Organic Chemistry A Survival Guide for Students
Ethan C. Levine Northeastern University December 2006
Table of Contents Intro………………………………………………………………………………………..i Letter to Students…………………………………………………………………..i Acknowledgments………………………………………………………………...ii References………………………………………………………………………..iii 1) It’s Back! General Chemistry Plus…...……………………………………………….1 Back to Basics: Atomic Structure and Bonding…………………………………..1 Drawing Organic Compounds…………………………………………………….5 Hybridization……………………………………………………………………...6 Molecular Orbital Theory…………………………………………………………8 Acids and Bases………………………………………………………………….13 Kinetics and Thermodynamics…………………………………………………..17 Final Note on Balancing Equations and Workup………………………………..21 2) Names, Structures, and Pictures………………………………………………………23 Carbon and Hydrogen……………………………………………………………23 Naming Alkanes………………………………………………………………….25 Functional groups………………………………………………………………...27 Some Structural Points…………………………………………………………...32 Rotating C—C Bonds……………………………………………………………33 Molecular Formula to Newman and Back Again………………………………..36 Cyclohexane and the Chair Conformation……………………………………….38 Isomers…………………………………………………………………………...41 Hydrocarbons and Formulas……………………………………………………..42 3) Alkenes………………………………………………………………………………...44 Nomenclature and Structure………………………………………………….….44 Basic Reaction Mechanism………………………………………………………47 HX (HBr, HCl, HI) and Some Complications…………………………………...48 Br2 and Avoiding Rearrangement………………………………………………..51 Water (HOH) and Alcohols (ROH)……………………………………………...53 Hydrogen……….………………………………………………………………...55 Breaking the Markovnikov Rule with HBr………………………………………56 4) Alkynes………………………………………………………………………………...57 Nomenclature and Structure……………………………………………………..57 Addition of HX (HBr, HCl, HI)………………………………………………….57 Water (HOH)…………………………………………………………………….58 Hydrogen…………………………………………………………………………59 Lengthening the Carbon Chain…………………………………………………..60 Multi-Step Synthesis……………………………………………………………..61 5) 3D Orgo: Stereochemistry…………………………………………………………….63 Drawing Chiral Centers………………………………………………………….63
R, S Nomenclature……………………………………………………………….64 Molecules with Two Chiral Centers……………………………………………..66 Creating Chiral Carbons…………………………………………………………68 A Note on Cumulenes……………………………………………………………70 Chiral Centers with Different Atoms…………………………………………….70 Light Rotation……………………………………………………………………71 6) Traveling Electrons: Resonance………………………………………………………75 Identifying Resonance Systems………………………………………………….75 Drawing Resonance Systems…………………………………………………….75 Stability…………………………………………………………………………..77 Acidity……………………………………………………………………………77 Molecular Orbitals……………………………………………………………….77 7) Two Double Bonds…………………………………………………………………….80 Two Types of Addition…………………………………………………………..80 Diels-Alder Reaction…………………………………………………………….81 Stereochemistry…………………………………………………………………..82 Line-up…………………………………………………………………………...84 Molecular Orbitals and Substituents……………………………………………..85 8) Radicals……………………………………………………………………………….87 Chlorination and Bromination of Alkanes.……………………………………...87 Which H Gets Replaced?………………………………………………………...88 Alkenes…………………………………………………………………………..89 Allylic Carbons…………………………………………………………………..90 Stereochemistry…………………………………………………………………..91 9) Substitution……………………………………………………………………………92 SN2 Reactions…………………………………………………………………….92 SN1 Reactions…………………………………………………………………….94 Competition Between SN2 and SN1 Reactions…………………………………...95 Switching it Up: OH and OR as Leaving Groups………………………………..96 Intramolecular Reactions………………………………………………………...97 Epoxides…………….….………………………………………………………...97 10) Elimination…………………………………………………………………………100 E2 Reactions….………………………………………………………………...100 E1 Reactions……………………………………………………………………101 Dehydration – Removing Water………………………………………………..102 Alkene Alkyne……………..………………………………………………..102 Competition Among E2, E1, SN2, and SN1……………………………………..103 11) Carbon and Metal…………………………………………………………………..104 C-Li: Organolithium Compounds………………………………………………104 Grignard Reagents……………………………………………………………...105
Gillman Reagents……………………………………………………………….105 12) Aromatic Compounds………………………………………………………………107 Antiaromatic Compounds………………………………………………………108 Stability…………………………………………………………………………109 H and C Only…………………………………………………………………...109 Atoms with Lone Pairs………………………………………………………….110 Molecular Orbital Theory Returns……………………………………………...111 13) All About Benzene…………………………………………………………………..114 Nomenclature: One Substituent………………………………………………...114 General Reaction Mechanism…………………………………………………..115 Halogenation……………………………………………………………………116 Nitration………………………………………………………………………...117 Sulfonation……………………………………………………………………..118 Friedel-Crafts Acylation………………………………………………………..119 Friedel-Crafts Alkylation……………………………………………………….120 Avoiding Rearrangement: Acylation and Reduction…………………………...121 Reactions of Benzene’s Substituents…………………………………………...122 Substituents and Reactivity……………………………………………………..124 Adding a Third Substituent……………………………………………………..127 Nucleophilic Aromatic Substitution……………………………………………129 Benzyne…………………………………………………………………………129 14) Carbonyls with Leaving Groups……………………………………………………131 General Mechanism…………………………………………………………….133 Acid-Catalyzed Mechanism…………………………………………………….134 Trouble with Acids……………………………………………………………..137 15) Aldehydes and Ketones……………………………………………………………..140 Freezing the Intermediate: H and C Nucleophiles……………………………...141 Freezing the Intermediate: O and S Nucleophiles……………………………...142 Replacing O: Nitrogen Nucleophiles…………………………………………...145 Replacing O: The Wittig Reaction……………………………………………...147 16) The α-Carbon of C==O……………………………………………………………149 Enols and Enolates…………………………….………………………………..149 Halogenation……………………………………………………………………151 Alkylation………………………………………………………………………152 Michael Reaction……………………………………………………………….153 Aldol Condensation…………………………………………………………….154 Claisen Condensation…………………………………………………………..157 17) Identifying Molecules: Mass Spectrometry, Infrared Spectroscopy, and NMR……158 Mass Spectrometry……………………………………………………………...158 Infrared Spectroscopy…………………………………………………………..161
NMR: Nuclear Magnetic Resonance Spectroscopy…………………………….163 Appendices………………………………………………………………………………..A I) Redox Reactions………………………………………………………………..A II) Hints for Biomolecules………………………………………………………...C III) Review Sheets………………………………………………………………...G Organic Compounds……………………………………………………...G Alkene Reactions…………………………………………………………H Alkyne Reactions………………………………………………………….J Substitution Reactions…………………………………………………….L Elimination Reactions……………………………………………………M Organometallic Compounds.……………………………………………..N Activating and Directing Substituents……………………………….…...O Glossary…………………………………………………………………………………..P Index……………………………………………………………………………………...I1
Copyright Information © 2006 by E. Levine Registration number: TXu-1-329-534
Dear students, Welcome to organic chemistry! Glad to have you here! That being said, you may have noticed that a number of people are less than thrilled to be in this class. It seems like everyone in the biological, chemical, and medical sciences needs to weather a year of organic chemistry. Some love it and wind up continuing on in the field. Others hate it from start to finish. Wherever you fall on that spectrum, my hope is that this study guide will make things a little bit better. I was one of those bizarre students who practically fell in love with the class. As a sociology major, I needn’t have been there anyway, but I went on to take a total of four courses in organic chemistry. I began tutoring during my sophomore year of college. Since then, I have seen close to 100 students, and one pattern has surfaced: students can succeed here. This class can be terrifying – but not necessarily because of the material. These seemed like the major sources of trouble: -
Pressure: organic chemistry is a filter class for med school. Passing is often necessary for continuing on in degree programs and getting internships.
-
Language: students were hit with a wealth of new vocabulary, coupled with teachers’ expectations that they would come to class fully aware of all terms’ and symbols’ meanings.
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Breadth and time constraints: tests covered as many as 120-140 pages of material. It was tough to find the major points in each chapter when preparing from exams. This was made all the more difficult by the fact that students were taking three or four other courses at the same time!
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Trouble approaching professors: it can be frightening to approach someone with a Ph. D., the power to pass or fail you, and who has spent years away from student life.
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Complicated explanations: material was often presented quickly and in confusing ways during lectures. It had to be possible to break topics down into simpler, more accessible concepts.
It was time to write a study guide! Time to break these seemingly complex ideas down into simpler ideas! Time to add some conversational language to the mix! Time to clarify which concepts were important and which were only of interest to those who wanted non-exam-related information! Best of luck in organic chemistry and beyond, Ethan Levine, 2006
[email protected]
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Acknowledgments Sincere thanks are due to Dr. Philip Le Quesne for his tireless support and assistance. His guidance throughout my directed study in organic chemistry helped me to acquire the confidence and communication skills necessary to begin this project. His suggestions and encouragement were essential to my completing it. Thanks also to Dr. Philip Warner, to whom I owe my very interest in organic chemistry. Additionally, I can not overemphasize my gratitude towards my tutees. I have worked with tens of students over the years. Their comments and dedication inspired me to continue teaching, to maintain my ties to chemistry, and to write this manual. Best of luck to all of you!
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References Text Bruice, Paula Yurkanis. Organic Chemistry, 4th Ed; Prentice Hall: New Jersey, 2003.
Syllabi and Lectures Brown, David M. CHE 202 – Introductory Organic Chemistry II. http://www.chm.davidson.edu/courses/202syl.pdf#search=%22organic%20chemistry%20 syllabus%22 (accessed October 15, 2006). Logan, Ralph H. CHM2423-2425 Organic Chemistry Syllabus. http://members.aol.com/profchm/org1syl.html (accessed August 8, 2006). Newland, Robert. Syllabus for Organic Chemistry I & II. http://users.rowan.edu/~newland/orgchem.htm (accessed 8 August 2006). Warner, Philip M. Lectures in Undergraduate Organic Chemistry, Northeastern University, September 2003 – April 2004.
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Chapter 1: It’s Back! General Chemistry Plus Welcome to organic chemistry! Glad to have you here! Before we jump into the world of carbon-containing compounds, we should review some general chemistry, along with a few new topics that are particularly important for this course. Back to Basics: Atomic Structure and Bonding Atoms can be broken down into two main parts: a nucleus, composed of positively-charge protons and neutral neutrons; and an electron cloud, composed of shells of negatively-charged electrons. The total number of protons is the atomic number of an atom/element, and the total number of (protons + neutrons) is the atomic weight. Although all atoms of the same element will have the same atomic number, different isotopes may exist – these atoms have different atomic weights because they have different numbers of neutrons. Shells in the electron cloud can be further broken down into orbitals, each of which can hold up to two electrons. There are four main kinds of orbitals: s, p, d, and f. To reference a specific orbital, include the shell and type – a 2p orbital is a p orbital in the second shell. As you move from one shell to the next (or one row to the next on the periodic table), you acquire more orbitals. This course deals primarily with s and p orbitals. The first electron shell contains one s orbital (so it can hold up to 2 electrons). Other shells contain one s and three p orbitals (room for a total of eight electrons in the valence shell). Elements in row three also contain five d orbitals and elements in row four and higher contain seven f orbitals. The outermost shell is called the valence shell. It may include lone pairs, or pairs of electrons that are not used in bonding (covalent bonding, at least – see below).
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The closer an electron is to the nucleus, the lower its energy. When an atom is in the ground state, all electrons are in the lowest-energy orbitals possible (begin filling orbitals starting with 1s, 2s, the 2p’s, etc.). In an excited state, one or more electrons will jump up to higher-energy orbitals. Ideally, atoms would like the s and p orbitals in their valence shell to be filled. This can be achieved by gaining, losing, or sharing electrons. Since atoms begin with an equal number of protons and electrons (for a net charge of zero), gaining/losing electrons means acquiring a charge – these atoms become ions. Ions with opposite charges attract each other, and can form ionic bonds. These are fairly weak bonds in which each atom keeps its electrons to itself, but is strongly enough attract to the other(s) that we can treat them as connected. When dissolved in solutions, ionically bonded compounds often break apart into separate ions. +
-
H ---Br (also written as HBr)
Atoms in covalent bonds share electrons. These can be best demonstrated using Lewis Structures¸ in which dots are used to represent the electrons in s and p orbitals. In these diagrams, the lower orbital is the s orbital, but you can arrange the s and p orbitals however you like.
Br *
**
:
** : Br .* Br
**
+
**
:
: Br .
**
:
:
: Br .
:
:
p : Le : p
:
:
p **
s
In the example above, bromine (Br), which has seven electrons in its valence shell, fills its last p orbital by sharing electrons with another bromine atom. Covalent bonds are stronger (and harder to break apart) than ionic bonds.
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As identical atoms, Br and Br share the electrons in the covalent bond evenly. Neither side of the bond is more positive/negative than the other – this is what’s called a nonpolar bond. Sometimes, covalent bonds form between atoms that have very different electronegativity, or ability to attract electrons. This can create a polar bond¸ in which one atom/side of the bond is partially negative and the other is partially positive. Chemists use the symbol δ to indicate a partial charge. Like covalent bonds, entire molecules can be polar or nonpolar – polar bonds pulling in opposite directions can cancel each other out! H3C—Cl δ+ δpolar bond polar molecule
O==C==O δ- δ+ δpolar bonds nonpolar molecule
Bonds are sometimes described in terms of dipole moment, which is a measure of polarity. More specifically, the dipole moment of a bond is equal to the magnitude of either atom’s charge (e) times the distance between the atoms (d) Each atom has the same magnitude of charge, so that the partial positive and negative charges balance each other out – you can therefore use either atom to find e. Distance between atoms is on the order of 10-8cm. The charge on one electron is 4.80 x 10-10 esu (electrostatic units). As such, dipole moment will be on the order of 10-18esu cm. To simplify this, we use a measurement called the debye (D). One debye is equal to 1 x 10-18esu cm. We represent dipole moment with the symbol μ: μ=exd A related concept is formal charge, or the charge on individual atoms within molecules. It can be calculated by adding the total charge of an atom’s protons with the total charge of it’s electrons (that is, the electrons we can assign to atom). We can also
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use this formula, in which ‘valence electrons’ mean the valence electrons in an uncharged atom, and ‘bonding electrons’ include all electrons in an atom’s bonds: formal charge = (valence electrons) – (lone-pair electrons + ½ bonding electrons) This can seem kind of daunting at first. Let’s practice with a couple of examples, using Lewis diagrams. 1) CH4 H
H x
C
H
x
.
.
H
.
H
H
x
C
.
H
x
H
Each H can be assigned one electron (the electron it contributes to its bond with carbon). Using the (proton charge + electron charge) method, we wind up with [(+1) + (-1)] = 0. With the longer formula, all electrons in H’s bond(s) are ‘bonding electrons’: formal charge = (valence electrons) – (lone-pair electrons + ½ bonding electrons) formal charge = (1) – (0 + 2/2) = 1 – 1 = 0 Carbon can be assigned 6 electrons: the electrons it contributes to its bonds as well as the two electrons in its first shell (the 1s shell). Carbon has 6 protons, which gives us a charge of zero: [(+6) + (-6)] = 0. Using the longer formula: (4) – (0 + 8/2) = 0. 2) HNO2 Let’s say that we are trying to assign charges to the atoms in this molecule.
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:: O
: :
:
O
O
: .
H
N
: :
N
H
+
N
O
-
O
:
:
O
: .: . .
H
H: Once again, H can be assigned one electron (balancing out its proton for a total charge of zero). Using the longer formula: formal charge = (1) – (0 + 2/2) = 0 N: We can assign a total of 6 electrons to N: the four from the diagram (every atom gets half of every bond it’s a part of, as well as any lone pairs if applicable). Check the periodic table – N has an atomic number of 7, meaning that it has 7 protons. If we add the electrons’ and protons’ charges, we get: [(-6) + (+7)] = +1. With the longer formula: Formal charge = 5 – (0 + 8/2) = +1 O in double bond: We can assign this O a total of 8 electrons, including the 6 electrons in the diagram and the 2 electrons in its first electron shell. Oxygen has 8 protons, giving us a formal charge of zero: [(-8) + (+8)] = 0. Using the longer formula: Formal charge = 6 – (4 + 4/2) = 6 – 6 = 0. O in single bond: We can assign this O a total of 9 electrons, including the 7 electrons in the diagram and the 2 electrons in its first shell. If we add the electrons’ and protons’ charges, we get: [(-9) + (+8)] = -1. With the longer formula: Formal charge = 6 – (6 + 2/2) = 6 – 7 = -1. Overall, two atoms should be charged: N+1, and O-1 in the single bond. Drawing Organic Compounds
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We have already seen several kinds of drawings. Lewis structures use dots to represent electrons (Li·), condensed structures are written like text and may include parentheses [CH3CH3, (CH3)3COH], and Kekulè structures use lines to represent covalent bonds and may or may not represent lone pairs with dots (Br—Br). We often combine condensed and Kekulè structures (CH3CH2CH==CH2). Chances are that you saw all of these drawing styles in general chemistry. Skeletal structures, which are probably new to you, use lines to indicate C—C bonds without specifically including the letters ‘C’ or ‘H.’ Every carbon is assumed to be bonded to enough H’s to fill its valence shell. Here are a few examples. Feel free to redraw skeletal structures (by writing the C’s in, for example) if that makes it easier to understand the images. Some teachers use mostly or exclusively skeletal structures in exams, but you are more than welcome to redraw molecules as long as you stay true to their structure! Molecule
Skeletal Structure
CH3CH2CH2CH3 O
O
HO
CH 2CH 3
HO
H
CH3
H
CH3
H2C==C(CH3)2
CH3CHBrCH2CH3
Br
Hybridization
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Carbon has four valence electrons, which would ordinarily arrange themselves .
:
.
C
this way:
This structure includes one full orbital (s), two half-full orbitals (p and p),
and one empty orbital (p). Far from ideal for an atom that would like to fill its valence shell! In order to have a better shot at 8 valence electrons, C atoms restructure their valence shells as hybrid orbitals. Instead of distinct s and p orbitals, atoms wind up with some orbitals that are part s and part p. Some teachers begin this topic with what is called the linear combination of atomic orbitals (LCAO): put simply, if we begin with a certain number of orbitals, we end with that number of orbitals! Carbon will always have four valence orbitals, some or all of which may wind up hybridized. The hybrid orbitals are named after the original orbitals used to make them. An sp3 orbital, for example, made from one s and three p orbitals. An sp2 orbital is made from one s and two p orbitals. Each type of hybridization allows C to bond to specific number of atoms at specific bond angles. Generally speaking, hybridized orbitals can bond to atoms while the leftover p orbitals form double/triple bonds or hold positive charges (more on that later!). Hybrid
Combination
sp3
(s + p + p + p) = sp3 + sp3 + sp3 + sp3
sp2
(s + p + p) + p = sp2 + sp2 + sp2 + p
sp
(s + p) + p + p = sp + sp + p + p
Structure C
C
C or C
(4 groups)
(3 groups)
(2 groups)
Bond Angle 109.5˚
120˚ 180˚
Multiple bonds are stronger and shorter than single bonds. To remember this, it can be helpful to imagine doing a pull-up at the gym. If you hold the bar with one hand, 1-7
you probably won’t have a terribly strong grip on it, and you may not manage to pull yourself up very far. If you hold the bar with both hands, you will have a stronger grip and be able to pull yourself farther (i.e. a stronger bond, and a shorter distance between you and the bar). Although we will be working mainly with carbon, several atoms can form hybrid orbitals. In order to recognize an sp3 hybridized atom, look for an atom that is bonded to four groups, in which a group can be either a bond, a lone pair, or a negative (-) charge. A positive (+) charge does NOT count as a group in this context. The following are examples of sp3 hybridized atoms. H H
-
C H
H
H
N
:
H
:
C
:
H
O
H
H
H H H
Molecular Orbital Theory This next concept is sort of like hybridization for an entire molecule. The orbitals of different atoms overlap to produce molecular orbitals. We will review two different bonds that result from this: sigma bonds, which are like combinations of s orbitals (and produce single bonds); and pi bonds, which are combinations of p orbitals and produce the second (or second and third) in double (and triple) bonds. When two orbitals overlap, they can either do so in an effective or ineffective way. An in-phase overlap produces a lower-energy bonding molecular orbital (bonding MO). An out-of-phase overlap produces a higher-energy anti-bonding molecular orbital (anti-bonding MO, represented with an asterisk*). Out-of-phase interactions are characterized by nodes, or spaces in which we will not find electrons.
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Let’s look at an H—H bond. A neutral H atom has only one electron, and that electron is usually found in the 1s orbital. When two H atoms combine, their sphere-like s orbitals overlap to form a sigma bond. Each H contributes one s orbital, and these can interact in-phase or out-of-phase. Here is a diagram of the possible combinations: (node)
electrons
In the lowest-energy or ground state, both electrons can be found in the bonding orbital (this diagram is in the ground state). In an excited state, one or both may jump up to the higher energy, anti-bonding orbital. Now let’s try a more complicated molecule: ethene (CH2==CH2). Each C atom has one spare p orbital (see earlier section on hybridization), and these can combine to form a pi bond. Whereas s orbitals were sphere-shaped, p orbitals have two sections that are out of phase with one another. As a result, there will be a horizontal node through every combination of orbitals that we create. To draw a molecular orbital diagram, follow these steps: 1)
Place half of the orbitals on each side: in this case, we’re working with 2 p orbitals, so we should place one on each side
2)
Draw one line for each possible combination: as a rule, expect one combination per orbital (we will have two combinations this time)
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3)
Combine the orbitals, beginning with the most in-phase (lowest energy) combinations possible: simply put, the more orbitals we have overlapping in-phase, the lower the energy. Line up as many similar porbital-lobes as possible. We should gain one node every time we move to a higher energy MO.
4)
Fill in the electrons: in the ground state (lowest energy), you should begin with the lowest orbital, adding two electrons
5)
Label bonding and anti-bonding orbitals if necessary. The lower half of the orbitals are bonding MO’s, the higher-energy half are antibonding. If we have an even number of molecular orbitals, the middle orbital is non-bonding. electrons: ground state
excited state
The lower orbital, in which the p orbitals are in-phase, is the bonding MO (π). The higher orbital, in which there is the orbitals are out-of-phase (and there is an additional node) is the antibonding MO (π*). All right, let’s try one more example: 1,3-butadiene, which has four p orbitals: CH2==CH—CH==CH2. We will set up our diagram, and then fill in the orbital combinations one at a time. 1-10
*Trick for exams! When analyzing these diagrams, teachers may ask you either to count the total nodes or vertical nodes (not counting the horizontal node) in a particular MO. Pay close attention to these questions!
2 orbitals on each side, expecting four combinations
all in phase, no vertical nodes
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2nd MO: one vertical node
3rd MO: two vertical nodes
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4th MO: three vertical nodes
One final concept that you can bet will come up later on in the course: with molecular orbitals, chemists may refer to the HOMO, or highest occupied molecular orbital; and LUMO, or lowest unoccupied molecular orbital. ‘Occupied’ here means containing electrons, and ‘highest’ refers to ‘highest energy.’ When two molecules react, the HOMO of one and the LUMO often interact. Acids and Bases When organic chemists speak of acids and bases, they generally use the Bronsted-Lowry definition: acids are proton (H+) donors, bases are proton acceptors. Acid-base reactions, in which an acid donates a proton to a base, are called protontransfer reactions. The products of these reactions are the conjugate base and conjugate acid of the reactants. HCl
+
-
HO
Cl
-
+
HOH
In the above reaction, HCl (an acid) donates a proton to HO- (a base). The products are the conjugate base of HCl (Cl-) and the conjugate acid of HO- (HOH). Note
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that this reaction can occur in either direction – it is possible for HOH to donate a proton to Cl-, producing our original reactants. To determine which reaction (or which direction of this reaction) will dominate, we need to look at the acidity, or acid strength, and basicity, or base strength, of the reactants on both sides of the equation. Let’s determine how to identify the stronger acid. A strong acid is an effective H+ donor. After donating a proton, it will form a stable product. That lost proton will stay lost! We might also call a strong acid unstable, as it is very interested in changing itself by donating a proton. In contrast, a strong acid’s conjugate base is weak and stable, content without getting the proton back. You can use a molecule’s structure to determine its strength as an acid, or the stability it will have in conjugate base form. Look for electronegative atoms and groups. The more electronegative groups a molecule has, and the closer they are, the stronger the acid (more stable the conjugate base). Consider the following trends: Electronegativity:
F>O>N>C
Acidity:
HF > HOH > NH3 > CH4
Base Stability:
F- > HO- > NH2- > CH3-
Now, for some acidity-comparing-style questions you are likely to see on an exam! 1) Which is the strongest acid? O C
a. CH4
b. CH3NH2
c. CH3OH
d. CH3COOH or H3C
OH
We want to find the molecule that has the most electronegative atoms possible. Since O is more electronegative than both H and N, we can rule out (a) and (b). Of the remaining choices, (c) has one O and (d) has two, very close to each other. The answer is (d).
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2) Which is the strongest base? a. CH3COO-
b. CH2FCOO-
c. CHF2COO-
d. CF3COO-
Clearly, there is a trend here – we either want more or fewer F’s. Let’s frame this as an acid question. If we were trying to find the weakest/most stable base (associated with the strongest acid), we would want as many electronegative atoms as possible. F is more electronegative than H, making (d) the most stable base. Yet this question asks for the strongest base, so we reverse our answer. The strongest base is (a). Using this type of reasoning, we can frame any acid/base strength question as an acid strength question, and then either keep or reverse our answer at the end! Acid strength can also be indicated by pKa. Chances are that your textbook explains the origin of this value, how to calculate it, etc. Chances are that you won’t really need to know all of that. Instead, remember that the lower the pKa, the stronger the acid. An acid’s pKa is determined in part by the solvent it’s in – unless otherwise stated, the values you will see should represent acid’s pKa in water. Strong acids:
pKa < 1
Moderately strong acids:
pKa = 1 – 5
Weak acids:
pKa = 5 – 15
Very weak acids:
pKa > 15
You may encounter exam questions that compare the pKa of an acid to the pH of the solution it’s in (pH is the measure of an entire solution’s acidity). In general chemistry, you probably learned that neutral solutions have a pH of 7, acidic solutions have a pH below 7, and basic solutions have a pH above 7. That’s good to know, but not
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all that important for these questions – just remember that, similar to pKa, solutions with lower pH values are more acidic. Compounds can have acidic and basic forms. One of these forms will be charged, depending on whether the compound is neutral or positively charged before donating a proton (or negative or neutral before accepting a proton). When the pH of a solution equals the pKa of a compound (pH = pKa), there are an equal amount of acidic and basic versions of that compound. When pH < pKa, we have a relatively acidic solution and you can expect the compound’s acidic form to dominate. When pH > pKa, we have a relatively basic solution and you can expect the basic form to dominate. Here are a couple of sample questions. 1) NH4+ has a pKa of 9.4. For the following pH values, will the acidic or basic form of this compound dominate? a. 3.0
b. 6.2
c. 9.4
d. 13.9
The first values are both lower than NH4+’s pKa, so we can accept the acidic form to dominate. At a pH of 9.4, which is equal to the pKa, neither form will dominate. At a pH of 13.9, which is higher than the pKa, we can expect the basic form of NH4+ to dominate. 2) CH3COOH has a pKa of 3.75. For the following pH values, will the charged or uncharged version of this compound dominate? a. 2.2
b. 3.75
c. 9.4
d. 11.0
First, we need to figure out which form of the compound (acidic/basic) is charged:
H3C
O
O
C
C OH
H3C
-
O
+
+
H
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As we can see, CH3COOH is neutral as an acid, then charged as a base (after it donates/loses a proton). When the acidic form dominates, the uncharged version dominates. Now for the pH values – 2.2 is lower than 3.75, so the acidic and uncharged molecule will dominate for (a). For choice (b), pH = pKa and neither form dominates. For choices (c) and (d), pH > pKa and the basic, charged version of CH3COOH dominates. One final note about defining acids and bases – your teachers and texts will sometimes refer to Lewis acids and bases, which are based on electron rather than proton donation/acceptance. Lewis acids accept pairs of electrons and Lewis bases donate them. Kinetics and Thermodynamics Follow your teacher’s lead on this one. This next section will address several topics in the broad field of kinetics and thermodynamics – some will come up in the beginning of the course, some may arise later on, some may never come up. As with all sections, if this review touches on something that doesn’t seem relevant to your particular curriculum, don’t spend much/any time on it. We’ll begin with a concept called Gibbs free energy change (ΔG°), which you may remember from general chemistry. The ° indicates standard conditions (concentration of 1M, temperature of 298K, pressure of 1atm). We can calculate ΔG using three different formulas. 1) ΔG° = (free energy of the products) – (free energy of the reactants) This means that ΔG° will be positive in an endergonic reaction (consumes energy, or consumes more than it releases) and negative in an exergonic reaction (releases
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energy, or releases more than it consumes). This is one of the better definitions to remember, as it clarifies what we are trying to calculate. 2) ΔG° = ΔH° – TΔS° ΔH° represents the change in enthalpy, or heat; ΔH° = (heat of the products) – (heat of the reactants). Endothermic reactions, which absorb heat, have a positive ΔH°. Exothermic reactions, which release heat, have a negative ΔH°. The ‘T’ in this equation represents the temperature. ΔS° represents the change in entropy, or disorder. The most favorable conditions for a reaction, or the conditions most likely to move a reaction forward, are a (-)ΔG°, (-)ΔH°, and (+)ΔS°. In other words, reactions prefer to release energy, release heat, and increase disorder. 3) ΔG° = –RTlnKeq This last formula pulls in lots of concepts from general chemistry. ‘R’ represents the gas constant (8.314 10-3 kJ mol-1K-1), T represents the temperature in K (K = °C + 273, so 10°C is 283K), and Keq is the equilibrium constant. Keq is a measure of the relative concentrations of reactants and products in a solution. To calculate Keq: wA + xB
Keq
yC + zD
[products]
[C]y[D]z
[reactants]
[A]w[B]x
Again, I would pay more attention to the first formula, as this will help you remember what ΔG° represents. The next major topic in thermodynamics is the reaction diagram. This is a graph that keeps track of the energy changes throughout a reaction or series of reactions. Every valley or low point in the diagram represents an intermediate (if it’s in the middle of the
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overall process) or final product (if it’s at the end). Every peak represents the transition state of a step in the overall reaction. One-Step Reaction
free energy
TS
products
reactants reaction progress
Two-Step Reaction TS2
free energy
TS1
intermediates reactants products
reaction progress
These diagrams reveal several things. First of all, we can tell whether the reactions are endergonic (energy-absorbing) or exergonic (energy-releasing) by comparing the products and reactants. In the first reaction, the products have higher energy than the reactants – this reaction is endergonic. In the second reaction, the
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products have lower energy than the reactants – this reaction is exergonic. We can also tell whether the transition state is more like the reactants or the products. According to Hammond’s Postulate, the transition state is similar to the molecule(s) with the highest energy. Remember that the TS is the highest-energy point in the whole reaction – it makes sense that it will be most similar to whichever side of a reaction has the highest energy! In an endergonic reaction, the TS resembles the products. In an exergonic reaction, the TS resembles the reactants. We can also identify the rate-determining step (RDS) of a multiple-step reaction. Look for the highest-energy transition state. The RDS is the slowest step of a reaction, which means that it requires the most energy to complete. The formulas we examined for ΔG° can also be used to determine the change in energy as the reaction reaches its transition state. Use the same types of formulas, but change ‘products’ to ‘transition state’ and use the symbol ‡ (ΔG‡, ΔH‡, ΔS‡). Diagrams can also be helpful in determining whether a reaction is under kinetic or thermodynamic control as well as its kinetic or thermodynamic stability. Essentially, thermodynamic control/stability depend on the products and kinetic control/stability depend on the reaction process. When a reaction is under thermodynamic control and thermodynamically stable, it forms the most stable product possible. This is indicated by a release in energy (exergonic reaction, negative ΔG°) and any other signs of stability in molecules that you will learn to look for. Kinetic control and stability are a bit more confusing. When a reaction is under kinetic control, it proceeds as quickly as possible. A reaction is kinetically stable if it proceeds slowly – this is indicated by a high ΔG‡.
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The rate at which a reaction proceeds is determined by the number of molecules in its RDS as well as a rate constant, indicted by ‘k.’ The rate of first-order reaction is determined by the concentration of one molecule (there is only one molecule necessary for the RDS). The rate of a second-order reaction is determined by the concentration of two molecules (two molecules are needed for the RDS). I First order:
A B; rate = k[A]
Second order: A + B C + D; rate = k[A][B] 2A B; rate = k[A]2 = k[A][A] Rate constants are not affected by the concentrations of reactants. They are, however, affected by temperature. You can calculate a rate constant using the Arrhenius Equation, which appears below. Memorize this equation if your teacher stresses it in your lecture or syllabus (or ask if you aren’t sure) – otherwise, chances are that you won’t have to use it in this course. k = Ae-Ea/RT A indicates the frequency factor, which addresses the frequency with which molecules collide in an orientation that allows the reaction to move forward. Ea is the experimental energy of activation, R is the gas constant, and T is the temperature. Final Note on Balancing Equations and Workup One key difference between general chemistry and organic chemistry is that we do not always, or even often, balance our equations in organic chemistry. We often merely represent the compounds we were trying to form/react in our products. Solvents and catalysts may not make it to the other side…In addition, putting reactants, catalysts,
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and solvents over and below arrows (expect to see this a lot!) makes it difficult to really “balance” both sides of a reaction. This is related to a concept called workup. At the end of a reaction, chemists may need to ‘clean up’ or ‘finish/fix’ their products – maybe adding an H+ to a negatively charged ion, or removing one from a positively charged ion. In these situations, you may see the term ‘workup’ written over an arrow, you may see H+ or –H+, or you/chemists/professors may simply assume that your molecule will gain or lose a proton somehow. Chemists may also need to separate their products from other materials.
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Chapter 2: Names, Structures, and Pictures Now that we’ve reviewed some material from general chemistry, it’s time to look at some organic compounds! This section will lay the ground rules for name, drawing, and describing some pictures of organic (carbon-containing) molecules. That means a lot of memorization of new words and rules. Depending on your learning style, you can either treat this section as strictly-memorization or as part-rote-memorization, partconcept-grasping. *Note about Drawings: In previous chemistry classes, you probably represented all atoms with their letter symbols (i.e. C for carbon, Li for lithium, etc.). Organic chemists often use lines and points to represent C’s and leave the H’s attached to carbons out of diagrams. For example: CH3CH2CH3 is written as written as
, CH3CH2CH2CH3 is
. On your exams, chances are that your teacher will allow you
to write molecules in either form (with letters or lines/points), but you should familiarize yourself with both styles. Carbon and Hydrogen Hydrocarbon groups, or carbon-and-hydrogen-groups, are called alkyl groups and are often labeled as ‘R’ groups. Multiple alkyl groups may be labeled as R, R I (R prime), RII (R double-prime), RIII and so on. Fragments/groups and whole molecules sometimes have multiple names, which can be divided into common names (which you may hear your teacher use, and which may not tell you much about the structure) and scientific or IUPAC names (which are based on the rules established and accepted by the International Union of Pure and
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Applied Chemistry). Expect to see IUPAC names in this review guide. When common names are used, they will be clearly labeled as such. Organic compounds/groups are assigned prefixes based on the number of carbons in a chain. If the C-chain is a substituent, or group attached to the larger molecule, end with the suffix ‘yl.’ If the C-chain is the parent chain, or longest group of consecutive compounds, we end the name with a suffix that tells us what kind of compound we have. Carbons 1 2 3 4 5 6 7 8 9 10
Prefix meth eth pro but pent hex sept oct non dec
As a Substituent methyl: CH3– ethyl: CH3CH2– propyl: CH3CH2CH2– butyl: CH3CH2CH2CH2– pentyl: CH3CH2CH2CH2CH2– hexyl: CH3CH2CH2CH2CH2CH2– septyl: CH3CH2CH2CH2CH2CH2CH2– octyl: CH3CH2CH2CH2CH2CH2CH2CH2– nonyl: CH3CH2CH2CH2CH2CH2CH2CH2CH2– decyl:CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2–
All of the above groups were drawn as straight chains. Some constitutional isomers of these groups, which have the same molecular formula (same number of each type of atom) but are connected differently, have common names that you may see in class or on an exam.
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Three Carbons CH3CH2CH2
CH3 H3C
propyl
CH iso-propyl
Four Carbons CH3
CH3 CH3CH2CH2CH2
H3C
butyl
CH
H3C
CH 2
CH 2
iso-butyl
CH
CH3 H3C
C
sec-butyl
CH3 tert-butyl
Naming Alkanes Alkanes are saturated hydrocarbons – they have all single bonds. Use the suffix ‘ane’ to name an alkane: a three-carbon alkane, for example, would be called propane. A two-carbon hydrocarbon with all-single bonds would be called ethane. As we have already seen, organic compounds may not be in straight-chain structures. Your first task in naming it to find the parent chain – feel free to follow branches/corners in diagrams, as long as you are looking at consecutive carbons. CH3 CH H3C 1
2
CH3 CH 2
parent chain: butane (4C)
4
3
CH3 7
CH2 H2 C
6
parent chain: heptane (7C)
5
H3C
CH CH2
4
CH2 CH 3
2
CH3 1
CH3
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The next step is to include any substituents on the molecule. Include alkyl substituents at the beginning of the molecule’s name, including a number to indicate the its location on the parent chain. CH3
- parent chain: propane (3C) - substituent: methyl on C2 - name: 2-methylpropane
CH 2
H3C 1
CH3 3
It is often possible to get different names for a molecule, depending on which side of the parent chain you begin with. As a rule, number in whichever direction gets you the lowest numbers for substituents. CH3
CH3
CH H3C 1
2
CH3 CH2 3
correct
4
CH H3C 4
3
CH3 CH2
1
correct: 2-methylbutane incorrect: 3-methylbutane
2
incorrect
Substituents and even sections of some parent chains may be written in parentheses sometimes. When you see this, draw the whole molecule before naming or reacting with it. Keep mind that carbon can handle four bonds – this can be useful in figuring out what the groups in parentheses are attached to! CH3(CH2)5CH3 ≡ CH3CH2CH2CH2CH2CH2CH3 H3C CHCH 2CH2CH3
(CH3)2CHCH2CH2CH3 ≡ H3C Parentheses also come in handy when trying to name complex substituents. You can treat like a whole molecule (or parent chain with one or more substituents) when naming it, and use parenthesis to indicate the whole group’s place on the molecule.
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CH3 common: 5-isobutyldecane
CH3
IUPAC: 5-(2-methylpropyl)-decane CH3 H3C
If a molecule has two or more identical substituents, separate the numbers with commas and use a prefix like ‘di’ (2) or ‘tri’ (3) to indicate the total count: CH3 CH3
H3C
2,3-dimethylhexane
CH3
Rings, or cyclic compounds and groups, can also be either parent chains or substituents. Just look for the longest chain of consecutive carbons – see whether it’s the ring or a group of carbons attached to the ring. To name a ring, put ‘cyclo’ before the usual alkane name – a five-C, straight-chain alkane is pentane; a five-C ring is cyclopentane. CH3
ethylcyclohexane (or 1-ethylcyclohexane)
H3C CH3 2-cyclopentyloctane
Functional Groups Many molecules are more complicated than alkanes, having either multiple bonds or functional groups composed of different atoms. Let’s take a look at some of them.
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Some general rules to keep in mind when naming (and different teachers value these rules differently): 1)keep looking for the parent chain, 2)number the chain in order to get the lowest numbers possible, 3)when naming multiple types of substituents, generally list them in alphabetical order (e.g. bromine and then methyl). Alkenes are hydrocarbons with one or more double bonds; alkynes are hydrocarbons with one or more triple bonds. Follow the rules we learned for alkanes when naming them – but change ‘ane’ to ‘ene’ or ‘yne,’ and use numbers to indicate the multiple bonds’ location. Write the number before ‘alkene/alkyne’ or in the middle (2pentene or pent-2-ene). Double/triple bond should have the lowest number possible.
cyclopentene
H3C
CH3
2-pentene or pent-2-ene
CH3
4-methyl-2-pentyne or 4-methylpent-2-yne
H3C H3C
Alkynes in which the triple bond is at the end of the molecule are called terminal alkynes. Alkynes in which the triple bond is somewhere in the middle of the molecule are called internal alkynes. HC
CH3 terminal
H3C
CH3 internal
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Alkyl halides have one or more halogens (7th column on the periodic table – F, Cl, Br, I). IUPAC names include the halogen’s name as a prefix (fluoro for fluorine/F, bromo for bromine/Br, chloro for chlorine/Cl, iodo for iodine/I) and a number to indicate the halogen’s location. If there are two or more of the same halogen, put the numbers together and add di, tri, etc. to indicate the number of halogens. Common names include two words: the alkyl group and the halogen with suffix ‘ide’ (e.g. ethyl bromide for CH3CH2Br). Cl CH3
2-chlorohexane
H3C
F 3-bromo-2-floropentane CH3
H3C
(careful to number in whichever direction gets you the lowest numbers for substituents!)
Br
Cl CH3
1,2,-dichlorobutane
Cl
Ethers include two hydrocarbon groups separated by an oxygen atom (ROR1), in which ‘R’ represents an alkyl group). IUPAC names treat the O and smaller group as a substituent – an alkoxy group, to be specific. Combine the alkane name with ‘oxy’ for the group, then use a number to indicate its position. Common names include both alkyl groups and then ‘ether,’ as three separate words (e.g. ethyl methyl ether for CH3CH2OCH3).
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CH3
O O H3C
CH3
CH3
H3C
methoxyethane
2-ethoxypentane
Epoxides are similar to ethers, in that they include a C—O—C bond. They have 3-membered C—O—C rings, and are named with the prefix ‘epoxy’ and 2 numbers to indicate the ring’s location.
O H3C
HC
CH
CH3
2,3-epoxybutane
Alcohols have one or more OH groups. Unlike halogens and single oxygens, OH groups change the suffix in a molecule’s name. Instead of ending in ane, ene, or yne, these names end in ‘ol.’ If there are multiple OH groups, change the suffix to diol, triol, etc. Numbers can either be written before the alkanol name (i.e. 2-pentanol) or right before the ‘ol’ suffix (i.e. pentan-2-ol). Common names include the alkyl group and ‘alcohol’ as separate words (e.g. CH3CH2CH2OH would be propyl alcohol). OH
OH OH
H3C propanol
CH3 H3C
CH3
2-propanol or propan-2-ol
H3C 3-hexanol or hexan-3-ol
In Amines, one or more C-groups (may also be called R-groups) are attached to nitrogen. They can be defined as primary (1), secondary (2), or tertiary (3) depending on
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how many C’s are attached to the N-atom. Once again, you should look for the longest chain of consecutive C’s to find your parent chain. Use the suffix ‘amine:’ CH3CH2NH2 is ethanamine. If there are other R groups attached to N, list them as you would any other alkyl substituent – but use ‘N’ instead of a number to show their position. H3C N CH3
H3C
H2N propanamine
NH
CH3
N-ethylbutanamine
H3C
CH3
N,N-dimethylpropanamine
Carboxylic acids are distinguished by carboxyl groups (COOH). These groups are a combination of an acyl or carbonyl group (C==O), and an alcohol group (OH). To name a carboxylic acid, find the parent chain that begins with (and therefore includes) the C in COOH, and add the suffix/phrase ‘oic acid.’ You can also treat the COOH group as a substituent. O
O C OH carboxy group
OH
H3C hexanoic acid or 1-carboxypentane
Carboxylic acid esters, often referred to simply as ‘esters,’ have one R-group and one OR group attached to a carbonyl. Their names include two words: the alkyl substituent attached to O (as it’s own word, simply ethyl, propyl, etc.) and then the C’s in the parent chain with the suffix ‘oate.’ Again, the parent chain includes the C in C==O.
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O
O
O CH2CH 3
H3C
O CH3
H3C methyl hexanoate
ethyl ethanoate
Amides include carbonyl groups attached to nitrogen. Their naming is similar to amines: use the suffix amide at the end of the parent chain (including the C in C==O), listing any groups attached to N as N-subtituents. O
O
H3C
H3C
NH2
NH
CH3
N-methylbutanamide
ethanamide
Ketones have two R-groups attached to a carbonyl. To name them, add the suffix ‘one’ and a number to indicate the carbonyl’s location. O
H3C
CH3 2-pentanone or pentan-2-one
Aldehydes are similar to ketones, except that the carbonyl groups are at the end of the molecule (similar to terminal alkynes). There is therefore no need to indicate the carbonyl group’s location – just use the suffix ‘al.’ O
H
CH3 ethanal
Some Structural Points
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Carbons can have a maximum of four bonds. We often refer to C’s in terms of the number of C’s they are bonded to: methyl (0°), primary (1°), secondary (2°), tertiary (3°), and quaternary (4°). The higher the degree (or the more C’s attached to), the more substituted the C-atom is. CH3 CH3 CH4 methyl
H3C
CH3
primary
H3C
* CH 2
CH3
secondary
(least substituted)
H3C
*CH
H3C CH3
tertiary
*C
CH3
CH3 quaternary
(most substituted)
C—H bonds are nonpolar. As a result, hydrocarbons (without any functional groups) should be insoluble in polar solvents and soluble in nonpolar solvents. Organic molecules with certain functional groups can sometimes dissolve in polar solvents, though. An OH group, for example, can pull two or three carbons into a polar solution. And a brief comment on boiling points. Larger and more highly substituted hydrocarbons have higher boiling points than smaller/less substituted ones. Rotating C—C Bonds! It’s time to look at some stereochemistry, or arrangements of atoms/molecules in space. Some people think of it as 3D chemistry. C—C bonds can rotate, which changes the relative positions of groups on both carbons. Different positions (assumed by rotating) are called conformers or conformations of bonds. We can represent different conformers with drawing called Newman Projections, which look directly down a bond. A dot is used to represent the atom (C) nearest you, and a circle represents the atom (C) directly behind it. Other substituents are represented with lines attached to the dot and circle. Here are some
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Newman projections for the C2—C3 bond in butane (CH3CH2CH2CH3). Look for changes in structure as the C—C bond rotates. H CH 3
H
H
H
H H CH 3
staggered anti
H CH 3
H CH H 3
H H CH 3
eclipsed
H CH 3
3
H CH 3
H CH 3
H H H
H CH H 3CH
H
H
H H
HH
H
H
H
H
staggered
H
eclipsed
staggered
Gauche
Gauche
*Note: Although H’s are written out in these pictures, they are left out of most Newman projections – you will see/write a bond to C with nothing on the end. As you can see, there are two main types of arrangements for bonds: staggered, in which groups are spaced as far apart as possible, and eclipsed, in which groups are directly in line with each other. Arrangements that fall between these categories are skewed. If you are good at visualizing molecules or like working hands-on to learn new concepts, consider making a model of butane and rotating the C2—C3 bond to see these conformers. Staggered conformers can be further broken down into Gauche and anti arrangements. Anti substituents are 180º apart from each other, on opposite ends of the bond. The substituents on a staggered conformer are either positioned on vertical lines (axial) or slanted lines (equatorial). Gauche substituents are 60º apart from each other, with one group in an axial position and the other in an equatorial position. If both groups are equatorial, they are not Gauche to one another (even though they may still be separated by 60º).
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This brings us to the concept of strain. When atoms/groups are in each other’s way or connected in an unstable manner, it can put strain on a molecule. There are three major types of strain. Angle strain occurs when C’s have bonds that are unstable or difficult to maintain. Ideally, tetrahedral C likes to have 109.5º bonds (this is the bond angle for an sp3 carbon). The farther away from 109.5º you get, the more strain. Threeand 4-membered rings are particularly unstable. Five-membered rings have very little angle strain; six-membered rings don’t have any. Torsional strain occurs when electrons in different atoms repel against each other. Ecplisped conformations suffer from this. Finally, steric strain occurs when atoms/groups are taking up the same space. The closer that groups are, and the larger those groups are, the more strain (for example, two CH3’s can cause steric strain if they’re close to each other, but more strain will be caused by two CH2CH3’s). Steric strain of this sort can be measured in Gauche interactions. If two carbon groups are spaced 60º apart, with one in an axial/vertical position and the other in an equatorial/slanted position, they will produce one Gauche interaction of strain (written as 1G). If two carbon groups are either anti to each other or both in equatorial positions, they will be too far apart to cause steric strain. H CH 3
H CH 3
H
H
H
H
H
H
H CH 3
0G (anti)
H H CH 3
H H
1G
H
CH3
H
CH3 H
0G (both equatorial)
Things get more complicated in rings. You will probably have to calculate the strain in cyclohexane rings. That means learning two additional rules: (1) every axial
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methyl (or carbon group) has 2G interactions with the ring, and (2) if a cyclohexane has two axial methyl groups spaced 2C apart (on carbons 1 and 3, for example), they will form what is called a U-pentane. This adds an additional 4G of strain. The name Upentane comes from the U-shaped, 5C ring that forms (including the two methyls and three C’s from the ring). CH3
cyclohexane
one CH3: 2G
CH3
CH3
U-pentane structure: 4G each individual CH3: 2G total: 8G
Molecular Formula to Newman and Back Again You may be given a molecular formula (e.g. CH3CH2CH2CH3) and be asked to draw a Newman projection from a specific bond, or vice versa. Here is one method that you can use to solve these problems: 1) Make a list of what each C in the bond is attached to 2) For Newman molecular formula, you can write out a structural formula at this point. Then, determine the longest chain and re-write your answer. 3) For molecular formula Newman: depends on what the question asks for specifically. If you are meant to draw the least stable conformer, draw an eclipsed conformer with the biggest groups in line with each other. To draw the most stable conformer, draw a staggered conformer with the biggest groups as far apart as possible. Example 1: Newman Projection Molecular formula
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CH3 CH3
Br
CH2CH3
First, let’s make a chart of what each C in the bond is attached to: Closer C (point) - CH3 - CH2CH3 - H
Farther C (circle) - CH3 - H - Br
Now, we can draw a structural formula of this molecule.
H
CH3
CH3
C
C
H2C CH3
H
Br
We were looking down the C2—C3 bond of a 5C chain:
H
CH3
CH3
C
C
3
2
H2C CH3 4
1
Br CH3 H
CH3CHCHCH2CH3
Br
5
Example 2: Molecular formula Newman Projection Draw the most stable conformer for the C2—C3 bond of this molecule: Br CH3CHCHCH2CH3 Br
Like before, let’s start with a chart for the C’s in our Newman projection:
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C2 - CH3 -H - Br
C3 - Br - H - CH2CH3
Now, step by step, let’s draw the most stable Newman Projection: CH3
CH3
CH3 Br
Br CH2CH3 staggered
CH2CH3
C-groups should be anti
Br's should be anti
H
Br
Br
H CH2CH3
done!
Cyclohexane and the Chair Conformation The most stable conformation for cyclohexane is called the chair conformation. To see it, we will draw cyclohexane in a sawhorse projection. Lines and points still represent bonds to/between carbons. As with the Newman projections, we may not always draw the H’s in our diagrams. H
H H
H
H
H
H
H H H
just C-C bonds
H H
with H's written out
just bonds to H's (most common)
The carbons in cyclohexane are sp3 hybridized, and can be bonded to four things: two other C’s in the ring and two substituents/H’s. Looking at these diagrams, we can continue to divide substituents into axial (vertical) and equatorial (slanted) positions. Additionally, H’s/groups can be described as either pointing up or pointing down. Each 2-38
C has one substituent pointing up and one pointing down. In other words, if the axial group on a C points up, the equatorial group points down and vice versa.
axial
equatorial
all
When two substituents are pointing in the same direction (up/down) – regardless of whether they are axial or equatorial – they are cis to each other. When two substituents are pointing opposite directions, they are trans. CH3
CH3
CH3
CH3
CH3
CH3
cis methyl groups
CH3 CH3
CH3
CH3
CH3
CH3 trans methyl groups
Cis/trans distinctions apply to fused rings as well. When two cyclohexanes are fused (attached and sharing a whole bond), they will be connected in cis or trans positions.
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trans-fused rings
cis-fused rings
(up/down in equatorial positions)
(down in one axial, one equatorial position)
In what is called a ring flip, cyclohexanes can move between two different chair conformations. They pass through several different stages during this process, but test questions are likely to focus on the beginning/ending chair conformations. That will also be our focus here. You may be asked to draw the ring-flipped version of a molecule. Here are some numbered drawings that my help. 1
5 6
3 4
2
Some find it helpful to see a vertical ring flip. You can visualize C1 flipping over/down, and then just be sure to number in a clockwise direction. 6
4 5
2
3
1
1
5
4
6
6
3 4
5 2
3
2 1
a few textbooks show the ring flip side-by-side, like this
When cyclohexane rings flip, axial groups become equatorial and equatorial groups become axial. The direction (up/down) in which groups are pointing does not change.
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CH3 1
5 3 4
3
2
Br
5
2
6
4
2
3
6
4
6
4
1
5
6
5 1
3
CH3
2 1
Br CH3 goes from axial to equatorial, but is still pointing up
Br goes from equatorial to axial, but is still pointing down
Isomers Time to look at a very popular test question. Teachers will often show you a pair of molecules and ask you to determine whether they are identical, (a specific type of) isomers, or neither. Isomers are molecules that have the same molecular formula but are structurally different. Constitutional isomers are connected to each other differently. CH3 H3C
C
CH3
CH3CH2CH2CH2CH3
CH3
Conformational isomers are connected in the same way, but in different conformations. You are most likely to see these in Newman projections.
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CH3
CH3
CH3
CH3 CH3 CH3
Stereomers are different in space, and include cis/trans isomers as well as a variety of other sterically/spatially different molecules (see section five for more information). H3C
CH3
H3C CH3
Be on the lookout for cyclohexanes here! Remember that substituents switch from axial to equatorial when rings flip – that means that the following molecules, for example, are identical and NOT stereoisomers. CH3 CH3 CH3 H3C
Hydrocarbons and Formulas This review book would be doing you a disservice if we didn’t cover the formula for calculating the number of C’s and H’s in hydrocarbons. Start with this formula: CnH(2n+2) Then remove 2H’s for every pi bond and ring. DO NOT remove any H’s for branches or alkyl groups. Let’s try some examples: 1) What is the molecular formula for octane? Octane is an alkane with no pi bonds or rings. We have 8 carbons total: 2-42
CnH(2n+2) = C(8) and H(2x8 + 2) = C8H18 2) What is the molecular formula for cyclopentane? No pi bonds, but one ring. We have 5 carbons total: CnH(2n+2) = C(5) and H[(2x5 + 2) – 2] = C5H12-2 = C5H10 3) What is the molecular formula for 3-hexyne? Alkynes have triple bonds – or 2 pi bonds. No rings. We have 6 carbons total: CnH(2n+2) = C(6) and H[(2x6 + 2)– 2x2] = C6H14-4 = C6H10 4) What is the molecular formula for 2,3-dimethyldecane? No pi bonds, no rings. We have 12 carbons total: decane and 2 methyl groups: CnH(2n+2) = C(12) and H(2x12 + 2) = C12H26 This addresses the concept of degree of unsaturation. Remember that saturated hydrocarbons include all single bonds, and also include as many H’s as possible. To calculate the degree of unsaturation, add the number of pi bonds and rings. Butane has 0, cyclobutane has 1, butene has 1, and butyne has 2 degrees of saturation.
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Chapter 3: Alkenes Now that we’ve learned how to name and describe some properties of organic compounds, it’s time to look at some reactions! Some texts and review books, including this one, jump straight into alkenes (with carbon-carbon double bonds). Others begin with alkanes (single bonds). Check out sections 9-11 for alkane reactions. Nomenclature and Structure Double bonds can be treated as functional groups in and of themselves, and they sometimes take priority over other groups when naming compounds. If a double bond is in the parent chain of a molecule, indicate the location of the double bond and use the suffix ‘ene.’ Be careful to use the lowest number possible! CH3CH2CH2CH3 butane
CH3CH2CH==CH2 1-butene
The carbons in and adjacent to double bonds have special names. These names can also refer to substituents attached to these carbons. allyl carbon
CH3 H2C
CH2
vinyl carbon H2C
vinyl group
allyl group
Cl H2C
Cl
vinyl chlorine
allyl chlorine
H2C
Structure plays a role in naming alkenes. If each carbon in a double bond is attached to similar groups (each C is attached to one H and one Br, for example), we can describe the bond with cis and trans nomenclature. In a cis configuration, similar groups
3-44
are on the same side of the bond. In a trans configuration, they are on opposite sides. This is similar to the cis/trans nomenclature of cyclohexane (see section two). To name the compound, add ‘cis-‘ or ‘trans-’ to the IUPAC name: Br
Br
H
Br
H
H
Br
H
cis-1,2,-dibromoethene
trans-1,2,-dibromoethene
Things get more complicated when the carbons are attached to three or four different groups. When this happens, we must use E/Z nomenclature. The first step is to assign priorities to the groups on each carbon, using atomic number (higher atomic number = higher priority). When the first atoms in each group are identical, we work our way through the chain, looking at every bond. In a Z configuration, groups with similar priorities are on the same side (similar to cis). In an E configuration, groups with similar priorities are on opposite sides (similar to trans). Example one: H3C
CH3
Br
Cl
Br has an atomic number of 35 and C has an atomic number of 6. This makes Br the high priority and C (CH3) the low priority group for the first carbon. Cl has an atomic number of of 17, making it the high priority for the second carbon. low
high
H3C
Br
CH3
Cl
low
high
3-45
This molecule is in a Z configuration. Example two: H3C
CH3
Br
CH3
When we look at the substituents on the second carbon, both begin with C (atomic number = 6). We need to go further into the substituents to assign priorities. In the CH3 group, carbon is attached to H(1), H(1), and H(1). In the CH2CH3, carbon is attached to H(1), H(1), and C(6). This gives CH2CH3 a higher priority. low
high
H3C
low
CH3
Br
CH3
high
This molecule is in a Z configuration. Example three: Br
CH3
H3C
CH2
Once again, the atoms directly attached to the second carbon are identical, making it necessary to go farther into the substituents. In CH2CH3, the carbon is attached to H(1), H(1), and C(6). In CH==CH2, we treat every bond as a separate group – that is, we describe the carbon as attached to H(1), C(6), and C(6). high
Br
low
H3C
CH3
CH2
low
high
3-46
This molecule is in the E configuration. *Note about stability: as we learned in section two, organic molecules like to keep bulky and electron-rich groups as far apart as possible. This means that trans and E isomers will generally be more stable than cis and Z isomers. The carbons involved in double bonds are sp2 hybridized (see section two). Remember that sp2 carbons’ bonds are planar! The bond angles in a double bond are approximately 120˚ (carbon is attached to three substituents, and divides the full circle of 360˚ among them). In ethene, the simplest alkene, bond angles are exactly 120˚. The C’s in double bonds, as well as groups attached to them, are called vinyl C’s and groups. The C’s adjacent to double bonds, as well as groups attached to them, are called allyl C’s and groups. allyl Cl
vinyl Cl allyl C
vinyl C
H2C
CH
CH3
H2C
Cl
Cl
C
CH 2
Basic Reaction Mechanism Throughout this course, most of the alkene reactions you encounter will change alkenes to alkanes, or convert the double bond into a single bond. Each carbon needs to find something else to hold on to for this to work. The basic reaction here is called electrophilic addition. Double bonds are made up of electrons, which are negatively charged. As a result, bonds are attracted to positively-charged (or at least electron-poor) atoms. The electrons of the double bond attack an electrophile, or electron-loving group, in the first step of this reaction:
3-47
El H2C
El
+ +
CH2
H2C
CH2
By the end of this step, one carbon still has four bonds (it’s simply moved the double bond into a different position). The other carbon, however, has lost a bond. Losing contact with electrons (negatively charged) leaves this carbon with a positive charge – it has become a carbocation. Conversely, he electrophile loses its charge when it bonds with electrons. Note that net charge is conserved – we began with one positive charge, and we end with one positive charge. Now, a nucleophile (attracted to positive charges, like the protons in an atom’s nucleus) steps in to finish the job:
Nu
El
El
Nu+
H2C
CH 2
H2C
CH 2
The rest of the reactions we explore in this section will follow this basic pattern (double bond attacks electrophile, C+ forms, nucleophile attacks C+). Mechanisms vary somewhat among reactions. HX (HBr, HCl, HI) and Some Complications HBr follows the basic reaction with H+ as an electrophile and Br- as a nucleophile: H
+
H H2C
CH2
+
H2C
CH2
Br
H2 C
H CH2
-
Br
3-48
This is all well and good when working with ethene, but what if the alkene isn’t symmetrical? What if the two carbons are different? To solve these problems, we need to look at carbocation stability. Carbocations are positively charged. Ideally, molecules want to share their charges among several atoms (see section six on resonance) or at least balance them out. Remember that like charges repel. A C+ surrounded by hydrogen atoms, which are basically protons (H’s electron is busy maintaining its bond to carbon), will be unstable. A C+ surrounded by a mixture of carbons and hydrogens, or all carbons, will fare better. Carbocations, like neutral carbons, are classified as methyl, primary, secondary, or tertiary, depending on how many carbons they are directly attached to. CH3 H3C
+
C
CH3
>
H3C
tertiary most stable
+
CH
>
CH3
+
H3C
CH2
primary
secondary
>
+
CH3
methyl least stable
Let’s apply this to a reaction. If I want to react HBr with CH3CH==CH2, the reaction can precede in one of two ways: H H3C A
+
H H3C
CH
-
+
Br
H3C
CH CH2 primary (less stable)
H
Br
CH
CH 2
minor product
CH2
H
B H3C
+
CH CH 2 secondary (more stable)
-
Br
H
CH
CH 2
Br
H3C
major product
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In reaction A, H+ adds to the secondary carbon. As a result, the primary carbon loses a bond and becomes positive. Primary carbocations are very unstable. In reaction B, H+ adds to the primary carbon. As a result, the secondary carbon loses a bond and forms a more stable, secondary carbocation. Sounds like reaction B is the winner! This being the case, H+ will show regioselectivity, or a preference for one particular carbon in the double bond (regio = region, H+ selectively adds to one particular region of the molecule). The product formed by the more stable reaction pathway, reaction B in this case, is the major product. The product formed by the less stable reaction pathway is the minor product. A chemist named Vladimir Markovnikov noticed this pattern, and produced what has become known as Markovnikov’s Rule: H adds to the carbon with the most H’s on it (the least substituted carbon). He was slightly mistaken. As we will see, the electrophile in a reaction adds to the carbon with the most H’s, and H is not always the electrophile. Reactions are described as either Markovnikov or anti-Markovnikov reactions, depending on whether they follow this rule. Sometimes, in a more dramatic effort to form the most stable C+ possible, molecules will move H’s and C-groups around. This is called rearrangement. If possible, it is best to move an H (along with its bond to C) – this is known as a hydride shift. Otherwise, it is possible to move a CH3 group (methyl shift), a larger carbon group, or even to add another carbon to a ring. In this last type of rearrangement, known as a ring expansion, five- and six-membered rings are typically formed. The atom that gains a bond (C+) becomes neutral, and the atom that loses a bond becomes positively
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charged. To determine whether rearrangement will happen, add H+ to form the most stable carbocation possible, and then see whether adjacent carbons are more substituted. 1) Hydride Shift +
H
CH3
CH3 H3C
CH
CH
H3C
CH2
+
CH
CH
tert
sec
CH 2
CH3
CH3 +
H3C
H
CH
CH3
+
H3C
C
CH
H
CH3
H
2) Methyl Shift +
H
CH3
CH3 H3C
CH
+
H3C
CH2
CH
quat
CH3
CH 2
sec
CH3 CH3
CH3 +
H3C
H
CH
CH3
+
H3C
C
CH
CH3
CH3
CH3
3) Ring Expansion (this is the hardest one!) H3C
H3C CH2
H3C
H3C +
CH
CH3
CH3 CH3
+
C
Br2 and Avoiding Rearrangement It is possible to add Br2 to a double bond. This may seem strange, as Br2 does not have a clear electrophile or nucleophile – the electrons are evenly split by both atoms!
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Remember that like charges repel. When the pi bond attacks, it pushes the electrons in the Br—Br bond away, creating partially positive (δ+) and partially negative (δ-) sides. The alkene then takes the δ+ atom, separating it from the δ- atom. This type of bondbreaking, in which one atom claims both electrons, is called heterolytic cleavage.
Br H2C
Br
and
CH CH3
δ+
δ-
Br
Br
Br +
H2C
CH
CH3 H2C
CH CH3
Br is a halogen, meaning that it has seven valence electrons. One of these is lost by the electrophile when the Br-Br bond breaks, but that still leaves six electrons (or three lone pairs). They are attracted to the C+, and attack it. This forms a triangle intermediate. Although the molecule will go back and forth between linear and triangular shapes, the triangle is the most stable (and thus dominates). This prevents rearrangement by keeping the positive charge away from carbon. +
Br H2 C
Br
Br +
CH
CH3
+
H2 C
CH CH3
H2C
CH CH3
Finally, the nucleophile attacks. If Br2 is the only nucleophile source in the mixture, then a Br- (left over from the beginning) will attack the most stable carbocation – in this case, the secondary carbon. If another nucleophile source, such as OH- or Cl- is present (often introduced via HOH and NaCl), that nucleophile will typically attack.
3-52
+
Br H2C
Br CH
H2C
CH3
CH
CH3
Nu-
Nu-
Water (HOH) and Alcohols (ROH) 1) With Acid Think back to our review of general chemistry – an acid is a proton (H+) donor. As such, when an acid is present in our solution, chances are that the donated H+ will be our electrophile. Commonly used acids include HCl (H+Cl-) and H2SO4 (H+HSO4-). Sometimes, the H+ is written on its own, rather than as part of a larger compound. These reactions follow the basic reaction mechanism, with the added step of removing an H+ at the end. Let’s figure out the mechanism and major product for this reaction: HOH H3C
CH
CH2
HCl
?
+
+
H H3C
CH
H OH H3C
CH2
+
H OH Cl H3C
CH
CH3
+
CH
HOH CH 2 H
OH
-
H3C
CH
CH3
H3C
CH
CH3
oxygen loses electrons in attacking, making it positive
Some teachers want you to show a base remove the extra H+ ("B," "base," or the base that was attached to H+) . Others will either accept a blank reaction arrow or the description "- H+"
To perform this reaction with an alcohol, simply substitute a carbon group for one of the H’s in water! 2) With Mercury: Avoiding Rearrangement
3-53
As we saw with HBr, any reaction that follows the basic mechanism risks rearrangement – but, as we saw with Br2, it is possible to avoid rearrangement by forming a triangle intermediate. Adding water or alcohol with mercury, a process called oxymercuration for water and alkoxymercuration for alcohols, accomplishes the same thing. Mercury acts as an electrophile at first, and must be replaced with an H. This process is called reduction. Overall: oxymercuration-reduction (what a mouth full!). Here is the process, again using water as an example. OH
1. Hg(OAc)2, HOH/THF H3C
CH
H3C
CH2
CH
CH3
*THF is a solvent
O
*Ac stands for Acetate:
C CH3
2. NaBH4
OAc
OAc Hg H3C
CH
Hg
OAc CH2
H3C
HC
CH OH
HgOAc H3C
CH 2 H OH
CH
CH 2
+
-
O Ac
HOH
OH
HgOAc H3C
+
CH 2
H3C
CH
CH3
H NaBH3
3) With Boron: Anti-Markovnikov! This last reaction is for water only. Every now and then, we might want to add H to the least substituted carbon. You can think of it as adding water backwards, the wrong way, an unusual way – or in an anti-Markovnikov reaction, if you’re feeling fancy. In order to pull this off, we need to find something else to act as an electrophile – and BH3 is more than willing to do so. This reaction is called hydroboration/oxidation. Generally, you will only need to know which reactants to use for this process:
3-54
1 BH3/THF H3C
CH
H3C
CH2
H
OH
*THF is a solvent
CH
CH 2
*In step 2, HOH is not always written out. HO- and H2O2 make HOH
2. HO-, H2O2, HOH
Some teachers may want you to know the mechanism. It’s long, but manageable. Boron (B) basically acts as an electrophile, and alkenes keep attacking until there isn’t any room left. Each time, a ring of partial bonds forms before B and H officially join the molecule.
H3C
CH
CH2
H--------BH2
----
BH2
----
H
CH-------CH 2
H3C
----
----
H--------BHCH2CH 2CH 3
CH-------CH 2
H3C
H3C
----
----
H--------B(CH 2CH 2CH 3)2
CH-------CH 2
H3C
H3C
H3C
H
BH2
CH
CH 2
H
BHCH2CH 2CH 3
CH
CH 2
H
B(CH 2CH 2CH 3)2
CH
CH 2
Then oxygen steps in. To save space, we will use ‘R’ to represent the carbon chains attached to B – feel free to do that during an exam as well, as long as you draw out the final product!
R
BH2
R
R O
B
-
OH
R
R
-
B
O
R
OH
B
O
R
R O R
(repeat for remaining R groups) OR RO
B
R
O
B
OR -
HO
(repeat for remaining OR groups)
RO
B
R OR
OR
OR -
O
OR
RO
B
-
OR
RO
HO R
B -
OH
OH
O
3HOR + BO33-
Hydrogen 3-55
This reduces a double bond to a single bond without adding functional groups. Chances are that you won’t need to learn the mechanism for this reaction during the first year of organic chemistry (and the only year, for some of you!), so we will just list the catalysts involved: Pd/C, Pt, Ni H2 H2C
CH2
H3C
CH3
Breaking the Markovnikov Rule with HBr It is possible to add HBr in a way that places the H on the most substituted carbon. The mechanism for this involves radicals, or unpaired electrons. For the time being, we will just learn which reactants are involved. See section eight for information on radicals, including the mechanism of this reaction. H
Br
CH
CH 2
HBr H3C
CH
CH2
H3C RO O R (peroxide)
*See section five for the stereochemistry associated with alkene reactions!
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Alkynes This is one of the shorter chapters and, as far as learning reactions goes, tends to be one of the easier ones – quite a few alkyne reactions are basically alkene reactions that happen twice (alkynes include triple bonds, and each pi bond can react). Before we get into that, let’s take a look at names at structures for alkynes. Chapter 4: Nomenclature and Structure If a triple bond is in the parent chain of a molecule, indicate the location of the triple bond and use the suffix ‘yne.’ Be careful to use the lowest number possible! Alkynes can be terminal (triple bond at the end of the molecule) or internal (triple bond in the middle of the molecule). Sometimes, alkynes are referred to as acetylenes, or substituted versions of acetylene. Acetylene is a common name for the simplest alkyne – two carbons with a triple bond between them. HC
C
CH 2
CH3
1-butyne terminal
H3C
C
C
CH3
2-butyne internal
HC
CH acetylene (ethyne)
The carbons in a triple bond are sp hybridized (see section two). The bond angles surrounding the triple bond are 180˚. Addition of HX (HBr, HCl, HI) This works just like the alkene HBr reaction (see section three), except that it is possible to react twice. To ensure two reactions per alkyne, use excess HX. To try and limit the number of reactions, use a 1:1 ratio of HX: alkynes. Look out for regioselectivity (preference for adding the electrophile to a specific C – the least substituted one) and rearrangement. This reaction follows Markovnikov’s Rule (H to least substituted C). When two reactions occur, the product is a geminal or vicinal dihalide.
4-57
H3C
C
CH
Br
H
C
CH
H
+
H
H3C
+
C
CH
H3C
-
Br
+
H H3C
Br
H
Br
Br C
CH2
H3C
+
C
CH 2 Br
H3C
C
CH3
-
Br
The vinyl cation (positive charge on an atom in a double bond, created by first step) is very unstable and reacts quickly. Br2 Once again, we have an alkene-style reaction happening twice. Adding Br2 to an alkene resulted in once Br on each carbon; adding to an alkyne will result in two Br’s on each carbon. We can use a ratio of 1:1, Br2:alkyne to try and limit each alkyne to one reaction, or use excess Br2 to ensure two reactions. Br Br
+
Br H3C
H3C
C
CH
Br
Br
C
CH
H3C
H3C
C
CH
Br
Br
C
CH +
Br Br
Br
H3C
H3C
Br
Br
C
CH
Br
Br
C
CH
Br
Br
Water (HOH) Unlike HBr and Br2 (and versions with different halogens), HOH adds to alkynes once. This addition is followed by what is called tautomerism – think of it as shifting the double bond and redistributing H’s to balance that out:
4-58
O
OH HC
tautomerism
HOH
CH
(catalysts, other reactants)
H2C
H3C
CH
CH ketone
enol
At first, water is divided up as we would expect it to be – one carbon gets an H, the other gets an OH. The product is called an enol. Then, the double bond moves between carbon and oxygen. The other carbon gains an H to make up for the lost bond. Oxygen loses an H to make room for the new bond. The product is called a ketone. Molecules tend to go back and forth between these states, with the ketone dominating. That process is called keto-enol tautomerism. See section X for the mechanism. In terms of regioselectivity, stick with the old rules for alkenes. Acid and mercury follow Markovnikov’s Rule; boron breaks it. If you use boron with a terminal alkyne (or use any reactants with ethyne), the C==O group winds up on the last carbon. This product is called an aldehyde: OH
O
+
tautomerism
HOH, H H3C
C
CH
H3C
C enol
CH2
H3C
OH H3C
C
CH
HOH, HgSO4, H2SO4
C ketone
CH3
O tautomerism
H3C
C enol
CH2
H3C
C ketone
CH3 O
OH
H3C
C
CH
1. BH3/THF 2. HO-, H2O2, HOH
tautomerism H3C
CH enol
CH
H3C
CH 2
CH
aldehyde
Hydrogen Adding hydrogen to a triple bond can achieve three possible results: 1) an alkane, 2) a cis alkene, and 3) a trans alkene. The reactants from section three on alkenes will produce an alkane; others are necessary to stop the reaction after making an alkene:
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H2 H3C
C
C
CH3
H3C
CH 2
Pt, Pd/C, or Ni
CH 2
alkane H
H3C
C
C
CH3
H
H2
Lindlar's Catalyst
cis
H3C
C
C
CH3
CH3
H3C
Li or Na H3C
CH3
NH3
H trans
H
CH3
Lengthening the Carbon Chain The more pi bonds between carbons, the more acidic the molecule. This makes alkynes better H+ donors than alkenes and alkanes. Although hydrocarbons rarely give up protons without a fight, strong nucleophiles can steal H+ from terminal alkynes. This leaves carbon with a negative charge. Carbon isn’t too fond of being negative, and will quickly attack another molecule to become neutral again. If it attacks an RX molecule, or a carbon chain attached to a halogen, a new carbon-carbon bond forms! Organic chemists are obsessed with making C-C bonds – be sure and remember this for your exam! Here is the process, using NH2- (likely from NaNH2) as the base:
H3C
C
CH
NH2-
H3C
C
-
C
H3C Br H3C
C
C
CH3
Chances are that your teacher will want you to learn several specific strong bases. In general, bases with a negative charge on nitrogen (including NaNH2 and LDA) will work. This reaction is called an SN2 reaction (S for substitution, as the alkyne group takes the place of Br; N for nucleophilic, as the negatively charged alkyne is a
4-60
nucleophile; 2 for a second-order reaction). See section nine for more information on this type of reaction. Multi-Step Synthesis At this point, we are prepared to do multi-step synthesis reactions. Synthesis questions require you to figure out how to make a specific molecule. Teachers may give you starting materials. To solve synthesis problems, you can either work forwards (from starting materials to final product) or backwards (from final product back to starting materials). Working backwards to solve synthesis problems is called retrosynthetic analysis, and is indicated with double-lined arrows:
. This technique is very
useful with multi-step problems, when it can be hard to see the connection between initial and final materials. *Whenever you think you might be working with alkynes, check and see whether the number of carbons in a chain increases. If it does, be sure to lengthen the chain first!* 1. Forwards: Starting Materials to Final Product H3C H3C
C
CH3 C
CH
changes: - from 3C chain to 4C chain - from alkyne to cis alkene
C
H
H
1) lengthen C chain -
NH2 H3C
C
CH
H3C
-
C
H3C
Br H3C
C
C
C
CH3
2) alkyne to cis alkene
H3C
C
C
CH3
H2 Lindlar's catalyst
H3C
CH3 C
H
C H
2. Backwards: Final Product to Starting Materials (Retrosynthetic Analysis)
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O HC
CH
H3C
CH 2
aldehyde: must have used BH3 to add HOH to an alkyne
CH
O H3C
CH 2
CH
H3C
C
CH
3C chain: must have added CH3 to acetlyene (ethyne)
-
(1) BH3/THF (2)HO , H2O2
H3C
C
CH
HC
CH
(1) NaNH2 (2) CH3Br
4-62
Chapter 5: 3D Orgo: Stereochemistry Chances are that you’ve met this concept before. Stereochemistry involves the spatial arrangements of molecules, including the placement of atoms within molecules and the interactions of different molecules/atoms/particles in space. Cis/trans and E/Z nomenclature in alkenes are a part of stereochemistry (see section three). When we looked at molecules in Newman projections (section two), we stumbled across the idea of steric strain, that is, groups of atoms sharing space and getting in each other’s way. This section will take 3D chemistry farther. Our main focus involves chiral centers, or atoms with four different groups attached to them. Chiral carbons, which are also called asymmetric carbons, can be abbreviated as C*.
Drawing Chiral Centers Remember that nifty model set your teacher keeps pressuring you to use? It’s time to break it out! This section of the course lends itself very well to model use, especially for students who have trouble visualizing things. Take a carbon and attach four different-colored groups to it. It doesn’t matter what the groups are, we are merely trying to create a chiral center to work with. 1) Perspective Formula Take the chiral center and arrange it so that two groups are in the same plane – one pointing straight up, one pointing down and to the side. Of the remaining groups, one should point away from you and one should point towards you. Your molecule is now arranged in line with the perspective formula, and should look something like this:
5-63
Y
*a dark triangle, or wedge, represents something pointing up or towards you
Z
*a dotted line, or hatch, represents something pointing down or away
X C W
Y
CH3
Z
Perspective formulas can be incorporated into rings as well: 2) Fischer Projections Time to move the model again! Rotate it so the molecule looks sort of like a + sign, with two groups are pointing up or towards you, and the remaining two groups are pointing down or away from you. The groups pointing towards you should be on the horizontal line of the + sign, the groups pointing away should be on the vertical line. Now, if we replace the dark triangles and dotted lines with straight lines – remembering that the groups on the vertical line point away, and that the groups on the horizontal line point towards us – we will have drawn a Fischer Projection of the molecule.
X Y
C
X
Z
W
is drawn as
Y
C
Z
W
R, S Nomenclature Like alkenes, chiral centers can be spatially arranged in two different ways. They are called R and S configurations. Here is the process of labeling: 1) Assign priorities to each group, using atomic number (same method as E,Z) 2) Draw an arrow from priority 1 2 3
5-64
3) Determine whether the arrow is clockwise or counterclockwise. If priority 4 points away (hatch or dotted line in perspective, vertical line in Fischer), a clockwise arrow means R and a counterclockwise arrow means S. If priority 4 does not point away, reverse the answer (then a clockwise arrow means S, and a counterclockwise arrow means R). 3
3
CH3
CH3 H C
H
:
C
H
OH
:
4
C
Cl
OH
Cl
OH
1
2
1
2
CH3 Br
H
:
C
OH
Cl
CH3
4
1 Br
3
3
CH3
CH3
C
H4
:
1 Br
C
OH
OH
2
2
H4
counterclockwise arrow. Priority 4 is pointing away/down, so the counterclockwise arrow means that the molecule is in an S configuration.
counterclockwise arrow. Priority 4 points up/towards us, so we reverse the answer: a counterclockwise arrow now means that the molecule is in an R configuration.
*Trick: To change any R molecule to an S (or vice versa), just switch any two groups! Try it out! Take the examples we just looked at and move the groups around! R and S versions of the same molecule are called enantiomers. They are a type of stereoisomers (see section two), with the same kinds of atoms connected in the same way, but arranged differently in space. Enantiomers are mirror images of each other, but non-superimposable. For a visual comparison, think of two hands or a pair of gloves. They are mirror images of one another; if you hold a right glove/hand up to a mirror, the reflection will look like the left hand/glove. Yet if you try to put your left hand on top of your right hand, they won’t match up. Enantiomers are the same way.
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Sometimes, you can tell simply by looking at a pair of molecules that they are enantiomers (chiral and clearly mirror images). Other times, you need to determine the R and S configurations. Cl H
Cl
C
CH3
H3C
C
H
vs.
H
H
C
CH3
Cl
C
OH
OH
OH
OH
R
S
R
S
H3C
CH3 H
R
H3C
Br
C HO
Cl
vs.
C Br
HO
S
H
CH3
H
H
Br
Br
C HO
R
CH3
C S
OH
To add R,S nomenclature to the name of a compound, add the appropriate letter(s) to the name of a the molecule in parenthesis: (R)1-bromo-ethanol. Molecules with Two Chiral Centers When molecules have two chiral centers, several configurations are possible: RR, SS, RS, and SR. As with individual centers, we can get the mirror image of a molecule by changing all of the chiral centers. RR and SS versions of the same molecule are enantiomers; RS and SR versions of the same molecule are enantiomers. OH
H
H3C
C
H
H3C
C
OH
Cl
C
H
Cl
C
CH3
CH3 S,S
H R,R
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If only one chiral center is different (e.g. RR and RS), then we have diastereomers or diastereoisomers. OH
H
H3C
C
H
H3C
C
OH
Cl
C
H
Cl
C
H
CH3 S,S
CH3 R,S
If both chiral centers are attached to identical groups (a methyl, an H, and a Cl, for example), then the R,S and S,R versions will be identical – the molecule will be its own mirror image! Such molecules are called meso compounds. Teachers absolutely love to force you to identify meso compounds (common questions ask students to distinguish enantiomers, diasteomers, identical compounds, and meso compounds from one another). Remember these characteristics in trying to identify meso compounds: 1) 2 chiral centers 2) each C is attached to the same groups 3) the molecule is in an S,R or R,S configuration There will be a line/plane of symmetry running through the compound (although you may not be able to tell by looking at the diagram!).
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H
CH3
CH3
C
C
- 2 chiral centers - each C* attached to Cl, CH3, H H
Cl
Cl
- R,S configuration (left-right) - Meso Compound!
OH H
CH 2CH 3
- 2 chiral centers - each C* attached to OH, CH2CH3, H
H
CH 2CH 3
- S,R configuration (top-bottom) - Meso Compound!
OH CH3
H3C
not a meso compound! teachers try to trick students with molecules like this...although there is a line of symmetry, there are no chiral centers in this molecule!
Creating Chiral Carbons 1) Replacing Hydrogen For now, we are going to keep this theoretical. You will likely encounter carbons that have two H’s and two other, different groups attached to them. If one H is replaced, we might create a chiral center! The type of chiral center we create will depend on which H is replaced. Consider the following example: CH3
CH3 Ha
replace an H with Cl
C Hb
CH 2CH 3
CH3
Br
+
C Hb
CH 2CH 3 R
Ha C Br S
CH 2CH 3
If we replace Ha, we create an S molecule. This makes Ha a pro-S hydrogen. If we replace Hb, we create an R molecule, making Hb a pro-R hydrogen. Since replacing these hydrogens with another atom/group will create enantiomers, they are called enantiotopic hydrogens. We can perform similar replacements on molecules that
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already have one C* (making a second C*). This creates diastereomers – as such, the hydrogens we replace to do this are called diastereotopic hydrogens. Ha H3C
Ha Hb
HO
replace an H with Cl
H3C
Cl
+
HO
CH3
Cl H3C
CH3
Hb
HO
CH3
H
H
H
R
R,R
S,R
2) Alkene Reactions Adding to alkenes can create stereocenters. Carbons that were initially bonded to only three groups (thanks to the double bond) are suddenly bonded to four. Sometimes, any combination of R and S chiral centers is possible. Other times, we are limited by the directions from which groups attack/add to the alkene. In syn addition, both groups add to the same side (or attack from the same direction). In anti addition, the groups add to different sides. Carbocations are planar, or flat. As such, whenever a carbocation is created, the nucleophile can add from either side of the molecule – both syn and anti addition are possible! To determine how many products to draw, pay attention to the number of chiral centers that are formed during a reaction. H
H
H HBr H
H3C
CH3
+
C
C
CH3
CH3
H
CH3
- 1 chiral center formed - Br- can attack from either side, making S or R chiral center - two products (enantiomers)
H
Br
C
C
CH3
CH3
H
CH3 Br
C H3CH 2C
H
H
+
Br C H
CH 2CH 3
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Reactions with triangular intermediates, such as adding Br2, are more controlled. Electrons repel one another. When the Br- gets ready to attack the molecule, it will want to do so as far away from the other Br atom as possible (although Br is positively charged in this intermediate, it still has two lone pairs to repel Br- with!). Br2 addition is therefore an anti addition. The other groups on the alkene do not move during the addition of Br-. To draw your final product, just keep the other groups’ relative configurations the same (I usually look at the groups that were pointing up, and have them point up again in the final product) and make sure that the Br’s are pointing in opposite directions! +
H
Br
CH3
H3C
H
Br2
CH3
H3C
CH 2CH 3
-
H - 2 chiral centers
CH 2CH 3
Br
H
CH3
H3C
Br
+
- Br's in opposite directions Br
- two possibilities: R,R and S,S
CH 2CH 3
CH3
Br
CH 2CH 3
H3C
Br
A Note on Cumulenes Cumulenes are molecules that have consecutive double bonds. The type of structural arrangement alternates between E,Z and R,S as you lengthen the chain: X
W
W C
Z
E,Z
Y
Z
X Y
R,S
X
W C Z
C E,Z
Y
Chiral Centers with Different Atoms Although most chiral atoms are carbons, other atoms – nitrogen in particular – can also be treated as chiral. If we treat the lone pair in nitrogen as a substituent, than we can treat a nitrogen atom bonded to three different groups as a chiral center. The
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configuration of N* is unstable. This is because the lone pair (which takes up less space than a real substituent) switches positions; the molecule continually flips inside-out, like
CH2CH3
N H3C
H
H3C
H N
:
:
an umbrella in a storm.
CH2CH3
Light Rotation Chiral molecules are optically active. This means that they can rotate a plane of polarized light. Molecules that rotate light in a clockwise or positive (+) direction are dextrorotatory; molecules that rotate light in a counterclockwise or negative (-) direction are levorotatory. To remember this, you can think of counterclockwise rotation as rotation to the left – ‘left’ and ‘levorotatory’ both begin with ‘l.’ Enantiomers rotate polarized light in different directions. As of now, however, there is no way to tell which enantiomer will rotate light in which direction without experimenting. Meso compounds (which lack enantiomers/are their own enantiomers) do not rotate light. Solutions with equal amounts of R and S molecules, called racemates, do not rotate light either. We can use data about light rotation to determine the relative amounts of S and R molecules in a solution. Every chiral compound has a specific rotation – a specific degree to which it rotates light. If the R enantiomer of a compound has a specific rotation of (+)25º, then the S enantiomer has a specific rotation of (-)25º. This is why racemates do not rotate light – the enantiomers’ specific rotations cancel each other out. If we know a solution’s observed specific rotation (the degree to which light rotates as it moves through a solution), we can determine the solution’s optical purity, or
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whether there is an enantiomeric excess of the R or S molecule. Both of these concepts tell us whether one enantiomer outnumbers the other, although they have slightly different formulas. When a sample is 60% optically pure, it is 60% R or S and 40% racemate; in other words, a 60% pure sample would actually include 80% of one enantiomer and 20% of the other. In the same solution, we would say that there is a 60% excess of one enantiomer. To calculate optical purity: specific rotation for the solution % Optical Purity = 100% x specific rotation for pure sample By ‘pure sample,’ we mean a solution in which there is only one enantiomer present. Let’s try some examples. 1) A 60% optically pure solution has a specific rotation of 33.4º. If the R enantiomer is levorotatory, what is it’s specific rotation? First, let’s fill in the details we have for this reaction: 33.4 Optical Purity = 60% = 100% x specific rotation for pure sample 33.4 0.6 = Y ; Y = (+/-) 92.78 Since levorotatory is counterclockwise and negative, the specific rotation of the R enantiomer is (-)92.78. 2) What is the observed rotation for a 30% optically pure solution of a chiral compound if the S enantiomer has a specific rotation of 104.0º? 30% = 100% x
specific rotation for the solution 104.0
Y 0.3 = 104.0 ; (0.3)(104.0) = Y = 31.2º
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The formula for enantiomeric excess, or ee, looks like this: [R] - [S] ee % = [R] + [S] Use the absolute (+) value of your answer, or reverse the numerator (use [S] – [R]) if S enatiomers are more prevalent than R enantiomers. You can use actual concentrations (in moles) or percentages for this formula. Let’s do some a sample problem! If an R enantiomer is in 80% excess, what is the total percentage of R enantiomers in the solution? Using percentages, ([R] + [S]) has to equal 100%. Let’s look at our formula: [R] - [S] R-S 80% = [R] + [S] = 100% We know that (0.8 = R – S) and that (R + S = 1.0). S = R – 0.8 R + S = R + (R – 0.8) = 1.0 2R – 0.8 = 1.0 2R = 1.8 R = 0.9 90% of this solution is R! Alternatively, we could have talked our way through that one. If the solution has an ee of 80%, then 20% of it is racemic. The racemic mixture breaks down into 10% R and 10% S. Our total of R = 80% + 10% = 90%.
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Some folks like to look at these problems and solve them without putting pen to paper, others prefer to work with the formulas. Figure out what works best for you, and go for it!
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Chapter 6: Traveling Electrons: Resonance Resonance, or electron delocalization is the movement of electrons within a molecule. Most of the electrons we have encountered have been stuck in one place (unless they were reacting), but some are able to move among bonds and atoms. It is important to remember that the net charge of the molecule stays the same. Individual atoms may acquire or lose charges, but resonance does not change the total count of electrons in a molecule.
Identifying Resonance Systems A resonance system is a group of atoms that share electrons. Some resonance systems encompass whole molecules, others only sections of molecules. To participate in resonance, bonds/atoms must either have electrons to donate or the ability to take on new electrons. They must also be a specific distance from other bonds/atoms that can share electrons. More specifically, bonds/atoms in resonance systems must be conjugated, or separated by one single bond.
Resonance Systems
Yes
No Charges (represent electrons that can move, or room for new electrons): C-, C+ Atoms with lone pairs: N, O, S Radicals: C· Conjugated double bonds: C==C—C==C
Neutral sp3 carbons (C with four single bonds): these are walls that electrons can not pass through! Cumulated double bonds: C==C==C Isolated double bonds: C==C—C—C==C
6-75
Drawing Resonance Structures Electrons in resonance are constantly moving, and it is impossible to really pinpoint their locations. How, then, can we draw the molecules? We can draw what are called resonance contributors, the structures we would find if electrons stopped moving and momentarily formed stable bonds, lone pairs, etc. Indicate electrons’ movement using single-headed arrows for individual electrons and double-headed arrows for pairs of electrons. Contributors are separated by these arrows:
.
Now that we’ve talked this into the ground, let’s look at some structures! +
H2C
CH2
+
H2C
CH 2
CH
+
+
H2C
H2C
CH2
CH2
+
H2C
CH 2
CH2
CH 2
+
CH2
CH
H2C
CH2
*treat sp3C like a wall!
*e-1s in negative charges can move to bonds *e-1s in bonds can move to atoms or to other bonds
-
H2C
-
H2C
or*: H C2
CH2
-
H2C
CH2
CH2
CH
H2C
CH2
-
H2C -
CH2
CH2
H2C
-
CH
*when a system includes more than three atoms, you can either break it down into three-atom sections or try and move electrons throughout the whole molecule. Either way, be sure to check and draw all possible structures!
________________________________________________________________________
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CH2
H N
H N
:
:
H+ N
H+ N
H+ N
-
HC
-
CH
-
:
:
:
CH
N
N
N
Generally, atoms only donate one pair of electrons at a time. Since N is donating its double bond, the lone pair stays put.
Stability Resonance is good! Molecules benefit from the ability to move their electrons around – otherwise, chances are that resonance would be rare or nonexistent. You may be asked to determine which resonance contributors are the most stable within a system. Here are some things to keep in mind: -
separated charges are less stable
-
positive charges: N+ > C+ > O+
-
negative charges: O- > C- > N-
-
more substituted (surrounded by C’s) double bonds are more stable
Acidity Acids, as we learned in section one, are H+ donors – and strong acids are molecules that are particularly good at this. In other words, a strong acid will create a stable molecule by donating that proton; otherwise, the resulting base might try and steal the proton back! As resonance makes molecules more stable, a base that can move its electrons around will be more stable than one that can not. An acid whose conjugate base has resonance (see below) will thus be a strong acid.
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H3C
OH
OH
H3C
O
O
-
negative charge stuck on O, less stable base (weaker acid)
-
negative charge can move into/through the ring, more stable base (stronger acid)
Molecular Orbitals We’ve already seen that resonance increases stability. Back in section two, we learned that molecular orbitals are arranged in a hierarchy of stability, with electrons usually occupying the most stable/lowest energy orbitals. Combining these two concepts – we wind up with greater stability for molecular orbitals, and a new structure for the hierarchy. Let’s look at two molecules that have six pi electrons (in three double bonds): 1,3,5-hexatriene and benzene. The first molecule’s orbitals will follow the same pattern that we encountered in section two. Benzene’s molecular orbitals will present a different pattern, thanks to resonance and greater stability. Of all molecules with resonance, benzene is the one you are most likely to see in an exam question about molecular orbitals – chances are that you’ll learn a lot about its reactions as well! See sections twelve and thirteen for more information.
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*LUMO (lowest unoccupied molecular orbital) CH2
H2C
*HOMO (highest occupied molecular orbital) 1,3,5-hexatriene
*tied for LUMO
*tied for HOMO benzene
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Chapter 7: Two Double Bonds! This section combines our knowledge of alkene reactions (section three) with our newly acquired knowledge of resonance (section six). We will focus on conjugated double bonds, which are separated by one single bond. When naming a compound with multiple double bonds in its parent chain, add a prefix before ‘ene’ to indicate the number of double bonds. As always, indicate the bonds’ locations with numbers – using the lowest numbers possible. Remember that each double bond can be either E or Z! isolated:
H2C
*conjugated: H2C
cumulated:
H2C
CH3
(1,4-hexadiene)
CH3
(1,3-hexadiene)
CH3
(1,2-hexadiene)
Two Types of Addition When working with a conjugated diene, it’s possible to react with one bond individually, just like we did in section three on alkenes. This is called 1,2 addition or direct addition (if we treat the diene system as a four-carbon chain, we are adding to carbons one and two). -
Br
H Br H2C
CH
H
H CH
CH2
H2C
+
CH
CH
CH2
H2C
Br CH
CH
CH2
Check this out, though – the carbocation we formed during this reaction is conjugated with the remaining double bond: we have formed a three-carbon resonance system! The electrons may move before we have a chance to add Br-, leaving us with a C+ on the fourth carbon. This gives us 1,4 addition or conjugate addition.
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-
Br H Br H2C
H
CH
CH
CH2
H H2C
H2 C
H +
CH
CH
CH2
H2 C
CH
+
CH
CH2
Br CH
CH
CH 2
How can we tell which reaction is going to dominate? That depends upon the conditions. Direct or 1,2 addition is a faster reaction: Br- doesn’t have to move very far to get to the C+ (this is referred to as proximity). That makes the 1,2 adduct the kinetic product (or the product favored by kinetic conditions). In this case – and in most cases – the conjugate or 1,4 adduct is the most stable product (even though the C+ in this reaction is less stable). The double bond is more substituted. As such, the 1,4 adduct is the thermodynamic product (or the product favored by thermodynamic conditions). For the most part, the major product in reactions will be the thermodynamic or more stable final product – but sometimes teachers will specify kinetic or thermodynamic conditions. Pay attention to this! Diels-Alder Reaction The rest of this section will involve a very important reaction that you can expect to see on all exams from now on! It’s called the Diels-Alder reaction, and looks like this: CH2
CH 2
CH2
dienophile
-----
CH2
CH 2
-
diene
CH2
CH 2
---
CH2
-
CH2 CH2
---
CH2
CH 2
Lots of new vocabulary with this one. The two molecules involved are a diene (a molecule with two double bonds, will always be conjugated) and a dienophile (a molecule with a double or triple bond – something that likes or attracted to the diene).
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When they come together, all three pi bonds move to form new single bonds. The result is a ring! The Diels-Alder reaction is a [4+2] cycloaddition, which is easy enough to remember – the diene donates four electrons, the dienophile donates two electrons, and the product is cyclic. It is also classified as a pericyclic reaction. Although this reaction is concerted (one-step, everything happens at once), it is similar to the 1,4 addition we just learned about. Try to draw it out like a conjugated addition to see the similarity – start with the diene attacking the dienophile. Now, we are going to look at some more complicated versions of this reaction. As confusing and elaborate as these molecules may look, we will always be doing the same basic reaction written above. On exams, it can be helpful to draw the basic DielsAlder mechanism to reference while working with more complex molecules. When working through reactions, some students have found it helpful to draw the basic product (6-member ring with one double bond) before adding any other substituents. Stereochemistry In order for this reaction to go forward, the diene must be arranged in space as though it were already part of a ring – this is called the s-cis configuration, as the double bonds are cis (pointing in the same direction) around the single bond. Most dienes will rotate to form the s-cis configuration. Teachers may show you a molecule with an s-trans structure and expect you to figure out that it will rotate into place. If the diene is stuck in a different position, Diels-Alder won’t happen. CH2
CH2 s-cis: ready for DA
H2C
CH2
CH2 s-trans: must rotate
s-trans: can't rotate
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Then there is the matter of substituents on the reactants. As this reaction is concerted, nothing has time to move – cis groups on the alkene stay cis, trans groups stay trans. This is called retention of configuration. In this example, the product is a meso compound and we only need to draw one version (otherwise, we’d also want a product with the substituents pointing down/away). CH2
Br
CH2
Br
Br
*cis-Br's are cis in final product CH2
Br
CH2
Br
Br
When there are groups on the diene and dienophile, we can have several possible products. Figure out all of the possible results for each molecule and then combine them (in this case, we have two versions of ‘cis’ for the alkene, and two directions that CH3 can wind up pointing). You may be asked to find the most stable product in these reactions. Think of steric hindrance. The farther away your substituents are, the better. In this example, the products in which CH3 is trans to the Br’s are more stable. CH3
CH3 Br
CH2
Br
CH2
CH3
CH3
Br
Br
Br
Br
Br
Br CH3
CH3 Br
Br
Br
Br
Things get trickier when dienes are in rings. The products in these reactions are drawn in ways that can appear confusing – we will work through an example, step by 7-83
step, and try to make these drawings easier to understand. When you are writing an exam, you can draw your products however you like, as long as you are clear about the stereochemistry of the reactions! That means that, although your teachers will likely use the sorts of drawings we’ll see here in our final product (in the sawhorse projection, which used for cyclohexane in section two), you don’t necessarily have to.
CH3
H3C CH3
CH3
CH3
CH3
CH3
CH3
CH3 *CH3's are cis *5th C in the ring points up
CH3
CH3
exo
CH3 endo
sawhorse projection, step by step:
CH3 CH3
CH3 CH3 cyclohexene
5th C from diene
CH3's: exo
CH3's: endo
The less stable product, in which the 5th C from the diene and the groups on the dienophile point in the same direction, is the exo product. The more stable product, in which the 5th C from the diene and the groups in the dienophile point in opposite directions, is the endo product. Line-Up
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Every now and then, you will see a Diels-Alder reaction in which neither reactant is symmetrical. That presents will present you with two possible reaction line-ups: OH
HO
O
OH
CH2
HO
O
H
H
vs. H
H
CH2
CH2
O
O
CH2
Use resonance to solve these problems. If we draw resonance contributors for each reactant, we find (+) and (-) charges. Remembering that opposite charges attract, line the (+) charge on the diene with the (-) on the dienophile (and vice versa). :OH
+
OH
+
CH2
CH2
H CH2
-
O
CH2
OH
O
line-up: *OH with O *CH2 with CH2
O
HO H
CH2
H -
O H
CH2
Molecular Orbitals and Substituents Diels-Alder reactions usually involve the HOMO of the diene and the LUMO of the dienophile. The diene is interested in releasing/donating electrons, which can be aided by electron releasing groups. Simultaneously, the dienophile is interested in attracting electrons, which can be aided by electron withdrawing groups. Here are some examples of each: 7-85
Electron Releasing Groups (ERG) (best for diene) OH OR NH2, NHR, NR2
Electron Withdrawing Groups (EWG) (best for dienophile) NH3+ C≡N SO3H NO2
7-86
Chapter 8: Radicals In previous sections, we explored alkene and alkyne reactions – in other words, reactions with molecules that had pi electrons with which to attack new groups. It’s time to work with alkanes. As these molecules are saturated (all single bonds), we must get rid of one or more atoms to make room for new substituents. The process involves radicals (free electrons, neither in a bond nor a lone pair). * Bonds break evenly in radical reactions – each atom gets one electron, or one half of the bond. This is called homolytic cleavage: Br
Br
2Br•
Chlorination and Bromination of Alkanes Alkanes undergo radical substitution reactions. We are using radicals to replace one atom (H) with another (Cl or Br). This involves three steps: initiation, in which the first radicals are produced; propogation, in which existing radicals produce more; and termination, in which radicals combine. Sometimes, peroxide (ROOR), heat (Δ) or light (hv) is used as a radical initiatior in the beginning. In fact, the presence of any of these means that you’re probably doing a radical reaction!
Example: Bromination of Ethane
CH3CH3 + Br2 CH3CH2Br
1) Initiation (no radicals radicals)
Br2 2Br•
2) Propagation (radicals radicals)
CH3CH3 + Br• CH3CH2• + HBr CH3CH2• + Br2 CH3CH2Br + Br•
3) Termination (radicals no radicals)
Br• + Br• Br2 CH3CH2• + Br• CH3CH2Br CH3CH2• + CH3CH2• CH3CH2CH2CH3)
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Take a look at the propagation phase. We needed a Br• for the first step, and wind up creating a Br• in the second step. That new Br• can then find another CH3CH3 to react with. This self-perpetuating part of the process is called a radical chain reaction. Another note about propagation – most of the CH3CH2Br is created here, not during termination. In this example, we replaced one H with a Br. The final product had 5H’s left, though – and any/all of them can be replaced! It is possible for radical substitution to continue until all H’s in an alkane have been replaced – look out for the word ‘excess’ (as in excess Br2 or Cl2), which indicates that several reactions will occur: excess Br2 CH4
CH3Br
CH2Br2
CHBr3
CBr4
ROOR
Which H Gets Replaced? Our last examples used alkanes in which all H’s were equivalent, but what if you have several different types of H’s to choose from? The first thing to consider is C• stability, which follows the same pattern as C+ stability: CH3 H3C
C
CH3
tertiary most stable
>
H3C
CH secondary
CH3
>
H3C primary
CH2
>
CH3 methyl least stable
The relative rates of substitution on primary, secondary, and tertiary carbons depend on whether you are working with Br or Cl. Bromination is a slow, endothermic (energy-absorbing) reaction. It is also highly selective, replacing different H• ‘s in the following ratios: (1600 for 3Cº) > (82 for 2ºC) > (1 for 1ºC). Chlorination is exothermic (releases energy). It is a much faster and much
8-88
less selective reaction, replacing different H• ‘s in these ratios: (5 for 3Cº) > (3.8 for 2ºC) > (1 for 1ºC). The difference between these two reactions is explained by the reactivityselectivity principle – as reactivity increases, selectivity decreases (and vise versa). 3º 1600 5
Bromination Chlorination
2º 82 3.8
1º 1 1
Time to apply this to a reaction! To predict the percent yield of each product, use number of H's x reactivity this formula: total for all types of H's . Br Br2 CH3CH2CH2CH3
CH2CH2CH2CH3 1-bromobutane
Br
+
CH3CHCH2CH3 2-bromobutane
1-bromobutane:
2-bromobutane:
(number of H’s)(reactivity)
(number of H’s)(reactivity)
(6)(1) = 6
(4)(82) = 328
6 6+328 = 0.018 = 2%
328 6+328 = 0.982 = 98%
These reactions are regioselective, as they prefer to react at a specific C (or remove a specific H). Alkenes Radicals reactions produce anti-Markovnikov products – H adds to the most substituted carbon! This is due to radical stability and the fact that H adds last (as opposed to first) in these reactions.
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H H3C
CH
RO OR
H3C
+
CH2
+
HBr
HBr
CH
RO OR
R OH
CH2
H3C
+
CH
CH
CH 2 Br
H
HBr H3C
CH 2 Br
Br
H3C
Br
CH
H3C
CH 2 Br
CH
CH 2 Br
+
Br
Allylic Carbons Allylic and benzylic (adjacent to benzene) radicals are particularly stable because of resonance. In fact, when we add them to our hierarchy of stable radicals, they beat out tertiary carbons:
>
H2C
H2C
benzylic
CH
>
CH2
CH3 H3C
C
CH3
tertiary
allylic
Adding to an allylic carbon can be tricky, as reactants usually just react with the double bond instead (as in the above reaction – the methyl group on the left was unchanged!). Using a reactant called NBS (N-bromosuccinimide), we can add to allylic carbons. Although HBr and Br2 are both formed in this reaction, neither forms at a high enough rate to react with many double bonds.. O Br N Br
+
H3C
CH
CH2
H2C
CH
CH2
O NBS
8-90
O
O hv N Br
N ROOR
O
Br
O
+
Br
+
H3C
CH
H2C
CH2
CH
CH2
+
HBr
CH2
+
O
O
N Br
+
NH
HBr
+
Br2
O
O
Br H2C
CH
CH2
+
Br2
H2C
CH
Br
Remember that resonance can happen during this reaction! The radical is conjugated with the double bond! In this example, resonance wouldn’t change the final product, but it will with other alkenes – and you may be asked to distinguish between the kinetic and thermodynamic products (see section seven). CH3
CH3 NBS
CH3
C
ROOR CH Br CH3
CH3
+ Br kinetic product
thermodynamic product
Stereochemistry When radical reactions form chiral centers, they form racemates (see section five). Draw S and R molecules when you are required to show stereochemistry.
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Chapter 9: Substitution Reactions This section focuses on SN reactions – substitution reactions involving nucleophiles. To review these terms, or introduce them (in case your textbook follows a different order than this review book), substitution means that we will be replacing one atom/group with another. A nucleophile is attractive to positively charged, or partially positive (often written as δ+) atoms and groups. Your teacher may use the words ‘nucleophile’ and ‘base’ interchangeably. In these reactions, a nucleophile (Nu) replaces a leaving group (Lv), which should be a weaker base than the nucleophile. Leaving groups are often halogens in basic (-) form, and are effective in this order: I- > Br- > Cl- > F Depending on the order of reactions (see section one, kinetics), they are either referred to as SN2 (second order, we begin with two molecules) or SN1 (first order, we begin with one molecule). This section will explore both types of reactions, looking at the following characteristics: mechanism, order of the reaction/number of molecules in the beginning, number of steps, type of nucleophile, solvent, stereochemistry, and carbon chain. SN2 Reactions These reactions are second order, and begin with two molecules: the nucleophile, and the molecule that gets attacked. Substitution takes one step – the mechanism appears below, using OH- as a nucleophile and Br- as a leaving group.
HO HO H3C
-
-
and CH 2
Br
H3C
CH 2
Br
H3C
CH 2
OH
+
-
Br
9-92
Let’s analyze that reaction. The nucleophile, OH-, needs to be strong enough to force the leaving group, Br-, out of the molecule. In other words, we need a strong nucleophile for an SN2 reaction. Strong nucleophiles are strong bases. They often have negative charges on oxygen, or contain a negative or neutral nitrogen atom. Ideally, we would like to carry this reaction out in a solvent that will 1) dissolve a molecule/atom with a negative charge and 2) keep our nucleophile strong. This means that we want a solvent that is both polar (like dissolves like) and aprotic (does not have protons to donate). If we try to carry this reaction out in a protic solvent, which does have protons to donate, the nucleophile may accept a proton and become weak. Protic solvents also sometimes surround charged atoms/groups:
-
O
H OH CH3
strong Nu
HO
CH3
(protic solvent) weak Nu
OH HO H H O CH3 H H OHHO
hydration shells form around the nucleophile
Polar, aprotic solvents include acetone, DMF, DMSO, and DMA (many are represented with abbreviations). Pay attention in class – your teacher will probably want you to know a small number of specific solvents. In an SN2 reaction, the nucleophile must attack an already crowded molecule. It can not attack from the leaving group side – the lone pairs on a halogen (Br has three lone pairs!) will repel the strong base. The other option is what’s called a back-side attack. The Nu attacks carbon from the side opposite the Lv. As a result, we wind up with what’s called inversion of configuration: if the leaving group was pointing up, the nucleophile will wind pointing down. An S molecule becomes an R, or vice versa.
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CH3O H3C
CH 2CH 3 H C Br
CH 2CH 3 OCH 3 C H
H3C
OCH3 points down/away
Br points up/out
Finally, in order to access the carbon, the nucleophile would prefer to attack the least crowded molecule possible: methyl > 1˚ > 2˚ >>> 3˚. Generally, 3˚ molecules do not undergo SN2 reactions.
CH3 NH3
H3C
C
Br
NH3
H3C
H3C
Br
NH2
accessible
CH3 not accessible
SN1 Reactions These reactions are first order, and begin with one molecule: the molecule that gets attacked. Substitution takes two steps – the mechanism appears below, using HOH as a nucleophile and Br- as a leaving group.
H3C
CH
OH
H OH
Cl CH3
*the last step can also be broken down into two steps, but you aren't obligated to show it this way:
H3C
+
CH
H3C
CH3
+
CH
CH3
OH
HOH H3C
CH
CH3
(base)
H3C
CH
CH3
The leaving group cleaves off of its own accord, leaving room for the nucleophile to attack. A weak nucleophile is preferable for SN1 – it will be slower to attack, giving the leaving group time to leave. Furthermore, the first step of this reaction makes a strong
9-94
Nu unnecessary, as there is nothing to ‘push out’ of the molecule. Weak nucleophiles are often neutral, although they will probably still contain either oxygen. In order to allow our polar nucleophile and charged carbocation to dissolve/stay dissolved, we need to work with a polar solvent (like dissolves like). We would additionally prefer to use a protic solvent, which will discourage SN2 reactions and keep the Nu weak. This reaction forms a carbocation (C+) intermediate. As such, we would prefer to work with highly substituted carbons in order to form the most stable C+ possible: 3˚ > 2˚ >> 1˚ >> methyl. In fact, 1˚ and methyl molecules rarely undergo SN1 reactions. Be on the lookout for rearrangement (see section four for a review, or if you have not yet covered carbocation rearrangement in class). Carbocations are planar (flat). This means that the Nu can attack from either side – we will get inversion and retention of configuration, or a racemate (equal amounts of R and S products) where applicable. OH H3C Cl H3C
C H
H OH CH 2CH 3
H3C
+
C
C
CH 2CH 3
H
CH 2CH 3 H
H H3C
C
CH 2CH 3
OH
Competition Between SN2 and SN1 Reactions It is extremely difficult (if not impossible) to create an environment in which 100% of reactions will be either SN2 or SN1. We can, however, create an environment in which one of these two reaction mechanisms will dominate by varying the nucleophile strength, solvent, and type of carbon. See Appendix III for a chart comparing the characteristics and ideal conditions for each type of SN reaction.
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Switching it Up: OH and OR as Leaving Groups Sometimes, your teachers may want you to synthesize an alkyl halide. They may also want you to replace OH with another strong leaving group – but attacking OH with a strong base will usually just result in proton donation, rather than substitution. In these situations, it is best to create an alkyl halide first, and then attack with the nucleophile: HO
CH3
+
-
NH3
O
CH3
+
+
NH4
There are two key methods for replacing OH/OR with a halogen. 1) Protonation (works for OH and OR) This method involves adding an H+ to the OH or OR, converting it from a poor leaving group into a good leaving group.
Afterwards, the SN can proceed as usual.
H Cl HO
CH 2CH 3
Cl
CH 2CH 3
+
H HO
CH 2CH 3
(prot.)
H+ H O CH 2CH 3 Cl
-
Cl
CH 2CH 3
(SN2)
2) OH only: fancy, good leaving groups The second technique follows this pattern: R-OH + XY R-OY + X- R-X. A new molecule is introduced, which breaks apart into a halogen atom (in basic form, soon to be our nucleophile) and a larger group that replaces the H in OH. Oxygen, now attached to this group, has been transformed into a good leaving group. Here is an example, using a group called Ts or tosylate (attached to the Cl that becomes our leaving group). Ts replaces the ‘H’ in ‘OH,’ and then Cl- moves in. As with aprotic solvents, chances are that your teacher has a particular set of groups/molecules in mind for you to remember.
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O H3C
OH
+
Cl
O
S
CH3
H3C O
S
O
CH3
+
Cl
-
O
tosylate (Ts) O
O Cl
-
H3C O
S
CH3
H3C
Cl
+
O
-
S
CH3
O
O
Intramolecular Reactions Some molecules have both leaving groups and nucleophiles. They can either react with other molecules or with themselves. The general rule is this: if an intramolecular reaction will produce a five or six-membered ring, go for it! If not, stick to intermolecular reactions. Intramolecular:
O -
O CH 2 CH 2 CH 2 CH 2
O CH 2 CH 2 CH 2 CH 2 Br
O -
O CH 2 CH 2 CH 2 CH 2 CH 2 Cl
O CH 2 CH 2 CH 2 CH 2 CH 2
Intermolecular -
O CH 2CH 2 CH 2
O
O CH 2CH 2 CH 2
Br
Br CH 2CH 2 CH 2 O
-
+
Br CH 2CH 2 CH 2 O
-
BrCH 2CH 2CH 2OCH 2CH 2CH 2O-
Epoxides
9-97
An expoxide is a molecule with a 3-membered ring comprised of two carbons and O
an oxygen:
. To name them, use the prefix ‘epoxy’ along with numbers to indicate the
location of the ring: O H3C
HC
CH
CH 2
CH3
2,3-epoxypentane
Three-membered rings are very unstable (see angle strain in section two). As a result, it is relatively easy for a nucleophile to attack the ring and open it up (in other words, angle strain turns OR into a leaving group). There are two mechanisms for this – an SN2-like mechanism that occurs under basic conditions, and an SN1-like mechanism that occurs under acidic conditions. The SN2-like reaction is one-step. The nucleophile attacks the epoxide in the very first step. As with regular SN2 reactions, the nucleophile prefers to attack the least substituted carbons possible – this may mean that it will prefer one particular carbon in the ring. OH
O HC H3C
CH 2
-
O CH 2CH 3
H3C
CH
CH 2 O CH 2CH 3
In the SN1-like reaction, the epoxide’s O atom is protonated, and begins to cleave off of the molecule. This begins to create a C+. As C+’s prefer to be highly substituted (C+: 3˚ > 2˚ > 1˚), the protonated O will detach from the most substituted carbon.
9-98
H+ O HC H3C
OH CH 2
H3C HO CH 2CH 3
CH
CH 2
O CH 2CH 3
9-99
Chapter 10: Elimination Reactions Elimination, or E reactions, compete with the nucleophilic substitution ( SN) reactions we explored in section nine. Instead of replacing atoms or groups, we remove them – along with an H+ from an adjacent carbon – to create an alkene. Like their SN counterparts, elimination reactions are categorized as either E2 or E1 according to reaction order. This section will explore both types of reactions, looking at the following characteristics: mechanism, order of the reaction/number of molecules in the beginning, number of steps, type of nucleophile, solvent, stereochemistry, and carbon chain.
E2 Reactions These reactions are second order, and begin with two molecules: the nucleophile, and the molecule that gets attacked. Elimination takes one step – the mechanism appears below, using OH- as a nucleophile and Br- as a leaving group. Rather than attacking the carbon attached to the leaving group, OH- grabs a β hydrogen, or an H on the carbon adjacent to the leaving group. Nucleophiles are bases, after all – it makes sense that they would accept a proton!
-
H H2 C
HO
-
HO CH Br
H CH3
H2C
CH
CH3
H2C
CH
CH3
Br
Ideal E2 conditions are very similar to ideal SN2 conditions. We would prefer a strong nucleophile, as the nucleophile’s attack must be strong enough to drive the reaction forward. Polar, aprotic solvents are best (will dissolve the nucleophile, won’t
10-100
weaken it or slow it down). Even stereochemistry is similar – the Nu will take an H+ that is anti to the leaving group (like a back-side attack, only on an adjacent carbon). E2 and SN2 differ in terms of their ideal carbon chains. The final product in an E2 reaction is an alkene, and alkenes are more stable when more substituted. This reaction will work with leaving groups on primary, secondary, and tertiary carbons – but tertiary is preferable. Let’s go back to the stereochemistry issue. In an E2 reaction, our spatial limitations sometimes prevent us from forming the most substituted alkene. -
HO H3C
H
H3C Br
Br
H
There are protons anti to Br on both adjacent carbons, so we form the most stable product possible
H
H
H
CH3
H
H
CH3
H Br
Br Only one of the adjacent protons is anti to Br, so the nucleophile takes that one
H H
CH3
CH3
H H HO
-
E1 Reactions These reactions are first order, and begin with one molecule: the molecule that gets attacked. Elimination takes two steps – the mechanism appears below, using HOH as a nucleophile and Br- as a leaving group.
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HOH H H3C
CH
CH3
H3C
+
CH
H3C
CH 2
CH
CH2
Br
Like SN1 reactions, E1 reactions work best with weak nucleophiles; polar, protic solvents; and highly substituted carbons. Molecules with leaving groups on primary carbons rarely undergo E1. Watch out for rearrangement with the carbocation intermediate! As the carbocation intermediate is planar/flat, the nucleophile can move in from either direction – take whichever H+ will make the most stable alkene.
Dehydration – Removing Water That term makes sense, right? Generally, HOH is used as a nucleophile in these reactions, but every now and then, we’d like to remove it! One of the most effective ways to remove water is protonation – this turns OH into a good leaving group.
+
H
H OH H3C
CH
CH 2
H3C
+
CH
Nu-
CH 2
H3C
CH
CH2
Alkene Alkyne! Using very strong bases and a combination of addition and elimination, we can convert an alkene into an alkyne. See section three to review the alkene reaction. Steps: 1) Add Br2 (bromination: Br2, possibly CH2Cl2 as a solvent) 2) Two eliminations (strong base: possibly LDA or NaNH2)
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Br Br2 H2C
H2C
CH 2
CH 2 Br
Br
Br
Br Na NH2
HC H
HC
CH
CH
Na NH2
HC
CH
H
H
Competition Among E2, E1, SN2, and SN1 All four of these reactions will usually occur simultaneously – but we can create conditions in which one or two of them will dominate. See Appendix III section for comparisons of E2 and E1. You can compare this to the SN review sheet (also in Appendix III) to get a better idea of what conditions favor what reaction(s). Generally speaking, SN2/E2 conditions are basic with aprotic solvents, and SN1/E1 conditions are acidic with protic solvents. Elimination is favored by bulky bases (too big to fit themselves in an already crowded molecule). Tert-butyl-oxide is a particularly good one for this. This compound can be drawn in several ways: Basic (E2) CH3 H3C
Acidic (E1) CH3
-
O CH3
t-buO(CH3)3O-
H3C
OH CH3
t-buOH (CH3)3OH
10-103
Chapter 11: Carbon and Metal This section deals with organometallic compounds, which have carbon-metal (C-M) bonds. Within a C—M bond, the metal is partially positive (δ+) and the carbon is partially negative (δ-). We can therefore treat the carbons in C—M bonds as nucleophiles/bases. They can attack other, positively charged/polarized carbons to form new carbon-carbon bonds! These reactions are called coupling reactions. Organic chemists get super excited about forming C—C bonds, so chances are high that this will show up on an exam. CH3—Li δδ+ C-Li: Organolithium Compounds Organolithium compounds have C-Li bonds and are formed by combining alkyl halides with lithium atoms: CH3CH2Br + Li CH3CH2—Li + LiBr These molecules can react with epoxides, much like the nucleophiles we explored in section nine on substitution reactions. Most teachers will accept several different versions of the final product: a negative charge on O, an ionic bond between O and Li (written as OLi or O-Li+), or an OH (we just assume that the molecule finds a proton from somewhere – refer back to workup in section 1). O H3C
Li
+
H2C
can also write as: H3C
CH 2
CH 2
CH 2
H3C
O Li
or
H3C
CH 2
CH 2
CH 2
CH 2
O
-
OH
11-104
Organolithium compounds can also react with acids – remember that the C in a C—M bond acts like a base, and should be willing to accept a proton from an acid! This converts the organolithium into an alkane. O CH3CH2Li
O
+
CH3CH3 H3C
+
OH
-
H3C
O
Grignard Reagents These compounds are formed by reacting an alkyl halide with magnesium. The Mg atom winds up between carbon and the halogen: CH3CH2Br + Mg CH3CH2—MgBr δδ+
Grignard reagents are not quite as reactive as organolithium compounds. They can react with epoxides, though. O H3C
MgBr
+
H2 C
CH2
H3C
CH2
CH2
-
O
Gillman Reagents The third and final type of organometallic compound up for review here is the Gillman reagent, which has a C—Cu bond. They are made by combining a pair of organolithium compounds with copper iodide (CuI): 2CH3CH2Li + CuI (CH3CH2)2CuLi + LiI Gillman reagents can replace a halogen – just like the nucleophiles we explored in section nine on substitution reactions. Only one of the R groups on Cu is used during these reactions. 11-105
CH2Cl
+
CH2
CH2CH2CH3
(CH3CH2CH2)2CuLi
In addition to these three organometallic compounds and their reactions, your teacher may want you to learn some specific organometallic reactions (and your textbook will almost certainly include some). Pay attention – or just ask – whether you need to learn any specific reactions beyond organolithium, Grignard, and Gillman reagents. Either way, be sure to know the reactions covered here for your exams!
11-106
Chapter 12: Aromatic Compounds Aromatic compounds are defined by a number of criteria. This material can be presented in a few different ways – read this section over, and see which makes the most sense to you. We’ll start with the language you are most likely to find in textbooks, then expand that with some more conversational languange. 1) Aromatic compounds have an uninterrupted cyclic cloud of pi electrons Think back to molecular orbital theory. Bonds between p orbitals were called pi bonds, which makes the electrons in double and triple bonds pi electrons. To carry that even further, think back to resonance. All electrons that can move around – lone pairs, negative charges, double bonds – are pi electrons. That should clue us in to expect resonance in aromatic compounds. If our system is cyclic, we need to have a ring. Finally, uninterrupted: the resonance system has to extend throughout the ring. sp3 C acts as a wall -
CH
Aromatic
-
CH
Not Aromatic
2) Aromatic compounds are planar. You won’t necessarily be able to tell whether a compound is planar (flat) by looking at it. You may, however, be expected to know that aromatic compounds are planar. 3) Huckel’s Rule: there should be (4n + 2) electrons in the resonance system
12-107
Some students have found it helpful to use Huckel’s rule. The ‘n’ can be any integer. As such, an aromatic system can have 2 electrons (4x0 + 2 = 2), 6 electrons (4x1 + 2 = 6), 10 electrons (4x2 + 2 = 10), and so on. Another way to check for this: aromatic systems must have an odd number of pairs of pi electrons. What counts as a pair? A double bond, a negative charge, or a lone pair. Radicals and positive charges do not count as pairs of electrons. -
CH
Aromatic
+
CH
Not Aromatic
When identifying aromatic compounds, it is important to remember that any compound with a ring that matches these criteria is aromatic. You don’t need to apply every criterion to the whole molecule, as long as it has an aromatic system somewhere.
Antiaromatic Compounds Take all of the criteria we just reviewed, trade an odd number of pairs of pi electrons for an even number of pairs, and you have antiaromatic compounds! Or, to put it in bullet-points, antiaromatics: 1) include a ring with an uninterrupted resonance system 2) are planar 3) do not follow Huckel’s rule, and have an even number of pairs of pi electrons Plenty of compounds fall short of both sets of criteria – these are non-aromatic compounds. You may be asked to identify compounds as aromatic, antiaromatic, or non-aromatic. Some teachers will divide compounds into aromatic and non-aromatic, including antiaromatics in the second category.
12-108
-
CH
+
CH
H3C
Aromatic
Anti-Aromatic
CH3 Non-aromatic
Stability We have already seen that resonance makes molecules more stable. The same is true of aromaticity. Aromatic compounds are more stable than non-aromatic compounds, which are more stable than anti-aromatic compounds
H and C Only First, let’s take a look at some aromatic and antiaromatic hydrocarbonds. Aromatic compounds should have rings with uninterrupted systems of resonance – and those systems should include an odd number of pairs of pi electrons. Anti-aromatics should have an even number of pairs of pi electrons, but are otherwise the same. Remember that a negative charge counts as one pair of electrons, while a positive charge counts as zero (positive charge indicates an absence of electrons).
12-109
+
CH -
CH Aromatic
-
CH +
CH Anti-Aromatic
Some molecules can gain or lose H’s to become aromatic or antiaromatic. Predictably, they are eager to become aromatic and resistant to becoming antiaromatic (they want to be as stable as possible!). This affects acidity: -
molecules that lose H’s to become aromatic are very acidic (strong acids)
-
molecules that lose H’s to become anti-aromatic are very basic (poor acids) CH 2
-
+
-H
CH
This reaction is favorable! Our reactant was very acidic. Aromatic
Non-aromatic
CH2
-
CH -H+
Non-aromatic
This reaction is unfavorable! Our reactant was very basic (or not very acidic).
Anti-Aromatic
Atoms with Lone Pairs To review – O, N, S (and other elements in their columns in the Periodic Table) have lone pairs. Cyclic compounds (rings) with non-C atoms in them are called heterocylic compounds.
12-110
Lone pairs can participate in resonance, but they won’t do this every time. As a general rule, each atom can contribute one pair of electrons to a resonance system. If an atom has a lone pair and is part of a double bond, chances are that the lone pair is not involved in resonance. If an atom has a lone pair and nothing else, the lone pair
:
participates in resonance. H N :
N
- N has a double bond - lone pair not in resonance system - aromatic: 3 pairs of pi electrons (all from double bonds)
- N can only donate lone pair - lone pair in resonance system - aromatic: 3 pairs of pi electrons (two double bonds, one lone pair)
*Note: remember that a negative charge represents two pi electrons. Sometimes negative
(
:
charges are written on their own (-), sometimes they are written along with their electrons ): either way, those electrons are not separate from the charge.
Molecular Orbital Theory Returns! That’s right. We need to look at some molecular orbital diagrams once more, this time in hopes of explaining the stability of aromatic compounds and instability of antiaromatic compounds. For the first task, take a look at these diagrams for 1,3,5 hexatriene (straight-chain molecule with three conjugated double bonds) and benzene (6membered aromatic ring, also with three conjugated double bonds).
12-111
H2C
CH2
12-112
Remember that lower-energy molecules are more stable. In aromatic benzene, molecular orbitals are closer in energy – the 2nd and 3rd orbitals are at equal energy levels, as are the 4th and 5th. All electrons in the molecule are still in bonding (lower energy) orbitals. All bonding orbitals are full.
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Chapter 13: All About Benzene Now that we’ve brushed up on resonance and aromaticity, we are going to look at one of the most important aromatic molecules: benzene. We’ll start out with some naming and reactions, then move on to see what substituted benzenes can do.
Nomenclature: One Substituent Some monosubstituted benzenes follow the same rules we reviewed in section two: list the substituent, then say ‘benzene’ for the parent chain. Br
bromobenzene
CH 2CH 3
ethylbenzene
Others have special names. Consult your teacher and textbook to see which ones you’ll need to know for your exam. Here are some examples: OH
CH3
toluene
phenol
NH2
aniline
CH2
styrene
Benzene, itself can also be a substituent. A benzene ring group is called a phenyl group (abbreviated: Ph); a benzene and one additional CH2 (in which the parent chain is attached to that additional carbon) is a benzyl group. All benzene-derived groups, including phenyl and benzyl, are collectively referred to as aryl groups (abbreviated: Ar).
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CH2
phenyl group
benzyl group
Ph CH2
benzyl group
General Reaction Mechanism Benzene is full of pi electrons (6 electrons in constant motion due to resonance, even though we often write them out as double bonds). Electrons are attracted to positive-charged groups, or electrophiles. We encountered double-bond reactions in section three. Let’s see what happens when one of the bonds in benzene tries to react like an alkene. After all, the pi electrons in alkenes are known for attacking electrophiles! ‘El’ represents an electrophile and ‘Nu’ represents a nucleophile.
NuEl+
+
CH
Nu
El
El
Notice anything wrong with that reaction? The mechanism followed alkene addition perfectly – pi bond graps El+, we create a C+, Nu- attacks – but something still wasn’t right. Think about stability. Prior to reacting, benzene was an aromatic compound. It had an odd number of pairs of pi electrons, all of which could move throughout its ring. After reacting, it became a non-aromatic compound (resonance doesn’t include the whole ring). That means a decrease in stability. Benzene is unlikely to stand for that.
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We need to find a way to add to benzene without losing any pi electrons. In a regular, unsubstituted benzene molecule, every C has room for one substituent. That means that, if we want to add a new group, we must eliminate a H. Let’s try that reaction again, this time using the Nu to remove an H. Be sure to take an H from the same carbon that you’re adding to!
NuEl+
+
CH
H El
El
That looks much better – we have added to benzene without sacrificing aromaticity. This was a substitution reaction in which an electrophile replaced an H, and it involved an aromatic compound. In other words, this reaction was an aromatic electrophilic substitution. Here’s the exciting part, as far as studying goes: every electrophilic substitution in this section (i.e. in your curriculum, most likely) has this mechanism! Really! We are going to look at several specific types of reactions now, all of which follow this pattern: 1) prepare the electrophile (often involves a catalyst) 2) general mechanism Halogenation We can replace an H on benzene with Cl or Br. As with alkenes, this process involves homolytic cleavage (all electrons go to one atom when the Cl—Cl or Br—Br bond breaks). This reaction won’t occur if we simply expose benzene to a halogen.
13-116
Instead, we need to use a catalyst: FeBr3 for Br2, FeCl3 for Cl2. The catalyst attracts the electrons in the halogens’ bond, creating δ+ and δ- sides. The symbol ‘B:’ is commonly used to represent a base (any base). 1) Preparing the electrophile Br
Br
+
FeBr3
Br ------- FeBr3
Br
δ+ δ2) General mechanism
:B +
H
CH
Br
Br
Br
Br ------- FeBr3
Here is the overall reaction: Br Br2 FeBr3
Nitration This reaction adds NO2 to a benzene ring. The electrophile (NO2+) is first created by reacting HNO3 with H2SO4. 1) Preparing the electrophile O
O +
N O
OH
+
H
O +
HSO 4
N
-
O
-
+ OH2
+
N O
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2) General mechanism
O
:B
+
+
N
H
CH
NO 2
NO 2
O
Here is the overall reaction: NO 2 HNO3 H2SO4
Sulfonation This reaction adds SO3H to benzene. The electrophile preparation is a bit more complicated for this – chances are that you will be tested on the overall reaction without needing to know the mechanism for creating the electrophile SO3, but we will review it just in case! 1) Preparing the electrophile O HO
O
S
OH
HO
O
S
OH
HO
O
H
O
+
S
O
S O
O H2O
O
O +
OH2
O
-
S
OH
O
O O
S O
2) General mechanism
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O :B S
O
+
H
CH
SO3H
SO3-
O
assume SO3- finds an H+ somewhere, or write this out as a separate step!
Here is the overall reaction: SO3H
H2SO4
Note that the sulfonation of benzene is reversible! Friedel-Crafts Acylation This reaction adds a group that begins with C==O to benzene. The RC==O (or ArC==O) group is called an acyl group; the C==O, itself is called a carbonyl. Friedel and Crafts were the scientists who discovered this process. The C==O group will begin with a Cl on one side and an R group (alkyl group, either H or a hydrocarbon group) on the other. React it with AlCl3 to remove the Cl and give the acyl group a positive charge! 1) Electrophile preparation O
O
+
C AlCl3
Cl
R
R
2) General mechanism
O +
+
C
R
H
O
:B
O
CH
R
R
13-119
Here is the overall reaction: O
O
Cl
R
R
AlCl3
Friedel-Crafts Alkylation This same process can be performed with alky (R or hydrocarbon) groups. Once again, you can begin with an alkyl chloride (R—Cl) and use AlCl3 to remove the Cl. 1) Preparing the electrophile
R
Cl
+
AlCl3
R+
2) General mechanism
:B R
+
+
CH
H R
R
The first step of this reaction (preparing the electrophile) forms a carbocation. This means that we need to watch out for rearrangement! If the R-group can shift a methyl or H to create a more stable C+, it will!
13-120
CH3
CH3
CH3CH2CH2Cl AlCl3
H CH 2 H3C
Cl
+
AlCl3 H3C
CH 2
+ CH2
H3C
CH
+ CH2
H3C
CH3
CH3
CH3
+
+
CH H3C
CH3
Avoiding Rearrangement: Acylation and Reduction What if we want to add a straight chain of three or more C’s to benzene? We need to find a way to add the desired number of C’s without letting the (+) move. This can be achieved through (1)acylation followed by (2) reduction of C==O to CH2. This can be accomplished using one of these methods: -
H2 and Pd/C
-
Wolff-Kishner Reduction: NH2NH2 and NaOH (or any source of OH-)
-
Clemmensen Reduction: Zn(Hg) in HCl
13-121
O
1. AlCl3,
Cl
CH3
CH3
2. NH2NH2, NaOH O O
AlCl3
+ Cl
CH3
CH3
O
CH3
NH2NH2, NaOH
CH3
Rearrangement is prevented here because resonance moves the (+) charge to oxygen. This is similar to the bromination of alkenes (see section three), in which we avoided rearrangement by moving the (+) charge to a bromine atom. O
O
+
+
C
C
R
R
Reactions of Benzene’s Substituents A monosubstituted benzene ring can react in two ways: 1) the substituent can change/react, or 2) we can add something else to the ring. Let’s begin with the reactions of substituents. You have already seen some of these reactions – carbon groups attached to benzene can undergo SN, E, coupling, and alkene reactions.
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CH 2 Br HO
SN (substitution)
CH 2 OH
-
Br
CH2
CH3 t-buO
E (elimination)
-
CH 2 CH3
CH 2 Br (CH3)2CuLi coupling (with Gillman reagent)
Br CH2
CH3 HBr
alkene reactions
You may also need to learn some new reactions for this part of the class – consult your teacher and/or textbook to see which specific reactions are relevant for your exams. The NO2 group on nitrobenzene can be changed into a diazonium group, which looks like this: N≡N+. The process involves several steps: NO 2
+
NH2
N
HNO3
Sn, HCl
HNO2
H2SO4
NaOH
HCl
N
The diazonium group is very reactive and can be replaced with many different groups (nucleophiles). Group
Reagent(s)
Reaction +
N
Br or Cl
CuBr or CuCl
Br
N CuBr
13-123
+
N
I
KI
KI
+
N
F
F
N
HBF4
HBF4
+
N
CN (C≡N)
I
N
CN
N
CuCN
CuCN
+
N
+
OH
N
H3O or OH
Cu2O
Cu2O, Cu(NO3)2, and HOH
Cu(NO3)2, HOH +
N
H
H
N
H3PO2
H3PO2
The N≡N+ group can also act as an electrophile. You may see examples in which a substituted benzene attacks a diazonium ion – this works best with activated benzenes (discussed later in this section!).
HO
N
+
N
HO
N
N
Substituents and Reactivity As mentioned earlier, it is also possible to add a new substituent to a benzene ring. The group that’s already on benzene will make the ring more or less likely to react
13-124
again – a group that makes benzene more reactive is an activating substituent; a group that makes benzene less reactive is a deactivating substituent. Remember that benzene forms a C+ intermediate when it attacks an electrophile. Electron donating groups (ED) will stabilize that positive charge, activating the ring (or at least making it more willing to become positive again!). Electron-withdrawing groups (EW) will destabilize that charge and deactivate the ring. Electrons can be donated/withdrawn through resonance, which literally adds or removes electrons, and induction, which is sort of like pulling on electrons and creating +/- regions without actually moving electrons to different bonds or atoms.
Substituent OH
Type of (De)Activator OR
NH2
Mechanism (EW/ED) ED by resonance --and--
Strongly activating O
O
R
R
O
R
NH
Moderately activating
Ar R
F
I
Cl
O
O
O
+
+
-
Weakly deactivating
ED by resonance --and-EW by induction
Moderately deactivating
EW via resonance and induction
Strongly deactivating
EW via resonance and induction
O N
HO S
N
O
ED by induction
Br
R
R2
Weakly activating
O
R
R3 N R1
EW by induction ED by resonance --and-EW by induction
O
*A note about Acidity: activators decrease acidity; deactivators increase acidity. Direction
13-125
In addition to making benzene more or less reactive, an existing substituent determines which C (or C’s) the next group will add to. In the end, the two groups can wind up in three different positions, relative to one another: CH3
CH3
CH3 CH3
CH3 CH3 para (anti, 3C away)
meta (2C away)
ortho (adjacent, 1C away)
Substituents that that donate electrons (whether by resonance or induction) are ortho-para directors. In other words, if an electron-donating group is already on benzene, the next group to be added will wind up in an ortho or para position. This
:
:
pattern can be explained if we look at some resonance diagrams: OH
OH
+
+
OH
+
OH -
OH -
CH
HC -
CH
Negative charges wound up at the ortho and para positions! This makes those positions very attractive to a positively-charged electrophile. Looking at these structures, it makes sense that most resonance donators (everything but halogens) are activating substituents – they create negative charges that pull electrophiles toward the benzene ring. Neutral Rgroups (which donate by induction only) are also ortho-para directors. Substituents that withdraw electrons by resonance are meta directors. Let’s see how the resonance diagrams look with one of these groups:
13-126
H
O
H
O
-
O
-
H
H
O
+
-
O
H
+
CH
HC +
CH
This time, we wound up with positive charges at the ortho and para positions. That will repel electrophiles. New groups will want to add to the meta positions instead! It makes sense that these groups are called deactivating substituents – they push electrophiles away. With ortho-para and meta directors, it is possible to have multiple reactions:
OH
OH
NO 2
CH3 Cl
Cl
Cl 2
Cl 2
FeCl 3
FeCl 3 Cl
Cl
Cl
See Appendix III for an expansion of the activator/deactivator chart that classifies groups as ortho-para or meta-directing! Adding a Third Substituent Sometimes, two groups on benzene will direct to the same place(s).
13-127
NO 2
NO 2
NO 2 Br
Br
Br
NO2: meta Br: ortho-para
NO2 direction
NO 2
Br direction
NO 2 Br
NO 2 Br
Cl 2
Cl
Br
FeCl 3
Cl
Other times, the two groups will direct to different places. Go with strength in these situations. Refer to the chart earlier in this section (or the review sheet at the end) that classified different groups as strongly, moderately, or weakly activating/deactivating. Those same patterns can be applied to direction. Go with the strongest group – if groups are equally strong, just write different products. OH
OH
OH
CH3
CH3
CH3
Conflict! OH: strong CH3: weak OH wins out!
OH: ortho-para CH3: ortho-para
CH3 direction
OH direction
OH
OH
OH Br
CH3
CH3
CH3
Br2 FeBr3
Br
13-128
Nucleophilic Aromatic Substitution These reactions, abbreviated SNAr, proceed slowly and require strong, electronwithdrawing substituents to lure a nucleophile towards the benzene ring. The incoming Nu must be a stronger base/Nu than the group being replaced (just like in SN1 and SN2). It is best to have EW-groups at the ortho and para positions, so that they can take on the extra electrons. Cl
OH
HO
-
Mechanism below: note that NO2 takes the electrons in the third structure!
+
O
+
N
-
O
O
Cl
HO
-
Cl
-
N
O
Cl
OH
OH
Cl
Cl
OH
-
OH
OH
-
CH
HC -
C +
+
O
-
N
O
O
-
+
N
O
O
-
+
N
O
O
-
+
N
O
-
O
-
+
N
O
O
-
N
O
Benzyne Almost done! The last topic for this section is benzyne, which looks like a benzene ring in which one double bond has been replaced with a triple bond. Benzyne is very unstable/reactive. It can be formed through an elimination reaction with bromobenzene or chlorobenzene.
benzyne
13-129
H3C
Br
H3C
Br
H3C -
H
NH2
H
Benzyne reacts almost immediately. The triple bond can perform the equivalent of an alkene reaction with NH3 (formed when NH2- took an H from bromobenzene). If the original benzene ring had any substituents aside from Br, we can wind up with two possible products: H3C
H3C
NH2
H3C
H
NH3
H
NH2
13-130
Chapter 14: Carbonyls with Leaving Groups The title of this section clues us in about its content: we will working with carbonyl or acyl compounds, which are characterized by a C==O group. These molecules will have leaving groups attached to the carbonyl – and leaving groups can be replaced by nucleophiles! This section explores substitution at an acyl group, also known as nucleophilic acyl substitution (SNAc).
Nomenclature and Reactivity Before reviewing reactions, let’s take a look at the types (and nomenclature) of the molecules we’ll be working with. The leaving groups on these molecules are nonhydrocarbon groups attached to C==O. Acyl halides have a halogen (F, Cl, Br, I) attached to the acyl group. As usual, the halogens function as leaving groups. These molecules are named as alkanoyl halides. The C in C==O should be C1 in the parent chain. O
H3C
CH3
Cl
ethanoyl chloride
H3C
O
Br
3-methylbutanoyl bromide
Anhydrides have two carbonyl groups separated by oxygen. Their leaving groups include oxygen, a C==O group, and the R group attached to C==O. Either acyl group can be attacked in a reaction. When naming anhydrides, name them as alkanoic1 alkanoic2 anhydride if the R groups are different, or as alkanoic anhydride if the R groups are identical.
14-131
O
H3C
O
O
O
H3C
CH3
ethanoic anhydride
O
H3C
-
leaving group(s)
O
O
O
O
H3C
CH3
O
O
-
O
-
CH3
leaving group(s)
butanoic ethanoic anhydride
Carboxylic acid esters, often referred to simply as esters, have an OR group attached to carbonyl. The OR groups are their leaving groups. They are named as alkyl alkanoates, in which the first alkyl group is the R from OR and the second alkyl group is the parent chain (including the C in C==O). O H3C O
CH3
ethyl propanoate
Carboxylic acids have an OH group attached to the carbonyl. This combined functional group (COOH, or an acyl and an OH) is called a carboxyl group. The OH groups are their leaving groups, and they are named as alkanoic acids. O
HO
CH3 butanoic acid
14-132
Amides have N-groups attached to the carbonyl. These N-groups can be primary (only attached to the C in C==O; NH2), secondary (NHR), or tertiary (NR2 or NRRI). The N-groups are the leaving groups in amides. They are named as alkanamides. O
O
CH3 H2N
CH3 butanamide
H3C
NH
N-methylethanamide
Now, for reactivity!. As we have learned, weaker bases are easier to replace in SN reactions. This makes molecules with weaker bases (a.k.a. better leaving groups) are more likely to undergo substitutions. We have just looked at different carbonyl compounds in order of decreasing reactivity: Compound Most Reactive
Acyl halides Leaving group: F, Cl, Br, F
Anhydrides Leaving group: OC(O)R Esters Leaving group: OR
-
Acids Leaving group: OH
-
Amides
-
Least Reactive
Reacts With OC(O)R: forms anhydride OR: forms ester OH: forms acid NR2: forms amide OR: forms ester OH: forms acid NR2: forms amide OH (with H+): forms acid ORI (with H+): forms a different ester NR2: forms amide OR (with H+): forms ester NR2: acid-base reaction Can also convert OH into a better leaving group! OH (with H+): forms acid OR (with H+): forms ester
General Mechanism Like benzene, these carbonyl compounds have a general reaction mechanism that will carry us through most of this section. It goes like this: 1) a nucleophile attacks, 2) a
14-133
tetrahedral intermediate forms (C is attached to 4 groups), 3) the weaker of the two bases gets kicked out. O
O H3C H3C
Lv
-
C
Nu-
O Lv H3C
Nu
Nu
tetrahedral intermediate
Note that both steps in this reaction are reversible. When the tetrahedral intermediate forms, the weaker base gets kicked out of the molecule. If the attacking Nu is a stronger base than the leaving group, we wind up with a new product. If the attacking Nu is a weaker base than the Lv, we wind up with no net reaction (the product and reactant are identical). O
O H3C H3C
Cl
-OCH 3
Nu- attacks
Cl
weaker base: Cl-
O
OCH 3
O
OCH 3
O
H3C
C
-
Cl
-
Nu- attacks
H3C
H3C
OCH 3
substitution
-
C
O OCH 3 H3C
OCH 3
Cl
weaker base: Cl-
no net reaction
Acid-Catalyzed Mechanism
14-134
Some reactions are sped up – or made possible – by acid catalysts (H+). Although these reactions are similar to the general mechanism, there are a few extra steps. Those steps involve adding and removing the H+ catalyst. Let’s look at the situations in which you’ll want to use an acid catalyst, and then take a look at the mechanism. All of these reactions have the same mechanism – just change your leaving groups and nucleophiles !
Reactant: Lv Carboxylic Acid: OH Ester: OR
Amide: NH2, NHR, NR2
Acid-Catalyzed Substitution Nucleophile OR (esterification) OH (hydrolysis) ORI (transesterification) OH (hydrolysis) OR (esterification)
-
Product Ester carboxylic acid different ester
-
carboxylic acid ester
Sample Reaction: Hydrolysis of Methyl Ethanoate O
O
HOH +
H3C
OCH3
H
H3C
OH
14-135
OH
+
+
O
OH
H
H3C H3C
H3C
OCH3
O takes H+
OCH3
HOH
Nu attacks
OCH3 +
HO H
tetrahedral intermediate
OH
OH
OH
+
H3C
C
C
OCH3
+
-H
H3C
C
OCH3
H
H3C
+
HO H
C
+
O CH3 H
OH
OH
H+ is lost and gained: (+) groups are better leaving groups than neutral ones!
+
OH H3C
C
O
OH
-H+
+
O CH3 H
OH HOCH3+ leaves
H3C
OH
lose H+
H3C
OH
final product!
Here is another example, without explanations of the steps. This time, an OR group replaces an NH2.
14-136
OH
+
+
O
OH
H
H3C H3C
H3C
NH2
HOCH 3
NH2
+
HO CH 3
OH
+
C
NH2
OH
OH H3C
C
+
-H
NH2
H3C
C
H
NH2
H3C
+
H O CH 3
H3C
C
+
O
OH
-H+
+ NH3
H3C
OCH 3
NH3
OCH 3
OCH 3
OH
+
C
H3C
OCH 3
OCH 3
Trouble with Acids Earlier in this book (and hopefully earlier in this course), we learned that OH makes a poor leaving group. This is perhaps even more true when OH is attached to an acyl group. Acids are known for donating protons, and that is exactly what happens when strong nucleophiles approach carboxylic acids. These reactions are irreversible (the base that forms is stabilized by resonance, and won’t want to take the H+ back). O
R
O
NH3 OH
R
-
O
+
NH4
That being said, chances are that your professor will give some synthesis problems that begin with carboxylic acids. You can begin these problems by converting
14-137
OH into a better leaving group. Then, replace the modified OH group with a halogen – and halogens are very easy to replace! Here are some techniques for this. *Note: the molecules that react with OH can be divided into 2 basic parts: the group that attaches to O, and a single halogen atom that attacks. 1) Replace OH with Br: PBr3 PBr3 PBr2 attaches to O, then Br- attacks Br P
O
Br
O
O
Br R
OH
R
OPBr2
R
Br
-
Br
2) Replace OH with Cl: PCl3 PCl3 PCl2 attaches to O, then Cl- attacks Cl P
O
Cl
O
O
Cl R
OH
R
OPCl2 Cl
R
Cl
-
3) Replace OH with Cl: SOCl2 SOCl2 SOCl attaches to O, then Cl- attacks This reaction is super favorable, as we begin with 2 molecules and end with 3 AND create a gas (increase in entropy/chaos).
14-138
O
O
O Cl S Cl
R
OH
O
-
R Cl
O R
O
O O
S
Cl
S
O
-
-
R Cl
O O
S
Cl
O SO2 (g)
Cl R
Cl
-
Cl
14-139
Cl
Chapter 15: Aldehydes and Ketones Ketones include a carbonyl surrounded by two R-groups, while aldehydes include a carbonyl surrounded by one H and one R-group or two H’s. O
R
O
O
R1
R
R
R
ketones
O
H
H
H
aldehydes
Like the molecules we looked at in section 14, ketones and aldehydes can be attacked by nucleophiles (at the C in C==O). Unlike the molecules we looked at in section 14, ketones and aldehydes don’t have leaving groups! Reactions can proceed one of two ways when the Nu attacks: 1) Freezing the Intermediate (Nu attacks, forms a tetrahedral intermediate, and it freezes in place) O
O H3C H3C
H
-
C
OH
+
H H
H3C
C
H
NuNu
Nu
2) Replacing O (instead of replacing a group attached to C==O, the Nu replaces O) O
Nu Nu-
H3C
H
H3C
H
The type of reaction will depend on the type of Nu involved. *A note about reactivity: these reactions depend on a nucleophile reaching the C in C==O. The most reactive molecules will be the easiest to reach:
15-140
O
O
O
> H
>
H
R
R
H
(most reactive)
R
(least reactive)
Freezing the Intermediate: H and C Nucleophiles When a C or H nucleophile attacks, we wind up with a tetrahedral intermediate that has no leaving groups. Instead of re-creating the C==O, the molecule protonates the O- to make an OH. The final product is an alcohol. It’s also possible to begin with a carbonyl that has a leaving group (see section 14 for examples), in which case it’s possible to get two reactions. 1) C Nucleophiles: RMgX, RC≡C-, CNO
O CH3
-
OH
H
H
C
CH 2CH 3
H
CH3
O
Cl
C
CH 2CH 3
CH 3CHCH 2CH 3
CH3
OH
O
CH 3MgBr H
OH
CH 3MgBr
CH 3MgBr H
CH3
H
C
CH3
CH3
2) H Nucleophiles: LiAlH4, R2LiAlH, NaBH4 -
O
O CH3
H
OH
OH
NaBH4 H
C H
CH 2CH 3
H
C
CH 2CH 3
CH 2CH 2CH 3
H
When reacting with a C==O that has a leaving group, use LiAlH4 to get two reactions:
15-141
OH
O
O LiAlH4 H3C
H3C H3C
Cl
C
H
H
CH 3CH 2OH
H
Use R2LiAlH to get only one reaction: O
O R2LiAlH H3C
H3C
Cl
H
Exception: If we begin with an amide, the C==O is reduced to CH2: O LiAlH4 H3C
CH3CH2NH2
NH2
Freezing the Intermediate: O and S Nucleophiles When an oxygen nucleophile attacks, the intermediate winds up including two oxygen groups – in other words, we have no obvious leaving group. It is possible to freeze the intermediate. OH
O ROH +
R
R
C
R
H
R
OR
These reactions are acid-catalyzed, reversible, and work best with highly active carbonyls (more effective with aldehydes than ketones). The simplest of these reactions form hydrates, which have two OH groups on the same carbon. They are formed by reacting an aldehyde or ketone with water and acid. OH
O HO H H3C
CH3
+
H
H3C
C
CH3
OH
15-142
+
O
OH
OH
+
H
H3C H3C
H3C
CH3
OH -H+
HO H
CH3
C
CH3
H3C
C
+
HOH
CH3
OH
Aldehydes and ketones also react with alcohols. This creates a hemiacetal or hemiketal, which has one OH and one OR group on the same carbon (hemiacetal from an aldehyde, hemiketone from a ketone).
+
O
H3C
H3C
CH3
OH
OH
OH
+
H
CH3
R OH
H3C
-H+
C
H3C
CH3
C
+
CH3
OR
H OR
hemiketal
If the hemiketal (or hemiacetal) reacts with another alcohol molecule, it can become a ketal (or acetal), which has two OR groups attached to the same carbon.
+
OH
H OH
H3C
+
H3C
C OR
CH3
H
H3C
C
R OH
CH3 OR
OR
OR
CH3 H3C
C
CH3
+
OR ketal
Ketals and acetals can act as protecting groups. This means that they can preserve a functional group (C==O) while the molecule undergoes reactions. The process involves three major steps/stages: 1) Add protecting group (ROH and H+) 2) React (non-protected part of the molecule) 3) Remove the protecting group (HOH and H+)
15-143
The best protecting groups are diols (molecules with two OH groups) that form 5- and 6membered rings with the C in C==O.
O
HO-CH2CH2-OH
O
O
O
HO-CH2CH2CH2-OH
O
+
H R
R
R
R
R
O
+
H
R
R
protected
R
protected
Protecting groups are incredibly important in synthesis (and likely to show up on a test!). Sample problem: synthesis O
H3C
O
O
Cl
H3C
OH
Let’s start by tracking the changes in our molecule: Changes: C1 (far right) - C=O becomes C-OH - lose Cl - 2 new H's
H
O C C
C
OH
Cl H
We can achieve all of the changes in C1 by adding LiAlH4 (see H and C nucleophiles, earlier in this section). There’s a problem, though – LiAlH4 will react with the other carbonyl group, reducing the C==O to a C—OH! To avoid this, we must protect the ketone C==O.
15-144
O
O
protection
H3C
O
reactions at C1
H3C
O
Cl
LiAlH4
O
O
O OH
H3C
H
O
HOH
O deprotection
H3C
O
O Cl
O
O
LiAlH4
H3C
O
H+
Cl
O
O
HOCH2CH2OH
H+ H3C
OH
H3C
OH
These reactions can also be carried out with sulfur nucleophiles (use SH and SR instead of OH and OR). Protection groups formed with SR groups are called thioketals and thioacetals. O
R
HSCH2CH 2CH 2SH H
R
+
S
S
R
R
Why use sulfur at all, when oxygen works so well? This gives us one extra option: if you expose a thioketal or thioacetal to H2 and Raney nickel, you can replace the C—S bonds with C—H bonds. This means that you can eliminate the carbonyl!
O
H2
HSCH2CH2CH2SH S
R
R
H
S
+
Raney Ni R
R
R
R
Replacing O: Nitrogen Nucleophiles
15-145
As we have seen, N-bases are generally stronger than O-bases. When an N-group attacks an aldehyde or ketone, it competes with O for a place on the molecule – and wins! Instead of freezing the intermediate, the molecule eliminates O (or OH). weaker base
OH O R R
C
R
NH3
R
stronger base
NH2
Here is the mechanism for ammonia and primary amines. It begins like the SNAc reactions from section 14 (Nu attacks, intermediate forms). Then, when the intermediate collapses, N lends its lone pair to the molecule and O leaves. The final product is an imine. OH
+
O
R
R
H R
R
HO+H
+
H
C
R
R
C
R -H
R
NH3
+
R
+
NH2
NH2
NH2
R
NH imine
This process works for any N-group that contains two H’s: O
R
NH 2CH 3 R
O
R
+
R
R
H
+
R
H
R
*Part of Wolff-Kishner Reduction! See section 13 on benzene*.
NOH
NH 2OH
R
R NNH 2
NH 2NH 2
O
R
H
N CH3
+
R
R
Secondary amines react similarly – they attack the carbonyl, a tetrahedral intermediate forms, and the O is forced out – but there is one extra step at the end. The secondary N
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wind up positive at the end of the imine reaction. An alpha (α) C, which is adjacent to the functional group (C==O or C==N, at this stage) then loses a proton to allow N to reclaim its electrons. The final product is an enamine: a double bond (ene) and N-group (amine) are attached to the same carbon. O
H3C
R2NH H
H
NR2
+
NR2
+
+
NR2
H H3C
H
CH2
H2C
H
H
enamine
All of these reactions are reversible. To recreate the C==O, react an imine or enamine with water and acid (HOH, H+). Replacing O: The Wittig Reaction The Wittig Reaction replaces the C==O bond with a C==C bond. We can create an alkene from a ketone or aldehyde, forming a new C—C bond! Organic chemists love this type of reaction! Before completing this reaction, we need to prepare a nucleophile. The Wittig Reaction involves an ylide, or a molecule with opposite charges on adjacent atoms. The ylide will be in resonance with an ylene, which has a double bond between those atoms. We can prepare the ylide using an alkyl halide, a base, and a molecule of Et3P (three ethyl groups attached to P).
Et3P
Et3P
CH 3CH 2 Br
Et3P
+
CHCH 3 ylide
+
CH 2CH 3
Et3P
Et3P
+
CHCH 3
CHCH 3 ylene
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The ylene and carbonyl group line up: C next to C, O next to P. The intermediate looks sort of like a box, as two new double bonds replace the old ones. O
R
PEt3
CHCH 3 R
R
O
R
R
CHCH 3
R
PEt3
O
PEt3
CHCH 3
C
CHCH 3
R
R
R CHCH 3
O
PEt3
R
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The α-Carbon of C==O This section explores reactions at the α carbon of carbonyl groups. We label the C’s adjacent to a functional group as alpha (α), beta (β), and so on. Most of these reactions will involve aldehydes and ketones. We will also explore some ester reactions towards the end of the section. *Throughout this section, ‘α C’ and ‘α carbon’ will refer to C’s adjacent to carbonyls!* O
H3C
H
Chapter 16: Enols and Enolates When the α C loses a proton, an enolate can form. We saw enols in section four on alkynes – they have a double bond (ene) and an OH group (ol) attached to the same carbon. Enolates are the negative ions of enols (a double bond and an O- attached to the same C). OH
O
R
-
R H
H
enol
enolate
Chances are that you’ll need to know three techniques for forming an enols/enolates. 1) Acid-catalyzed: enols H HOH
O
+
OH
OH
+
R
R
R H
H
H
H HOH
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This is the mechanism for keto-enol tautomerism, which we first saw in section four. H3O+ is a great acid to use in this reaction, as water (its conjugate base) is unlikely to change the ketone or aldehyde (see sections fourteen and fifteen). 2) Base-catalyzed: enolates
R
R
O R
-
CH
H H
-
O
O
H
H
-
HO
This reaction is very similar to the acid-catalyzed version. Once again, OHmakes a great base for this reaction because it is unlikely to change the ketone or aldehyde. You may also see stronger bases used for this. LDA, or lithium diisopropylamide, is a common choice (negative charge on N makes it very basic). +
Li
-
H3C
N
CH3
CH3
CH3
3) From α,β-unsaturated compounds: either These compounds are in resonance with an enolate-like structure to begin with: Nu
-
O
O
O Nu
Nu
H3C
H -unsaturated
H3C
H
enolate
H3C
H
aldehyde (or ketone)
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If a nucleophile attacks the β C, resonance will create an enolate. To create an enol instead, react in acidic conditions. This works best with weaker nucleophiles (groups that are too weak to successfully add to the carbonyl of an aldehyde or ketone). Stronger nucleophiles will react at the carbonyl. Br
-
O
Br
R
H
O
-
R
H
OH
O
R
H3C
H
H C
-
HC
C
CH
This pattern applies to carbonyls with leaving groups as well – if the nucleophile can successfully attack the carbonyl, it will. Otherwise, it will attack the β carbon. The α C of an enol or enolate acts as a nucleophile – it can attack things! Expect to see some synthesis problems in which you have to add something to the α C of a ketone or aldehyde. Halogenation This reaction places one or more halogens on the α C. Under acidic conditions, one hydrogen will be replaced with Br, Cl, F, or I. O
H3O
H3C H
+
O
OH H3C
H3C
H
H Br Br Br
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Afterwards, you can do SN reactions to place a nucleophile at the α C. Remember this for synthesis problems! O
O HO
H3C
-
H3C H
H OH
Br
Halogenation also occurs in basic solutions. This process replaces all α H’s (all hydrogens on the α carbons). -
-
O H3C
O
O
HO
H3C
H3C H
H
H Br Br Br
If you the ketone or aldehyde has a methyl group attached to C==O, the haloform reaction may occur. Replacing all α H’s creates a CX3 group. Even though it includes a carbon and resembles methyl, CX3 is an excellent leaving group. The OH used to create an enolate will attack the C==O. The reaction proceeds like a section-14 (carbonyls with leaving groups) reaction: a tetrahedral forms, and the weaker base (CX3) is eliminated. O
O
I3
O
-
HO -
H
CH3
HO
H
CI 3
H
OH
Alkyation It is possible to add alkyl groups to the α C. This creates a new C—C bond, so you can expect to see it show up on your exam! These reactions work best with bases. If you alkylate a ketone or aldehyde with LDA, you may get multiple reactions. The LDA will simply keep creating enolates, which in turn will attack more alkyl groups.
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O
O
-
O CH3
CH 3Br
LDA
O
O
O CH3
CH3
LDA
H3C
H3C
CH3
CH3
CH 3Br
If you only want to add one alkyl group, you’ll want to bring in some reactions from section 15. Change the ketone or aldehyde to an enamine, add the alkyl group, and then change the enamine back into a carbonyl compound. O
NR 2
HNR 2
H3C H
NR 2
H
+
H3C H
NR 2
CH 3Cl
+
O HOH
H3C
H3C
H
H CH3
H
+
H3C H CH3
Michael Reaction In this reaction, an enolate attacks an α,β-unsaturated compound. The molecules are often big – this may make the reaction look a bit overwhelming or confusing. Remember that we are performing the same reaction we looked at in the beginning of this section. Using a enolate for our Nu won’t change the mechanism. 1)
General reaction: adding to the β C of an α,β-unsaturated compound
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-
O
O
Nu
O Nu
Nu
H3C
H3C
H
H
-unsaturated
2)
H3C
enolate
H
aldehyde (or ketone)
Michael Reaction: enolate as a nucleophile O
O
H3C
H3C
H O
O
H3C
H
H O
O
-
H2C H2C
-
C
H2C
H
C
H
H
O 3
rewrite final product:
5 4
H3C
O
CH3
1
H
H
O
3 2
5 4
H
O H2C 2
C 1
H
The trickiest part of a Michael reaction is often drawing the final product. While writing out a synthesis, feel free to draw the product however you like as long as it is accurate. While working through a multiple choice question (or trying to see whether your product is the correct), you may need to redraw the final product. It can be helpful to practice redrawing (as in the example above). Aldol Condensation The Aldol Condensation has two parts: 1) an enolate attacks an aldehyde or ketone, and 2) water is eliminated. The first step is called the Aldol Addition, the second is dehydration. 1) Addition
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O OH
O
(base) H3C
H H3C
OH
O
O
H3C
H
H3C
-
H2C
H
C
H
H O H2C
OH redraw product:
H3C
H
2
4 3
H
1
C
H
4
3
O
2
H3C
O H2C
H
OH
C
1
C
This is just like the reactions we looked at in section fifteen. A C Nu attacks, a tetrahedral intermediate forms, and we freeze the intermediate in place. 2) Dehydration OH
H3C
O
O
H3C
H
OH base-catalyzed
H
O
O
H3C
H H
+
H OH
HO
H3C
H
-
O
O
O +
CH
acid-catalyzed H3C
H
H3C
H
H3C
H
H HOH
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If possible (if there are enough H’s), water is eliminated after the addition. This can be done under acidic or basic conditions. The final product is an α,β-unsaturated compound!
Remember that any carbonyl’s α C can be used as a nucleophile, and that any carbonyl can be attacked. This means that you may have multiple reaction pathways if you are using 1) different molecules (non-identical) or 2) molecules with 2 α C’s. You may need to work backwards through Aldol Condensations. In these questions, you will be presented with a final product (after dehydration) and need to write out the reactant(s). To do this, keep in mind the fact that the new bond is the α,β bond next to the carbonyl! Break the molecule up there, and add water again. O CH3 H3C CH3
O
(new bond)
O
(O) (2H's)
CH3 H3C
H3C
HC
C
CH3
CH3
CH3
O
H3C
O CH 2
CH3 H
CH3
Aldol reactions can also be intramolecular. This occurs when it is possible to form a 5- or 6-membered ring.
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O
O
1
3 2
H
- C2 attacks C5: 4-membered ring (NO) - C4 attacks C1: 4-membered ring (NO)
5 4
CH3
- C6 attacks C1: 6-membered ring (YES)
6
OH O
O
OH -
H
CH2
O
H
CH 2 O
Claisen Condensation An ester attacks an ester in the Claisen Condensation. More specifically, the α C of one ester attacks the carbonyl group of another. This is an SNAc reaction (see section fourteen): a tetrahedral intermediate forms and the weaker nucleophile (OR) is eliminated. O
H3C
O
CH3OOCH 3
O
CH3OH H3C
O
OCH 3
O
O H3C
H3C
OCH 3
-
H2C
C
OCH 3
-
O OCH 3 H3C
O H2C
O
OCH 3
OCH 3
***Note on bases: esters can react with other bases. To avoid changing your OR group, use a base identical to it (in this example, we used –OCH3 in a molecule with an ester that had an OCH3). That way, even if substitution happens, it won’t affect your products. The Claisen Condensation, like the Aldol Condensation, can involve two identical molecules or two different ones (which increases the number of possible products). It can also be intramolecular, if that will result in a five- or six-membered ring.
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Chapter 17: Identifying Molecules: Mass Spectrometry, Infrared Spectroscopy, and NMR Chemists need to be able to identify the molecules that they are working with. This is essential for determining the compounds present in unknown materials. Chemists also need some way of determining the product(s) of a reaction (and whether a reaction has occurred at all). This section reviews three methods for identifying compounds.
Mass Spectrometry This technique uses molecular mass to identify a compound. The process goes like this: 1) An energy beam hits the molecule 2) One electron breaks off from the molecule, leaving a radical cation (• +)
CH3CH2CH3
e-1 beam
CH3CH2CH3+.
3) The radical cation can break into fragments: one radical and one cation. Other times, the radical cation remains intact. CH3CH2CH3+. CH3CH2CH3+.
+
CH3CH2CH3+.
.CH3
CH3
.CH2CH3 +
CH2CH3
4) The mass spectrometer records the m/z value of the (+) fragments and intact radical cations. This translates to mass/charge and, since the charge is +1, is equal to the mass (m/1 = m). Mass is measured in atomic mass units.
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CH3CH2CH3+. +
+
(m/z = 44)
CH3
(m/z = 15)
CH2CH3
(m/z = 29)
The results are displayed in a mass spectrum, which is a graph of the relative amounts of each fragment (indicated by mass). The amounts of each fragment are indicated by their relative abundance. The highest peak, or base peak, represents the most abundant fragment and is set at an arbitrary value of 100. Molecular ions can fragment in many ways. They are the most likely to break off in ways that will create the most stable C+’s (and radicals) possible. As such, the base peak should represent the most stable C+ fragment.
Source: Bruice 2002 The above graph is a mass spectrum for pentane. The base peak, with a mass of 43, represents a three C fragment: CH3CH2CH2+ (3C + 7H = 36 + 7 = 43). Roughly 1/5 of the molecules were intact (non-fragmented) radical cations, with a mass of 71. There 17-159
are also peaks for fragments with one, two, and four C’s: CH3+ (m/z = 15), CH3CH2+ (m/z = 29), and CH3CH2CH2CH2+ (m/z = 57). Now, let’s look at the part of mass spectrometry that involves some math! Carbon atoms have isotopes. They appear most frequently as 12C (98.9%), but can also appear as 13C (1.1%). This means that for every fragment of mass M, you will have a small peak at (M + 1), in which one of the C atoms is a 13C isotope. You can use this information to determine how many C’s are in a fragment. Let’s say that fragment M has 4 C atoms. The relative frequency of the (M + 1) fragment should be (4 x 1.1%), multiplied by the relative frequency of the fragment M. We wind up with this equation: relative frequency of (M + 1) peak Number of C atoms = 0.011 x (relative frequency of M peak) If you have the values for the M and (M + 1) peaks, you can calculate the number of C’s in a fragment! Isotopes are also important when working with functional groups. You can expect to see specific ratios of peaks with different isotopes. Atom Br Cl
Isotopes and Ratio 79 Br and 81Br – 1:1 35 Cl and 37Cl – 3:1
Results for Graph M and (M + 2) peaks, equal frequency M and (M + 2) peaks, 3:1 difference
Molecules with functional groups often fragment in predictable ways. Functional Group Br and Cl Cl
O in Ethers
Fragmentation Br/Cl breaks off α cleavage: bond between α C and another C breaks, (+) goes to Cl group C—O cleavage, (+) goes to R group α cleavage, (+) goes to
Results for Graph Peak at (M – Br/Cl) 2 sets of M and (M + 2) peaks with a 3:1 height ratio: one set for the intact molecule, one after α cleavage Peaks at (M – OR) and (M – OR1) Peaks at (M – α group), C group
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O in Ethers O in Alcohols O in Alcohols C==O ketones H
having been attached to α C Peaks at (M – α group), C group having been attached to α C
group with O α cleavage, (+) goes to group with O OH and γH leave (see diagram below) α cleavage, (+) goes to group with O
Peak at (M – HOH), or (M – 18) Peaks at (M – α group), C group having been attached to α C
.+
OH
CH3CHCH2CH2CHCH3
.
+
CH3CHCH2CH2CHCH3
HOH
Infrared Spectroscopy Covalent bonds are in constant motion; they vibrate at specific frequencies. A bending vibration changes a bond’s angle. A stretching vibration changes a bond’s length. Molecules absorb energy when exposed to radiation that matches the frequency of one of their bonds’ vibrations. Infrared spectroscopy exposes molecules to infrared radiation, producing a graph of absorbed energy. The y-axis consists of the percentage of transmission (how much radiation passes through the molecule). The x-axis indicates wavelenth and wavenumber. Every downward-pointing spike in the graph indicates that energy has been absorbed.
Source: Bruice 2002
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The IR spectrum can be divided into two sections: the functional group region and the fingerprint region. The functional group region ranges from 4000-1400cm-1. As the name suggests, this is where most functional groups show absorption bands. You will likely be expected to recognize the absorption bands of a few different functional groups. The fingerprint region ranges from 1400-600cm-1 (the left-hand third of the graph). Like fingerprints, which are unique to individuals, the absorption bands in a molecule’s fingerprint region are unique to that molecule. You will not be expected to recognize a molecule simply by looking at its ‘fingerprint.’ Here are some groups that you will likely be expected to recognize. Pay attention in class to see whether you are specifically required to memorize the positions of other groups’ absorption bands. Functional Group
Wavenumber
Description
1800-1600 cm-1
VERY sharp peak
3650 – 3200 3300 – 2500
broad peak broad peak
O C
OH (alcohol) OH (in COOH)
Again, you will never be expected to recognize an entire molecule by looking at its fingerprint!
Source: Bruice 2002
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Looking at this graph, you should be able to identify signs of two functional groups: a sharp peak around 1700 that represents C==O and a broad peak around 3400 that represents an OH group. NMR: Nuclear Magnetic Resonance Spectroscopy This tends to be the most test-question-producing method for identifying compounds. NMR applies a magnetic field to a compound, producing a spectrum of signals from a particular type of atom. NMR can produce signals from 1H, 13C, and 31P among others. Throughout this section, and most of organic chemistry discourse, ‘NMR’ means 1
H NMR unless otherwise specified. Every set of chemically equivalent protons produces a signal. Chemically
equivalent protons are either (1)on the same C or (2) in symmetrical positions on a molecule.
a b a CH3CH2CH3
a a CH3OCH3
2 signals
1 signal
c d b a CH3CHCH2CH3
Br 4 signals
a b d c CH3CH2CH2OCH3
4 signals
The relative positions of signals depend on their chemical shift (δ), which is a measure of their frequency compared to a reference compound. The reference compound, which is usually TMS (tetramethylsilane or (CH3)4Si) is set at a chemical shift of 0. Signals at higher frequencies appear to the left, or downfield, of TMS’s signal.
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Source: Bruice 2002 What determines a signal’s location on the graph? This has to do with shielding. Electrons can get in the way of the magnetic field, which shields the protons and weakens their signal. In other words, the signal will have a lower frequency. If a proton is near functional groups or bonds that pull the electrons away, the signal will be at a higher frequency. Electronegative groups (OH, COOH, carbonyl) and multiple bonds (aromatics, alkenes, alkynes) pull electrons away. In the above graph, we can see two different sets of H’s: those in the methyl groups and those in the CH2. This is why there are two signals. The H’s in the CH2 are de-shielded by the electronegative Br atom. As a result, their signal is at a higher frequency and appears to the left of the methyl protons’ signal. NMR Spectra: chemical shift (ppm) when δTMS = 0 12 ppm
10 – 8 ppm
O
O
8 – 7 ppm
6 – 5 ppm
H
CH2
R OH
R
H
R
4 – 3 ppm
H3C
O
H3C
X
2 – 0 ppm
CH, CH2, CH3 (in that order)
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The height of a signal is determined by the number of H’s it represents. A signal that represents 4 protons will be twice as high as a signal that represents two protons. Be careful, though – the heights of signals tell us the relative, not the absolute numbers of H’s making each signal! Take another look at the last graph. There are 9 methyl protons (from 3 methyl groups) and 2 protons in CH2. As a result, the methyl protons’ signal is approximately 9H 4.5x as high as the CH2 protons’ signal: 2H = 4.5. The final major concept in NMR is signal splitting. Protons on adjacent C’s affect each other’s signals. A signal will be split into (N + 1) separate lines, where N is the number of H’s on an adjacent carbon. Signals can appear as singlets (not split), doublets (split into two lines), triplets (split into three lines), etc.
a b CH3CH2Cl
signal a: split into a triplet by group b (N + 1) = (2 + 1) = 3 signal b: split into a quartet by group a (N + 1) = (3 + 1) = 4
Sometimes, a signal can be split by the protons on 2 or more carbons. If the protons on these carbons are equivalent, simply take the total number of H’s for your Nvalue. You will simply wind up with more lines – perhaps a septet or quintet. a b a CH3CH2CH3
signal b: split into a septet by groups a (N + 1) = (3 + 3 + 1) = 7
If the protons are not equivalent, multiply your different (N + 1) values. This will produce multi-layed coupling: you may wind up with a triplet of doublets, or a doublet of doublets, etc. 17-165
signal b: split into a quartet by group a (N + 1) = (3 + 1) = 4 a b c CH3CH2CH2Br
signal b: split into a triplet by group c (N + 1) = (2 + 1) = 3 overall: quartet of triplets
Source: Bruice 2002 In this graph, we can see two signals: one from the proton in CH and one from the protons in CH3. The signal from CH should be to the left of the signal from CH3, as the 1 CH proton is deshielded by Cl2. The signal from CH should also be 3 the height of the 1H 1 signal from CH3 (3H = 3 ). The proton in CH is split into a quartet by CH3: (N + 1) = (3 + 1) = 4. The protons in CH3 are split into a doublet by CH: (N + 1) = (1 + 1) = 2.
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Chapter 18: Appendices Appendix One: Redox Reactions Redox is an umbrella term for reduction and oxidation reactions. Chances are that you’ll hear these terms over and over in this class. You may hear vague definitions, different definitions at different times, or find that you are simply assumed to know what these words mean. If you’re confused, don’t worry! These concepts can be reframed in a more accessible way. Better still, you may already know how to do a number of these reactions – the only part left is to classify them! Below are some charts to clarify the meanings of reduction and oxidation. Several of the terms overlap (e.g. H donation and replacing a functional group with H).
REDUCTION Equivalent Terms
Examples O
OH NaBH4
C
H Donation (increasing number of C—H bonds)
H3C
CH3
H2 H2C
CH2
CH H3C
CH3
CH 3CH 3
Pt
O
Replacement of functional groups by H H3C
Addition of electrons
NH2NH2
C CH3 2+
Mg
-
CH3CH2CH3
HO
e-1
Mg
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OXIDATION Equivalent Terms
Examples O
H3C
H
CH3
Loss of H or R groups
CH3
1. O3 2. Zn, HOH H
O
CH3 H3C
CH3
Br Br2 H2C
H2C
CH2
CH 2 Br
Increase in C—O, C—N, or C—X bonds OH
H3C
Loss of electrons
O
H2CrO 4 H3C
CH3
-e-1
Li
Li
CH3
+
Carbons can be described in terms of their oxidation state. This is the number of bonds to O, N, and halogens that each C has. Reduction reactions decrease oxidation states, while oxidation reactions increase them.
0
1
CH4
CH3F
CH3
oxidation reactions 2 O
CH3OCH3 CH3
H
H
H3C
OH
H
CCl4 O
OH
O CH3
Cl
4
O
Cl
CH3NH2
H3C
3
H
Br
NH2
Br
O==C==O
O
CH3 H3C H
Br
18-168
NH
reduction reactions
Again, most of these reactions should look familiar to you after spending some time in organic chemistry. It may not be necessary to classify every redox reaction you encounter, but it can be very helpful to understand these terms!
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Appendix Two: Hints for Biomolecules
Some professors require that you study biomolecules towards the end of this course. They may have lectures on carbohydrates, lipids, and proteins. Many of the reactions involved should be reactions you’ve already learned – although this may not be said outright. While we are not going to delve into the details of biomolecules here, we will take a look at the relevant reactions. Carbohydrates Most of the carbohydrate reactions you learn will be carbonyl reactions. As the carbonyl groups are surrounded by C’s and H’s, you should take another look at aldehydes and ketones for hints! Here are some examples: O H 2C
HC H HO
H
OH NaBH 4
H
H
OH
H
OH
H
HO
OH OH H
+
OH
H
OH
H
OH
OH
Example 1: D-glucose is reduced (see reduction of aldehydes!)
O 1
H HO
2 3
OH H
H
4
OH
H
5
OH
H H HO H H
1 2 3 4
OH OH H O OH
OH
H 4
HO HO
5 3
5
CH 2OH
H
H
O
H 2
1
OH
H
OH
OH
Example 2: D-glucose forms a ring (the ring is a hemiacetal!)
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That last example – converting a straight-chain 6C sugar into a ring – may have looked a bit scary. Don’t worry! Here is how you do it: 1)
Draw out a couple of chair conformations for reference: one axial, one equatorial. *axial lines alternate between pointing up and down *axial and equatorial lines on the same carbon should point in opposite directions equatorial
axial
2) Have the OH in C5 attack C1 – the mechanism is straight out of the carbonyl chapters! 1
H H HO H
2 3 4 5
O 1
OH OH H OH
H H HO H
2 3 4 5
OH
1
O OH OH H OH
H H HO H
2 3 4 5
OH OH OH H O OH
OH
3) Draw out your rings. O should be in the top right, with C’s numbered clockwise. 4) Add the substituents! The CH2OH should be in the equatorial position (more stable). The OH on C1 can be axial or equatorial. With the others, keep directions consistent. If two OH’s both pointed left in the chain, they should point in the same direction (up or down) on the ring. In this example, the OH’s on C2 and C3 point in one direction, and the OH on C4 points the other way. OH OH
OH 4 5 3
O
HO
2 1
O OH
O
HO
OH C2 and C3 up, C4 down
OH
OH OH
C2 and C3 down, C4 up
*the more stable molecule is the one with more equatorial bonds (first ring in this example)*
Lipids A review of resonance and substitution reactions will help you with lipid reactions. The material primarily involves double bonds and radicals. Remember that:
Radicals and double bonds can both participate in resonance 18-171
Double bonds can split up into 2 separate radicals
If two radicals combine, they can form a new bond
Substitution reactions are particularly relevant to terpene synthesis. A review of SN and E reactions might be useful if you need to learn about terpene synthesis. Isoprene units combine through a combination of an SN2-style attack (electrons in a double bond act as a nucleophile) and an E1-style elimination (a proton is removed, creating a double bond).
O
CH3
O
P H3C
O O
CH3
P -
O O
O
CH3
-
O
P
-
H2C
O
+
C
O
P
H3C
O O
H
O O
CH3
O
-
O
O O
-
O
-
CH3
P PH
P -
-
O
O
CH3
O
P
-
H3C
O O
P -
PH O
-
O
-
:B
Proteins Amino acids have at least two functional groups: a N-group and a carbonyl group: O H2N OH R
As such, many amino acid reactions are similar to the amine and carbonyl reactions you learned earlier in the class. Keep these general concepts in mind:
N can attack (review SN reactions!)
carbonyl can be attacked (review carbonyl reactions, acids!)
α C of carbonyl can attack (review reactions at the α C!)
18-172
You may also want to take another look at acidity/basicity. In section one (review of general chemistry), we reviewed some potential pH questions. We used a molecule’s structure and pKa to determine whether it would be charged or neutral at different pH values. We also determined whether it’s acidic or basic form would dominate. Proteins have two functional groups, each with their own pKa values. You can do the same sorts of questions with them – determining whether each functional group will be neutral/charged or acidic/basic at certain pH values. You may also be asked to calculate the pI value, also called the isoelectric point. This is the pH at which the amino acid’s net charge is zero. To calculate this, simply take the average of the pKa values of the N and carbonyl groups. O
pKa = 2.34
H2N OH
pKa = 9.69
CH3
pI =
2.34 + 9.69 = 6.02 2
If the R-group of an amino acid also has a functional group with a pKa, take the average of the two similar groups (i.e. the two that are positively charged as ions or the two that are negatively charged as ions). O
pKa = 2.19 H2N OH
pKa = 9.67
HO
O
use the COOH groups: pI =
pKa = 4.25
4.25 + 2.19 = 3.22 2 18-173
Appendix Three: Review Sheets Organic Compounds
Compound
Structure
Alkane
R
Alkene
RCH==CHRI
Alkyne Alkyl Halide
RC≡≡CRI RX X ≡ I, Br, Cl, F
Ether
RORI
Example with Name CH3CH2CH3 propane CH3CH==CH2 propene CH3C≡≡CH Propyne CH3CH2CH2CH2Cl cholorobutane CH3OCH2CH3 methoxyethane
O
O
Epoxide R HC
CH R1
Alcohol
ROH
Amine
RNH2
H3C HC
CH CH3
2,3-epoxybutane CH3CH2OH ethanol CH3CH2NHCH3 N-methylethanamine O
O
Carboxylic Acid
H3C R
OH
OH
butanoic acid O
O
Carboxylic Acid Ester
O CH3
H3C O R1
R
methyl butanoate
O
Amide
R
O
NH2
H3C
NH2
butanamide
O
Ketone
R
O
R1
H3C
CH3
propan-2-ol or 2-propanol O
Aldehyde
R
O
H
H3C H
propanal 18-174
Alkene Reactions Adding
Reaction (see section text for mechanisms)
Cl
HBr
H2C
CH
HCl
CH3
H3C
CH
CH3
Br
H2C
Br2, Cl2
CH
Br2
CH3
H2C
CH
CH3
Com H to least subs Cl or Br to mo Look out for r Rearrangemen Forms triangu Possible to ha attack – see be
Br
Br
H2C
Br2, Cl2 + Nu
CH
Br2
CH3
H2C
CH
Same mechan OH- (or any ot intermediate in Nu attacks mo
Acid-catalyzed H to least subs OH to most su Look out for r Acid-catalyzed H to least subs OR to most su Look out for r Oxymercurati NO rearrangem THF is a solve
Ac stands for
Alkoxymercur NO rearrangem
THF is a solve
Ac stands for Hydroboration H to most sub OH to least su HOH only, no THF is a solve Step 2 may be
CH3
Na OH OH
OH
HOH and H+
H2C
CH
CH3
HOH, H+
H2C
CH
CH3
OR +
ROH and H
+
H2C
CH
CH3
ROH, H
H2 C
CH
CH3
OH
HOH and Hg
H2C
CH
CH3
1. Hg(OAc)2, HOH/THF 2. NaBH4
H2 C
CH
CH3
OR
ROH and Hg
H2C
CH
CH3
1. Hg(OAc)2, ROH/THF 2. NaBH4
H2C
CH
CH3
OH
HOH via BH3
H2C
CH
CH3
1. BH3/THF 2. HOH, HO-, H2O2
H2 C
CH 2 CH3
18-175
Alkyne Reactions Adding
Reaction (see section text for mechanisms)
Br HBr
HBr
HC
C
CH3
H3C
C
Cl
Br Cl
CH
C
Cl
Cl
Cl 2
Br2, Cl2
HC
C
CH3
OH
HOH and H+
HOH, H+ H3C
C
CH
H3C
C
CH2
H3C
H3C
C
CH
HOH, HgSO4 H2SO4
HOH via BH3
H3C
CH3
CH2
H3C
C
CH3
O
OH H3C
C
CH
1. BH3/THF 2. HO-, H2O2, HOH
H2: to alkane
C
O
C
H3C
CH
H3C
CH
CH2
CH
H2 H3C
C
C
CH3
H3C CH 2
Multiple react an alkene, whi
Keto-enol taut Ketone usually H to least subs OH to most su Keto-enol taut Ketone usually H to least subs OH to most su Aldehyde form H to most sub OH to least su Step 2 may be Multiple react the alkane stag additional or d
CH3
O
OH
HOH and Hg
CH3
CH 2 CH3
Pt, Pd/C, or Ni
H2: cis alkene
H2: trans alkene
H2 H3C
C
C
CH3
H3C
CH3
‘Poisons’ reac reaching the a
H H3C
H H
H’s come from this reaction
Terminal alky Do this first in involving chai
Lindlar's Catalyst
Li or Na H3C
C
C
CH3
NH3 H
R-groups (chain-lengthening)
CH3
1. LDA or NaNH2 H3C
C
Co Multiple react an alkene, whi H’s to least su Br’s to most s
H3C
CH
C
C
R
2. RBr
18-176
Substitution (SN) Reactions Reaction
SN2
SN1 Cl
-
H3C
HO
Mechanism
CH3CH2
Br
CH3CH2
CH
CH3
H3C
+
CH
CH
OH OH
H3C
Order Steps Nucleophile
nd
2 order, begins with 2 molecules One-step reaction Strong Nu (negatively charged or including N) Solvent Polar, aprotic (DMSO, DMF, acetone) Stereochemistry Back side attack! We have inversion of configuration: S R and R S C-chain methyl > 1º > 2º > > > 3º
+
CH
HOH H3C
CH3
CH
C
st
1 order, begin with 1 molecule Two-step reaction Weak Nu (neutral)
Polar, protic (anything acidic) No preference for either side/direction. We win up with a racemate (even mix of R and S). 3º > 2 º > > > 1º and methyl
18-177
Elimination (E) Reactions
Reaction
E2
E1 Cl
Cl
H3C
Mechanism
H3C
CH
CH 2 H
H3C HO
CH
CH
CH3
-
H H3C
Order Steps Nucleophile Solvent Stereochemistry C-chain
nd
H3C
CH2
2 order, begins with 2 molecules One-step reaction Strong Nu (negatively charged or including N) Polar, aprotic (DMSO, DMF, acetone) Leaving group and H must be anti to one another 3º > 2 º > 1º, but will work with 1º
+
CH
CH 2
st
HOH
H3C
1 order, begin with 1 molecule Two-step reaction Weak Nu (neutral) Polar, protic (anything acidic) No preference – just form the most s 3º > 2 º > > > 1º , won’t work with 1
18-178
Organometallic Compounds Compound
Formation 1)
Reactions Acids: takes H, makes an alkane O
CH3CH2Li
+
CH3CH3 H3C
Organolithium: RLi
O
OH
RBr + Li 2) Expoxides: opens ring R—Li O + LiBr (need 2Li H3C Li + H2C CH2 per RBr, but it may not be written this way on an exam!) Epoxides: opens ring
H3C
+
H3C
CH2
O
CH2
-
O
-
O
Grignard: RMgX
RBr + Mg RMgBr
H3C
Mg Br
+
H2 C
CH2
H3C
CH2
CH2
O
-
Coupling; Replaces a Halogen (similar to SN) Gilliman: R2CuLi
2RLi + CuI R2CuLi + LiI
(CH3)2CuLi
+
CH3CH2CH2Br
CH3CH2CH2
18-179
CH3
Appendix One: Redox Reactions Redox is an umbrella term for reduction and oxidation reactions. Chances are that you’ll hear these terms over and over in this class. You may hear vague definitions, different definitions at different times, or find that you are simply assumed to know what these words mean. If you’re confused, don’t worry! These concepts can be reframed in a more accessible way. Better still, you may already know how to do a number of these reactions – the only part left is to classify them! Below are some charts to clarify the meanings of reduction and oxidation. Several of the terms overlap (e.g. H donation and replacing a functional group with H). REDUCTION Equivalent Terms
Examples O
OH NaBH4
C
H Donation (increasing number of C—H bonds)
H3C
CH3
H2 H2C
CH2
CH H3C
CH3
CH 3CH 3
Pt
O
Replacement of functional groups by H
NH2NH2
C H3C
CH3 2+
Addition of electrons
Mg
-
CH3CH2CH3
HO
e-1
Mg
OXIDATION Equivalent Terms
Examples O
H3C
H
CH3
Loss of H or R groups
CH3
1. O3 2. Zn, HOH H
O
CH3 H3C
CH3
A
Br Br2 H2C
H2C
CH2
CH 2 Br
Increase in C—O, C—N, or C—X bonds OH
H3C
Loss of electrons
O
H2CrO 4 H3C
CH3
-e-1
Li
Li
CH3
+
Carbons can be described in terms of their oxidation state. This is the number of bonds to O, N, and halogens that each C has. Reduction reactions decrease oxidation states, while oxidation reactions increase them.
0
1
CH4
CH3F
CH3
oxidation reactions 2 O
H
CH3OCH3 CH3
H
H3C
OH
H
CCl4
CH3
O
OH
O
Cl
CH3
4
O
Cl
CH3NH2
H3C
3
H
Br
NH2
Br
O==C==O
O
NH
H3C H
Br
reduction reactions
Again, most of these reactions should look familiar to you after spending some time in organic chemistry. It may not be necessary to classify every redox reaction you encounter, but it can be very helpful to understand these terms!
B
Appendix Two: Hints for Biomolecules Some professors require that you study biomolecules towards the end of this course. They may have lectures on carbohydrates, lipids, and proteins. Many of the reactions involved should be reactions you’ve already learned – although this may not be said outright. While we are not going to delve into the details of biomolecules here, we will take a look at the relevant reactions. Carbohydrates Most of the carbohydrate reactions you learn will be carbonyl reactions. As the carbonyl groups are surrounded by C’s and H’s, you should take another look at aldehydes and ketones for hints! Here are some examples: O H 2C
HC H HO
H
OH NaBH 4
H
H
OH
H
OH
H
HO
OH OH H
+
OH
H
OH
H
OH
OH
Example 1: D-glucose is reduced (see reduction of aldehydes!)
O 1
H HO
2 3
OH H
H
4
OH
H
5
OH
H H HO H H
1 2 3 4
OH OH H O OH
OH
H 4
HO HO
5 3
5
CH 2OH
H
H
O
H 2
1
OH
H
OH
OH
Example 2: D-glucose forms a ring (the ring is a hemiacetal!)
C
That last example – converting a straight-chain 6C sugar into a ring – may have looked a bit scary. Don’t worry! Here is how you do it: 1) Draw out a couple of chair conformations for reference: one axial, one equatorial. *axial lines alternate between pointing up and down *axial and equatorial lines on the same carbon should point in opposite directions equatorial
axial
2) Have the OH in C5 attack C1 – the mechanism is straight out of the carbonyl chapters! 1
H H HO H
2 3 4 5
O 1
OH OH H OH
H H HO H
2 3 4 5
OH
1
O OH OH H OH
H H HO H
2 3 4 5
OH OH OH H O OH
OH
3) Draw out your rings. O should be in the top right, with C’s numbered clockwise. 4) Add the substituents! The CH2OH should be in the equatorial position (more stable). The OH on C1 can be axial or equatorial. With the others, keep directions consistent. If two OH’s both pointed left in the chain, they should point in the same direction (up or down) on the ring. In this example, the OH’s on C2 and C3 point in one direction, and the OH on C4 points the other way. OH OH
OH 4 5 3
O
HO
2 1
O OH
O
HO
OH C2 and C3 up, C4 down
OH
OH OH
C2 and C3 down, C4 up
*the more stable molecule is the one with more equatorial bonds (first ring in this example)*
Lipids A review of resonance and substitution reactions will help you with lipid reactions. The material primarily involves double bonds and radicals. Remember that:
Radicals and double bonds can both participate in resonance D
Double bonds can split up into 2 separate radicals
If two radicals combine, they can form a new bond
Substitution reactions are particularly relevant to terpene synthesis. A review of SN and E reactions might be useful if you need to learn about terpene synthesis. Isoprene units combine through a combination of an SN2-style attack (electrons in a double bond act as a nucleophile) and an E1-style elimination (a proton is removed, creating a double bond).
O
CH3
O
P H3C
O O
CH3
P -
O O
O
CH3
-
O
P
-
H2C
O
+
C
O
P
H3C
O O
H
O O
CH3
O
-
O
O O
-
O
-
CH3
P PH
P -
-
O
O
CH3
O
P
-
H3C
O O
P -
PH O
-
O
-
:B
Proteins Amino acids have at least two functional groups: a N-group and a carbonyl group: O H2N OH R
As such, many amino acid reactions are similar to the amine and carbonyl reactions you learned earlier in the class. Keep these general concepts in mind:
N can attack (review SN reactions!)
carbonyl can be attacked (review carbonyl reactions, acids!)
α C of carbonyl can attack (review reactions at the α C!)
You may also want to take another look at acidity/basicity. In section one (review of general chemistry), we reviewed some potential pH questions. We used a molecule’s E
structure and pKa to determine whether it would be charged or neutral at different pH values. We also determined whether it’s acidic or basic form would dominate. Proteins have two functional groups, each with their own pKa values. You can do the same sorts of questions with them – determining whether each functional group will be neutral/charged or acidic/basic at certain pH values. You may also be asked to calculate the pI value, also called the isoelectric point. This is the pH at which the amino acid’s net charge is zero. To calculate this, simply take the average of the pKa values of the N and carbonyl groups. O
pKa = 2.34
H2N OH
pKa = 9.69
CH3
pI =
2.34 + 9.69 = 6.02 2
If the R-group of an amino acid also has a functional group with a pKa, take the average of the two similar groups (i.e. the two that are positively charged as ions or the two that are negatively charged as ions). O
pKa = 2.19 H2N OH
pKa = 9.67
HO
O
use the COOH groups: pI =
pKa = 4.25
4.25 + 2.19 = 3.22 2
F
Appendix Three: Review Sheets Organic Compounds Compound
Structure
Alkane
R
Alkene
RCH==CHRI
Alkyne Alkyl Halide
RC≡≡CRI RX X ≡ I, Br, Cl, F
Ether
RORI
Example with Name CH3CH2CH3 propane CH3CH==CH2 propene CH3C≡≡CH Propyne CH3CH2CH2CH2Cl cholorobutane CH3OCH2CH3 methoxyethane
O
O
Epoxide R HC
CH R1
Alcohol
ROH
Amine
RNH2
H3C HC
CH CH3
2,3-epoxybutane CH3CH2OH ethanol CH3CH2NHCH3 N-methylethanamine O
O
Carboxylic Acid
H3C R
OH
OH
butanoic acid O
O
Carboxylic Acid Ester
O CH3
H3C O R1
R
methyl butanoate
O
Amide
R
O
NH2
H3C
NH2
butanamide
O
Ketone
R
O
R1
H3C
CH3
propan-2-ol or 2-propanol O
Aldehyde
R
O
H
H3C H
propanal
G
Alkene Reactions Adding
Reaction (see section text for mechanisms)
Cl
HBr
H2C
CH
HCl
CH3
H3C
CH
CH3
Br
H2C
Br2, Cl2
CH
Br2
CH3
H2C
CH
CH3
Com H to least subs Cl or Br to mo Look out for r Rearrangemen Forms triangu Possible to ha attack – see be
Br
Br
H2C
Br2, Cl2 + Nu
CH
Br2
CH3
H2C
CH
Same mechan OH- (or any ot intermediate in Nu attacks mo
Acid-catalyzed H to least subs OH to most su Look out for r Acid-catalyzed H to least subs OR to most su Look out for r Oxymercurati NO rearrangem THF is a solve
Ac stands for
Alkoxymercur NO rearrangem
THF is a solve
Ac stands for Hydroboration H to most sub OH to least su HOH only, no THF is a solve Step 2 may be
CH3
Na OH OH
OH
HOH and H+
H2C
CH
CH3
HOH, H+
H2C
CH
CH3
OR +
ROH and H
+
H2C
CH
CH3
ROH, H
H2 C
CH
CH3
OH
HOH and Hg
H2C
CH
CH3
1. Hg(OAc)2, HOH/THF 2. NaBH4
H2 C
CH
CH3
OR
ROH and Hg
H2C
CH
CH3
1. Hg(OAc)2, ROH/THF 2. NaBH4
H2C
CH
CH3
OH
HOH via BH3
H2C
CH
CH3
1. BH3/THF 2. HOH, HO-, H2O2
H2 C
CH 2 CH3
H
Alkyne Reactions Adding
Reaction (see section text for mechanisms)
Br HBr
HBr
HC
C
CH3
H3C
C
Cl
Br Cl
CH
C
Cl
Cl
Cl 2
Br2, Cl2
HC
C
CH3
OH
HOH and H+
C
CH
H3C
C
CH2
H3C
H3C
C
CH
HOH, HgSO4 H2SO4
H3C
C
CH3
O
C
CH2
H3C
C
CH3
O
OH H3C
C
CH
1. BH3/THF 2. HO-, H2O2, HOH
H3C
CH
H3C
CH
CH2
CH
H2 H3C
C
C
CH3
H3C CH 2
Multiple react an alkene, whi
Keto-enol taut Ketone usually H to least subs OH to most su Keto-enol taut Ketone usually H to least subs OH to most su Aldehyde form H to most sub OH to least su Step 2 may be Multiple react the alkane stag additional or d
CH3
HOH, H+ H3C
HOH via BH3
H2: to alkane
O
OH
HOH and Hg
CH3
CH 2 CH3
Pt, Pd/C, or Ni
H2: cis alkene
H2: trans alkene
H2 H3C
C
C
CH3
H3C
CH3
‘Poisons’ reac reaching the a
H H3C
H H
H’s come from this reaction
Terminal alky Do this first in involving chai
Lindlar's Catalyst
Li or Na H3C
C
C
CH3
NH3 H
R-groups (chain-lengthening)
CH3
1. LDA or NaNH2 H3C
C
Co Multiple react an alkene, whi H’s to least su Br’s to most s
H3C
CH
C
C
R
2. RBr
I
Substitution (SN) Reactions Reaction
SN2
SN1 Cl
-
H3C
HO
Mechanism
CH3CH2
Br
CH3CH2
CH
CH3
H3C
+
CH
CH
OH OH
H3C
Order Steps Nucleophile
nd
2 order, begins with 2 molecules One-step reaction Strong Nu (negatively charged or including N) Solvent Polar, aprotic (DMSO, DMF, acetone) Stereochemistry Back side attack! We have inversion of configuration: S R and R S C-chain methyl > 1º > 2º > > > 3º
+
CH
HOH H3C
CH3
CH
C
st
1 order, begin with 1 molecule Two-step reaction Weak Nu (neutral)
Polar, protic (anything acidic) No preference for either side/direction. We win up with a racemate (even mix of R and S). 3º > 2 º > > > 1º and methyl
J
Elimination (E) Reactions
Reaction
E2
E1 Cl
Cl
H3C
Mechanism
H3C
CH
CH 2 H
H3C HO
CH
CH
CH3
-
H H3C
Order Steps Nucleophile Solvent Stereochemistry C-chain
H3C
CH2
2nd order, begins with 2 molecules One-step reaction Strong Nu (negatively charged or including N) Polar, aprotic (DMSO, DMF, acetone) Leaving group and H must be anti to one another 3º > 2 º > 1º, but will work with 1º
+
CH
CH 2
HOH
H3C
1st order, begin with 1 molecule Two-step reaction Weak Nu (neutral) Polar, protic (anything acidic) No preference – just form the most s 3º > 2 º > > > 1º , won’t work with 1
K
Organometallic Compounds Compound
Formation 1)
Reactions Acids: takes H, makes an alkane O
CH3CH2Li
+
CH3CH3 H3C
Organolithium: RLi
O
OH
RBr + Li 2) Expoxides: opens ring R—Li O + LiBr (need 2Li H3C Li + H2C CH2 per RBr, but it may not be written this way on an exam!) Epoxides: opens ring
H3C
+
H3C
CH2
O
CH2
-
O
-
O
Grignard: RMgX
RBr + Mg RMgBr
H3C
Mg Br
+
H2 C
CH2
H3C
CH2
CH2
O
-
Coupling; Replaces a Halogen (similar to SN) Gilliman: R2CuLi
2RLi + CuI R2CuLi + LiI
(CH3)2CuLi
+
CH3CH2CH2Br
CH3CH2CH2
L
CH3
Glossary Many textbooks have an extensive glossary that can be helpful when learning new concepts and reviewing for exams. They are a wonderful, but underused resource! This list is limited primarily to the words that appeared in bold in this book. 1,2 addition 1,4 addition Acetal Acetylene Acid Acidity Activating substituent Acyl Acyl halide Alcohol Aldehyde Aldol Addition Aldol Condensation Alkane Alkene Alkoxymercuration Alkyl groups Alkyl halide Alkyne Allyl Alpha (α) C Amide Amine Angle strain
Anhydride Anion Anti Anti addition Anti-aromatic compound Anti-bonding orbital Anti-Markovnikov
see 'direct addition' see 'conjugate addition' two OR groups one the same C. formed when alcohols attack aldehydes simplest alkyne: HC≡≡CH. alkynes are sometimes referred to as substituted acetylenes. generally interpreted as 'Bronsted-Lowry acid,' i.e. a proton donor acid strength (determined ability to donate a proton, stability after donating that proton) makes a compound more reactive C==O group attached to C and/or H C==O attached to a halogen ROH molecule, an alkyl group attached to an OH C==O group attached to one H and one alkyl group first step of the Aldol Condensation two steps: 1) the α C of a ketone/aldhyde attacks the C==O of another ketone/aldehyde, 2) dehydration HC molecule with exclusively single bonds HC molecule with at least one double bond technique for adding an alcohol (ROH) to a double or triple bond without allowing for rearrangement. followed by reduction. see hydrocarbon groups RX molecule, an alkyl group is attached to a halogen HC molecule with at least one triple bond adjacent to a double bond (either an adjacent C or a substituent attached to it) adjacent to a C with a functional group contain carbonyl groups attached to N alkyl group attached to a N-group. Can be primary, secondary, tertiary, or quaternary depending on how many C's are attached to N the result of unstable bond angles. 3-membered rings have considerable angle strain. 5-membered rings have very little angle strain, and 6membered rings have none (when in their most stable arrangements) two C==O groups separated by O negatively-charged atom substituents are 180° apart from one another two groups add to opposite sides of a molecule cyclic, planar, contains an even number of pairs of pi electrons. resonance can occur throughout the ring. very unstable higher-energy molecular orbitals, produced by out-of-phase overlap violates Markovnikov's rule
P
Reaction Aprotic Aromatic compound Aromatic electrophilic substitution Arrhenius Equation Aryl group Asymmetric carbon Atomic number Atomic weight Axial Base Base peak Basicity Bending vibration Benzyl group Benzyne Bonding orbital Bronsted-Lowry acid Bronsted-Lowry base Carbocation Carbonyl Carboxyl group Carboxylic acid Carboxylic acid ester Catalyst Cation Chair conformation Chemical shift Chemically equivalent protons Chiral center Cis
Claisen Condensation Concerted Condensed structures Conformational isomers Conformations
does not contain protons that it can donate, an aprotic solvent does not have protons that will interact with molecules in the solution cyclic, planar, contains an odd number of pairs of pi electrons. resonance can occur throughout the ring. very stable an electrophile replaces an atom/group in an aromatic compound calculates the rate constant of a reaction benzene or substituted benzene chiral carbon, attached to four different groups the number of protons in an atom combined total of protons and neutrons in an atom vertical, groups are pointing straight up or straight down generally interpreted as 'Bronsted-Lowry base,' i.e. a proton accepter in a mass spectrum: represents the most abundant fragment. set at an arbitray value of 100 (relative to the other peaks) base strength (determined by ability to attract a proton, stability after accpeting that proton) a bond's angle changes when it is exposed to radiation benzene ring and one adjacent C benzene ring in which one of the double bonds is replaced by a triple bond lower-energy molecular orbitals, produced by in-phase overlap proton donor proton accepter positively-charged C C==O group COOH (a carbonyl and an OH) RCOOH molecule, an alkyl group attached to a carboxyl group see 'ester' increases a reaction's speed without being consumed or used up positively-charged atom most stable conformation for cyclohexane in NMR: a measure of each signal's frequency, relative to a reference compound attached to the same C or in symmetrical positions on a molecule. produce the same signal in an NMR graph molecule attached to four different groups pointing in the same direction (both up/towards you or both down/away). in cis double bonds, similar groups are pointing in the same direction. the α C of an ester attacks the C==O of another ester all-at-once written like text, no bonds drawn out bonds are arranged differently due to rotation (e.g. staggered vs. eclipsed) see 'conformers'
Q
Conformers Conjugate acid Conjugate addition
Conjugate base Conjugated Constitutional isomers Coupling Covalent bonds Cumulated Cumulene Cyclic compound Deactivating substituent Debye (D) Degree of unsaturation Dehydration Dextrorotatory Diastereomers Diastereotopic hydrogens
Dipole moment Direct addition
E configuration Eclipsed Electron delocalization Electron releasing groups (ERG) Electron withdrawing groups (EWG) Electronegativity Electrophile Electrophilic addition Elimination Enamine Enantiomeric excess
different spatial arrangements of individual bonds, determined by rotation. see staggered, eclipsed, and skewed. a base becomes its conjugate acid by accepting a proton electrophilic addition to conjugated double bonds, in which resonance occurs between the electrophile adding and the nucleophile adding. the groups wind up at C1 and C4, if the first C in the conjugated system is labeled C1 an acid becomes its conjugate base by donating a proton separated by one single bond connected differently (e.g. straight-chain vs. branched) reaction of organometallic compounds that creates new C-C bonds bonds formed by sharing electrons. stronger than ionic bonds. adjacent - not separated by one single bond contains consecutive double bonds contains a ring makes a compound less reactive units of measurement for dipole moment, 1D = 1 x 10-18esu cm total of pi bonds (one pi bond per double bond, two per triple bond) and rings within a molecule removal of HOH rotates polarized light in a clockwise direction versions of the same chiral molecule that are different, but not enantiomers (e.g. RR and RS). include at least 2 chiral centers two H's on the same C - replacing one will produce an S chiral center, replacing the other will produce an R stereocenter. overall results will be 2 diastereomers (there must be another chiral center in the molecule already!) a measure of bond polarity determined by multiplying the charge on either atom (e) by the distance between them (d) electrophilic addition to conjugated double bonds, in which the electrophile and nucleophile add to the same bond. the groups wind up at C1 and C2, if the first C in the conjugated system is labeled C1 in double bonds: high-priority groups are pointing in opposite directions, similar to trans least stable arrangement for bonds, in which substituents are lined up with one another see 'resonance' donate electrons to a system or molecule, either through resonance or induction pull electrons away from a system or molecule, either through resonance or induction tendency of an atom to attract electrons electron-loving, a molecule or atom that is attracted to negative charges an electrophile is attacked by the molecule it's being added to. in alkenes, this is followed by a nucleophile attacking the molecule. removal of atoms/groups, creates a double bond double bond (ene) and N-group (amine) attached to the same C excess of one enantiomer that is present in a mixture of 2 enantiomers
R
Enantiomers Enantiotopic hydrogens
Endergonic Endo product Endothermic Enol Enthalpy (H) Entropy (S) Epoxide ERG Equatorial Ester Ether EWG Excited state Exergonic Exo product Exothermic Fingerprint region
First-order reaction Formal charge Functional group Functional group region Gauche Gauche interactions (G) Geminial dihalide Gibbs free energy change (ΔG°) Gillman reagent Grignard reagent Ground state Haloform reaction
chiral molecules that are non-superimposable mirror images of each other. S and R versions of the same molecule are enantiomers. two H's on the same C - replacing one will produce an S chiral center, replacing the other will produce an R stereocenter. overall results will be 2 enantiomers. a reaction that consumes more energy than it releases. will have a positive ΔG° placement for substituents: closer to the longer or less saturated bridge a reaction that absorbs heat one double bond (ene) and one OH group (ol) attached to the same C heat, ΔH° = (heat of the products) – (heat of the reactants) disorder or chaos, increased by going from solid to liquid to gas or by increasing the number of molecules contains a 3-membered ring with 2C and 1O see 'electron releasing groups' diagonal, groups are slanting upwards or downwards RCOOR' molecule in which an alkyl group is attached to a COOR' group (like a carboxyl group with R' instead of H) ROR molecule, two alkyl groups with one O between them see 'electron withdrawing groups' higher-energy state, electrons may jump up to higher-energy orbitals a reaction that releases more energy than it consumes. will have a negative ΔG° placement for substituents: closer to the shorter or more saturated bridge a reaction that releases heat unique to each molecule, it is next to impossible to recognize a molecule simply by looking at its fingerprint region (without the help of a computer!) one molecule in the rate-determining step the charge on an individual atom in a molecule reactive part(s) of a molecule; refers to anything except for saturated alkyl groups region in an IR spectroscopy chart in which most functional groups appear substituents are 60° apart from one another strain from groups being close together. see section two for instructions on measuring this. two halogens attached to the same C overall change in energy after a reaction: (free energy of the products) – (free energy of the reactants) R2CuLi, an organometallic compound containing 2 R-groups, copper, and lithium RMgX, an organometallic compound containing magnesium and a halogen lowest-energy state, all electrons in lowest-energy orbitals the α H's in a methyl ketone/aldehyde are all replaced by halogens. the CX3 group can then be replaced by an OH
S
Halogen Hammond's Postulate
Hemiacetal
Hemiketal Heterocyclic compound Heterolytic cleavage Highest occupied molecular orbital (HOMO) HOMO Homolytic cleavage Huckel's Rule Hybrid orbitals
Hydrate Hydride shift Hydroboration Hydrocarbon (HC) groups Imine Infrared spectroscopy Iniation In-phase Internal alkyne Inversion of configuration Ion Ionic bonds Isolated Isomers Isotopes IUPAC names
column 7 in the periodic table: F, Cl, Br, I compares the transition state to reactants and products. The transition state (TS) represents the highest-energy point of a reaction. Therefore, the TS is similar to the reactants in reactions that releases energy (exergonic) and similar to the products in reactions that consume energy (endergonic). one OH group and one OR group on the same C. the C is attached to an H and an R-group. formed when an alcohol attacks an aldehyde one OH group and one OR group on the same C. the C is attached to an H and an R-group. formed when an alcohol attacks a ketone contains a ring that is not exclusively composed of C's (may include an N or O) a bond breaks, with both electrons going to one atom
highest-energy molecular orbital that contains electrons see 'highest occupied molecular orbital' a bond breaks, with one electron going to each atom aromatic compounds must have (4n + 2) pi electrons in their resonance systems, where n = any integer combinations of s and p orbitals that allow molecules to bond in certain ways. sp orbitals, for example, allow C to have two double bonds or one triple and one double bond C attached to two OH groups, forms when HOH attacks a ketone or aldehyde H, along with its bond, moves during a reaction to create a more stable carbocation technique for anti-Markovnikov addition of water, in which the H goes to the most substituted C molecules, or sections of molecules, composed of C and H. often labeled 'R' groups contains C==N bond method for identifying molecules that involves exposing them to infrared (IR) radiation and measuring bonds' vibrations first step in a radical substitution reaction: no radicals to radicals orbitals overlap favorably to produce bonding orbitals triple bond is somewhere in the middle of the molecule stereochemistry changes during a reaction: S becomes R, R becomes S charged atom bonds between oppositely-charged atoms. weaker than covalent bonds. bond is based on attraction between ions, not shared electrons. separated by more than one single bond molecules with the same chemical formula that are structurally different. see constitutional, conformational, and stereoisomers. atoms of the same element that have different numbers of neutrons and, therefore, different atomic weights scientific names for molecules, based on the rules established by the International Union of Pure and Applied Chemistry
T
Kekule structures Ketal Keto-enol tautomerism Ketone Kinetic control Kinetic product Kinetic stability LCAO LDA Leaving group Levorotatory Lewis acid Lewis base Lewis structure Lindlar's catalyst Linear combination of atomic orbitals (LCAO) Lone pairs Lowest occupied molecular orbital (LUMO) LUMO m/z Major product Markovnikov Reaction Markovnikov's Rule Mass spectrometry
Mass spectrum Meso compound
Meta Meta director Methyl shift Minor product
use lines to represent covalent bonds, may include dots to represent lone pairs two OR groups one the same C. formed when alcohols attack ketones interconversion of ketone/aldehyde and enol tautomers C==O group attached to two alkyl group reactions proceed as quickly as possible, without regard to how stable the products will be forms the fastest reactions are kinetically stable when they proceed slowly, and have high energy difference between the reactants and transition states see 'linear combination of atomic orbitals' lithium-diisopropyl amide, a very strong base replaced in a substitution reaction rotates polarized light in a counterclockwise direction accepts a pair of electrons donates a pair of electrons diagrams in which the valence electrons of an atom are drawn around its chemical symbol stops hydrogenation of an alkyne to form a cis double bond essentially, we need to end with the same number of orbitals we begin with. For C, we begin with 4 orbitals (s p p p), and we need to end with 4 orbitals after hybridization pairs of electrons that aren't used in bonding
lowest-energy molecular orbital that does not contain electrons see 'lowest unoccupied molecular orbital' mass/charge most likely to form, either because it is the most stable product (thermodynamic conditions) or the fastest to form (kinetic conditions) obeys Markovnikov's rule in electrophilic addition, H adds to the C with the most H's (the least substituted C). method for identifying molecules; involves exposing a molecule to an energy beam, which causes it to break into radical and cation fragments. the cations' masses and relative abundance are measured a graph of the relative amounts of each fragment obtained through mass spectroscopy chiral molecule that is its own mirror image, and has no enantiomer. must include at least 2 chiral centers, each of which is attached to the same groups, and be either RS or SR. has a plane of symmetry, which may or may not be obvious from the picture. arrangement of substituents on benzene: 2 groups are 2 C's away from each other substituent on benzene that causes the next atom/group to add to a meta position methyl group, along with its bond, moves during a reaction to create a more stable carbocation less likely to form, either because it is less stable (thermodynamic control) or forms more slowly than the major product (kinetic control)
U
Molecular orbitals Newman projections Nodes Non-aromatic compound Nonpolar bonds Nuclear magnetic resonance spectroscopy (NMR) Nucleophile Nucleophilic Acyl Substitution (SNAc) Nucleophilic Aromatic Substitution (SNAr) Nucleophilic Substitution (SN) Observed specific rotation Optical activity Optical purity Orbitals Organic compounds Organometallic compound Ortho Ortho-para director Out-of-phase Oxidation Oxymercuration Para Parent chain Pericyclic Phenyl group Pi bonds pKa Polar bonds Propogation
formed when the orbitals of different atoms within one molecule overlap look directly down an individual bond, useful in studying different arrangements that bonds can get into when they rotate regions where we do not find electrons, created by out-of-phase interactions neither aromatic nor anti-aromatic covalent bonds in which electrons are shared equally (or almost equally) between atoms method for identifying molecules that involves exposing them to a magnetic field; specific atoms will show signals. 'NMR' usually means 'proton NMR' nucleus-loving, a molecule or atom that is attracted to positive charges (the nucleus includes positive-charged protons!). nucleophiles are bases. a nucleophile replaces an atom/group attached to a C==O. a nucleophile replaces an atom/group in an aromatic compound nucleophile replaces an atom or group degree to which a sample rotates polarized light ability of a molecule to rotate polarized light; chiral molecules are optically active excess of one enantiomer that is present in a mixture of 2 enantiomers region around the nucleus in which an electron is most like to be present contain C contains C and a metal arrangement of substituents on benzene: 2 groups are on adjacent C's substituent on benzene that causes the next atom/group to add to an ortho or para position orbitals overlap unfavorably to produce anti-bonding orbitals several equivalent terms: losing H; increase in CO, CN, and CX bonds, loss of electrons technique for adding HOH to a double or triple bond without allowing for rearrangement. followed by reduction. arrangement of substituents on benzene: 2 groups are 3 C's away from each other longest chain of consecutive carbons in a molecule. used to determine the name of a compound. concerted reaction that involves the cyclic rearrangement of electrons (e.g. Diels-Alder) benzene ring combinations of p orbitals, produce double and triple bonds a measure of acid strength - the lower the pKa, the stronger the acid covalent bonds in which electrons are not shared equally between atoms. these bonds have partially negative and partially positive ends. middle of a radical substitution reaction: radicals to radicals. this is where the radical chain reaction occurs
V
Pro-R hydrogen Pro-S hydrogen Protecting group
Protic Proton-transfer reactions Racemate Racemic mixture Radical chain reaction Radical initiator Radical substitution reactions Rate constant Rate-determining step (RDS) Reactivity-selectivity principle Rearrangement Reduction Regioselectivity
Resonance Resonance contributors Retention of configuration Retrosynthetic analysis R-groups Ring expansion Ring flip
Saturated Sawhorse projection S-cis configuration Second-order reaction Shielding
replacing this H with a functional group will produce an R chiral center replacing this H with a functional group will produce an S chiral center preserve a functional group while a molecule undergoes reactions that would otherwise change it. afterwards, the functional group can be restored by removing the protecting group. contains protons that it can donate, a protic solvent had protons that can interact with other molecules in solution acid-base reactions in which one molecule (acid) donates a proton to another (base) equal mixure of 2 different enantiomers see 'racemate' during propogation, radicals produce new radicals that can start the whole process over again substance used to create radicals necessary for a reaction. the initiator does not participate in the reaction. one atom or group replaces another through a radical reaction determines the speed at which reactions proceed (along with the number of molecules in the rate-determining step) slowest step of a reaction, determines the rate at which the whole reaction can proceed as reactivity increases, selectivity decreases H's and C's shift in order to produce a more stable carbocation several equivalent terms: gaining H, replacing a funcitonal group with H, addition of electrons preference for which region a reaction (or a step in a reaction) occurs in. when adding to a double bond, an electrophile may have a preference for a particular C. electrons moving within a molecule, not trapped between 2 atoms different theoretical structures a molecule can acquire through resonance stereochemistry does not change during a reaction: S stays S, R stays R, cis stays cis, trans stays trans a method of solving synthesis problems that involves working backwards from the product any part of a molecule that isn’t drawn out; often represents hydrocarbon groups rearrangement in which a C that was outside of a ring becomes part of a ring. most likely to occur when it can create a 5- or 6-membered ring cyclohexanes can transition between 2 different chair conformations. causes axial groups to become equatorial and equatorial groups to become axial, but does NOT change cis and trans arrangements molecule with exclusively single bonds lines used to represent all bonds, including C-H cis-arrangement around a single bond. in the Diels-Alder reaction, the double bonds in the diene need to be pointing in the same direction two molecules in the rate-determining step in NMR: electrons can get in the way of the magnetic field, which
W
Sigma bonds Skeletal structures Skewed SN SNAc SNAr Specific rotation Staggered Stereochemistry Stereoisomers Steric strain Strain Stretching vibration Substituent Syn addition Tautomerism Tautomers
Terminal alkyne Termination Tetrahedral Intermediate Thermodynamic control Thermodynamic product Thermodynamic stability Thioacetal Thioketal Torsional strain Trans
Unsaturated Valence shell Vicinal dihalide Vinyl
weakens the protons' signals. weaker signals have lower frequencies. combinations of s orbitals, produce single bonds (and the first bond in double and triple bonds) lines indicate C-C bonds, H's are not specifically written out (zig-zaglike structures) somewhere in between staggeed and eclipsed see 'nucleophilic substitution' see 'nucleophilic acyl substitution' see 'nucleophilic aryl substitution' degree to which a pure sample of an enantiomer will rotate polarized light comparatively stable arrangement for bonds, in which substituents and H's are as far apart as possible 3D chemistry, the arrangement of molecules in space arranged differently in space, but connected in the same way (e.g. cis vs. trans) occurs when substituents are close together. in general, groups and atoms like to have as much space to themselves as possible. stress on molecules: see angle, steric, and torsional strain a bond's length changes when it is exposed to radiation any group, other than H, that is attached to C two groups add to the same side of a molecule rapid interconversion between tautomers isomers with different arrangement of bonding electrons, different structures a molecule can take on through tautomerism (e.g. ketone and enol) triple bond is at the end of the molecule end of a radical substitution reaction: radicals to no radicals forms when a carbonyl is attacked by a nucleophile. temporarily, the C has four bonds (tetrahedral structure) reactions produce the most stable products possible
most stable reactions produce the most stable products possible acetal formed with RSH groups instead of ROH groups ketal formed with RSH groups instead of ROH groups electrons in different bonds (generally bonds to H or substituents) repel each other when they get too close - occurs in eclipsed conformations pointing in opposite directions (one up/towards you and one down/away). in trans double bonds, similar groups are pointing in opposite directions. molecule with at least one double bond, possibly multiple double/triple bonds. outermost shell of electrons see 'geminal dihalide' attached to a double bond (either a C in the bond or a substituent attached to it)
X
Workup Ylene Ylide Z configuration
clean-up' that occurs at the end of reactions - separates the products from other materials, may protonate or de-protonate final products molecule in which a double bond could move onto one of its atoms, creating an ylide molecule with opposite charges on adjacent atoms in double bonds: high-priority groups are pointing in the same direction, similar to cis
Y
Index 1,2-addition 1,4-addition α C of C==O Acetals Acetylene Acidity aryl compounds (benzene) resonance Acids and bases Bronsted-Lowry conjugate acids and bases Lewis proton-transfer reactions see also 'carboxylic acids' Activating substitutents Acyl halides Friedel-Crafts acylation (benzene) nomenclature nucleophilic acyl substitution synthesis Alcohols acetals/ketals (reacting with aldehydes/ketones) alkenes dehydration (elimination) esters (reacting with carbonyls with leaving groups) hemiacetals/hemiketals (reacting with aldehydes/ketones) making OH a leaving group nomenclature protecting groups synthesis: benzene (nucleophilic aryl substitution) synthesis: alkenes (electrophilic addition) synthesis: alkyl halides (nucleophilic substitution) synthesis: reduction (aldehydes/ketones) Aldehydes and ketones acetals/ketals H and C nucleophiles hemiacetals/hemiketals hydrates N nucleophiles O and S nucleophiles protecting groups thioketals and thioacetals
80 80 see 'alpha C of carbonyl' 143 see 'alkynes' 14 125 77 13 13 13 17 13 see 'benzene' 119, 121 131 133 137 143 52, 53 102 133, 134 143 96 30 143 129 53 92 141 140 143 141 143 142 145 142 143 145 I-1
Wittig reaction Aldehydes nomenclature reactions Aldol Condensation Alkanes nomenclature nucleophilic substitution radical substitution Alkenes alcohols (alkoxymercuration) alkynes (elimination) allyl C's and groups bond angle bromination cis and trans nomenclature creating chiral centers E and Z nomenclature electrophilic addition HOH HX HX and peroxide hydrogenation Markovnikov's Rule nomenclature stability synthesis (elimination) vinyl C's and groups Alkyl halides nomenclature nucleophilic substitution synthesis: alcohols (nucleophilic substitution) synthesis: alkanes (radical substitution) synthesis: alkenes (electrophilic addition) Alkoxymercuration Alkynes bond angle bromination HOH HX hydrogenation internal lengthening the chain nomenclature synthesis (elimination) terminal
147 32 see 'aldehydes and ketones' 154 25 92 87 44 53 102 47 47 51 44 69 45 47 53 48 56, 89, 90 55 50 28, 44 47 100, 102 47 29 60, 92, 152 96 87 48 53 57 57 58 58 57 59 57 60 28, 57 102 57, 60 I-2
Allyl groups alpha C of carbonyl Aldol Condensation alkylation Claisen Condensation enolates haloform reaction halogenation intramolecular reactions keto-enol tautomerism Michael Reaction Amides HOH and ROH (nucleophilic acyl substitution) nomenclature synthesis Amines aldehydes/ketones nomenclature synthesis (nucleophilic substitution) Angle strain Anhydrides nomenclature nucleophilic acyl substitution Anti addition Anti-aromatic compounds Anti-bonding orbitals Aprotic solvents Aromaticity anti-aromatic compounds criteria heterocyclic compounds Huckel's Rule molecular orbital theory non-aromatic compounds stability Arrhenius equation Asymmetric carbons Atomic number Atomic structure Atomic weight Axial and equatorial Back-side attack Base peak Basicity see also 'acids and bases'
149 154 152 157 149 152 151 156, 157 149 152 134 32, 133 131 145, 153 30 92 35 131 133 69 see 'aromaticity' see 'molecular orbital theory' 93 107 108 107 110 108 111 108 109 21 63 1 1 1 34, 38 see 'nucleophilic C356substitution' see 'mass spectroscopy' 14
I-3
Brending vibrations Benzene acidity activation/deactivation acylation (Friedel-Crafts) adding a third substituent alkylation (Friedel-Crafts) benzyl groups benzyne reactions diazonium electrophilic substitution halogenation meta directors nitration nomenclature nucleophilic aryl substitution ortho-para directors phenyl groups reactions of substituents sulfonation Benzyl groups Benzyne Bond angle Bonding orbitals Bronsted-Lowry acids and bases Carbocations resonance stability Carbonyls with leaving groups Carboxylic acid esters Carboxylic acids nomenclature nucleophilic acyl substitution synthesis Chair conformation Chemical shift Chemically equivalent protons Chirality Cis and trans nomenclature alkanes (including cyclohexane) alkenes Claisen Condensation Cleavage Concerted reactions Conformational isomers
see 'infrared spectroscopy' 114 125 125 119 127 120, 121 114 129 123 115 116 126 117 114, 126 129 126 114 122 118 see 'benzene' 129 35, 47, 57 see 'molecular orbital theory' see 'acids and bases' 47 75, 76 49 see 'nucleophilic acyl substitution' see 'esters' 31, 132 134, 137 131, 134 see 'cyclohexane' see 'NMR' see 'NMR' 63 39 44 157 51,87 82 33,41 I-4
eclipsed, skewed, staggered strain Conjugated acids and bases Conjugate addition Conjugation Constitutional isomers Coupling Covalent bonds Cyclohexane axial and equatorial chair conformation cis and trans nomenclature ring flip strain Cumulenes Deactivating groups Degree of unsaturation Dehydration Delocalization Diastereomers Diastereotopic hydrogens Diels-Alder mechanism lining molecules up S-cis configuration stereochemistry Dipole moment Direct addition Drawing organic compounds E and Z nomenclature Eclipsed configurations Electron delocalization Electronegativity Electron-releasing groups (ERG) Electron-withdrawing groups (EWG) Electrophiles Electrophilic addition see also 'alkenes' Electrophilic aromatic substitution see also 'benzene' Elimination alkene to alkyne competition with substitution dehydration mechanism
34 35 see 'acids and bases' see 1,4-addition 75,80 41 see 'organometallic compounds' 2 34, 38 39 39 40 35 70 see 'benzene' 43 102 see 'resonance' 67 69 81 81 84 82 82 3 see 1,2-addition 5 see 'alkenes' see 'conformational isomers' see 'resonance' 3 85 85 47 47 115 100 102 103 102 100, 102 I-5
nucleophile strength solvents stereochemistry Enamines alkylation of alpha C synthesis (N nucleophiles and aldehydes/ketones) Enantiomeric excess Enantiomers Enantiotopic hydrogens Endergonic reactions Endo products Endothermic reactions Enolates Enthalpy Entropy Epoxides nomenclature nucleophilic substitution organometallic compounds Equatorial ERG Esters Claisen Condensation nomenclature substitution Ethers EWG Excited state Exergonic reactions Exo products Exothermic reactions Fingerprint region First-order reactions Fischer projections Formal charge Functional group region Functional groups (general, nomenclature) Gauche interactions Geminal dihalides Gibbs free energy Gillman reagents Grignard reagents Ground state Haloform reaction
100, 102 100, 102 101, 102 153 146 see 'optical activity' 65,66 68 17 18, 84 18 149 18 18 30 97 104, 105 see 'axial and equatorial' 85 157 31, 132 134 29 85 2 17 18, 84 18 see 'infrared spectroscopy' 21, 94, 101 64 3 see 'infrared spectroscopy' 27 35 57 17 see 'organometallic compounds' see 'organometallic compounds' 2 152 I-6
Hammond's postulate Hemiacetals Hemiketals Heterocyclic compounds Heterolytic cleavage Highest occupied molecular orbital (HOMO) HOMO Homolytic cleavage Huckel's rule Hybridization linear combination of atomic orbitals (LCAO) orbitals Hydrates Hydride shift Hydroboration-oxidation Imines Infrared spectroscopy bending vibrations fingerprint region functional group peaks functional group region stretching vibrations Initiation Inversion of configuration Ionic bonds Ions Isomers Isotopes IUPAC nomenclature Ketals Keto-enol tautomerism Ketones reactions nomenclature Kinetics and Thermodynamics control stability products LCAO Leaving groups halides OH and OR Lewis acids and bases Lewis structures Linear combination of atomic orbitals (LCAO Lone pairs
20 143 143 110 52 see 'molecular orbital theory' see 'molecular orbital theory' 87 107 6 7 7 142 see 'rearrangement' 54 146 161 161 162 162 162 161 see 'radicals' 93 2 2 41 1 23 143 59, 149 see 'aldehydes and ketones' 32 17, 81 20 20 81 see 'hybridization' 92 96 see 'acids and bases' 2 see 'hybridization' 1, 75, 76 I-7
M + 1 fragments Markovnikov's Rule Mass spectrometry base peak Br and Cl isotopes fragmentation functional groups M + 1 fragments m/z values process Meso compounds Meta directors Methyl shift Michael Reaction Molecular orbital theory anti-bonding orbitals aromaticity bonding orbitals highest occupied molecular orbital (HOMO) in-phase lowest unoccupied molecular orbital (LUMO) nodes out-of-phase pi bonds resonance sigma bonds m/z values Newman projections NMR chemical shift chemically equivalent protons functional groups shielding signal height signal splitting Nodes Nomenclature acyl halides alcohols aldehydes alkanes alkenes alkynes amides amines anhydrides
see 'mass spectroscopy' 50 158 159 160 158 160 160 158 158 67 see 'benzene' see 'rearrangement' 153 8 8, 9 111 8, 9 13 8, 9 13 8, 11 8, 9 8, 9 77 8, 9 see 'mass spectroscopy' 33 163 163 163 164 164 165 165 see 'molecular orbital theory' 23 131 30 32 25 28, 44, 45 28, 57 32, 133 30 131 I-8
benzene (aryl compounds) carboxylic Acids epoxides esters ethers ketones prefixes Non-aromatic compounds Non-polar bonds Nuclear magnetic resonance spectroscopy Nucleophiles Nucleophilic acyl substitution acid-catalyzed mechanism mechanism reacting with acids reactivity tetrahedral intermediate Nucleophilic aromatic substitution see also 'benzene' Nucleophilic substitution back-side attack competition between reactions competition with elimination epoxides intramolecular mechanisms nucleophile strength replacing OH and OR solvents stereochemistry Observed specific rotation Optical activity enantiomeric excess observed specific rotation optical purity racemates specific rotation Orbitals atomic hybrid molecular Order of reactions Organometallic compounds Coupling reactions Gillman reagents Grignard reagents
114, 126 31, 132 30 31, 132 29 32 24 see 'benzene' see 'polarity' see 'NMR' 48 131 134 134 137 133 134 129 92 93 95 103 97 97 92,94 93,94 96 93,95 93,95 see 'optical activity' 71 73 71 72 71 71 1 6 see 'molecular orbital theory' 21 104 104 105 105 I-9
Organolithium Ortho-para directors Oxymercuration-reduction Out-of-phase Pericyclic reactions Perspective formulas pH Phenyl groups pKa Polarity Pro-R and Pro-S hydrogens Propogation Protecting groups Protic solvents Proton-transfer reactions R,S nomenclature Racemates Radicals allylic bromination chain reaction initiation nomenclature propogation regioselectivity stability substitution reactions termination Rate-determining step Reaction order Reactivity-selectivity principle Rearrangement hydride shift methyl shift ring expansion Regioselectivity Resonance acidity molecular orbital theory stability Retention of configuration Retrosynthetic analysis Ring expansion Ring flip Rotation of bonds S-cis configuration Saturation
104 see 'benzene' 54 see 'molecular orbital theory' 82 63 15 see 'benzene' 15 3, 33 68 see 'radicals' 143 93 see 'acids and bases' 64 71 90 88 87 56,89,90 87 88 88 87 87 20 21 89 50 51 51 51 49,88 75 77 77 77 83, 95 61 see 'rearrangement' see 'cyclohexane' 33 see 'Diels-Alder' 25, 43 I-10
Second-order reactions Skewed conformation Specific rotation Staggered conformation Stereoisomers (stereomers) Steric strain Strain Stretching vibrations Substitution Syn addition Tautomerism Termination Tetrahedral intermediate Thermodynamics Thioketals and thioacetals Torsional strain Trans Valence shells Vicinal dihalide Vinyl groups Wittig Reaction Workup Ylene Ylide Z configuration
21, 92, 100 see 'conformational isomers' see 'optical activity' see 'conformational isomers' 42 35 35 see 'infrared spectroscopy' see 'nucleophilic substitution' 69 58, 149 see 'radicals' see 'nucleophilic acyl substitution' see 'kinetics and thermodynamics' 145 35 see 'cis and trans' 1 57 see 'alkenes' 147 22 147 147 see 'E and Z nomenclature'
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