1-sec-butyl-2-ethylcyclopentane. (d). 1-cyclobutyl-3-methylpentane e). 3-
isopropyl-2,5-dimethylhexane. (f) cis-1-isopropyl-3-pentylcyclohexane ...
Chemistry 217, Problem Set 5 Recommended Problems from the text: 4.2—4.4, 4.8, 4.11-4.17, 4.20-4.21, 4.23-4.24, 4.26-4.32, 4.37-4.38, 4.40-4.44, 4.47-4.49, 4.53-4.60, 4.70, 5.2-5.11 (1st ed.: 4.2-4.5, 4.10-4.16, 4.19-4.20, 4.22-4.24, 4.26-4.31, 4.36-4.37, 4.39-4.42, 4.454.47, 4.51-4.56, 4.57-4.58, 4.68, 5.2-5.11) Supplemental Problems: Nomenclature: Klein, Ch. 5; Newman projections: Klein, 6.16.2 Cyclohexanes: Klein, Ch. 6.3-6.7; Stereochemistry: Klein, Ch. 7 1.
2.
Draw structures of the following molecules.
(a)
1-tert-butyl-2-methylcyclobutane
(b)
(c)
2-cyclohexyl-4,4-dimethylpentane
(d)
1-isopropyl-3-methylcyclopentane
1,1,2,2,3,3,-hexaethylcyclohexane
3-ethyl-2,4,6-trimethylheptane (e) Give the IUPAC names for the following compounds:
a)
1-ethyl-1-methylcyclohexane
c)
1-sec-butyl-2-ethylcyclopentane
e)
3-isopropyl-2,5-dimethylhexane
(b)
(d)
(f)
isopropylcyclodecane
1-cyclobutyl-3-methylpentane
cis-1-isopropyl-3-pentylcyclohexane
3-tert-butyl-1,5-diethylcycloheptane
g)
(h) 3-ethyl-2,4,4,5-tetramethylheptane
3. Draw an energy diagram for all conformations of the following molecules when they are rotated about the bond indicated with an arrow. You may have drawn your Newman diagrams in different way, but the relative energies should be the same and the number or ecliped/gauche interactions should be the same. (a) 1 Me-Me eclipsed 1 Me-H eclipsed 1 H-H eclipsed
1 Me-Me eclipsed 1 Me-H eclipsed 1 H-H eclipsed H3C CH3
H3C CH3 H H3C
H H
H
H
CH3 H
3 Me-H eclipsed H CH3
2 Me-Me gauche 2 Me-H gauche 2 H-H gauche H3C
H H 3C
CH3 CH3
H
H
H
1 Me-Me gauche 4 Me-H gauche 1 H-H gauche H3C H
CH3 H H CH3
1 Me-Me gauche 4 Me-H gauche 1 H-H gauche H H
CH3 CH3 H CH3
CH3 H
(b) 1 Et-Me eclipsed 1 Me-H eclipsed 1 H-H eclipsed H3CH2C CH3
1 Et-H eclipsed 1 Me-Me eclipsed 1 H-H eclipsed
H
H
CH3 H
H3C CH3 H H3CH2C
1 Et-H eclipsed 2 Me-H eclipsed
H H
1 Et-Me gauche 1 Me-Me gauche 1 Et-H gauche 1 Me-H gauche 2 H-H gauche H3CH2C
CH3 CH3
H
H H
2 Et-H gauche 1 Me-Me gauche 2 Me-H gauche 1 H-H gauche H3C H
CH3 H H CH2CH3
H CH3 H H3C
1 Et-Me gauche 1 Et-H gauche 3 Me-H gauche 1 H-H gauche H H
CH3 CH2CH3 H CH3
CH2CH3 H
(c) 1 iPr-Me eclipsed 1 Me-H eclipsed 1 H-H eclipsed
1 iPr-Me eclipsed 1 Me-H eclipsed 1 H-H eclipsed H3C H
H
H3C
H
1 iPr-H eclipsed 2 Me-H eclipsed H CHH3
H H3C
1 Me-iPr gauche 1 iPr-H gauche 3 Me-H gauche 1 H-H gauche
H
H
H3C
H
H
H
CH3
4.
2 iPr-Me gauche 2 Me-H gauche 2 H-H gauche
H
CH3
Draw the flat structure of the following compounds. H H
H
(a)
H3C
CH3
CH3
H
(c)
H
H
H
CH3
CH3
CH3
H
H H
H
H
(b)
CH2CH2CH3
H
(d)
H
H CH3
1 Me-iPr gauche 1 iPr-H gauche 3 Me-H gauche 1 H-H gauche
H3C
H
CHH3
CH3
H3C H
H
H
H H3C
H CH3
CH3
H
H
H3C
(e) (f) 5. Name each of the compounds in question 5 (when a benzene ring is a substituent as it is in part b, it is called a phenyl). (a) 2-methylbutane (b) 2-phenylbutane (c) hexane (d) cyclopentane (e) 1,1-dimethylcyclopentane (f) 1,2-dimethylcyclopentane 6. For each compound below, convert the chair structure into a flat drawing. 6. For each compound below, convert the chair structure into a flat drawing. Br
Br
HO
OH
(a)
Br
(b)
OH
Br
Br Br
(c)
OH HO
HO Br
OH
OH Br
7. Draw both chair conformations of the following molecules, and determine which one is most stable.
(a)
(b)
(c)
8. For each compound below, draw the opposite chair conformation and determine which of the two is most stable.
(a)
2 ax Me 1 eq. Et
1 ax Et 2 eq. Me
1 ax t-Bu 1 eq. Me
1 eq. t-Bu 1 ax. Me
(b)
3 eq. Me 1 ax. Me
(c)
3 ax Me 1 eq. Me
9. The structures of five sugars are drawn below. Rank them from least to most stable. Some of them may have very similar stabilities. OH
OH O
OH
HO
OH
O
OH
OH
OH O OH
OH
OH
HO HO
least stable
galactopyranose HO
O OH
1 axials
OH O OH
HO HO
HO OH
no axials
OH OH
OH
O OH
OH
HO
gulopyranose
HO
2 axials
OH OH
mannopyranose
OH
O
OH
HO
OH
O OH
HO HO
OH
HO
glucopyranose
OH
10.
OH
OH
OH O
OH
altropyranose
HO
OH
HO
OH
OH
OH O
2 axials
altropyranose = gulopyranose < mannopyranose = galactopyranose < glucopyranose
1 axials
most stable
Define the following terms, and give an example of each. a. conformers: orientations of a molecule which result from rotation about single bonds. Three examples: CH3 Cl
Cl
H
and HO
HO
Cl
H
HO CH3
CH3 CH3 and H HO
and
H Cl
(notice that these Newman diagrams are the same conformations as the compounds shown on the left).
b. enantiomers: two stereoisomers which are mirror images of each other, but are non-superimposable. Examples:
and
and
d. chiral molecule: a molecule which has an enantiomer. Examples:
e. stereocenter (stereogenic center): a carbon which has four different things attached to it. Examples (all stereocenters are starred). Cl HO
11.
*
* *
*
*
Locate all stereocenters, if any, in the following molecules: HO
O
* *
O O
O
O F3 C
*
O
*
* HN
O
*
*
*
N H
Prozac
* *
Zocor
(a)
Ritalin
12.
Determine if the following molecules are chiral or achiral. If it is chiral, draw the enantiomer of the compound. If it is achiral, locate the mirror plane.
a)
(b)
(b)
The mirror image of this compound is the same compound. Therefore, it is achiral. The mirror plane bisects the molecule, and contains the methyl group and the hydrogen.
The symmetry of this molecule is now broken, and the mirror images are not superimposable. There is no longer a mirror plane, so the molecule is chiral. The enantiomer is the mirror image. The enantiomer can also be drawn like this:
(c)
same
redraw to see mirror plane:
H carbon has: 1 methyl, 1 hydrogen, 2 ethyl groups Since it does not have four different groups, it is achiral. The mirror image is superimposable. The mirror plane contains the hydrogen and the methyl group.
(c)
non-superimposable
The symmetry of this molecule is now broken, and the mirror images are not superimposable. There is no longer a mirror plane, so the molecule is chiral. The enantiomer is the mirror image. The enantiomer can also be drawn like this: not a stereocenter stereocenter
(d)
Br
Br
Br
Br
Even though this molecule contains two chiral carbons, the molecule is achiral because there is a mirror plane (it is meso).
(e)
(f)
There is no longer a mirror plane, so this molecule is chiral. The enantiomer is:
Br
Br
13. The 72 students taking organic chemistry right now and their parents represent 16 different countries. The flags of those countries are shown below. Examine each flag for planes of symmetry. Some of them have no symmetry plane, some have a vertical plane, some have horizontal plane, and some have both a vertical and horizontal plane (disregard the symmetry plane in the plane of the page that all flags have).
Canada (vertical)
China (no plane)
Germany (vertical)
India (vertical)
Iraq (vertical w/o writing incl. writing: no plane)
Mauritius (vertical)
Nepal (no plane)
Pakistan (no plane)
Poland (vertical)
Singapore (no plane)
Sweden (horizontal)
United Arab Emirates (no plane)
Vietnam (vertical)
Zambia (no plane)
United Kingdom (no plane)
United States (no planee)
