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
(1 st 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, 4.45-
4.47, 4.51-4.56, 4.57-4.58, 4.68, 5.2-5.11)
Supplemental Problems: Nomenclature: Klein, Ch. 5; Newman projections: Klein, 6.1-
6.2 Cyclohexanes: Klein, Ch. 6.3-6.7; Stereochemistry: Klein, Ch. 7
1. Draw structures of the following molecules.
(a) 1-tert-butyl-2-methylcyclobutane (b) 1-isopropyl-3-methylcyclopentane
(c) 2-cyclohexyl-4,4-dimethylpentane (d) 1,1,2,2,3,3,-hexaethylcyclohexane
2.
(e) 3-ethyl-2,4,6-trimethylheptane
Give the IUPAC names for the following compounds: a)
1-ethyl-1-methylcyclohexane
(b) isopropylcyclodecane c)
1sec -butyl-2-ethylcyclopentane
(d)
1-cyclobutyl-3-methylpentane e)
3-isopropyl-2,5-dimethylhexane
(f) cis -1-isopropyl-3-pentylcyclohexane

g)
3tert -butyl-1,5-diethylcycloheptane
(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
H
3
C
CH
3
H
H
H
H
3
C
1 Me-Me eclipsed
1 Me-H eclipsed
1 H-H eclipsed
H
3
C
CH
3
H
H
CH
3
H
2 Me-Me gauche
2 Me-H gauche
2 H-H gauche
H
3
C
CH
3
CH
3
H H
H
3 Me-H eclipsed
H
H
H
CH
3
CH
3
H
3
C
1 Me-Me gauche
4 Me-H gauche
1 H-H gauche
H
3
C
CH
3
H
H
CH
3
H
1 Me-Me gauche
4 Me-H gauche
1 H-H gauche
H
CH
3
CH
3
H
CH
3
H

(b)
2 Et-H gauche
1 Me-Me gauche
2 Me-H gauche
1 H-H gauche
H
3
C
CH
3
H
H
CH
2
H
CH
3
1 Et-H eclipsed
1 Me-Me eclipsed
1 H-H eclipsed
H
H
3
C
CH
3
H
3
CH
2
C
H
H
1 Et-Me gauche
1 Me-Me gauche
1 Et-H gauche
1 Me-H gauche
2 H-H gauche
H
3
CH
2
C
H
CH
3
CH
3
H
H
1 Et-Me eclipsed
1 Me-H eclipsed
1 H-H eclipsed
H
3
CH
2
C
CH
3
H
H
CH
H
3
1 Et-H eclipsed
2 Me-H eclipsed
1 Et-Me gauche
1 Et-H gauche
3 Me-H gauche
1 H-H gauche
H
CH
3
CH
2
CH
3
H
CH
3
H
H
H
3
H
CH
3
CH
2
H
CH
3
C

(c)
1 iPr-Me eclipsed
1 Me-H eclipsed
1 H-H eclipsed
H
3
C
H
H
H
CH
3
1 iPr-Me eclipsed
1 Me-H eclipsed
1 H-H eclipsed
H
H
3
C H
CH
3
1 iPr-H eclipsed
2 Me-H eclipsed
H
3
C
H
CH
3
H
4.
1 Me-iPr gauche
1 iPr-H gauche
3 Me-H gauche
1 H-H gauche
H
3
C
H
H
CH
3
H
1 Me-iPr gauche
1 iPr-H gauche
3 Me-H gauche
1 H-H gauche
H
3
C
H
H
CH
3
H
Draw the flat structure of the following compounds.
H
3
C
H
H
CH
3
H
2 iPr-Me gauche
2 Me-H gauche
2 H-H gauche
H
H H
(a)
H
3
C
CH
3
CH
3
H
(c)
H
H H
CH
3
CH
2
CH
2
CH
3
(b)
H
H
H
CH
3
CH
3
(d)
H
H
H
H

H H
H
3
C
CH
3
H
CH
3
H H
3
C
(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
Br
Br Br
HO
OH
HO OH
(a) Br
OH (b) OH HO (c)
Br
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

(b)
1 eq. t-Bu
1 ax. Me
1 ax t-Bu
1 eq. Me
(c)
3 eq. Me
1 ax. Me
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 OH
OH
OH
O OH O OH O OH
O OH O OH
HO
OH altropyranose
OH HO OH
OH glucopyranose
HO OH
OH mannopyranose
HO
OH
OH
O
OH
OH
2 axials
HO
HO
HO no axials
OH
O
OH
HO
HO
OH
OH
O
OH
1 axials
HO
OH gulopyranose
OH
OH
OH
O
OH
HO OH
2 axials
HO OH
OH galactopyranose
HO
HO
OH
O
OH
HO
1 axials least stable altropyranose = gulopyranose < mannopyranose = galactopyranose < glucopyranose
Cl and most stable
10. 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:
CH
3 CH
3
CH
3
Cl
H and and
HO HO Cl HO H
H HO
H
Cl
CH
3
(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
O
F
3
C
*
HN
*
O
*
N
H
*
*
*
*
*
(a) Ritalin (b) Prozac (c) Zocor
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)
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.
(b)
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:

(e)
Br
(f)
(c)
(d)
Br 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.
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
Br
Even though this molecule contains two chiral carbons, the molecule is achiral because there is a mirror plane (it is meso).
Br 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)
Mauritius
(vertical)
China
Nepal
(no plane)
Germany
(no plane) (vertical)
Pakistan
(no plane)
India
(vertical)
Poland
(vertical)
Iraq
(vertical w/o writing incl. writing: no plane)
Singapore
(no plane)
Sweden
(horizontal)
Vietnam
(vertical)
United Arab
Emirates
(no plane)
Zambia
(no plane)
United Kingdom
(no plane)
United States
(no planee)