Name __________________________
Exam 1 SOLUTION
All of the questions were the same, but rearranged on the different exams.
1. [8 pts] For the following molecule,
a. Complete the Lewis structure by filling in missing bonds and lone pairs (use the skeleton
structure given, do not rearrange the atoms).
b. Assign formal charges, and list non-zero charges on the molecule.
c. Draw resonance structures for the molecule.
d. If one resonance structure is better than the other, indicate this, and give a reason for this
indication. If the structures are equivalent, indicate that.
There should be 24 electrons in the molecule (CH3NO2) - it is neutral. To draw the resonance
structures, note that the electrons move, but not the atoms!
H
H
H
O
C
N
H
C
O
H
H
H
O
N
H
O
C
O
N
O
H
Since the negative charge goes from one oxygen to the
next, there is no difference in resonance structures.
2. [8 pts] The structure of caffeine is shown below. Give the molecular formula for caffeine, and
provide the hybridization for the indicated atoms. Place your answers in the boxes provided.
O
Molecular Formula
N
N
O
C8H12N4O2
N
N
sp2
sp3
Don’t forget the lone pair on N!
3.
[10 pts] Forskolin, a natural plant extract, has recently been shown to stimulate melanin production
in mice, possibly leading to a topical “sunless tanning” lotion that can chemically ”tan” skin. (Nature
2006, 443, 340) Identify the chiral centers in forskolin by marking each one with an asterisk (*).
How many stereoisomers of forskolin are possible?
Number of Isomers? 2
O
OH
*
8
= 256
*
*
*
*
OH
*
O
O
*
*
O
OH
Exam 1
Fall 2006
Page 1 of 4
4. [8 pts] Draw the line structure for 2-methylbutane. Looking down the C2–C3 bond, draw Newman
projections for the lowest energy conformation and the highest energy conformation.
All unique Newman projections, in order of increasing energy:
H
H
2-methylbutane
CH3
H
CH3
CH3
H
CH3
CH3
H
CH3
H
H CH3
H
CH3
H
H3C
Staggered
H
CH3
CH3
H
H
Eclipsed
Lowest Energy
H
H CH3
Highest Energy
H CH3
CH3
H
H
CH3
CH3
Staggered
H
CH3
CH3
Eclipsed
All staggered conformations are lower in energy than eclipsed conformations. The lowest energy
conformer minimizes the number of gauche interactions, and the highest energy conformer maximizes
the number of eclipsed methyl groups.
5.
[12 pts] Draw both chair conformers of each molecule below, and circle the more stable conformer.
CH3
H
H
H3C
HO
OH
menthol
H
H
H
CH3
H
H
CH3
H
CH3
CH3
Br
Exam 1
OH
More stable - two
groups equitorial (esp.
the isopropyl group!)
CH3
Br
H
H
Fall 2006
H3C
CH3
H
H
Br
More stable - two
groups equitorial
Page 2 of 4
6. [12 pts] Nomenclature. Give the IUPAC (or common) names for the following molecules. Be sure to
indicate stereochemistry, where appropriate.
2
4 3
5
7
1
F
2-fluoro-3,4,5-trimethylheptane
No stereochemistry is indicated, so none is given in the name.
6
Br
4
6
3
5
2
OH
1
(2S, 3S)-2-bromohex-4-en-3-ol or (2S, 3S)-2-bromo-4-hexen-3-ol
Non-carbon functional groups take priority in numbering, so the
alcohol group should have the lowest number possible.
cis-1-t-butyl-3-methylcyclopentane or (1S, 3R)-1-t-butyl-3-methylcyclopentane
You can indicate stereochemistry using either method.
Short answer. Briefly explain the following observations or answer the questions (1-3 sentences).
Use drawings or figures when appropriate. Full credit will be given for correct vocabulary usage.
7. [8 pts] Explain why cyclopentyne has never been isolated.
Cyclopentyne
8.
The carbons on the triple bond are sp-hybridized, and would have an ideal bond
angle of 180°. However, to close up into a 5-membered ring, the angle would have to
“squeeze” down to 120° (That’s the internal angle of a pentagon). Doing so would
create a great deal of ring strain, so much that the molecule has never been isolated!
[8 pts] Draw the stable Newman projection for propane. Label the dihedral angle. Explain why the
staggered configuration is more stable than the eclipsed configuration.
Dihedral angle, !
H
H
H
H
H
CH3
Staggered
9.
In the staggered conformation, the dihedral angle is 60°, whereas the
dihedral angle is 0° in the eclipsed conformation. This allows for two
stabilizing “factors”. (1) In the staggered conformation, there is less steric
hinderance, where the side groups can run into each other and cause
torsional strain. (2) In the staggered conformation, the bonding orbitals
on one group overlap directly with an empty antibonding orbital off of the
neighboring carbon, allowing for electron delocalization.
[8 pts] Use Fisher projections to draw all isomers of 2,3-difluoropentane and label the enantiomeric
pairs.
F
CH3
CH3
CH3
CH3
F
H
H
F
H
F
F
H
H
F
F
H
H
F
F
F
H
H
H
H
H
H
H
The moelcule is
H
H
asymmetric, so
CH3
CH3
CH3
CH3
there are no meso
compounds!
enantiomers
enantiomers
Exam 1
Fall 2006
Page 3 of 4
Multiple Choice. Circle the answer that best answers the question. [3 pts each, 18 pts]
10. Which of the following represents a cis isomer? For a cis isomer, the two groups must either
both be “up” or both be “down”.
CH3
CH3
CH3
CH3
CH3
CH3
11. How many structural isomers exist for the formula C2H6O?
a. 1
b. 2  one alcohol and one ether (C-C-O and C-O-C)
c. 3
d. 4
12. Select the most energetically stable molecule. A greater degree of branching yields greater
stability and a more negative ∆H.
13. Which molecule has the lowest boiling point? Boiling points increase when chains have less
branching and more surface area (greater dispersion, or London forces).
14. Which of the following statements is true?
a. All mirror images are enantiomers. No! Some are meso compounds!
b. All molecules that have stereocenters are chiral. No! Not those pesky meso compounds!
c. Isomers that are not superimposable on their mirror images are enantiomers.
d. Superimposable structural isomers are enantiomers. Structural isomers have different
connectivity- it can’t be a mirror image!
15. What are the correct stereochemistry configurations for the following molecule?
3 CH2OH
F1
H
H
2
OH
CH2CH3
CH2OH
H
F
2
b. 2R, 3S
1
H
OH
CH2CH3
3
Exam 1
a. 2R, 3R
c. 2S, 3R The alcohols are at 1 and 3, the F at carbon 2.
d. 2S, 3S
Fall 2006
Page 4 of 4