CHEM 3390 2016R
Problem Set #1 ANSWERS
Friday September 25 2015 due Friday October 2 2015. 20 points total, 5 for each question.
1. For each of the circled groups, indicate the appropriate stereochemical relationship.
a.
b.
c.
d.
e.
Enantiotopic
Diastereotopic
Homotopic
Enantiotopic
Diastereotopic
2. The molecule shown is chiral. Briefly explain how this is possible. How many distinct
stereoisomers are theoretically possible for this structure?
The individual phenyl rings of the biphenyl units cannot be coplanar, so they twist about the
central C-C bond. This creates a spiral-like geometry that is chiral. There are FOUR possible
stereoisomers as shown. Structures 1 and 4 are enantiomers of one another. Their
diastereomers 2 and 3 are also enantiomers of each other.
NOTE that the CH2 groups cannot be stereogenic centres!
In fact, it turned out that the barrier to rotation around the right-hand biphenyl unit was quite
small so it flipped back and forth rapidly at room temperature. Only the left-hand biphenyl unit
was configurationally stable. Thus, only a pair of enantiomers (shown as the intermediate step
in the scheme above) could be observed.
Bringmann, G.; Mühlbacher J.; Reichert, M.; Dreyer, M.; Kolz, J.; Speicher, A. J. Am. Chem. Soc.,
2004, 126, 9283–9290.
3. Use Newman projections to draw all the conformations possible for meso-2,3-butanediol.
Which conformer will have the lowest energy? Why?
On purely steric grounds we might expect the anti conformer to be most stable. However, in the
gauche conformation there is H-bonding between the adjacent OH groups. This adds stability
and overcomes the energy-raising due to the CH3/CH3 gauche relationship. Note that the gauche
energy is about 0.8 to 0.9 kcal mol-1, but a typical H-bond between hydroxyls lowers energy by
close to 5 kcal mol-1.
NOTE: eclipsed geometries are not “possible conformations”. As we noted in class, they are
TRANSITION STATES and not local minima. A conformation must be a stable structure with a
finite lifetime.
See http://chemistry.stackexchange.com/questions/6647/which-is-most-stable-conformationof-optically-inactive-butane-2-3-diol.
4. Draw a specific stereoisomer of 1-bromo-1-phenyl-2,3-dimethylbutane that could form
exclusively Z-1-phenyl-2,3-dimethylbut-1-ene on treatment with DBU in warm CH2Cl2. (DBU =
1,8-diazabicyclo[5.4.0]undec-7-ene, see Klein p 385).
This is right out of second-year introductory organic! E2 eliminations are stereospecific,
requiring an antiperiplanar alignment of the C-H and C-X bonds. The (S,S) enantiomer would also
work but the (S,R) and (R,S) compounds would lead to the E alkene product.