SAMPLE EXAM

***** Answer Key *****

Question 1:

Part A: 2,4-Dimethylglutaric acid is a component of inhibitor of the angiotensin-converting enzyme(ACE), pentopril. Is this compound chiral? Why? Why not?

ANSWER: The isomer of 2,4-dimethylglutaric acid shown is chiral, since it does not have the plane of symmetry. It has however, a two-fold axis of symmetry running vertically in the plane of the drawing dissecting the C₂-C₃-C₄ and H_c-C-H_d planar angle in half.

Part B: In the synthesis of pentopril from the above dicarboxylic acid, one of the carboxylic group is converted into an amide. Would conversion of one carboxylate give a different product than the conversion of the other one? Why?

ANSWER: Regardless of which carboxylic acid residue is converted into amide, only one compound is formed. This can be shown by converting each carboxyl group separately, and superimposing the product amides upon each other.

Part C: Define stereochemical descriptors(R,S) for racemic 2,4-dimethylglutaric acid.

ANSWER: Both chiral center are of R configuration.

Question 2:

Assign configurations(R, S descriptors and cis/trans geometry) in the enantiomer of H-1 antihistamine agent levocabastine, show below.

ANSWER: The configuration of levocabastine is 3S, 4R, r-1', c-4' (since nitrile group has a higher precedence than the phenyl group). The configuration in the piperidine ring is more precisely defined with RS descriptors, than with cis/trans descriptors. The later would only define the relative configuration at the carbon atoms 3 and 4. Both sites at 1' and 4'-positions are achiral due to the presence of two identical substituents(CH₂CH₂). The nitrogen site is achiral due to a rapid pyramidal inversion.

Question 3:

Mark(encircle) chemical residues of higher precedence in the following pairs according to Cahn-Ingold-Prelog rules:

ANSWER:

Ethynyl > t-butyl;
ethylamino > methyldeuterioamino;
phosphonate > sulfate(P > O);

Question 4:

Determine the relationship between the two compounds below whether they are:

Superimpossable without bond rotation
Diastereomers
Enantiomers
Conformers

ANSWER: Diastereomers( the same configuration at the top chiral center and a reversed configuration at the bottom chiral center). The best way to solve this problem is to assign RS descriptors in each structure, after making sure they have the same constitution.

Consider a step-by step process below:

Step 1: Convert structure A to a Wedge projection similar to structure B.
Step 2: Rotate the NHCH₃ around C₂of structure A upward. The following video demonstrates step 2. Note that hydrogen atoms of the methyl groups and the benzen ring are not shown on the video and image below.
Step 3: Examine both A and B, the hydrogen and the methyl group are displaced between the two compounds.

Question 5

Part A: Draw an approximate energy profile for 1,2-dibromoethane molecule as a function of the torsion angle about the C-C bond.

ANSWER: Anologous to the energy profile of n-butane. Please see Potential Energy of Butane.

Part B: Label the graph with the Newman projections of conformers corresponding to energy minima and maxima.

Note: You need not to know numerical values of energy differences between conformers to draw the graph.

ANSWER: Please see Potential Energy of Butane.