EXAM II
***** Answer Key *****
Problem 1 (total 20 points, each question worth 2
points)
Which of the following statements are true (write "true or "false", as applicable, in front of
the question on the same line):
- All compounds which have carbon atom with four different ligands are chiral.
- Enantiomers are stereoisomers of a chiral compound.
- Enantiomers of the same compound have different physical properties.
- Tereoisomers with different configurations are enantiomers.
- Diastereomers are compounds of the same constitution, but not related through the
mirror reflection operation.
- Meso-compounds have chiral carbon atoms and are therefore chiral.
- Compounds which only have an axis of symmetry are always chiral.
- cis- and trans-isomers are diastereomers.
- All structures which are related to each other as an object to its mirror image are enantiomers.
- The d-prefix means the compound has the same configuration as d-glyceraldehyde.
ANSWER:
- False, because meso compounds which are achiral have chiral carbon
atoms( for different ligands).
- True, by definition of enantiomers.
- False, the physical properties of enantiomers such as melting point,
boiling point, reflective index ..etc are the same only fluroptical
properties are different.
- False, for example, diastereomers have different configurations, yet
they are not related by mirror plane reflection.
- True, by definition of diastereomers.
- False, meso compounds have a plane of symmetry and are hence achiral.
- True, only compounds which have a plane or center of symmetry are
achiral; all others are chiral.
- True, cis/trans isomers are compounds of the same constitution, but
are not related by mirror plane of reflection according to the definition
of diastereomer.
- False, because to be enantiomers such structures must also be
nonsuperimposable.
- False, the lower case of d-refers to the sign of optical rotation not
configuration. Compounds which have the same optical rotation do not
necessarily have the same configuration.
Problem 2 (total 20 points, 4 points each)
- List symmetry elements (n-fold axis, a plane, or a center of symmetry) of the structures shown below.
- Draw where approximately in the structure the element may be applied.
- State in what relation it is to the structure (e.g. in plane of the structure, vertical in plane of
projection, horizontal perpendicular to the plane of projection, etc.).
- If no symmetry is detected, state "NONE", however, in such case identify the chiral center by an asterisk (*).
ANSWER:
Problem 3
Part I: Designate the RS configurations of each chiral centers in
the structures shown below (12 points, 3 for each chiral center). Be sure
to label each ligand of the chiral carbon atoms with the appropriate precedence
numbers (from 1 to 4). If two chiral centers share the same ligand, redraw
the structure to show separately precedence at each center.
ANSWER:
Part II: Draw the Fisher projection of
2S,3R-dibromobutane (10 pts).
ANSWER:
Problem 4
Part I: (12 points, 3 each) Define the relation of each of the the structures B through E to the structure A shown below as:
- Identical (ID);
- Enantiomeric (ENT);
- Diastereomeric (DIA);
ANSWER:
- Determine configuration of each compound A to E.
- Compounds C and E have the same configuration. Thus C and E are ID.
- Compound B has both configurations opposite to those of A. Thus A and
B are ENT.
- Compound D has one configuration opposite and one the same as A. Thus
A and D are DIA.
Part II: (8 points total, 2 for each
problem). Which of the four pairs of the structures shown below are:
- Identical (ID);
- Enantiomeric (ENT);
- Diastereomeric (DIA);
- Constitutional isomers (CON);
ANSWER:
Problem 5
Define which of the conformations in each row shown below is more stable
(18 points, 6 points for each problem).
ANSWER:
For Part I
- A is most stable due to H-bonding between hydroxyl group and a large
distance between methyl groups.
- B is less stable due to repulsion between methyl groups.
- C is least stable because of loss of H-bond and existence of repulsion
between methyl groups.
For Part II