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100. (A) The number of meso forms does not affect the number of stereoisomers, just the 
number of chiral centers. The number of isomers may be calculated by the formula 2n.
101. (B) Diastereomers may have different physical properties, but their molecular formu-
las must be the same.
102. (D) Enantiomers have opposite configurations and optical rotations. Therefore, com-
pound C would have an S absolute configuration and a negative optical rotation.
103. (D) The two compounds are cis/trans isomers (or diastereomers), and as such, they 
are expected to have different physical properties altogether. There are differences in one 
chiral center, so there would be a difference in the optical activity.
104. (C) To determine the absolute configuration about a carbon atom, note the lowest-
priority group (in both carbons in the ring, it is a hydrogen) and assign priority to the other 
groups. In the case of both carbons, it is the nitrogen group, followed by the isopropyl. 
Beginning with the highest-priority group, count toward the next highest group and end 
with the lowest group. Counting to the right (clockwise) is R, and counting to the left 
(counterclockwise) is S. The configuration about each carbon atom is independent of the 
configuration about the other carbon atom. Doing this, the direction of rotation is to the 
left (counterclockwise), and so it is S for both.
105. (D) In order to have hydrogen bonding, there must be a hydrogen atom attached 
to an extremely electronegative element: O, N, or F. The only structures that meet that 
criterion are the alcohol and the amine, III and IV. 
106. (A) To determine the absolute configuration about a carbon atom, note the lowest-
priority group (in both of these cases an H) and assign priority to the other groups. In 
this case, the highest-priority group on each C is the Br group, followed by the ethyl, and 
then the methyl group. Beginning with the highest-priority group (Br here), count toward 
the next highest group (ethyl here) and end with the lowest group. Counting to the right 
(clockwise) is R, and counting to the left (counterclockwise) is S. The configuration about 
each carbon atom is independent of the configuration about the other carbon atom. Doing 
this indicates that structure I is R and structure II is S.
107. (B) Simply examining the chlorine atoms is sufficient to eliminate most of the 
answers. The chlorine atoms are in different positions relative to the aldehyde group, as 
one is 3-chloro and the other is 4-chloro. Therefore, the two compounds are not identical, 
conformers, or enantiomers, which leaves regioisomers.
108. (A) Meso compounds have stereocenters, and there is a plane of symmetry splitting 
the molecule. In I and IV, there is no plane of symmetry present. In II, there is a plane of 
symmetry between the two chiral centers. In III, there is a plane of symmetry cutting the 
molecule in two. 
109. (D) Compound I can form two hydrogen bonds, compound II cannot form any 
hydrogen bonds, and compound III can form one hydrogen bond. The compound with 
the greatest number of hydrogen bonds will have the highest boiling point, and so on.