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55. (C) To determine the absolute configuration about a carbon atom, note the lowest-
priority group and assign priority to the other groups. Beginning with the highest-priority 
group (Cl in this case), 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, both carbons turn out to be S.
56. (A) Achiral compounds are those that are superimposable on their mirror images. I and 
II cannot be superimposed, but III can. It is a meso compound, as the chlorine atoms are 
cis, creating a symmetry plane between them.
57. (D) You are told that both have a chiral center; there is no way to determine the rela-
tionship (+ or –) of either compound. Changing atoms (F for Cl) may or may not retain the 
+ or –. There is no information given or predictable about the type (direction) of the rotation.
58. (D) The glycine is achiral; therefore, it is possible to produce pure glycine. The other 
natural amino acids occur as D and L enantiomers. Laboratory synthesis gives a mixture of 
equal amounts of the D and L enantiomers. The mixture of equal amounts of the D and L 
enantiomers is called a racemic mixture. It would not be optically active (eliminating answer 
A); mesomeric and diastereometric are not terms used to describe enantiomers.
59. (B) A chiral center has four different groups attached to it. In 2,3-hydroxypentane, 
there are two chiral centers, the carbon atoms with hydroxy groups. Each of those carbons 
has two different R groups, a hydrogen atom and a hydroxy group, attached to it.
60. (A) It is impossible to know what the optical rotation of the product will be. The 
mechanism does cause inversion of the configuration; however, this does not mean inver-
sion of optical rotation.
61. (A) The presence of the lone pair on the nitrogen atom leads to the formation of the 
double bond to it (N=X), and the nitrogen is best able to accommodate the positive charge.
62. (D) Assign the priorities to each atom (Br > F > -CH2-C- > CH3). Then see if the 
Br →F → and so on rotation is to the right (R) or to the left (S). Answer D has the proper 
absolute configuration on each atom.
63. (B) These are enantiomers. 
64. (D) The trans fluorine atoms cancel, while in all other cases the fluorine atoms do 
not cancel.
65. (A) The compounds are isomers; therefore, their molar masses are the same. Only ethyl 
methyl amine is capable of forming hydrogen bonding, while trimethyl amine cannot.
66. (A) To determine the absolute configuration about a carbon atom, note the lowest-
priority group and assign priority to the other groups. 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