review-questions-for-MCAT-OC-BC-pdf_pag239

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78. (B) Sodium sulfate is a soluble ionic compound, so its solubility will be greater than 
those of the other compounds, even though they can all hydrogen-bond to water.
79. (D) The boiling point of a compound is highly influenced by the intermolecular forces 
between the compounds. The stronger the intermolecular forces, the higher the boiling 
point. Compound I can hydrogen-bond to other molecules, while compound II can use 
only weaker dipole-dipole forces.
80. (A) These two structures are diastereomers of each other, not mirror images. The left 
one is S and S around its stereocenters, while the right is S and R.
81. (C) These two compounds are mirror images of each other, making them enantiomers.
82. (A) N1 has only single bonds (three of them); therefore, it is sp3 hybridization. N2 has 
a double bond and a single bond, which indicates sp2 hybridization. Each nitrogen atom 
has a lone pair (not shown). 
83. (A) There are three chiral carbon atoms (n = 3) on the ring containing the oxygen, so 
there are 2n = 8 stereoisomers.
84. (C) 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 OH group, followed by the COOH group. 
Beginning with the highest-priority group (OH here), count toward the next highest group 
(COOH 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 
both carbons are R.
85. (B) The side chain is a sec-butyl group, and alkyl groups are nonpolar.
86. (A) The optical rotation might be changed or might not be changed; therefore, it is 
impossible to determine the effect without more information.
87. (A) To determine the absolute configuration about a carbon atom, note the lowest-
priority group (in the first and third carbons, it is an H, and in the second case, it is an ethyl 
group) and assign priority to the other groups. In this case, the highest-priority group on 
each C is the R group containing the ether group followed by the OCH3 group. Beginning 
with the highest-priority group, count toward the next highest group (OCH3 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 both carbons 1 and 2 
(counting from right to left) are R and the third carbon is S.
88. (D) The molecules are the same. Rotating the right-hand structure until the –OCH3 
is vertical yields the same structure as the left-hand structure.