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226  Answers
31. (A) The longest chain has six carbon atoms and only single bonds, which makes it a 
hexane. The two substituents (bromo and chloro) alphabetically give bromochlorohexane. 
Numbering from the end that gives the substituents the lowest numbers gives the answer.
32. (C) An acid anhydride consists of two carbonyl groups linked by an oxygen atom.
33. (C) The isomers are n-hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbu-
tane, and 2,3-dimethylbutane. 
34. (D) An isobutyl group is a 2-methylpropyl.
35. (A) The nitrogen adjacent to a carbonyl is an amide. Two carbon atoms make this a 
substituted acetamide. The –OH group is para (number 4) on the phenyl ring. The ring is 
attached to the N, making it an N-.
36. (B) The chain splits at the oxygen. There are five carbon atoms to the left (pent) and 
three carbon atoms to the right (prop). An oxygen atom adjacent to a carbonyl can be an 
ester (-oate suffix). The acid side of the ester has the carbonyl group. The part of an ester 
derived from a five-carbon chain is a pentanoate, with the carbonyl carbon being 1. If the 
carbonyl carbon is 1, then the fluorine is on 4. Combining these gives a 4-fluoropentanoate. 
The propyl group, from the alcohol, remains and, according to the nomenclature rules, is 
the first part of the name.
37. (D) Every ring is one degree of unsaturation, and every π bond is one degree of unsatu-
ration. Answer A gives 3, answer B gives 3, answer C gives 3, and answer D gives 4. 
38. (A) This is the only reduction (the only answer where the oxidation state is reduced).
39. (D) The longest chain has seven carbon atoms and only single bonds; therefore, this 
is a heptane. The branches are methyl, bromo, and hydroxyl (alcohol). The alcohol has 
highest priority; therefore, this is an alcohol, with the –OH being closest to the end of the 
chain. This gives heptan-2-ol. Following through on the numbering will yield 4-bromo and 
5-methyl. Adding these to the root name (alphabetically) gives the answer.
40. (D) An acetal is R2C(OR)2. The Rs may be the same or different.
41. (C) The compound has an eight-carbon backbone with a double bond (octene) and a 
carboxylic acid group (octenoic acid). Numbering starts at the end closest to the acid end 
(2-octenoic acid), and the R groups are oriented on opposite sides of the double bond, 
making it an (E) conformation.
42. (A) There are two basic structures, propene and cyclopropane. There are three cyclo-
propane isomers, (1,1), (cis-1,2), and (trans-1,2). There are seven propene isomers, (1,1), 
(2,3), (3,3), (cis-1,2), (trans-1,2), (cis-1,3), and (trans-1,3).
43. (D) If the compound were saturated, it would have 2(12) + 2 = 26 hydrogen atoms, 
instead of 14. The degree of unsaturation is (26 − 14)/2 = 6.