Solutions to Review Problems

1. 1. A
      N-methyl-3-butenamide
   2. B
      Propanoyl bromide
   3. C
      Pentanamide
   4. D
      Propyl propanoate
   5. E

      Ethanoic anhydride (common name: acetic anhydride)

2. D
   Treating a carboxylic acid with thionyl chloride results in the production of an acyl chloride. In this reaction, butanoic acid is converted to butanoyl chloride, which is choice (D). Since none of the other choices are acyl chlorides, they can be eliminated.
3. C
   Hydrolysis of an acid chloride results in the formation of a carboxylic acid and HCl. Since pyridine can act as a base, it serves to neutralize the HCl that is formed. The reaction does not result in the formation of hydroxide ions, so choice (A) is incorrect. Pyridine does not react with the carboxylic acid product, it reacts with the HCl, so choice (D) is incorrect. Finally, choice (B) is incorrect because the acyl chloride is very reactive. The correct answer is choice (C).
4. B
   In this question, an acid chloride is treated with an alcohol, and the product will be an ester. However, the esterification process is affected by the presence of bulky side-chains on either reactant. It is easier to esterify an unhindered alcohol than a hindered one. In this reaction, the primary hydroxyl group is less hindered and will react with benzoyl chloride more rapidly, so choice (B) is correct. Choice (A) is incorrect because the hydroxyl group is a hindered secondary hydroxyl, and the reaction rate will be slower. Choice (C) is incorrect because it is not an ester. Choice (D) is incorrect because steric hindrance would prevent this product from being formed.
5. A
   Acid chlorides react with ammonia or other amines to form amides. Since the amine is replacing the hydroxyl group of the carbonyl, there must be at least one hydrogen on the amine. Therefore, only ammonia and primary or secondary amines can undergo this reaction. In order to obtain a primary amide, ammonia, choice (A), must be used. The reaction of an alcohol with an acid chloride produces an ester, so choice (B) is incorrect. A primary amine reacting with an acid chloride would give an N-substituted amide; thus, choice (C) is incorrect. Choice (D) is incorrect because tertiary amines will not react with acid chlorides.
6. A
   Methyl propanoate is an ester, which can be synthesized by reacting a carboxylic acid with an alcohol in the presence of acid: choice (A). Reacting ethanol with propanoyl chloride, choice (B), will also result in the formation of an ester, but because ethanol is used, ethyl propanoate will be formed, not methyl propanoate. This is also the case for choice (D), since ethanol is used here as well. Therefore, choices (B) and (D) are incorrect. Choice (C) is incorrect because propanoyl chloride will not form an ester in the presence of base alone. Therefore, choice (A) is the correct response.
7. D
   This question asks for the products when ammonia reacts with acetic anhydride. Recall from the text that an amide and an ammonium carboxylate will be formed. The only choice showing such a pair is (D), acetamide and ammonium acetate.
8. C
   This question gives a reaction scheme for the interconversion of propanoic acid to various derivatives, and asks what intermediate products are formed. The first reaction involves the formation of an acid chloride using thionyl chloride. Acid chlorides are made by replacing the hydroxyl group with chlorine. Thus, choices (A) and (B), which depict intact hydroxyl groups, can be eliminated. The second reaction is nucleophilic attack by ammonia on propanoyl chloride. The product should be propanamide, since ammonia will replace the chloride on the carbonyl carbon, which is correctly shown in both answer choices (C) and (D). The final reaction involves amide hydrolysis in the presence of base; therefore, the resulting carboxylic acid will exist in solution as a carboxylate salt. Thus, choice (C) is the correct answer, since it has sodium propanoate as the product of the third reaction.

9. This question asks about preparation of pentanoic acid derivatives. Where there is more than one way to make the products, the most efficient method will be given.

   1. A

      1-Pentanol

      Carboxylic acids are easily reduced by LAH to produce the corresponding primary alcohol.

   2. B

      Pentanoyl bromide

      To form pentanoyl bromide, pentanoic acid is reacted with PBr₃. The bromide replaces the hydroxyl on the carbonyl carbon.

   3. C

      N-Methylpentanamide

      N-Methylpentanamide can be prepared by first producing the acid chloride using thionyl chloride, and then reacting it with methylamine to yield the amide.

   4. D

      Ethyl pentanoate

      The ethyl ester of pentanoic acid can be formed by reacting it directly with ethanol in the presence of hydrochloric acid.

   5. E

      Pentanoic anhydride

      The most common method of preparing anhydrides is the reaction between an acid chloride and a carboxylate anion. To form pentanoic anhydride, one mole of pentanoic acid must be treated with thionyl chloride to yield pentanoyl chloride. This reacts with one mole of sodium pentanoate to form pentanoic anhydride.

10. Saponification is the process whereby fats are hydrolyzed under basic conditions to produce soaps.