Answers and Explanations

DISCRETE PRACTICE QUESTIONS

1 . D

Glycine’s   R   group   is  a   hydrogen    atom;   this    amino   acid    is  therefore   achiral because the central

carbon  is  not bonded  to  four    different   substituents.   The other   amino   acids   are all chiral  and

therefore   have    both    L-  and D-enantiomers.

2 . B
This reaction is similar to the Gabriel synthesis. Phthalimide acts as a nucleophile, the methyl
carbon acts as an electrophile, and bromide acts as the leaving group. Therefore, the reaction
between methyl bromide and phthalimide results in the formation of methyl phthalimide.
Subsequent hydrolysis then yields methylamine.

3 . B
Cysteine is well known for containing a sulfur atom because it is able to form disulfide bridges;
however, methionine also contains a sulfur atom in its R group.

4 . B
An amide is formed from an amine and a carboxyl group or its acyl derivatives. In this question,
an amine is already given; the compound to be identified must be an acyl compound. The only
acyl compound among the choices given is a carboxylic acid.

5 . C
During the Strecker synthesis, ammonia attacks a carbonyl, forming an imine, choice (B). This
imine is attacked by cyanide, forming an amine, choice (D), and a nitrile, choice (A). Amide bonds
are formed between amino acids, but do not appear during the Strecker synthesis.

6 . C