Question

Draw the product formed when the following amino acid is treated with each reagent: (a)${\mathbf{CH}}_{\mathbf{3}}\mathbf{OH}\mathbf{;}{\mathbf{H}}^{\mathbf{+}}$; (b)${\mathbf{CH}}_{\mathbf{3}}\mathbf{COCl}$, pyridine; (c) HCl (1 equiv); (d) NaOH (1 equiv); (e)${\mathbf{C}}_{\mathbf{6}}{\mathbf{H}}_{\mathbf{5}}\mathbf{N}\mathbf{=}\mathbf{C}\mathbf{=}\mathbf{S}$.

Short answer

The ball and stick model represents the amino acid phenylalanine. The reaction of phenylalanine with methanol in a mild acid occurs as an esterification reaction.

• With acetyl chloride in pyridine, the amine group of the amino acid will react.

• With HCl, the basic group that is the amine group, will react.

• With the base, NaOH, the carboxylic acid group of the amine will react.

Step-by-step solution

The given amino acid

The ball and stick model represents the amino acid phenylalanine. The reaction of phenylalanine with methanol in a mild acid occurs as an esterification reaction.

• With acetyl chloride in pyridine, the amine group of the amino acid will react.

• With HCl, the basic group that is the amine group, will react.

• With the base, NaOH, the carboxylic acid group of the amine will react.

The reaction (a)

The reaction (a) is an example of an esterification reaction. The carboxylic acid group of the phenylalanine reacts with methanol under mildly acidic conditions. The reaction is as shown:

The reaction (b)

In reaction (b), the compound acetyl chloride reacts with the amine group of the phenylalanine and forms the following product:

The reaction (c)

In reaction (c), HCl reacts with the amine group of the phenylalanine and forms the following product:

Step 5:The reaction (d)

In reaction (d), the base, NaOH, reacts with the carboxylic acid group of the amine.

The reaction is as shown below:

Step 6:The reaction (e)

In reaction (e), benzylisothiocyanate reacts with phenylalanine and forms the following product: