Question

Draw a structural formula for the product formed when alanine is treated with the following reagents. (a) Aqueous \(\mathrm{NaOH}\) (b) Aqueous \(\mathrm{HCl}\) (c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}, \mathrm{H}_{2} \mathrm{SO}_{4}\) (d) \(\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O}, \mathrm{CH}_{3} \mathrm{COONa}\)

Short answer

Answer: The products formed are as follows: (a) CH3CH(NH2)COO-Na+ (b) CH3CH(NH3+)COOH (c) CH3CH(NH2)COOCH2CH3 (d) CH3CH(NH2)COOCH3

Step-by-step solution

Write the structure of alanine

The structure of alanine can be written as follows: \(\mathrm{CH}_{3}\mathrm{CH}\left(\mathrm{NH}_{2}\right)\mathrm{COOH}\)

React with aqueous NaOH

Since aqueous \(\mathrm{NaOH}\) is a strong base, it will deprotonate the acidic \(\mathrm{COOH}\) group in alanine. The reaction can be represented as: \(\mathrm{CH}_{3}\mathrm{CH}\left(\mathrm{NH}_{2}\right)\mathrm{COOH} + \mathrm{NaOH} \rightarrow \mathrm{CH}_{3}\mathrm{CH}\left(\mathrm{NH}_{2}\right)\mathrm{COO}^{-}\mathrm{Na}^{+} + \mathrm{H}_{2}\mathrm{O}\)

React with aqueous HCl

Since aqueous \(\mathrm{HCl}\) is a strong acid, it will protonate the basic \(\mathrm{NH}_{2}\) group in alanine. The reaction can be represented as: \(\mathrm{CH}_{3}\mathrm{CH}\left(\mathrm{NH}_{2}\right)\mathrm{COOH} + \mathrm{HCl} \rightarrow \mathrm{CH}_{3}\mathrm{CH}\left(\mathrm{NH}_{3}^{+}\right)\mathrm{COOH} + \mathrm{Cl}^{-}\)

React with CH3CH2OH and H2SO4

The presence of \(\mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH}\) and concentrated \(\mathrm{H}_{2}\mathrm{SO}_{4}\) indicates a Fischer esterification reaction. In this reaction, the \(\mathrm{COOH}\) group in alanine will react with the \(\mathrm{OH}\) group in ethanol to form an ester. The reaction can be represented as: \(\mathrm{CH}_{3}\mathrm{CH}\left(\mathrm{NH}_{2}\right)\mathrm{COOH} + \mathrm{CH}_{3}\mathrm{CH}_{2}\mathrm{OH} \xrightarrow[\mathrm{H}_{2}\mathrm{SO}_{4}]{} \mathrm{CH}_{3}\mathrm{CH}\left(\mathrm{NH}_{2}\right)\mathrm{COOCH}_{2}\mathrm{CH}_{3} + \mathrm{H}_{2}\mathrm{O}\)

React with (CH3CO)2O and CH3COONa

In this reaction, the reagent \((\mathrm{CH}_{3}\mathrm{CO})_{2}\mathrm{O}\) in the presence of \(\mathrm{CH}_{3}\mathrm{COONa}\) performs an esterification of the carboxylic acid group in alanine. This reaction will form an ester linkage between the alanine and acetate. The reaction can be represented as: \(\mathrm{CH}_{3}\mathrm{CH}\left(\mathrm{NH}_{2}\right)\mathrm{COOH} + (\mathrm{CH}_{3}\mathrm{CO}){2}\mathrm{O} \xrightarrow[\mathrm{CH}_{3}\mathrm{COONa}{}]{} \mathrm{CH}_{3}\mathrm{CH}\left(\mathrm{NH}_{2}\right)\mathrm{COOCH}_{3} + \mathrm{CH}_{3}\mathrm{COOH}\)