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

a) The reaction with aqueous NaOH forms a carboxylate ion. The product's structural formula is H2NCH(CH3)COO- Na+. b) The reaction with aqueous HCl forms an ammonium ion. The product's structural formula is H3N+CH(CH3)COOH. c) The reaction with CH3CH2OH and H2SO4 forms an ester. The product's structural formula is H2NCH(CH3)COOCH2CH3. d) The reaction with (CH3CO)2O and CH3COONa forms an amide. The product's structural formula is H3N+CH(CH3)COOH · CH3CO- .

Step-by-step solution

Identifying the Reaction

In this reaction, alanine is treated with aqueous \(\mathrm{NaOH}\), which is a strong base. The base will react with the acidic carboxyl group in alanine and form a carboxylate ion.

Drawing the Structural Formula

After the reaction with \(\mathrm{NaOH}\), the carboxyl group will lose a proton and become an ion. The structural formula for the product is: \(\mathrm{H}_{2}\mathrm{N}\mathrm{CH}(\mathrm{CH}_{3})\mathrm{COO}^{-} \mathrm{Na}^{+}\). #b) Aqueous \(\mathrm{HCl}\)

Identifying the Reaction

In this reaction, alanine is treated with aqueous \(\mathrm{HCl}\), which is a strong acid. The acid will react with the basic amino group in alanine and form an ammonium ion.

Drawing the Structural Formula

After the reaction with \(\mathrm{HCl}\), the amino group will gain a proton and become an ion. The structural formula for the product is: \(\mathrm{H}_{3}\mathrm{N}^{+}\mathrm{CH}(\mathrm{CH}_{3})\mathrm{COOH}\). #c) \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}, \mathrm{H}_{2}\mathrm{SO}_{4}\)

Identifying the Reaction

In this reaction, alanine is treated with ethanol \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\) and sulfuric acid \(\mathrm{H}_{2}\mathrm{SO}_{4}\). The carboxyl group in alanine will react to form an ester.

Drawing the Structural Formula

After the reaction with ethanol and sulfuric acid, the carboxyl group forms an ester with the ethanol molecule. The structural formula for the product is: \(\mathrm{H}_{2}\mathrm{N}\mathrm{CH}(\mathrm{CH}_{3})\mathrm{COO}\mathrm{CH}_{2}\mathrm{CH}_{3}\). #d) \(\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O}, \mathrm{CH}_{3}\mathrm{COONa}\)

Identifying the Reaction

In this reaction, alanine is treated with acetic anhydride \(\left(\mathrm{CH}_{3} \mathrm{CO}\right)_{2} \mathrm{O}\) and sodium acetate \(\mathrm{CH}_{3}\mathrm{COONa}\). The amino group in alanine will react to form an amide.

Drawing the Structural Formula

After the reaction with acetic anhydride and sodium acetate, the amino group forms an amide with an acetyl group. The structural formula for the product is: \(\mathrm{H}_{3}\mathrm{N}^{+}\mathrm{CH}(\mathrm{CH}_{3})\mathrm{COOH} \cdot \mathrm{CH}_{3}\mathrm{CO}^{-}\).