Question

Write the zwitterion form of alanine and show its reaction with the following. (a) \(1 \mathrm{~mol} \mathrm{NaOH}\) (b) \(1 \mathrm{~mol} \mathrm{HCl}\)

Short answer

Answer: The zwitterion form of alanine is \(\mathrm{CH_3CH(NH_3^+)COO^-}\). When reacting with 1 mole of NaOH, it forms \(\mathrm{CH_3CH(NH_2)COO^- + Na^+ + H_2O}\). When reacting with 1 mole of HCl, it forms \(\mathrm{CH_3CH(NH_3^+)COOH + Cl^-}\).

Step-by-step solution

Identify alanine's structure

Alanine is an amino acid with the following chemical formula: \(\mathrm{CH_3CH(NH_2)COOH}\). It has an amino group \(\mathrm{(NH_2)}\), a carboxyl group \(\mathrm{(COOH)}\), and a methyl group \(\mathrm{(CH_3)}\).

Write the zwitterion form of alanine

In the zwitterion form, the amino group \(\mathrm{(NH_2)}\) will be protonated and become \(\mathrm{(NH_3^+)}\), and the carboxyl group \(\mathrm{(COOH)}\) will be deprotonated and become \(\mathrm{(COO^-)}\). The zwitterion form of alanine is: \(\mathrm{CH_3CH(NH_3^+)COO^-}\).

Reaction of alanine zwitterion with 1 mole NaOH

When reacting with NaOH, a strong base, the alanine zwitterion will lose the extra proton from its protonated amino group. The reaction will be: \(\mathrm{CH_3CH(NH_3^+)COO^- + NaOH \rightarrow CH_3CH(NH_2)COO^- + Na^+ + H_2O}\)

Reaction of alanine zwitterion with 1 mole HCl

When reacting with HCl, a strong acid, the alanine zwitterion will gain a proton on its deprotonated carboxyl group. The reaction will be: \(\mathrm{CH_3CH(NH_3^+)COO^- + HCl \rightarrow CH_3CH(NH_3^+)COOH + Cl^-}\)