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

Question: Describe the zwitterion form of alanine and its reactions with 1 mol of NaOH and 1 mol of HCl. Answer: The zwitterion form of alanine is \(\mathrm{H_3N^+}-\mathrm{CH(CH_3)COO^-}\), which has a positively charged amino group and a negatively charged carboxyl group. With 1 mol of NaOH, the amino group loses a proton and the molecule becomes \(\mathrm{H_2N-CH(CH_3)COO^-} \thinspace + \thinspace \mathrm{Na^+} \thinspace + \thinspace \mathrm{H_2O}\). With 1 mol of HCl, the carboxyl group gains a proton, resulting in \(\mathrm{H_3N^+-CH(CH_3)COOH} \thinspace + \thinspace \mathrm{Cl^-}\).

Step-by-step solution

Write the zwitterion form of alanine

Alanine is an amino acid with the molecular formula \(\mathrm{C_3H_7NO_2}\). It has an amino group (\(\mathrm{-NH_2}\)), a carboxyl group (\(\mathrm{-COOH}\)), and a methyl group (\(\mathrm{-CH_3}\)) attached to a central carbon atom. At physiological pH, alanine exists as a zwitterion, which means it has both positive and negative charges on different parts of the molecule. The zwitterion form of alanine has the amino group protonated as \(\mathrm{-NH_3^+}\), while the carboxyl group is deprotonated as \(\mathrm{-COO^-}\). The zwitterion form of alanine is written as: \(\mathrm{H_3N^+}-\mathrm{CH(CH_3)COO^-}\)

Reaction with 1 mol NaOH

When alanine reacts with \(\mathrm{1 \thinspace mol \thinspace NaOH}\), the hydroxide ion (\(\mathrm{OH^-}\)) from NaOH will abstract a proton from the protonated amino group (\(\mathrm{-NH_3^+}\)), resulting in a neutral amino group (\(\mathrm{-NH_2}\)). The reaction is as follows: \(\mathrm{H_3N^+}-\mathrm{CH(CH_3)COO^-} \thinspace + \thinspace \mathrm{NaOH} \thinspace \rightarrow \thinspace \mathrm{H_2N-CH(CH_3)COO^-} \thinspace + \thinspace \mathrm{Na^+} \thinspace + \thinspace \mathrm{H_2O}\)

Reaction with 1 mol HCl

When alanine reacts with \(\mathrm{1 \thinspace mol \thinspace HCl}\), the chloride ion (\(\mathrm{Cl^-}\)) from HCl donates a proton to the deprotonated carboxyl group (\(\mathrm{-COO^-}\)), thus, converting it to a protonated carboxyl group (\(\mathrm{-COOH}\)). The reaction is as follows: \(\mathrm{H_3N^+}-\mathrm{CH(CH_3)COO^-} \thinspace + \thinspace \mathrm{HCl} \thinspace \rightarrow \thinspace \mathrm{H_3N^+-CH(CH_3)COOH} \thinspace + \thinspace \mathrm{Cl^-}\)