JEE Main 2025 Apr 8 Shift 2, Chemistry Q14: Proteins

At pH = 2, the solution is strongly acidic, and excess H${}^{+}$ protonates the basic amino group while suppressing ionisation of the carboxylic acid, so the amino group becomes $-N{H}_3^{+}$ while $-COOH$ stays undissociated.

This gives the cationic species A: $H_3{N}^{+}-CH(R)-COOH$, where R is the isopropyl side chain.

At pH = 10, the solution is strongly basic, and excess OH${}^{-}$ deprotonates the carboxylic acid to $-CO{O}^{-}$ while the amino group remains neutral, since protonation of the amine is disfavoured under basic conditions.

This gives the anionic species B: $H_{2}N-CH(R)-CO{O}^{-}$.

These structures for A and B correspond exactly to option (1).