Choose the correct option for structures of A and B, respectively.

Step 1: Understand amino acid ionisation

The amino acid shown is valine: $\ce{H2N-CH(CH(CH3)2)-COOH}$

It has two ionisable groups:

• $\ce{-COOH}$ (acidic, $pK_a \approx 2.3$)
• $\ce{-NH2}$ (basic, $pK_a \approx 9.7$)

Step 2: At pH = 2 (strongly acidic)

Both groups are protonated:

• $\ce{-COOH}$ remains as $\ce{-COOH}$ (pH < $pK_a$)
• $\ce{-NH2}$ is protonated to $\ce{-NH3+}$

$A = \ce{H3N+-CH(CH(CH3)2)-COOH}$ (cationic form)

Step 3: At pH = 10 (strongly basic)

Both groups are deprotonated:

• $\ce{-COOH}$ → $\ce{-COO-}$ (pH > $pK_a$)
• $\ce{-NH3+}$ → $\ce{-NH2}$ (pH > $pK_a$)

$B = \ce{H2N-CH(CH(CH3)2)-COO-}$ (anionic form)

Answer: Option (a)

Key Points to Remember:

• At low pH: amino group protonated ($\ce{-NH3+}$), carboxyl group neutral ($\ce{-COOH}$)
• At high pH: amino group neutral ($\ce{-NH2}$), carboxyl group deprotonated ($\ce{-COO-}$)
• Zwitterion exists at isoelectric point (pI)