Selective functional group transformations with NaOH and LiAlH4

Steps:

• Compound X contains -COOH and -OH groups.
• NaOH(aq) deprotonates -COOH to -COONa; -OH groups remain unchanged, yielding Y.
• LiAlH4 reduces -COOH to -CH2OH; -OH groups remain unchanged, yielding Z as a polyol.
• Option A depicts Y as the carboxylate salt with intact -OH and Z as the corresponding reduced diol.

Why A is correct:

• NaOH selectively forms the sodium carboxylate from -COOH per acid-base reaction; LiAlH4 reduces -COOH to -CH2OH via hydride addition without affecting alcohols.

Why the others are wrong:

• B: Z shows -COH (aldehyde), but LiAlH4 fully reduces -COOH to -CH2OH, not aldehyde.
• C: Y incorrectly has -ONa on alcohol; NaOH(aq) does not deprotonate alcohols under these conditions.
• D: Combines B and C errors, with -ONa on alcohol and incomplete reduction in Z.

Final answer: A