See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: LiAlH4 is a strong, non-selective reducing agent that reduces all reducible functional groups, including both C=O (aldehyde/ketone/ester) and C=C double bonds conjugated with carbonyl groups. Step 1: Identify the functional groups in C6H5CH=CHCHO (cinnamaldehyde). The molecule contains: - An alpha,beta-unsaturated aldehyde (conjugated C=C and CHO) - The aromatic ring (not reduced by LiAlH4 under normal conditions) Step 2: LiAlH4 reduces both the C=C double bond and the CHO group in alpha,beta-unsaturated carbonyl compounds. It delivers hydride to both the carbonyl carbon and the beta carbon (via 1,2- and 1,4-reduction pathways), but with LiAlH4, both the alkene (conjugated) and the aldehyde are fully reduced. Step 3: Full reduction of C6H5CH=CHCHO with LiAlH4 gives C6H5CH2CH2CH2OH (3-phenyl-1-propanol), where the C=C double bond is reduced to C-C single bond and the CHO is reduced to CH2OH. Step 4: Evaluate other options: - (b) C6H5CH=CHCH2OH: only the CHO is reduced, C=C remains - this would be the product with a milder, selective reagent like DIBAL-H or NaBH4 under controlled conditions, not LiAlH4. - (c) C6H5CH2CH2CHO: only the C=C is reduced, CHO remains - LiAlH4 would not leave an aldehyde unreduced. - (d) C6H5CH2CHOHCH3: incorrect carbon skeleton, does not arise from this substrate. LiAlH4 being a strong hydride donor reduces both the conjugated double bond and the aldehyde completely, giving the saturated primary alcohol. Therefore, the correct answer is A.