See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: The reaction sequence involves aldol condensation followed by dehydration and then hydrogenation. Step 2 - Aldol Condensation (10% NaOH, 5°C): CH3CHO undergoes aldol condensation with another molecule of CH3CHO. The alpha-carbon of one acetaldehyde attacks the carbonyl carbon of the second acetaldehyde. This gives the aldol product: CH3CH(OH)CH2CHO (3-hydroxybutanal), a beta-hydroxy aldehyde with 4 carbons. Step 3 - Dehydration (Delta/heating): The beta-hydroxy aldehyde undergoes dehydration upon heating to give an alpha,beta-unsaturated aldehyde: CH3CH=CHCHO (but-2-enal, also called crotonaldehyde), still with 4 carbons. Step 4 - Hydrogenation (H2/Ni): The unsaturated aldehyde (crotonaldehyde) is reduced by H2 over Ni catalyst. Both the C=C double bond and the CHO group are hydrogenated, giving CH3CH2CH2CH2OH, which is butan-1-ol (butanol), a 4-carbon alcohol. Step 5 - Why other options fail: - (a) propanol: 3 carbons, impossible since both starting material and product have even carbon counts here. - (b) ethanol: only 2 carbons, same as acetaldehyde starting material, no condensation. - (d) pentanol: 5 carbons, would require a different starting material or reaction. The product with 4 carbons after aldol condensation (2+2), dehydration, and full hydrogenation is butanol. Therefore, the correct answer is C.