Match List-I with List-II.
| List-I Compound | List-II Product in Basic Medium (in NaOH + Heat) |
| (A) Ethanal | (I) Benzoic acid + Phenylmethanol |
| (B) Methanal | (II) 3-Hydroxybutanal + But-2-enal |
| (C) Benzenecarbaldehyde | (III) 4-Hydroxy-4-methylpentan-2-one + 4-Methylpent-3-en-2-one |
| (D) Acetone | (IV) Formic acid + Methanol |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(II), (B)-(IV), (C)-(I), (D)-(III)
What a carbonyl compound gives with alkali depends on whether it has an α-hydrogen. With an α-hydrogen it undergoes aldol condensation: the aldol is formed and, on heating, loses water to give the α,β-unsaturated carbonyl compound. Without an α-hydrogen, an aldehyde undergoes the Cannizzaro reaction with concentrated alkali.
(A) Ethanal — This matches (II) 3-Hydroxybutanal + But-2-enal. Ethanal has α-hydrogens, so it gives the aldol, which dehydrates on heating.
\( \displaystyle \ce{2CH3CHO} \xrightarrow{\text{dil. NaOH}} \ce{CH3CH(OH)CH2CHO} \xrightarrow[-\ce{H2O}]{\Delta} \ce{CH3CH=CHCHO} \)
(B) Methanal — This matches (IV) Formic acid + Methanol. HCHO has no α-hydrogen; it disproportionates (Cannizzaro), giving methanol and the formate salt.
\( \displaystyle \ce{2HCHO + NaOH} \xrightarrow{\text{conc.}} \ce{CH3OH + HCOONa} \)
(C) Benzenecarbaldehyde — This matches (I) Benzoic acid + Phenylmethanol. Benzaldehyde has no α-hydrogen, so it also gives the Cannizzaro reaction.
\( \displaystyle \ce{2C6H5CHO + NaOH} \xrightarrow{\text{conc.}} \ce{C6H5CH2OH + C6H5COONa} \)
(D) Acetone — This matches (III) 4-Hydroxy-4-methylpentan-2-one + 4-Methylpent-3-en-2-one. Acetone has α-hydrogens; its aldol (the ketol) dehydrates on heating.
\( \displaystyle \ce{2CH3COCH3} \xrightarrow{\ce{Ba(OH)2}} \ce{(CH3)2C(OH)CH2COCH3} \xrightarrow[-\ce{H2O}]{\Delta} \ce{(CH3)2C=CHCOCH3} \)
Shortcut: (A)–(II) appears only in Option (2).