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)
Reactions in NaOH + heat depend on presence of alpha hydrogen.
• Compounds with α-H → Aldol condensation
• Compounds without α-H → Cannizzaro reaction
(A) Ethanal → (II) 3-Hydroxybutanal + But-2-enal
Ethanal contains α-hydrogen, so in NaOH + heat it undergoes aldol condensation.
First it forms 3-hydroxybutanal (aldol), which on heating dehydrates to but-2-enal (crotonaldehyde).
(B) Methanal → (IV) Formic acid + Methanol
Methanal does not contain α-hydrogen, so it undergoes the Cannizzaro reaction in basic medium.
Two molecules react to give formic acid (or formate in base) and methanol.
(C) Benzenecarbaldehyde → (I) Benzoic acid + Phenylmethanol
Benzaldehyde also lacks α-hydrogen, so it undergoes the Cannizzaro reaction with NaOH to form benzoic acid and benzyl alcohol (phenylmethanol).
(D) Acetone → (III) 4-Hydroxy-4-methylpentan-2-one + 4-Methylpent-3-en-2-one
Acetone has α-hydrogen, so it undergoes aldol condensation.
It first forms diacetone alcohol (4-hydroxy-4-methylpentan-2-one) which on heating dehydrates to mesityl oxide (4-methylpent-3-en-2-one).
Therefore the correct matching is:
(A)-(II), (B)-(IV), (C)-(I), (D)-(III).