Match List-I with List-II
|
List-I (Reaction) |
List-II (Name) |
|
(A) |
(I) Aldol Reaction |
|
(B) |
(II) Cannizzaro Reaction |
|
(C) |
(III) Haloform Reaction |
|
(D) |
(IV) Hell Volhard Zelinski Reaction |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(I), (B)-(IV), (C)-(II), (D)-(III)
|
List-I (Reaction) |
List-II (Name) |
|
(A) |
(I) Aldol Reaction |
|
(B) |
(IV) Hell Volhard Zelinski Reaction |
|
(C) |
(II) Cannizzaro Reaction |
|
(D) |
(III) Haloform Reaction |
(A) CH₃CHO + Dil. NaOH → CH₃CHOHCH₂ CHO
Acetaldehyde undergoes self-condensation in presence of dilute base to give β-hydroxy aldehyde. This is Aldol reaction. (A) → (I)
(B) CH₃CH₂CH₂COOH + Br₂ / Red P / H₂O → CH₃CH₂CHBrCOOH
This is α-bromination of carboxylic acid. This reaction is Hell–Volhard–Zelinski reaction. (B) → (IV)
(C) C₆H₅CHO + Conc. KOH → C₆H₅COONa + C₆H₅CH₂OH
Benzaldehyde (no α-hydrogen) undergoes disproportionation in strong base. This is Cannizzaro reaction. (C) → (II)
(D) CH₃COCH₃ + NaOCl → CH₃COONa + CHCl₃ Methyl ketone gives chloroform.
This is Haloform reaction. (D) → (III)