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Match List-I with List-II
Choose the correct answer from the options given below: |
(A)-(I), (B)-(IV), (C)-(II), (D)-(III) (A)-(I), (B)-(III), (C)-(II), (D)-(IV) (A)-(I), (B)-(II), (C)-(IV), (D)-(III) (A)-(III), (B)-(IV), (C)-(I), (D)-(II) |
(A)-(I), (B)-(IV), (C)-(II), (D)-(III) |
The correct answer is Option (1) → (A)-(I), (B)-(IV), (C)-(II), (D)-(III)
(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) |
