Match List-I (Name) with List-II (Structure)
|
List-I (Name) |
List-II (Structure) |
|
(A) Cinnamaldehyde |
(I) |
|
(B) Vanillin |
(II) |
|
(C) Salicylaldehyde |
(III) |
|
(D) Benzophenone |
(IV) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
All four compounds are aromatic carbonyl compounds; their common names tell where the C=O group is and what else is on the ring. An aldehyde has a –CHO group, while a ketone has a C=O group between two carbon groups.
(A) Cinnamaldehyde — This matches (III). It is 3-phenylprop-2-enal, \( \ce{C6H5-CH=CH-CHO} \): the –CHO group sits at the end of a C=C side chain on the ring. It gives cinnamon its smell.
(B) Vanillin — This matches (I). It is 4-hydroxy-3-methoxybenzaldehyde: a –CHO group, an –\( \ce{OCH3} \) group at position 3 and an –OH group at position 4, the flavour of vanilla.
(C) Salicylaldehyde — This matches (IV). It is 2-hydroxybenzaldehyde: the –OH group is ortho to the –CHO group.
(D) Benzophenone — This matches (II). It is diphenyl ketone, \( \ce{C6H5-CO-C6H5} \): a C=O group joining two benzene rings.
Shortcut: (B)-(I) appears only in Option (3).