Question: 79: Match the reagents (List-I) with the product (List-II) obtained from phenol
| List – I | List – II | ||
|---|---|---|---|
| (a) | (i) NaOH (ii) \mathrm{CO}_{2} (iii) \mathrm{H}^{+} | (i) | Benzoquinone |
| (b) | (i) Aqueous \mathrm{NaOH}+\mathrm{CHCl}_{3} (ii) \mathrm{H}^{+} | (ii) | Benzene |
| (c) | Zn dust, \Delta | (iii) | Salicyl aldehyde |
| (d) | \mathrm{Na}_{2}\mathrm{Cr}_{2}\mathrm{O}_{7}, \mathrm{H}_{2}\mathrm{SO}_{4} | (iv) | Salicylic acid |
Choose the correct answer from the options given below :
(1) (a) – (iii), (b) – (iv), (c) – (i), (d) – (ii)
(2) (a) – (ii), (b) – (i), (c) – (iv), (d) – (iii)
(3) (a) – (iv), (b) – (iii), (c) – (ii), (d) – (i)
(4) (a) – (iv), (b) – (ii), (c) – (i), (d) – (iii)
Answer: Option (3)
Explanation:
In reaction (a), phenol reacts with sodium hydroxide to form sodium phenoxide,
which on treatment with carbon dioxide followed by acidification undergoes Kolbe–Schmitt reaction
to give salicylic acid. Hence, (a) corresponds to (iv).
In reaction (b), phenol reacts with aqueous sodium hydroxide and chloroform followed by acidification.
This is the Reimer–Tiemann reaction, which introduces a formyl group at the ortho position,
forming salicylaldehyde. Therefore, (b) corresponds to (iii).
In reaction (c), heating phenol with zinc dust removes the hydroxyl group and converts phenol into benzene. Thus, (c) corresponds to (ii).
In reaction (d), phenol undergoes oxidation with acidified sodium dichromate to form benzoquinone. Hence, (d) corresponds to (i).
Thus, the correct matching is (a)-(iv), (b)-(iii), (c)-(ii), (d)-(i),
which corresponds to option (3).