Match List - I with List - II. List-I (Reactants) (A) Phenol, \(\mathrm{Zn} / \Delta\) (B) Phenol, \(\mathrm{CHCl}_{3}, \mathrm{NaOH}, \mathrm{HCl}\) (C) Phenol, \(\mathrm{CO}_{2}, \mathrm{NaOH}, \mathrm{HCl}\) (D) Phenol, Conc. \(\mathrm{HNO}_{3}\) List-II (Products) (I) Salicylaldehyde (II) Salicylic acid (III) Benzene (IV) Picric acid Choose the correct answer from the options given below.

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Alcohols Phenols and Ethers

Match List - I with List - II.

List-I (Reactants)
(A) Phenol,

Zn /Δ

(B) Phenol,

CHCl_{3}, NaOH, HCl

CO_{2}, NaOH, HCl

(D) Phenol, Conc.

HNO_{3}

List-II (Products)
(I) Salicylaldehyde
(II) Salicylic acid
(III) Benzene
(IV) Picric acid

Choose the correct answer from the options given below.

Answer: A → III, B → I, C → II, D → IV

Solution

Phenol heated with zinc dust undergoes reductive removal of the

- O H

group (deoxygenation) to give benzene.

Phenol treated with

C H C l_{3}

and NaOH, then acidified with HCl (Reimer–Tiemann reaction), introduces a

- C H O

group at the position ortho to

- O H

, giving salicylaldehyde.

Phenol treated with

C O_{2}

and NaOH under pressure, then acidified with HCl (Kolbe–Schmidt reaction), carboxylates the ortho position, giving salicylic acid.

Phenol treated with concentrated

H N O_{3}

nitrates all three activated positions to give 2,4,6-trinitrophenol, picric acid.

So (A) Phenol/Zn → Benzene, (B) Phenol/CHCl₃,NaOH,HCl → Salicylaldehyde, (C) Phenol/CO₂,NaOH,HCl → Salicylic acid, (D) Phenol/conc. HNO₃ → Picric acid.

JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Match the columns

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JEE Main · 2024 · 01 Feb 2024 · Shift 2 · Alcohols Phenols and Ethers

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