JEE Main 2014 · 6 Apr · Shift 1 · Chemistry Q55

← JEE Main 2014 · 6 Apr · Shift 1

ChemistrySingle correct+4 / -1Question 55 of 90

Sodium phenoxide when heated with $\ce{CO2}$ under pressure at 125°C yields a product which on acetylation produces C.

$\ce{C6H5ONa + CO2 ->[125^\circ][5 Atm] B ->[H+][Ac2O] C}$

The major product C would be :

Sodium phenoxide + CO₂ (125°, 5 atm) gives B; B with H⁺ and then Ac₂O gives C.

2-Acetoxybenzoic acid (OCOCH₃ and COOH ortho to each other)

Benzene ring with OCOCH₃ and, on the adjacent carbon, COOH.

2,4-Diacetylphenol (OH with COCH₃ at the ortho and para positions)

Benzene ring with OH, COCH₃ on the adjacent carbon and a second COCH₃ para to OH.

Methyl salicylate (OH with COOCH₃ ortho to it)

Benzene ring with OH and, on the adjacent carbon, COOCH₃.

3-Acetoxybenzoic acid (OCOCH₃ and COOH meta to each other)

Benzene ring with OCOCH₃ and COOH meta to it.

Answer and worked solution

Official answer: A

Kolbe-Schmitt carboxylation gives sodium salicylate, and $\ce{H+}$ gives salicylic acid (B). Acetylation with $\ce{Ac2O}$ gives C, 2-acetoxybenzoic acid (aspirin).

Topics:Phenols Carboxylic acids

← Previous question Next question →

JEE Main 2014 · 6 Apr · Shift 1, Chemistry question 55