JEE Advanced 2023 · Paper 1 · Chemistry Q7

← JEE Advanced 2023 · Paper 1

ChemistrySingle correct+3 / -1Question 41 of 51

In the given reaction scheme, P is a phenyl alkyl ether, Q is an aromatic compound; R and S are the major products.

$\mathbf{P} \xrightarrow{\ce{HI}} \mathbf{Q} \xrightarrow[\text{(iii) } \ce{H3O+}]{\text{(i) NaOH; (ii) } \ce{CO2}} \mathbf{R} \xrightarrow[\text{(ii) } \ce{H3O+}]{\text{(i) } \ce{(CH3CO)2O}} \mathbf{S}$

The correct statement about S is

It primarily inhibits noradrenaline degrading enzymes.

It inhibits the synthesis of prostaglandin.

It is a narcotic drug.

It is ortho-acetylbenzoic acid.

Answer and worked solution

Official answer: B

The sequence.

P is anisole (a phenyl alkyl ether). HI cleaves it to phenol, Q, and $\ce{CH3I}$ .
R: the Kolbe reaction (NaOH, $\ce{CO2}$ , then acid) gives salicylic acid.
S: acetic anhydride acetylates the OH, giving acetylsalicylic acid (aspirin).

The options.

(B) True. Aspirin relieves pain by inhibiting the synthesis of prostaglandins.
(A) False: inhibiting noradrenaline-degrading enzymes describes antidepressants.
(C) False: aspirin is a non-narcotic analgesic.
(D) False: it is o-acetoxybenzoic acid, not o-acetylbenzoic acid.

Topics:Phenols Ethers Drugs

← Previous question Next question →

JEE Advanced 2023 · Paper 1, Chemistry question 7