In the following reactions, P, Q, R, and S are the major products.
The correct statement(s) about P, Q, R, and S is(are)
← JEE Advanced 2023 · Paper 2![(1) 1-tert-Butyl-4-ethylbenzene → [(i) KMnO4, KOH, Δ; (ii) H3O+] → P. (2) Methyl 3-(chlorocarbonyl)benzoate (benzene with COOMe and COCl meta to each other) → [(i) NaOH, H2O; (ii) H3O+] → Q. (3) Methyl 3-(3-cyanophenyl)-3-oxopropanoate (a β-keto ester, ArCOCH2COOMe with CN meta on the ring) → [(i) H3O+, Δ; (ii) H2CrO4] → R. (4) 2-(4-Bromophenyl)-1,3-dioxolane (acetal-protected 4-bromobenzaldehyde) → [(i) Mg, dry ether; (ii) CO2, then H3O+; (iii) ammoniacal AgNO3, H3O+] → S.](/figures/ja-2023-p2/ja-2023-p2-che-q07-f1.png)
ChemistryMultiple correct+4 / -2Question 41 of 51
JEE Advanced 2023 · Paper 2, Chemistry question 7
Answer and worked solution
Official answer: C, D
P. Hot alkaline oxidises the ethyl group to COOH. The tert-butyl group has no benzylic H and survives. P = 4-tert-butylbenzoic acid, a monocarboxylic acid.
Q. NaOH hydrolyses both the acid chloride and the ester. Acidification gives isophthalic acid (benzene-1,3-dicarboxylic acid).
R.
- Hot acid hydrolyses the nitrile to COOH and the ester to a -keto acid.
- The -keto acid decarboxylates, giving 3-acetylbenzoic acid.
- Chromic acid oxidises the aryl methyl ketone to COOH.
R = isophthalic acid, so (C) is true.
S.
- The acetal protects the aldehyde while the Grignard reagent forms and is carboxylated.
- Acid work-up removes the acetal, giving 4-formylbenzoic acid.
- Tollens' reagent oxidises the CHO, giving terephthalic acid.
The options.
- (A) False: P is not terephthalic acid (the dacron monomer), and the glyptal monomer is phthalic acid, not isophthalic acid.
- (B) False: P is a monocarboxylic acid.
- (D) True: R has no carbonyl -H (it is a diacid), so it cannot undergo aldol condensation. S has no aldehyde, so it cannot undergo the Cannizzaro reaction.