JEE Advanced 2022 · Paper 2 · Chemistry Q13

ChemistryMultiple correct+4 / -2Question 49 of 54

Considering the following reaction sequence, the correct statement(s) is(are)

Compounds

\mathbf{P}

and

\mathbf{Q}

are carboxylic acids.

Compound

\mathbf{S}

decolorizes bromine water.

Compounds

\mathbf{P}

and

\mathbf{S}

react with hydroxylamine to give the corresponding oximes.

Compound

\mathbf{R}

reacts with dialkylcadmium to give the corresponding tertiary alcohol.

Answer and worked solution

Official answer: A, B, C

Sequence.

Friedel-Crafts acylation with succinic anhydride gives $\mathbf P=\ce{C6H5CO(CH2)2COOH}$ , 4-oxo-4-phenylbutanoic acid.
Clemmensen reduction gives $\mathbf Q=\ce{C6H5(CH2)3COOH}$ .
$\ce{SOCl2}$ gives the acid chloride $\mathbf R$ .
Intramolecular Friedel-Crafts acylation gives $\mathbf S$ , α-tetralone.
A second Clemmensen reduction gives tetralin.

Check each option.

(A) P and Q are both carboxylic acids. True.
(B) S is an aryl ketone, which does not decolorise bromine water. False.
(C) P and S both have a ketone C=O, so both form oximes. True.
(D) Acid chloride with $\ce{R'2Cd}$ gives a ketone, not a tertiary alcohol. False.

The official key also accepts (A), (B), (C). (A), (C).

JEE Advanced 2022 · Paper 2, Chemistry question 13