JEE Advanced 2023 · Paper 2 · Chemistry Q3

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ChemistrySingle correct+3 / -1Question 37 of 51

In the following reactions, P, Q, R, and S are the major products.

The correct statement about P, Q, R, and S is

Four reactions of 1-chloro-2-methylpropane (isobutyl chloride): (1) [(i) Mg, dry ether; (ii) H2O] → P. (2) [(i) Mg, dry ether; (ii) CO2, dry ether; (iii) H3O+; (iv) NaOH] → Q. (3) [(i) Mg, dry ether; (ii) CH3CHO, then H2O; (iii) CrO3] → R. (4) [(i) ethanolic NaCN; (ii) H2/Ni; (iii) CHCl3/KOH, Δ; (iv) LiAlH4, then H2O] → S.

P is a primary alcohol with four carbons.

Q undergoes Kolbe's electrolysis to give an eight-carbon product.

R has six carbons and it undergoes Cannizzaro reaction.

S is a primary amine with six carbons.

Answer and worked solution

Official answer: B

The substrate is isobutyl chloride, $\ce{(CH3)2CHCH2Cl}$ (call the $\ce{(CH3)2CHCH2-}$ group R′).

P. The Grignard reagent is protonated by water to give isobutane, $\ce{(CH3)2CHCH3}$ . This is an alkane, not an alcohol, so (A) is false.

Q. Carboxylation gives $\ce{R'COOH}$ (3-methylbutanoic acid). NaOH turns it into the sodium salt Q. Kolbe electrolysis couples two R′ radicals:
$\ce{2R'COO- -> R'-R' + 2CO2},$
giving 2,5-dimethylhexane, $\ce{C8H18}$ . (B) is true.

The Grignard reagent adds to acetaldehyde, giving $\ce{(CH3)2CHCH2CH(OH)CH3}$ .
$\ce{CrO3}$ oxidises it to 4-methylpentan-2-one ( $\ce{C6}$ ).

A ketone with $\alpha$ -H does not undergo the Cannizzaro reaction, so (C) is false.

$\ce{R'CN}$ is reduced to $\ce{R'CH2NH2}$ , a $\ce{C5}$ primary amine.
The carbylamine reaction gives the isocyanide $\ce{R'CH2NC}$ .
$\ce{LiAlH4}$ reduces the isocyanide to the secondary amine $\ce{R'CH2NHCH3}$ ( $\ce{C6}$ ).

(D) is false.

Topics:SN1 and SN2 Carboxylic acids Cyanides and isocyanides Oxidation and reduction

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JEE Advanced 2023 · Paper 2, Chemistry question 3