JEE Advanced 2023 Paper 2 Q03 Chemistry Organic Reactions & Mechanisms Carboxylic Acids & Derivatives Hard

JEE Advanced 2023 Paper 2 · Q03 · Carboxylic Acids & Derivatives

In the following reactions, $P$, $Q$, $R$, and $S$ are the major products. Isobutyl chloride, (CH$_3$)$_2$CHCH$_2$Cl, undergoes the following four reaction sequences: (a) (i) Mg, dry ether; (ii) H$_2$O $\rightarrow P$. (b) (i) Mg, dry ether; (ii) CO$_2$, dry ether; (iii) H$_3$O$^+$; (iv) NaOH $\rightarrow Q$. (c) (i) Mg, dry ether; (ii) CH$_3$CHO, then H$_2$O; (iii) CrO$_3$ $\rightarrow R$. (d) (i) ethanolic NaCN; (ii) H$_2$/Ni; (iii) CHCl$_3$/KOH, $\Delta$; (iv) LiAlH$_4$, then H$_2$O $\rightarrow S$. The correct statement about $P$, $Q$, $R$, and $S$ is

  1. A. $P$ is a primary alcohol with four carbons.
  2. B. $Q$ undergoes Kolbe's electrolysis to give an eight-carbon product.
  3. C. $R$ has six carbons and it undergoes Cannizzaro reaction.
  4. D. $S$ is a primary amine with six carbons.
Reveal answer + step-by-step solution

Correct answer:B

Solution

Starting material throughout is isobutyl chloride, (CH$_3$)$_2$CHCH$_2$Cl, which forms the Grignard reagent (CH$_3$)$_2$CHCH$_2$MgCl on treatment with Mg/dry ether. (a) Aqueous work-up of the Grignard gives the alkane: $P =$ isobutane, (CH$_3$)$_2$CHCH$_3$, a 4-carbon alkane (not an alcohol). Option (A) is FALSE. (b) Carbonation of the Grignard gives isovaleric acid (3-methylbutanoic acid), (CH$_3$)$_2$CHCH$_2$COOH; NaOH converts it to its sodium salt: $Q = $ sodium 3-methylbutanoate. Kolbe's electrolysis: $2\,(CH_3)_2CHCH_2COO^- \rightarrow (CH_3)_2CHCH_2{-}CH_2CH(CH_3)_2 + 2\,CO_2$, giving 2,5-dimethylhexane, an 8-carbon product. Option (B) is TRUE. (c) Addition of the Grignard to acetaldehyde gives the secondary alcohol (CH$_3$)$_2$CHCH$_2$CH(OH)CH$_3$ (4-methylpentan-2-ol); CrO$_3$ oxidises it to the methyl ketone $R = $ 4-methylpentan-2-one, (CH$_3$)$_2$CHCH$_2$COCH$_3$. $R$ has 6 carbons, but it has $\alpha$-hydrogens, so it does NOT undergo Cannizzaro reaction (Cannizzaro requires no $\alpha$-H). Option (C) is FALSE. (d) Ethanolic NaCN gives the nitrile (CH$_3$)$_2$CHCH$_2$CN; H$_2$/Ni reduces it to the primary amine (CH$_3$)$_2$CHCH$_2$CH$_2$NH$_2$ (3-methylbuta…[truncated]

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