JEE Advanced 2023 Paper 1 Q03 Chemistry Organic Reactions & Mechanisms Carboxylic Acids & Derivatives Hard

JEE Advanced 2023 Paper 1 · Q03 · Carboxylic Acids & Derivatives

Consider the following reaction scheme starting from styrene: (i) hydroboration–oxidation gives $P$ = $PhCH_2CH_2OH$; (ii) $P$ to $Q$ via $CrO_3/H^+$ then $Cl_2/\text{red P}$ gives $PhCH(Cl)COOH$; (iii) $P$ to $R$ via $SOCl_2$, then KCN, then $H_3O^+/\Delta$ gives $PhCH_2CH_2COOH$; (iv) $R$ to $S$ via conc.$H_2SO_4/\Delta$ gives an anhydride. Choose the correct option(s) for the major products $Q$, $R$ and $S$.

  1. A. $Q = PhCH(Cl)COOH$; $R = PhCH_2COOH$; $S = $ acid
  2. B. $Q = PhCH(Cl)COOH$; $R = PhCH_2CH_2COOH$; $S = $ anhydride
  3. C. $Q = PhCH_2COCl$; $R = PhCH_2CN$; $S = $ amide
  4. D. $Q = PhCH(Cl)COCl$; $R = PhCH_2COOH$; $S = $ acid chloride
Reveal answer + step-by-step solution

Correct answer:B

Solution

$P = PhCH_2CH_2OH$ (anti-Markovnikov from $B_2H_6/H_2O_2$). $P \xrightarrow{CrO_3/H^+} PhCH_2COOH \xrightarrow{Cl_2/\text{red P}} Q = PhCH(Cl)COOH$ (Hell-Volhard-Zelinsky). $P \xrightarrow{SOCl_2} PhCH_2CH_2Cl \xrightarrow{KCN} PhCH_2CH_2CN \xrightarrow{H_3O^+/\Delta} R = PhCH_2CH_2COOH$. $R \xrightarrow{\text{conc}\,H_2SO_4/\Delta} S = $ anhydride. Matches option (B). [Note: section labeled MCQ_MULTI but official key has only one correct option, so this record is shipped as MCQ_SINGLE per pipeline precedent.]

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