JEE Advanced 2024 Paper 2 Q06 Chemistry Organic Reactions & Mechanisms Named Reactions & Reagents Hard

JEE Advanced 2024 Paper 2 · Q06 · Named Reactions & Reagents

The option(s) with correct sequence of reagents for the conversion of P to Q is(are)

(P is a cyclopentanone-based intermediate bearing an acetate ester (-OC(=O)CH3), a -CO2Et ester, two C≡C alkynes and a -CN nitrile. Q is the corresponding cyclopentane-diol bearing two cis-alkenes, a -CO2H carboxylic acid and a -CHO aldehyde.)

  1. A. i) Lindlar's catalyst, $H_2$; ii) $SnCl_2/HCl$; iii) $NaBH_4$; iv) $H_3O^+$
  2. B. i) Lindlar's catalyst, $H_2$; ii) $H_3O^+$; iii) $SnCl_2/HCl$; iv) $NaBH_4$
  3. C. i) $NaBH_4$; ii) $SnCl_2/HCl$; iii) $H_3O^+$; iv) Lindlar's catalyst, $H_2$
  4. D. i) Lindlar's catalyst, $H_2$; ii) $NaBH_4$; iii) $SnCl_2/HCl$; iv) $H_3O^+$
JEE Advanced multi-correct — pick every correct option, then check.
Reveal answer + step-by-step solution

Correct answer:C, D

Solution

Required transformations from P to Q: (a) reduce ketone to secondary alcohol ($NaBH_4$); (b) reduce nitrile $-CN$ to aldehyde $-CHO$ (Stephen reduction with $SnCl_2/HCl$); (c) hydrolyse the ester groups (acetate and $-CO_2Et$) to alcohol/carboxylic acid (acidic $H_3O^+$); (d) selectively reduce the two internal alkynes to cis-alkenes (Lindlar's catalyst, $H_2$). Lindlar reduction must be carried out last so that earlier reagents do not over-reduce the alkenes; conversely Lindlar must come after $SnCl_2/HCl$ in the alternative pathway. The two valid orderings are (C) NaBH$_4 \to$ SnCl$_2$/HCl $\to$ H$_3$O$^+$ $\to$ Lindlar/H$_2$ and (D) Lindlar/H$_2$ $\to$ NaBH$_4$ $\to$ SnCl$_2$/HCl $\to$ H$_3$O$^+$. (A) places NaBH$_4$ after SnCl$_2$/HCl, which would reduce the freshly formed aldehyde back to alcohol; (B) places H$_3$O$^+$ before SnCl$_2$/HCl, hydrolysing the imine intermediate prematurely.

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