The option(s) with correct sequence of reagents for the conversion of $\mathbf{P}$ to $\mathbf{Q}$ is(are) :

(C) i) $\mathrm{NaBH}_{4}$; ii) $\mathrm{SnCl}_{2} / \mathrm{HCl}$; iii) $\mathrm{H}_{3} \mathrm{O}^{+}$; iv) Lindlar's catalyst, $\mathrm{H}_{2}$

(A) i) Lindlar's catalyst, $\mathrm{H}_{2}$; ii) $\mathrm{SnCl}_{2} / \mathrm{HCl}$; iii) $\mathrm{NaBH}_{4}$; iv) $\mathrm{H}_{3} \mathrm{O}^{+}$

(B) i) Lindlar's catalyst, $\mathrm{H}_{2}$; ii) $\mathrm{H}_{3} \mathrm{O}^{+}$; iii) $\mathrm{SnCl}_{2} / \mathrm{HCl}$; iv) $\mathrm{NaBH}_{4}$

(D) i) Lindlar's catalyst, $\mathrm{H}_{2}$; ii) $\mathrm{NaBH}_{4}$; iii) $\mathrm{SnCl}_{2} / \mathrm{HCl}$; iv) $\mathrm{H}_{3} \mathrm{O}^{+}$

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · Aldehydes Ketones and Carboxylic Acids

The option(s) with correct sequence of reagents for the conversion of

P

Q

is(are) :

(A)i) Lindlar's catalyst, $H_{2}$ ; ii) $SnCl_{2} / HCl$ ; iii) $NaBH_{4}$ ; iv) $H_{3} O^{+}$
(B)i) Lindlar's catalyst, $H_{2}$ ; ii) $H_{3} O^{+}$ ; iii) $SnCl_{2} / HCl$ ; iv) $NaBH_{4}$
(C)i) $NaBH_{4}$ ; ii) $SnCl_{2} / HCl$ ; iii) $H_{3} O^{+}$ ; iv) Lindlar's catalyst, $H_{2}$ ✓ Correct
(D)i) Lindlar's catalyst, $H_{2}$ ; ii) $NaBH_{4}$ ; iii) $SnCl_{2} / HCl$ ; iv) $H_{3} O^{+}$

Answer: (C)

The conversion requires:

Reduction of the ketone group with $NaBH_{4}$ to form an alcohol
Reduction of the nitro group to amine using $SnCl_{2} / HCl$
Hydrolysis with $H_{3} O^{+}$ to remove any protecting groups or complete the reaction
Reduction of the triple bond to a double bond using Lindlar's catalyst with $H_{2}$

This sequence is given in option (C).

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · One or more correct options