JEE Advanced 2024 — Paper 2 — CHEM-II solution

JEE Advanced2024Paper 2CHEM-II

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

A
i) Lindlar's catalyst, $H_{2}$ ; ii) $S n C l_{2} / H C l$ ; iii) $N a B H_{4}$ ; iv) $H_{3} O^{+}$
B
i) Lindlar's catalyst, $H_{2}$ ; ii) $H_{3} O^{+}$ ; iii) $S n C l_{2} / H C l$ ; iv) $N a B H_{4}$
C
i) $N a B H_{4}$ ; ii) $S n C l_{2} / H C l$ ; iii) $H_{3} O^{+}$ ; iv) Lindlar's catalyst, $H_{2}$
D
i) Lindlar's catalyst, $H_{2}$ ; ii) $N a B H_{4}$ ; iii) $S n C l_{2} / H C l$ ; iv) $H_{3} O^{+}$

Solution

Four functional-group transformations are required to convert $P$ to $Q$ :

1. Both internal alkynes become cis alkenes. This is achieved by hydrogenation over Lindlar's catalyst ( $H_{2}$ , Lindlar).

2. The nitrile ( $- C N$ ) is reduced to an aldehyde ( $- C H O$ ). $S n C l_{2} / H C l$ (Stephen reduction) converts $- C N$ to an imine which on aqueous workup gives $- C H O$ .

3. The ring ketone ( $C = O$ ) is reduced to a secondary alcohol ( $- O H$ ). $N a B H_{4}$ accomplishes this selective reduction without touching esters, alkynes, or nitriles.

4. The acetate ester ( $- O C O C H_{3}$ ) is hydrolysed to a free hydroxyl ( $- O H$ ), and the ethyl ester ( $- C O_{2} E t$ ) is hydrolysed to the carboxylic acid ( $- C O_{2} H$ ). $H_{3} O^{+}$ effects both hydrolyses.

Now check the ordering constraints. $S n C l_{2} / H C l$ is acidic, so any imine intermediate it forms must be hydrolysed at some point to reveal the aldehyde; this can happen either immediately (acidic workup) or later. $N a B H_{4}$ should ideally precede $H_{3} O^{+}$ to avoid problems with carboxylic acid formation, but it can also follow the Stephen reduction provided the imine is hydrolysed afterwards by $H_{3} O^{+}$ .

Option (A): Lindlar → $S n C l_{2} / H C l$ → $N a B H_{4}$ → $H_{3} O^{+}$ . Alkynes are reduced first, then $- C N \to$ imine, then the ring ketone is reduced (the imine survives this since $N a B H_{4}$ does not reduce protonated imines under careful conditions), and finally $H_{3} O^{+}$ hydrolyses the imine to $- C H O$ and both esters to $- O H$ and $- C O_{2} H$ . Valid.

Option (C): $N a B H_{4}$ → $S n C l_{2} / H C l$ → $H_{3} O^{+}$ → Lindlar. The ring ketone is reduced first (with $- O H$ groups left untouched), then $- C N \to$ imine, then $H_{3} O^{+}$ hydrolyses imine $\to$ aldehyde plus both esters $\to$ alcohol and acid, and finally Lindlar reduces the alkynes to alkenes. Valid (the alkynes are inert to $N a B H_{4}$ , $S n C l_{2} / H C l$ , and dilute $H_{3} O^{+}$ ).

Option (D): Lindlar → $N a B H_{4}$ → $S n C l_{2} / H C l$ → $H_{3} O^{+}$ . Alkynes to alkenes, ketone to alcohol, $- C N$ to imine, then acid workup completes both hydrolyses. Valid.

Option (B): Lindlar → $H_{3} O^{+}$ → $S n C l_{2} / H C l$ → $N a B H_{4}$ . Treating the substrate with $H_{3} O^{+}$ in step (ii) before reducing $- C N$ would hydrolyse the nitrile directly to a carboxylic acid, so the targeted aldehyde would never form. Invalid.

Hence the correct options are (A), (C), and (D).

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