Mindyard | The option(s) with correct sequence of reagents for the conversion of P to Q is(are) i) Lindlar’s catalyst, H2 ; ii) SnCl2 / HCl ; iii) NaBH4 ; iv) H3O+ , i) Lindlar’s catalyst, H2 ; ii) H3O+ ; iii) SnCl2 / HCl ; iv) NaBH4 , i) NaBH4 ; ii) SnCl2 / HCl ; iii) H3O+ ; iv) Lindlar’s catalyst, H2 , i) Lindlar’s catalyst, H2 ; ii) NaBH4 ; iii) SnCl2 / HCl ; iv) H3O+

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

Options

$i) Lindlar’s catalyst, H_{2}; ii) Sn {Cl}_{2} / HCl; iii) NaBH 4; iv) H_{3} O^{+}$

$i) Lindlar’s catalyst, H_{2}; ii) H_{3} O^{+}; iii) Sn {Cl}_{2} / HCl; iv) NaBH 4$

$i) NaBH 4; ii) Sn {Cl}_{2} / HCl; iii) H_{3} O^{+}; iv) Lindlar’s catalyst, H_{2}$

$i) Lindlar’s catalyst, H_{2}; ii) NaBH 4; iii) Sn {Cl}_{2} / HCl; iv) H_{3} O^{+}$

Correct Answer :

Option C

$i) NaBH 4; ii) Sn {Cl}_{2} / HCl; iii) H_{3} O^{+}; iv) Lindlar’s catalyst, H_{2}$

Option D

$i) Lindlar’s catalyst, H_{2}; ii) NaBH 4; iii) Sn {Cl}_{2} / HCl; iv) H_{3} O^{+}$

Solution :

The correct sequences of reagents for the conversion of compound P to compound Q are:

i) NaBH₄ ; ii) SnCl₂/HCl ; iii) H₃O⁺ ; iv) Lindlar’s catalyst, H₂
and
i) Lindlar’s catalyst, H₂ ; ii) NaBH₄ ; iii) SnCl₂/HCl ; iv) H₃O⁺

Analysis of Functional Group Transformations from P to Q:

1. Reduction of Ketone to Alcohol & Hydrolysis of Acetate:
Compound P contains a ketone carbon (C=O) on the cyclopentane ring and an acetate group (-OCOCH₃). Treatment with NaBH₄ selectively reduces the ketone group to a secondary alcohol (-OH). Subsequent acidic hydrolysis (H₃O⁺) cleaves the ester/acetate group to yield a second alcohol group (-OH) on the ring and also hydrolyzes the ester (-CO₂Et) to a carboxylic acid (-CO₂H).

2. Stephen Reduction (Nitrile to Aldehyde):
The cyano group (-CN) on one of the side chains in compound P needs to be converted into an aldehyde group (-CHO) in compound Q. Using SnCl₂/HCl followed by hydrolysis (H₃O⁺) converts the nitrile (-CN) into an aldehyde (-CHO) via Stephen reduction.

3. Partial Reduction of Alkynes to cis-Alkenes:
Compound P has two carbon-carbon triple bonds (alkynes, C≡C) in its side chains, whereas compound Q contains two cis-double bonds (alkenes, C=C). Hydrogenation using Lindlar’s catalyst and H₂ selectively reduces the triple bonds to cis-alkenes without over-reducing them to alkanes.

Step-by-Step Sequence Evaluation:

• Sequence 1:
Step i) NaBH₄ reduces the ketone to an alcohol.
Step ii) SnCl₂/HCl converts the nitrile (-CN) to an aldimine intermediate.
Step iii) H₃O⁺ hydrolyzes the aldimine to an aldehyde (-CHO), hydrolyzes the ethyl ester (-CO₂Et) to a carboxylic acid (-CO₂H), and hydrolyzes the acetate group to an alcohol (-OH).
Step iv) Lindlar’s catalyst with H₂ selectively reduces both alkynes to cis-alkenes, yielding product Q.

• Sequence 2:
Step i) Lindlar’s catalyst with H₂ first reduces both alkynes in P to cis-alkenes.
Step ii) NaBH₄ reduces the ketone to an alcohol.
Step iii) SnCl₂/HCl converts the nitrile (-CN) to an aldimine.
Step iv) H₃O⁺ hydrolyzes the aldimine to aldehyde (-CHO), ester to carboxylic acid (-CO₂H), and acetate to alcohol (-OH), yielding product Q.

Thus, both sequence options are correct ways to achieve the conversion of P to Q.

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