Sequence of Reagents for Conversion of Functional Groups in Polyfunctional Compound

To determine the correct sequence of reagents for converting compound $\mathbf{P}$ into compound $\mathbf{Q}$, let us first analyze the structural changes and functional group transformations required:

1. Reduction of Alkynes to $cis$-Alkenes:

   • Compound $\mathbf{P}$ contains two internal carbon-carbon triple bonds ($-\text{C}\equiv\text{C}-$).
   • Compound $\mathbf{Q}$ contains two $cis$-double bonds ($-\text{CH}=\text{CH}-$).
   • Reagent: Lindlar's catalyst ($\text{Pd}/\text{CaCO}_3/\text{quinoline}$) with $\text{H}_2$ selectively reduces alkynes to $cis$-alkenes without affecting ketones, esters, nitriles, or aldehydes.

2. Reduction of Ketone to Secondary Alcohol:

   • Compound $\mathbf{P}$ contains a ring ketone ($>\text{C}=\text{O}$).
   • Compound $\mathbf{Q}$ contains a secondary alcohol ($-\text{CH-OH}$) at this position.
   • Reagent: Sodium borohydride ($\text{NaBH}_4$) selectively reduces ketones and aldehydes to alcohols, leaving esters ($-\text{CO}_2\text{Et}$ and $-\text{OCOCH}_3$) and nitriles ($-\text{CN}$) unreacted under standard conditions.

3. Conversion of Nitrile to Aldehyde (Stephen Reduction):

   • Compound $\mathbf{P}$ contains a nitrile group ($-\text{CN}$).
   • Compound $\mathbf{Q}$ contains an aldehyde group ($-\text{CHO}$).
   • Reagent: Tin(II) chloride with hydrochloric acid ($\text{SnCl}_2/\text{HCl}$) reduces the nitrile to an aldimine intermediate ($-\text{CH}=\text{NH}\cdot\text{HCl}$), which upon subsequent acidic hydrolysis ($\text{H}_3\text{O}^+$) yields the aldehyde ($-\text{CHO}$).

4. Hydrolysis of Esters:

   • Compound $\mathbf{P}$ contains an ethyl ester ($-\text{CO}_2\text{Et}$) and an acetate ester ($-\text{OCOCH}_3$).
   • Compound $\mathbf{Q}$ contains a carboxylic acid ($-\text{CO}_2\text{H}$) and a secondary alcohol ($-\text{OH}$).
   • Reagent: Acidic hydrolysis ($\text{H}_3\text{O}^+$) hydrolyzes both ester groups to their corresponding carboxylic acid and alcohol.

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Evaluation of Options:

• Option (D):

  1. Step i ($\text{Lindlar's catalyst, H}_2$): Converts both $-\text{C}\equiv\text{C}-$ triple bonds into $cis$-alkenes ($-\text{CH}=\text{CH}-$).
  2. Step ii ($\text{NaBH}_4$): Selectively reduces the ring ketone ($>\text{C}=\text{O}$) to a secondary alcohol ($-\text{CH-OH}$). The esters and nitrile remain unaffected.
  3. Step iii ($\text{SnCl}_2/\text{HCl}$): Reduces the nitrile ($-\text{CN}$) to an aldimine intermediate ($-\text{CH}=\text{NH}\cdot\text{HCl}$).
  4. Step iv ($\text{H}_3\text{O}^+$): Hydrolyzes:
     • The aldimine intermediate into an aldehyde ($-\text{CHO}$).
     • The ethyl ester ($-\text{CO}_2\text{Et}$) into a carboxylic acid ($-\text{CO}_2\text{H}$).
     • The acetate ester ($-\text{OCOCH}_3$) into a secondary alcohol ($-\text{OH}$).

  Result: Yields compound $\mathbf{Q}$ successfully. Hence, (D) is correct.

• Option (C):

  1. Step i ($\text{NaBH}_4$): Reduces the ketone to a secondary alcohol.
  2. Step ii ($\text{SnCl}_2/\text{HCl}$): Reduces the nitrile to an aldimine intermediate.
  3. Step iii ($\text{H}_3\text{O}^+$): Hydrolyzes the aldimine to an aldehyde, the ethyl ester to a carboxylic acid, and the acetate ester to an alcohol.
  4. Step iv ($\text{Lindlar's catalyst, H}_2$): Selectively hydrogenates the two alkynes to $cis$-alkenes without affecting the carboxylic acid, aldehyde, or alcohol groups.

  Result: Yields compound $\mathbf{Q}$ successfully. Hence, (C) is correct.

• Option (A) & Option (B):

  • In Option (A), treating the intermediate with $\text{NaBH}_4$ after $\text{SnCl}_2/\text{HCl}$ would reduce the unhydrolyzed aldimine intermediate ($-\text{CH}=\text{NH}$) to a primary amine ($-\text{CH}_2\text{NH}_2$) rather than an aldehyde.
  • In Option (B), performing $\text{H}_3\text{O}^+$ hydrolysis before $\text{SnCl}_2/\text{HCl}$ means there is no acidic hydrolysis step after $\text{SnCl}_2/\text{HCl}$ to convert the aldimine intermediate into the aldehyde group ($-\text{CHO}$).
  • Therefore, (A) and (B) are incorrect.

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Conclusion:

The correct sequence of reagents for the conversion of $\mathbf{P}$ to $\mathbf{Q}$ is given by options C and D.

Correct Options: C, D