Question

What are \(X\) and \(Y\) in the following reactions? 

• A. \(X = {B}_2{H}_6;\) \(Y = {H}_2/{Pd}\)
• B. \(X = {B}_2{H}_6;\) \(Y = {H}_2, {Pd-BaSO}_4\)
• C. \(X = {(i) NaBH}_4 {(ii) H}_2{O};\) \(Y = {H}_2, {Pd-BaSO}_4\)
• D. \(X = {(i) NaBH}_4 {(ii) H}_2{O};\) \(Y = {H}_2/{Pd}\)

Hint:

- Diborane (\(B_2H_6\)) is commonly used for the reduction of benzylic alcohols to hydrocarbons. - Rosenmund reduction (H\(_2\), Pd-BaSO\(_4\)) is selective for reducing acyl chlorides to aldehydes without further reduction to alcohols.

Answer 1

The Correct Option is B

1. Step 1: Reduction of -OH to -CH₂OH (X Formation)
- The first step involves the reduction of the benzylic hydroxyl (-OH) group to a methylene (-CH₂OH) group.
- Diborane (B₂H₆) is commonly used for this transformation, reducing benzylic alcohols to corresponding hydrocarbons.

2. Step 2: Conversion to -COOH and then -CHO (Y Formation)
- SOCl₂ (Thionyl chloride) converts the carboxyl (-COOH) group to an acyl chloride (-COCl).
- The Rosenmund reduction using H₂, Pd-BaSO₄ selectively reduces acyl chloride (-COCl) to an aldehyde (-CHO), giving the final product.

Thus, the correct sequence of reactions follows:
X = B2H6 (reduction of -OH to -CH2OH)
Y = H2, Pd-BaSO4 (Rosenmund reduction)

Since this matches option (2), it is the correct choice.