Question

What are X and Y in the following set of reactions?

• A. \( X = \text{(i) DIBAL-H, (ii) H}_2\text{O}, \quad Y = \text{(i) DIBAL-H, (ii) H}_2\text{O} \)
• B. \( X = \text{H}_2 / \text{Catalyst}, \quad Y = \text{H}_2 / \text{Catalyst} \)
• C. \( X = \text{H}_2 / \text{Catalyst}, \quad Y = \text{(i) DIBAL-H, (ii) H}_2\text{O} \)
• D. \( X = \text{(i) DIBAL-H, (ii) H}_2\text{O}, \quad Y = \text{H}_2 / \text{Catalyst} \)

Hint:

For selective reduction:
DIBAL-H reduces esters to aldehydes at low temperatures.

Answer 1

The Correct Option is C

Step 1: Identifying the transformations.
- The first reaction involves the reduction of the ester to an aldehyde, which is best achieved using Diisobutylaluminum hydride (DIBAL-H) at low temperatures.
- The second reaction involves catalytic hydrogenation, which reduces the aldehyde to a primary alcohol. 

Step 2: Assigning X and Y.
- The intermediate Y corresponds to the aldehyde, which is obtained by the partial reduction of the ester using DIBAL-H.
- The final product is a primary alcohol, which suggests the use of catalytic hydrogenation after the formation of the aldehyde.