Question

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

• A. $H_{2}/cat; (CH_{3})_{2}Cd$
• B. $Na/C_{2}H_{5}OH; CH_{3}MgBr, H_{2}O$
• C. $DIBAL-H, H_{2}O; CH_{3}MgBr, H_{2}O$
• D. $LiAlH_{4}, H_{2}O; (CH_{3})_{2}Cd$

Hint:

DIBAL-H stops reduction at the aldehyde stage for nitriles.

Answer 1

The Correct Option is C

Step 1: Identify the reaction type.
This sequence reduces a nitrile to an aldehyde first, and then uses a Grignard reagent on that aldehyde. We must find reagents X and Y.

Step 2: Choose the nitrile reducer.
To stop the reduction at the aldehyde stage, we need a mild and selective reducing agent. DIBAL-H is the correct choice for turning a nitrile into an aldehyde.

Step 3: Add the water step.
After DIBAL-H reacts, we add water to hydrolyse the intermediate and release the aldehyde. So X is DIBAL-H followed by water.

Step 4: Choose reagent Y.
The aldehyde then reacts with a Grignard reagent. Here the Grignard reagent is methyl magnesium bromide, $CH_3MgBr$.

Step 5: Add the work-up.
After the Grignard adds to the carbonyl, water is added to give the alcohol product. So Y is $CH_3MgBr$ followed by water.

Step 6: State the answer.
So X is DIBAL-H with water, and Y is $CH_3MgBr$ with water. \[ \boxed{X = DIBAL\text{-}H,\,H_2O;\ Y = CH_3MgBr,\,H_2O} \]