Question

X and Y in the following reactions are:
(CH₃–CH₂)₂C=O → HCN/KCN → X → LiAlH₄ → (CH₃)₂C(OH)CH₂NH₂

• A. X = (CH$_3$)$_2$CHCH$_2$CN, Y = NaBH$_4$
• B. X = (CH$_3$)$_2$CCOONH$_2$, Y = CH$_3$NH$_2$
• C. X = (CH$_3$)$_2$C(CN)CH$_2$OH, Y = LiAlH$_4$
• D. X = (CH$_3$CH$_2$)$_2$C(CN)CH$_2$OH, Y = CH$_3$NH$_2$

Hint:

HCN/KCN adds a -CN group and -OH to a ketone to form cyanohydrins. LiAlH$_4$ reduces nitriles to amines.

Answer 1

The Correct Option is C

 

Step 1: Reaction of (CH₃–CH₂)₂C=O with HCN/KCN

The starting compound is a ketone, (CH₃–CH₂)₂C=O (acetone). When this compound reacts with hydrogen cyanide (HCN) or potassium cyanide (KCN), a nucleophilic addition occurs.

The cyanide ion (CN^-) attacks the carbonyl carbon, which is electrophilic, leading to the formation of a cyanohydrin intermediate. This intermediate has both a hydroxyl group (-OH) and a nitrile group (-CN) attached to the same carbon atom. The product of this reaction is labeled as X.

Step 2: Reduction of the Cyanohydrin Intermediate (X) with LiAlH₄

The cyanohydrin intermediate, X, formed in Step 1 is then subjected to reduction using LiAlH₄ (Lithium aluminum hydride). LiAlH₄ is a strong reducing agent that is capable of reducing nitriles (-CN) to amines (-NH₂).

The nitrile group (-CN) is reduced to an amine group (-NH₂), while the hydroxyl group (-OH) remains intact. As a result, the product obtained is (CH₃)₂C(OH)CH₂NH₂, which contains both an alcohol group (-OH) and an amine group (-NH₂) on the same carbon chain. This final product is labeled as Y.

Final Answer:

X is the cyanohydrin intermediate formed after the reaction with HCN/KCN, and Y is the final reduced product, which is (CH₃)₂C(OH)CH₂NH₂.