Consider the given reaction, Identify 'X' and 'Y' :
Consider the given reaction, Identify 'X' and 'Y' :
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Remember the key reactions:
Carbonyl + HCN (base cat.) \(\rightarrow\) Cyanohydrin.
Nitrile (-CN) + LiAlH\(_4\) \(\rightarrow\) Primary amine (-CH\(_2\)NH\(_2\)).
LiAlH\(_4\) is a powerful reducing agent that reduces nitriles, esters, carboxylic acids, and carbonyl compounds.
- A
- B
- C
- D
The Correct Option is C
Solution and Explanation
The reaction involves two main steps:
1. Reaction of a carbonyl compound with HCN in the presence of a catalyst 'X' to form a cyanohydrin.
2. Reduction of the cyanohydrin with LiAlH\(_4\) followed by hydrolysis (H\(_3\)O\(^+\)) to give the final product 'Y'.
Step 2: Identifying the Starting Material and Reagent 'X'
The first step is the formation of a cyanohydrin. This is a nucleophilic addition of CN\(^-\) to a carbonyl group. Since HCN is a weak acid, a basic catalyst is required to generate a sufficient concentration of the nucleophile CN\(^-\). Therefore, 'X' must be a base, such as NaOH or KCN. This eliminates options (B) and (D).
Let's analyze the structure of the product 'Y' given in the correct option (C) to deduce the starting material. The structure of 'Y' in option (C) is 1-amino-2-methylbutan-2-ol:
\[ \begin{array}{c} \text{OH} \\ | \\ \text{CH}_3\text{CH}_2 - \text{C} - \text{CH}_2\text{NH}_2 \\ | \\ \text{CH}_3 \end{array} \]
This product is formed by the reduction of a nitrile group (-CN) to an amino group (-CH\(_2\)NH\(_2\)). Reversing this step, the intermediate cyanohydrin must have been:\[ \begin{array}{c} \text{OH} \\ | \\ \text{CH}_3\text{CH}_2 - \text{C} - \text{CN} \\ | \\ \text{CH}_3 \end{array} \]
This cyanohydrin is formed by the addition of HCN to a ketone. Reversing the cyanohydrin formation, the original carbonyl compound must have been Butan-2-one (or methyl ethyl ketone):\[ \begin{array}{c} \text{O} \\ || \\ \text{CH}_3\text{CH}_2 - \text{C} - \text{CH}_3 \end{array} \]
So, the reaction starts with Butan-2-one, even if the diagram in the question is ambiguous.Step 3: Verifying the Reaction and Identifying 'Y'
Reaction 1: Formation of cyanohydrin.
\[ \text{Butan-2-one} + \text{HCN} \xrightarrow{\text{NaOH (X)}} \text{2-hydroxy-2-methylbutanenitrile} \]
Reaction 2: Reduction of the nitrile group.\[ \text{2-hydroxy-2-methylbutanenitrile} \xrightarrow{\text{1. LiAlH}_4 \text{ 2. H}_3\text{O}^+} \text{1-amino-2-methylbutan-2-ol (Y)} \]
The reducing agent LiAlH\(_4\) reduces the -CN group to a -CH\(_2\)NH\(_2\) group.Step 4: Final Answer
The catalyst 'X' is NaOH, and the final product 'Y' is 1-amino-2-methylbutan-2-ol. This matches option (C).
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