In the reaction, `CH_(3)*CH_(2)*Cloverset(KCN)to(A) overset(LiAlH_(4))to(P)` end product (P) is

Options

A. `CH_(3)CH_(2)*NO_(2)`
B. `CH_(3)CH_(2)CH_(2)NO_(2)`
C. `CH_(3)CH_(2)NH_(2)`
D. `CH_(3)*CH_(2)*CH_(2)NH_(2)`

Explanation

To solve the given question step by step, we will analyze the reactions involved and identify the end product (P).

### Step 1: Identify the starting compound
The starting compound is ethyl chloride, which is represented as \( CH_3CH_2Cl \).

### Step 2: Reaction with KCN
When ethyl chloride reacts with potassium cyanide (KCN), a nucleophilic substitution reaction occurs. The cyanide ion (\( CN^- \)) from KCN replaces the chloride ion (\( Cl^- \)) in ethyl chloride.

**Reaction:**
\[ CH_3CH_2Cl + KCN \rightarrow CH_3CH_2CN + KCl \]

Here, the product formed is ethyl cyanide (or propionitrile), represented as \( CH_3CH_2CN \).

### Step 3: Reduction with LiAlH4
The next step involves the reduction of ethyl cyanide using lithium aluminium hydride (LiAlH4). Lithium aluminium hydride is a strong reducing agent that can convert nitriles into primary amines.

**Reaction:**
\[ CH_3CH_2CN + LiAlH_4 \rightarrow CH_3CH_2CH_2NH_2 \]

In this reaction, the cyanide group (\( CN \)) is reduced to an amine group (\( NH_2 \)), resulting in the formation of propylamine, which is represented as \( CH_3CH_2CH_2NH_2 \).

### Step 4: Identify the end product (P)
The final product (P) after the complete reaction sequence is propylamine, \( CH_3CH_2CH_2NH_2 \).

### Final Answer:
The end product (P) is propylamine, represented as \( CH_3CH_2CH_2NH_2 \).

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