In the following sequence of reaction, the major product ( C)
`CH_(3)CH_(2)I overset(NaCN)to(A) underset("Partial hydrolysis")overset(OH^(-))to(B)overset(Br_(2)//NaOH)to(C)`

Options

A. `CH_(3)CH_(2)NH_(2)`
B. `CH_(3)CH_(2)-overset(O)overset(||)(C)-NHBr`
C. `CH_(3)CH_(2)COONH_(4)`
D. `CH_(3)CH_(2)-overset(O)overset(||)(C)-NBr_(2)`

Answer

To solve the given question step by step, we will follow the sequence of reactions as described.

### Step 1: Formation of Compound A
The first reaction involves the nucleophilic substitution of ethyl iodide (CH₃CH₂I) with sodium cyanide (NaCN).

**Reaction:**
\[ \text{CH}_3\text{CH}_2\text{I} + \text{NaCN} \rightarrow \text{CH}_3\text{CH}_2\text{CN} + \text{NaI} \]

**Product A:**
The product formed here is ethyl cyanide (or propionitrile), which is represented as:
\[ \text{A} = \text{CH}_3\text{CH}_2\text{CN} \]

### Step 2: Partial Hydrolysis to Form Compound B
Next, we perform partial hydrolysis of compound A (ethyl cyanide) using water and a base (OH⁻).

**Reaction:**
\[ \text{CH}_3\text{CH}_2\text{CN} + \text{H}_2\text{O} \overset{\text{OH}^-}{\rightarrow} \text{CH}_3\text{CH}_2\text{C(=O)NH}_2 \]

**Product B:**
The product formed from partial hydrolysis is ethyl carbamate (or ethyl carbamoyl), which is represented as:
\[ \text{B} = \text{CH}_3\text{CH}_2\text{C(=O)NH}_2 \]

### Step 3: Formation of Compound C
Finally, we treat compound B with bromine (Br₂) and sodium hydroxide (NaOH). This is a reaction known as the Hofmann degradation, where we convert the carbamate to an amine.

**Reaction:**
\[ \text{CH}_3\text{CH}_2\text{C(=O)NH}_2 + \text{Br}_2 + \text{NaOH} \rightarrow \text{CH}_3\text{CH}_2\text{NH}_2 + \text{NaBr} + \text{H}_2\text{O} + \text{CO}_2 \]

**Product C:**
The product formed here is ethylamine, which is represented as:
\[ \text{C} = \text{CH}_3\text{CH}_2\text{NH}_2 \]

### Final Answer:
The major product C is ethylamine (CH₃CH₂NH₂).

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