Consider the following sequence of reactions :
`H_(2) C = CH - CH = CH_(2) underset("CCI"_4) overset(Br_(2) ("1 mole") ) (to) A underset(2. H_(2), Ni) overset(1. "KCN (excess)") (to) B`
The end product (B) is :

Options

A. `H_(2) N - (CH_(2))_2 - CH = CH - (CH_2)_2 - NH_2`
B. `H_(2) N - (CH_2)_6 - NH_2`
C. `NC - CH_2 - CH = CH - CH_2 - CN`
D. `H_(2) C = CH - underset(underset(NH_2)(|))CH - (CH_2)_2 - NH_2`

Explanation

To solve the given sequence of reactions, we will break down each step and identify the products formed at each stage.

### Step 1: Identify the starting compound
The starting compound is a diene, which can be represented as:
\[ \text{H}_2\text{C} = \text{CH} - \text{CH} = \text{CH}_2 \]

### Step 2: Reaction with Bromine
The diene reacts with bromine (Br₂) in the presence of carbon tetrachloride (CCl₄). This reaction leads to the addition of bromine across the double bonds. The product formed (let's call it A) will have bromine atoms added to the carbon atoms where the double bonds were present.

The reaction can be represented as:
\[ \text{H}_2\text{C} = \text{CH} - \text{CH} = \text{CH}_2 + \text{Br}_2 \rightarrow \text{A} \]

The product A will be:
\[ \text{Br}-\text{CH}_2-\text{CH}-\text{CH}-\text{Br} \]

### Step 3: Reaction with KCN
Next, the product A undergoes a reaction with excess potassium cyanide (KCN). In this step, the bromine atoms are replaced by cyanide (CN) groups through a nucleophilic substitution reaction.

The reaction can be represented as:
\[ \text{A} + 2 \text{KCN} \rightarrow \text{B} + 2 \text{KBr} \]

The product B will now have cyanide groups attached to the carbon atoms where the bromine atoms were present:
\[ \text{CN}-\text{CH}_2-\text{CH}-\text{CH}-\text{CN} \]

### Step 4: Hydrogenation
The final step involves the hydrogenation of product B using hydrogen gas (H₂) in the presence of a nickel catalyst (Ni). This reaction will convert the cyanide groups into amine groups (NH₂) and will also saturate the carbon chain.

The reaction can be represented as:
\[ \text{B} + 2 \text{H}_2 \rightarrow \text{C}_4\text{H}_8\text{N}_2 \]

The final product B will be:
\[ \text{NH}_2-\text{CH}_2-\text{CH}-\text{CH}-\text{NH}_2 \]

### Final Product
Thus, the end product B is:
\[ \text{NH}_2\text{CH}_2\text{CH}(\text{NH}_2)\text{CH}_2 \]

### Summary of the Steps
1. Start with the diene: \(\text{H}_2\text{C} = \text{CH} - \text{CH} = \text{CH}_2\).
2. Add Br₂ in CCl₄ to form product A: \(\text{Br}-\text{CH}_2-\text{CH}-\text{CH}-\text{Br}\).
3. React with KCN to form product B: \(\text{CN}-\text{CH}_2-\text{CH}-\text{CH}-\text{CN}\).
4. Hydrogenate to form the final product: \(\text{NH}_2-\text{CH}_2-\text{CH}-\text{CH}-\text{NH}_2\).