A chlorocompound (A) on reduction with Zn-Cu and ethanol gives the hydrocarbon (B) with five carbon atoms. When (A) is dissolved in dry ether and treated with sodium metal it gave 2,2,5,5,-tetramethylhexane. The treatment of A as
`A overset(alc. KCN)to C`
The reaction of C with Na, `C_(2)H_(5)OH` gives

Options

A. `(CH_(3))_(3)C CH_(2)CONH_(2)`
B. `(CH_(3))_(3)CNH_(2)`
C. `(CH_(3))_(3)C CH_(2)NH_(2)`
D. `(CH_(3))_(2)CHCH_(2)NH_(2)`

Answer

To solve the problem step by step, we will analyze the information given and deduce the required reactions and products.

### Step 1: Identify Compound A
The problem states that a chlorocompound (A) on reduction with Zn-Cu and ethanol gives a hydrocarbon (B) with five carbon atoms.

- Since the hydrocarbon (B) has five carbon atoms, we can deduce that compound A must be a chlorinated alkane that can yield a five-carbon hydrocarbon upon reduction.

### Step 2: Determine the Structure of Compound A
The reduction of A gives a hydrocarbon (B) with five carbon atoms. The reduction of a chlorinated compound typically removes the halogen and adds hydrogen.

- A reasonable assumption for A could be **1-chloro-2,2-dimethylpropane** (C5H11Cl). This compound has a chlorine atom that can be removed during reduction, yielding a hydrocarbon with five carbon atoms.

### Step 3: Reaction of Compound A with Sodium in Dry Ether
When compound A is treated with sodium metal in dry ether, it produces **2,2,5,5-tetramethylhexane**.

- This reaction is known as the Wurtz reaction, which couples two alkyl halides. The structure of 2,2,5,5-tetramethylhexane confirms that A could indeed be 1-chloro-2,2-dimethylpropane, as it can couple to form the larger hydrocarbon.

### Step 4: Reaction of Compound A with Alcoholic KCN
The next part of the problem states that compound A is treated with alcoholic KCN to give compound C.

- The reaction of 1-chloro-2,2-dimethylpropane with KCN will lead to nucleophilic substitution, where the cyanide ion (CN-) replaces the chlorine atom. This gives us **2,2-dimethylbutanenitrile (C)**.

### Step 5: Reaction of Compound C with Sodium Ethanol
Finally, we need to determine the product when compound C (2,2-dimethylbutanenitrile) is treated with sodium in ethanol.

- The reaction of nitriles with sodium in ethanol typically leads to a reduction known as the Mendius reduction. In this case, the nitrile group (C≡N) is reduced to an amine group (C-NH2).

- The product will be **3,3-dimethylbutanamine**.

### Summary of Reactions
1. **A (1-chloro-2,2-dimethylpropane) + Zn-Cu/EtOH → B (hydrocarbon with 5 carbons)**
2. **A + Na (dry ether) → 2,2,5,5-tetramethylhexane**
3. **A + alcoholic KCN → C (2,2-dimethylbutanenitrile)**
4. **C + Na (ethanol) → 3,3-dimethylbutanamine**

### Final Answer
The final product obtained from the reaction of C with sodium in ethanol is **3,3-dimethylbutanamine**.

---