A chloro compound(A) on reduction with Zn-Cu and ethanol give hydrocarbon (B) with 5 carbon atoms. When(A) is dissolved dry ether and treated with sodium metal it gave 2, 2, 5, 5 tetra methyl hexane. The treatment of A as A'C (in presence of alcoholic KCN)
The reaction of A with aq. KOH will preferably favour:

Options

A. `SN_(1)` Mechanism
B. `E_(1)` Mechanism
C. `SN_(2)` Mechanism
D. `E_(2)` Mechanism

Explanation

To solve the problem step-by-step, we need to analyze the information given about compound A and the reactions it undergoes.

### Step 1: Identify Compound A
We know that compound A is a chloro compound that, when reduced with Zn-Cu and ethanol, gives a hydrocarbon (B) with 5 carbon atoms. The hydrocarbon formed must be a straight-chain alkane, likely pentane (C5H12).

### Step 2: Analyze the Reaction with Sodium Metal
When compound A is treated with sodium metal in dry ether, it produces 2,2,5,5-tetramethylhexane. This suggests that A has a structure that allows for the formation of this product through a Wurtz reaction, which typically involves coupling of alkyl halides.

### Step 3: Determine the Structure of A
Given that the product is 2,2,5,5-tetramethylhexane, we can deduce that compound A must have a branched structure that can lead to this product. A plausible structure for A could be 2-chloro-2-methylpentane (C5H11Cl), which would allow for the formation of the desired hydrocarbon through the coupling of two alkyl radicals.

### Step 4: Reaction with Aqueous KOH
When compound A is treated with aqueous KOH, we need to consider the mechanism of the reaction. The chlorine atom (Cl) will leave, forming a carbocation. The stability of the carbocation will determine the pathway of the reaction:

1. **SN1 Mechanism**: The formation of a carbocation suggests that an SN1 mechanism is favored, especially since we have a tertiary carbocation from the structure of A.
2. **Rearrangement**: If the carbocation formed is stable, it may undergo rearrangement to form a more stable carbocation before the nucleophile (OH-) attacks.

### Step 5: Final Product Formation
After the carbocation rearrangement, the hydroxide ion (OH-) from aqueous KOH will attack the carbocation, leading to the formation of an alcohol. The final product will be a tertiary alcohol due to the structure of compound A.

### Conclusion
The reaction of compound A with aqueous KOH will preferably favor the formation of a tertiary alcohol through an SN1 mechanism.