Alkyl halides on reduction with Zn-Cu couple and alcohol give -

Options

A. Alkanes
B. Alkanes
C. Alkynes
D. Cyclic compounds

Answer

To solve the question regarding the reduction of alkyl halides with a zinc-copper couple in the presence of alcohol, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- We start with an alkyl halide, represented as R-X, where R is the alkyl group and X is the halogen (e.g., Cl, Br, I).

2. **Understand the Reducing Agent**:
- The reducing agent in this reaction is a zinc-copper couple (Zn-Cu). This combination is known for facilitating the reduction of alkyl halides.

3. **Reaction Mechanism**:
- When the alkyl halide (R-X) reacts with the zinc-copper couple in an alcohol solvent, zinc donates electrons. The zinc metal (Zn) is oxidized to Zn²⁺ while releasing two electrons:
\[ \text{Zn} \rightarrow \text{Zn}^{2+} + 2e^- \]

4. **Formation of Free Radicals**:
- The alkyl halide undergoes homolytic bond fission, where the bond between R and X breaks, resulting in the formation of a free radical (R•) and a halide ion (X⁻):
\[ \text{R-X} \rightarrow \text{R}• + \text{X}⁻ \]

5. **Nucleophilic Attack**:
- The free radical (R•) can then react with another molecule of alkyl halide (R-X). The free radical (R•) acts as a nucleophile and attacks the carbon of another alkyl halide, leading to the formation of a new bond:
\[ \text{R}• + \text{R-X} \rightarrow \text{R-R} + \text{X}⁻ \]

6. **Product Formation**:
- The final product of this reaction is an alkane (R-R), as the two alkyl groups combine to form a new carbon-carbon bond:
\[ \text{Product} = \text{Alkane (R-R)} \]

7. **Name of the Reaction**:
- This reaction is known as the **Franklin Reaction**, which is similar to the Wurtz Reaction but uses zinc instead of sodium.

### Final Answer:
The product formed from the reduction of alkyl halides with a zinc-copper couple in alcohol is an **alkane (R-R)**.