Alkyl halide on reduction with Zn + HCl gives

Options

A. alcohol
B. alkene
C. alkane
D. ether

Explanation

To solve the question "Alkyl halide on reduction with Zn + HCl gives", we will follow these steps:

### Step 1: Understand the Reactants
- The reactant is an alkyl halide, which can be represented as R-X, where R is an alkyl group and X is a halogen (like Cl, Br, or I).

**Hint:** Alkyl halides are compounds where a halogen atom is bonded to an alkyl group.

### Step 2: Identify the Reducing Agent
- The reducing agent in this reaction is zinc (Zn) in the presence of hydrochloric acid (HCl). This combination produces nascent hydrogen (H), which is a highly reactive form of hydrogen.

**Hint:** Zn + HCl generates nascent hydrogen, which is crucial for reduction reactions.

### Step 3: Understand the Reduction Process
- The reduction of an alkyl halide (R-X) involves the removal of the halogen (X) and the addition of hydrogen (H). This process converts the alkyl halide into an alkane (R-H).

**Hint:** Reduction typically involves the gain of hydrogen or the loss of halogen.

### Step 4: Write the Reaction
- The general reaction can be written as:
\[ R-X + Zn + HCl \rightarrow R-H + ZnXCl \]
- Here, R-H is the alkane formed, and ZnXCl is a byproduct.

**Hint:** The alkyl halide loses the halogen and gains hydrogen to form an alkane.

### Step 5: Conclude the Product
- The final product of the reaction is an alkane (R-H).

**Hint:** Remember that the alkane is formed by replacing the halogen in the alkyl halide with hydrogen.

### Final Answer:
The alkyl halide on reduction with Zn + HCl gives an alkane (R-H).

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