Ethyl methyl ketone on reduction with zinc amalgam and Conc. HCl gives :

Options

A. Sec. alcohol
B. `1^(th)` alcohol
C. `3^(th)` alcohol
D. n-butane

Explanation

To solve the question regarding the reduction of ethyl methyl ketone with zinc amalgam and concentrated HCl, we will follow these steps:

### Step-by-Step Solution:

1. **Identify the Compound**:
Ethyl methyl ketone is also known as 3-pentanone. Its structure can be represented as:
\[ CH_3-CO-CH_2-CH_3 \]
where the carbonyl group (C=O) is located between the methyl (CH3) and ethyl (C2H5) groups.

**Hint**: Recognize the structure of ethyl methyl ketone and identify the carbonyl group.

2. **Understand the Reaction**:
The reaction in question is a Clemensen reduction, which involves the reduction of a ketone to an alkane using zinc amalgam and concentrated hydrochloric acid (HCl). The general reaction can be represented as:
\[ R_1C(=O)R_2 + Zn(Hg) + HCl \rightarrow R_1R_2C + H_2O \]
Here, the carbonyl group (C=O) is reduced to a methylene group (C-H2).

**Hint**: Recall that Clemensen reduction converts carbonyl compounds to alkanes.

3. **Apply the Reaction to Ethyl Methyl Ketone**:
When ethyl methyl ketone undergoes Clemensen reduction, the carbonyl group (C=O) is removed, and two hydrogen atoms are added. The reaction can be written as:
\[ CH_3-CO-CH_2-CH_3 + Zn(Hg) + HCl \rightarrow CH_3-CH_2-CH_2-CH_3 + H_2O \]
This results in the formation of butane (C4H10).

**Hint**: Write the balanced equation for the reduction, ensuring to account for the formation of water.

4. **Final Product Identification**:
The product of the reaction is butane, which is a saturated hydrocarbon (alkane) with the formula C4H10. The structure of butane can be represented as:
\[ CH_3-CH_2-CH_2-CH_3 \]

**Hint**: Identify the final product based on the number of carbon atoms and hydrogen atoms.

### Conclusion:
The reduction of ethyl methyl ketone with zinc amalgam and concentrated HCl gives butane.

### Final Answer:
**Butane (C4H10)**