An organic liquid of the composition `C_4 H_8 O_2` yields a sodium salt of an acid `C_3 H_6 O_2` and methanol on boiling with `NaOH` solution. The gives liquid is :

Options

A. `CH_3 CH_2 COOCH_3`
B. `CH_3 COOC_2 H_5`
C. `HCOOC_3 H_7`
D. `CH_3 CH_2 CH_2 COOH`

Explanation

To solve the question, we will follow a systematic approach to identify the organic liquid with the composition \( C_4H_8O_2 \) that yields a sodium salt of an acid \( C_3H_6O_2 \) and methanol upon boiling with sodium hydroxide (NaOH) solution.

### Step 1: Determine the Degree of Unsaturation
The degree of unsaturation can be calculated using the formula:
\[ \text{Degree of Unsaturation} = \frac{2C + 2 - H}{2} \]
Where \( C \) is the number of carbon atoms and \( H \) is the number of hydrogen atoms.

For \( C_4H_8O_2 \):
- \( C = 4 \)
- \( H = 8 \)

Substituting the values:
\[ \text{Degree of Unsaturation} = \frac{2(4) + 2 - 8}{2} = \frac{8 + 2 - 8}{2} = \frac{2}{2} = 1 \]

### Step 2: Identify Functional Groups
A degree of unsaturation of 1 indicates the presence of either a double bond (like a carbonyl group) or a ring. Given that the compound contains two oxygen atoms, it suggests the presence of a carbonyl group (C=O).

### Step 3: Analyze the Reaction with NaOH
The organic liquid reacts with NaOH to form a sodium salt of an acid \( C_3H_6O_2 \) and methanol. The sodium salt of the acid suggests that the original compound is likely an ester.

### Step 4: Identify the Acid and Its Salt
The acid \( C_3H_6O_2 \) can be identified as propanoic acid, which has the structure:
\[ \text{Propanoic Acid: } CH_3CH_2C(=O)OH \]
When the ester reacts with NaOH, it forms the sodium salt of propanoic acid and methanol.

### Step 5: Determine the Structure of the Ester
Since the ester must have one more carbon than the acid (4 carbons in total), the structure of the ester can be deduced as:
\[ \text{Ester: } CH_3CH_2C(=O)OCH_3 \]
This is methyl propanoate.

### Step 6: Confirm the Reaction
When methyl propanoate reacts with NaOH, it undergoes hydrolysis:
\[ CH_3CH_2C(=O)OCH_3 + NaOH \rightarrow CH_3CH_2C(=O)O^-Na^+ + CH_3OH \]
This confirms that the products are the sodium salt of propanoic acid and methanol.

### Conclusion
The organic liquid with the composition \( C_4H_8O_2 \) is methyl propanoate.

### Final Answer
The given liquid is methyl propanoate.

---