An organic compound `(a)` reacts with sodium metal and forms `(b)`. On heating with conc. `H_(2)S0_(4)` (a) gives diethyl ether. `(a)` and `(b)` are respectively

Options

A. `C_(2)H_(5)OH and C_(2)H_(5)O Na`
B. `C_(3)H_(7)OH and C_(3)H_(7)O Na`
C. `CH_(3)OH and CH_(3)O Na`
D. `C_(4)H_(9)OH and C_(4)H_(9)O Na`

Explanation

To solve the problem step by step, we will identify the organic compounds `(a)` and `(b)` based on the given reactions.

### Step 1: Identify Compound `(a)`
The problem states that compound `(a)` reacts with sodium metal to form compound `(b)`. In organic chemistry, when an alcohol reacts with sodium metal, it forms sodium alkoxide and hydrogen gas.

The general reaction is:
\[ \text{R-OH} + \text{Na} \rightarrow \text{R-O}^- \text{Na}^+ + \frac{1}{2} \text{H}_2 \]

Given that the final product after heating with concentrated sulfuric acid is diethyl ether, we can deduce that compound `(a)` must be an alcohol.

Since diethyl ether is formed, we can assume that the alcohol is ethanol (ethyl alcohol), which has the formula:
\[ \text{C}_2\text{H}_5\text{OH} \]

Thus, we conclude:
\[ (a) = \text{C}_2\text{H}_5\text{OH} \]

### Step 2: Identify Compound `(b)`
From the reaction of ethanol with sodium, we form sodium ethoxide. The reaction can be represented as follows:
\[ \text{C}_2\text{H}_5\text{OH} + \text{Na} \rightarrow \text{C}_2\text{H}_5\text{O}^- \text{Na}^+ + \frac{1}{2} \text{H}_2 \]

Here, sodium ethoxide is the product `(b)`, which has the formula:
\[ (b) = \text{C}_2\text{H}_5\text{O}\text{Na} \]

### Step 3: Confirm the Formation of Diethyl Ether
When we heat the alcohol (ethanol) with concentrated sulfuric acid, it leads to the formation of diethyl ether through a dehydration reaction. The reaction can be summarized as:
\[ 2 \text{C}_2\text{H}_5\text{OH} \xrightarrow{\text{conc. } H_2SO_4} \text{C}_2\text{H}_5\text{O}\text{C}_2\text{H}_5 + \text{H}_2\text{O} \]

Thus, the final product is indeed diethyl ether, confirming our earlier deductions.

### Final Answer
Therefore, the compounds `(a)` and `(b)` are:
- `(a) = \text{C}_2\text{H}_5\text{OH}` (Ethanol)
- `(b) = \text{C}_2\text{H}_5\text{O}\text{Na}` (Sodium ethoxide)