Product C in the reaction
`C_(2)H_(5)Br underset(("aqueous"))overset(NaOH)to A overset(Na) to B overset(CH_(3)I)toC` will be

Options

A. Propane
B. Ethyl iodide
C. Ethane
D. Ethyl methyl ether

Explanation

To determine the product C in the given reaction sequence, let's break down the steps involved:

### Step-by-Step Solution:

1. **Identify the Starting Material**:
The starting material is ethyl bromide, which can be represented as CH₃CH₂Br.

2. **Reaction with Aqueous NaOH**:
Ethyl bromide (CH₃CH₂Br) reacts with aqueous sodium hydroxide (NaOH). This is a nucleophilic substitution reaction (SN2) where the hydroxide ion (OH⁻) acts as a nucleophile and attacks the electrophilic carbon atom bonded to bromine.
- The bromine atom (Br) is a good leaving group and leaves as Br⁻.
- The product formed is ethanol (CH₃CH₂OH).

3. **Reaction with Sodium (Na)**:
The ethanol (CH₃CH₂OH) then reacts with sodium (Na). Sodium reacts with alcohols to form alkoxides. In this case, sodium ethoxide (CH₃CH₂O⁻Na⁺) is formed.
- The reaction can be represented as:
\[ CH₃CH₂OH + Na \rightarrow CH₃CH₂O⁻Na⁺ + \frac{1}{2} H₂ \]

4. **Reaction with Methyl Iodide (CH₃I)**:
Sodium ethoxide (CH₃CH₂O⁻Na⁺) now reacts with methyl iodide (CH₃I). In this reaction, the ethoxide ion (CH₃CH₂O⁻) acts as a nucleophile and attacks the electrophilic carbon in methyl iodide.
- The iodide ion (I⁻) is a good leaving group and leaves.
- The product formed is methoxyethane (CH₃CH₂OCH₃), also known as ethyl methyl ether.

5. **Final Product**:
The final product C is methoxyethane (CH₃CH₂OCH₃).

### Summary of Products:
- A (Ethanol): CH₃CH₂OH
- B (Sodium Ethoxide): CH₃CH₂O⁻Na⁺
- C (Methoxyethane): CH₃CH₂OCH₃

### Final Answer:
The product C in the reaction is **methoxyethane (CH₃CH₂OCH₃)**.

---