IsoprophI bromide was treated separately with sodium ethoxide under two different conditions.
Reaction I:
Treatment of isopropyI bromide with `(Me_(3)CONa)` at `40^(@)C` gave almost exclusively compound `(A) (C_(3)H_(6))`.
Reaction II:
Treatment of `(i-PrBr)` with `NaOC_(2)H_(5)` at `30^(@)C` yileled compound `(A) (C_(3)H_(6))` along with a small amount of an ether `(B) (C_(5)H_(12)O)`.
Compound `(A)` was readily oxidised by a neutral solution of cold dil. `KMnO_(4)` to give a brown preciptate.
Which of the following is an accurate representations of compound `(B)` ?

Options

A. MeOMe`
B. `Et-O-Et`
C. `i-Pr-O-Et`
D. `EtOMe`

Explanation

To solve the problem, we need to analyze the reactions of isopropyl bromide with sodium ethoxide under two different conditions and determine the structure of compound B.

### Step-by-Step Solution:

1. **Identify Isopropyl Bromide Structure:**
- Isopropyl bromide (i-PrBr) has the structure:
\[ \text{(CH}_3\text{)}_2\text{CHBr} \]
- This means it has a bromine atom attached to a carbon that is also bonded to two other methyl groups.

2. **Reaction I:**
- Isopropyl bromide is treated with \( \text{(Me}_3\text{C)Na} \) at \( 40^\circ C \).
- This reaction likely leads to the formation of compound A, which is \( \text{C}_3\text{H}_6 \).
- Given the conditions, this reaction favors elimination (E2 mechanism) leading to the formation of propene (C3H6).

3. **Reaction II:**
- In this reaction, isopropyl bromide is treated with sodium ethoxide (\( \text{NaOC}_2\text{H}_5 \)) at \( 30^\circ C \).
- This reaction yields compound A (C3H6) and a small amount of compound B (C5H12O).
- The sodium ethoxide acts as a nucleophile and can perform an \( \text{S}_N2 \) reaction.

4. **Mechanism of Reaction II:**
- The ethoxide ion (\( \text{C}_2\text{H}_5\text{O}^- \)) attacks the carbon attached to bromine, leading to the formation of an ether.
- The bromine leaves as \( \text{NaBr} \), and the ethoxide group attaches to the isopropyl group.
- The product formed will be isopropyl ethyl ether, which has the molecular formula \( \text{C}_5\text{H}_{12}\text{O} \).

5. **Identify Compound B:**
- Compound B is the ether formed from the reaction, which is isopropyl ethyl ether.
- The structure can be represented as:
\[ \text{(CH}_3\text{)}_2\text{CHOCH}_2\text{CH}_3 \]
- This matches the molecular formula \( \text{C}_5\text{H}_{12}\text{O} \).

6. **Conclusion:**
- The accurate representation of compound B is isopropyl ethyl ether, which corresponds to option C in the provided choices.

### Final Answer:
- The accurate representation of compound B is isopropyl ethyl ether, which corresponds to option C.