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.
Referring to Q. No. 67 which of the following represents the intermediate `T.S` for the formation of compound `(B)` ?

Options

A.
B.
C.
D.

Explanation

To solve the problem, we need to analyze the reactions of isopropyl bromide with sodium ethoxide under two different conditions and identify the transition state for the formation of compound B.

### Step-by-Step Solution:

**Step 1: Understand the Reactants and Products**
- Isopropyl bromide (i-PrBr) has the structure:
\[ \text{(CH}_3)_2\text{CHOBr} \]
- When treated with sodium ethoxide (NaOC₂H₅), it can yield compound A (C₃H₆) and compound B (C₅H₁₂O).

**Step 2: Analyze Reaction I**
- In Reaction I, isopropyl bromide is treated with (Me₃CONa) at 40°C, leading to the formation of compound A. This suggests a straightforward substitution reaction, likely an SN2 mechanism, where the bromine leaves and a nucleophile replaces it.

**Step 3: Analyze Reaction II**
- In Reaction II, isopropyl bromide is treated with NaOC₂H₅ at 30°C. This reaction yields compound A along with a small amount of ether (compound B). The ether formation indicates that there is an additional step occurring here.

**Step 4: Identify the Mechanism**
- The formation of compound B (an ether) suggests that the reaction involves an SN2 mechanism where the ethoxide ion (C₂H₅O⁻) attacks the isopropyl bromide. The bromine atom acts as a leaving group.

**Step 5: Draw the Transition State**
- In the transition state, the nucleophile (C₂H₅O⁻) is attacking the carbon atom that is bonded to the bromine. The bromine is partially leaving, resulting in a high-energy state.
- The structure of the transition state can be represented as:
\[ \text{C} \text{(CH}_3)_2\text{C} \text{(O}^- \text{C}_2\text{H}_5) \text{(Br)} \text{(partial bond)} \]
- In this state, the carbon is bonded to both the ethoxide and the bromine, which is about to leave.

**Step 6: Identify the Correct Option**
- Based on the structure of the transition state, we can conclude that option C represents the correct transition state for the formation of compound B.

### Final Answer:
The transition state for the formation of compound B is represented by option C.

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