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 most accureately repreasents the activated complex formed in reaction `(II)` that leads to compound `(A)` ?

Options

A.
B.
C.
D.

Explanation

To solve the question regarding the activated complex formed in reaction II that leads to compound A, we will follow a systematic approach.

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactant in reaction II is isopropyl bromide (i-PrBr) and sodium ethoxide (NaOC2H5).

2. **Understand the Reaction Conditions**:
- The reaction is conducted at 30°C, which suggests that we are likely dealing with a reaction that can proceed via an elimination mechanism rather than substitution due to the presence of a strong base (NaOC2H5).

3. **Determine the Product**:
- The product A is identified as an alkene (C3H6). Given that isopropyl bromide has two methyl groups, the likely alkene product is propene (C3H6).

4. **Mechanism of Reaction**:
- The formation of compound A (alkene) from isopropyl bromide and sodium ethoxide suggests an E2 elimination mechanism. In this mechanism, the base (ethoxide ion) abstracts a proton from the β-carbon while the leaving group (bromine) departs from the α-carbon.

5. **Identify the Transition State**:
- In the E2 mechanism, the transition state involves the simultaneous removal of a hydrogen atom and the departure of the bromine atom. This transition state can be depicted as having partial charges: the ethoxide ion will have a partial negative charge as it is abstracting a proton, and the bromine will have a partial negative charge as it is leaving.

6. **Draw the Transition State**:
- The transition state can be represented with the isopropyl structure, showing the ethoxide ion attacking the hydrogen on the β-carbon, while the bromine is leaving. The transition state will have a partially formed double bond between the α and β carbons, and the ethoxide ion will be positioned close to the hydrogen being removed.

7. **Conclusion**:
- Based on the above analysis, the activated complex formed in reaction II that leads to compound A can be accurately represented by the transition state structure that shows the ethoxide ion, the leaving bromine, and the developing double bond.