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.
The formations of `(A)` and `(B)` are best explained by:

Options

A. `SN^(2)` reaction and `E2` reaction, respectively
B. `E2` reaction and `SN^(2)` reaction, respectively
C. `E1` reaction and `SN^(1)` reaction, respectively
D. `E2` reaction and `SN^(1)` reaction, respectively

Explanation

To solve the question regarding the reactions of isopropyl bromide with sodium ethoxide under different conditions, we can break it down into a step-by-step analysis of each reaction and the mechanisms involved.

### Step 1: Identify the Reactants
The reactant in both reactions is isopropyl bromide, which can be represented as:
\[ \text{(CH}_3\text{)}_2\text{CHBr} \]

### Step 2: Reaction I with (Me₃CONa) at 40°C
In the first reaction, isopropyl bromide is treated with a bulky base, tert-butoxide (Me₃CONa), at a higher temperature (40°C).

- **Mechanism**: The bulky base favors an elimination reaction (E2 mechanism) rather than substitution due to steric hindrance.
- **Process**:
- The base abstracts a hydrogen atom from one of the methyl groups adjacent to the carbon bonded to bromine.
- The bromine atom leaves, resulting in the formation of a double bond.
- **Product**: The product formed is propene (C₃H₆), which is compound A.

### Step 3: Reaction II with NaOC₂H₅ at 30°C
In the second reaction, isopropyl bromide is treated with sodium ethoxide (NaOC₂H₅) at a lower temperature (30°C).

- **Mechanism**: Sodium ethoxide is a less bulky base, which allows for both elimination and substitution reactions. However, the primary pathway here is substitution (SN2 mechanism).
- **Process**:
- The bromine atom leaves, forming a carbocation intermediate.
- The ethoxide ion (C₂H₅O⁻) attacks the carbocation, leading to the formation of an ether.
- **Products**: The products formed are:
- Propene (C₃H₆) again (compound A).
- An ether (compound B), which has the molecular formula C₅H₁₂O.

### Step 4: Summary of Products
- **Compound A**: Propene (C₃H₆) is formed in both reactions.
- **Compound B**: An ether (C₅H₁₂O) is formed only in the second reaction.

### Step 5: Oxidation of Compound A
The question mentions that compound A (propene) is readily oxidized by cold dilute KMnO₄, which indicates that it can undergo oxidation reactions typical for alkenes.

### Conclusion
The formation of compounds A and B can be explained as follows:
- **Compound A (C₃H₆)** is formed by an E2 mechanism in the first reaction and by an SN2 mechanism in the second reaction.
- **Compound B (C₅H₁₂O)** is formed by the SN2 mechanism in the second reaction.

### Final Answer
The formations of compounds A and B are best explained by the E2 mechanism for compound A in the first reaction and the SN2 mechanism for compound B in the second reaction.

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