The compound that is most reactive with alcoholic KOH is

Options

A. `CH_2=CH-Br`
B. `CH_3CH_2Br`
C. `(CH_3)_2CH-Br`
D. `CH_3COCH_2CH_2Br`

Answer

To determine which compound is most reactive with alcoholic KOH, we need to analyze the reactivity of each compound provided in the options and the stability of the products formed after the reaction.

### Step-by-Step Solution:

1. **Identify the Reaction Type:**
Alcoholic KOH is a strong base and typically promotes elimination reactions (dehydrohalogenation) to form alkenes from alkyl halides.

2. **Analyze Each Option:**
- **Option 1: CH2=CHBr (Allyl bromide)**
- Reaction: The base (OH-) abstracts a hydrogen atom, and Br- leaves, forming CH≡CH (ethyne).
- Stability: Ethyne is less stable than alkenes due to the presence of a triple bond, which is more strained and has a shorter bond length.
- Conclusion: This compound is not the most reactive.

- **Option 2: CH3CH2Br (Bromoethane)**
- Reaction: The base abstracts a beta hydrogen, and Br- leaves, forming CH2=CH2 (ethylene).
- Stability: Ethylene is a stable alkene, but we need to compare it with other options.
- Conclusion: This compound is reactive but not the most reactive.

- **Option 3: CH(CH3)2Br (2-Bromopropane)**
- Reaction: The base abstracts a hydrogen from one of the methyl groups, and Br- leaves, forming CH3C(CH3)=CH2 (propene).
- Stability: Propene is more stable than ethylene due to having more alkyl substituents (hyperconjugation).
- Conclusion: This compound is more reactive than the previous ones.

- **Option 4: CH3C(=O)CH2CH2Br (Bromoacetone)**
- Reaction: The base abstracts a hydrogen from the beta carbon, and Br- leaves, forming CH2=CH2 (ethylene) with resonance stabilization due to the adjacent carbonyl group.
- Stability: The product benefits from resonance stabilization, making it more stable than the products from the previous options.
- Conclusion: This compound is likely the most reactive due to the stability of the product formed.

3. **Final Conclusion:**
After analyzing all options, **Option 4 (CH3C(=O)CH2CH2Br)** is the most reactive with alcoholic KOH because it leads to the formation of a more stable product due to resonance stabilization.