Which of the following is fast de-brominated ?

Options

A.
B.
C.
D.

Answer

To determine which of the given compounds is fast de-brominated, we need to analyze the stability of the products formed during the dehydrohalogenation reaction. The process involves the elimination of HBr in the presence of an alcoholic KOH. Let's break down the analysis step-by-step:

### Step 1: Identify the Compounds
We have four compounds with bromine substituted. We need to analyze each compound to see how the de-bromination will occur.

### Step 2: Understand the Mechanism
The de-bromination reaction is a type of dehydrohalogenation, where HBr is eliminated. This typically occurs through the abstraction of a hydrogen atom adjacent to the carbon atom bonded to bromine.

### Step 3: Analyze Each Compound
1. **Compound 1**: Contains a double bond and a bromine atom. The hydrogen can be removed from either of the two adjacent carbons. However, one carbon is sp2 hybridized, making it less favorable to remove hydrogen from there due to higher energy requirements.

2. **Compound 2**: This compound has bromine attached to a carbon in a benzene ring. The removal of HBr here leads to the formation of a stable benzene ring, which is very stable due to resonance.

3. **Compound 3**: This is a cyclohexane with bromine. The hydrogen can be removed from any position, leading to the formation of a cyclohexene. This compound is stable, but not as stable as the benzene formed in Compound 2.

4. **Compound 4**: This compound also has a double bond and bromine. The hydrogen removal would lead to a cumulated diene, which is less stable compared to the benzene ring formed in Compound 2.

### Step 4: Determine the Fastest De-bromination
Based on the analysis:
- Compound 2 (the one forming benzene) is the most stable product due to resonance stabilization.
- Therefore, it will undergo de-bromination the fastest.

### Conclusion
The compound that is fast de-brominated is **Compound 2**, which leads to the formation of a stable benzene ring.

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