Arrange the following alkyl halides in decreasing order of the rate or `beta`-elimination reaction alcoholic KOH.
A. `CH_(3)-underset(CH_(3))underset(|)overset(H)overset(|)(C)-CH_(2)Br`
B. `CH_(3)-CH_(2)-Br`
C. `CH_(3)-CH_(2)-CH_(2)-Br`

Options

A. `A gt B gt C`
B. `C gt B gt A`
C. `B gt C gt A`
D. `A gt C gt B`

Explanation

To arrange the given alkyl halides in decreasing order of the rate of β-elimination reaction with alcoholic KOH, we need to analyze the structure of each compound and the stability of the alkenes formed after the elimination reaction.

### Step-by-step Solution:

1. **Identify the Alkyl Halides**:
- A: `CH3-CH(CH3)-CH2Br` (1-bromo-2-methylpropane)
- B: `CH3-CH2Br` (ethyl bromide)
- C: `CH3-CH2-CH2Br` (1-bromopropane)

2. **Understand β-Elimination**:
- β-elimination involves the removal of a hydrogen atom from the β-carbon and the halogen (Br) from the α-carbon, resulting in the formation of an alkene. The stability of the alkene formed is crucial.

3. **Determine the Substitution of the Alkenes**:
- For compound A (1-bromo-2-methylpropane):
- The elimination will yield 2-methylpropene, which is a disubstituted alkene (two alkyl groups attached to the double bond).
- For compound B (ethyl bromide):
- The elimination will yield ethene, which is a non-substituted alkene (no alkyl groups attached to the double bond).
- For compound C (1-bromopropane):
- The elimination will yield propene, which is a monosubstituted alkene (one alkyl group attached to the double bond).

4. **Compare the Stability of the Alkenes**:
- Disubstituted alkenes are more stable than monosubstituted alkenes, which are more stable than non-substituted alkenes. Thus, the order of stability is:
- 2-methylpropene (disubstituted) > propene (monosubstituted) > ethene (non-substituted).

5. **Rate of β-Elimination Reaction**:
- The rate of β-elimination reaction is directly related to the stability of the alkene formed. Therefore, the order of the rate of reaction will be:
- A (1-bromo-2-methylpropane) > C (1-bromopropane) > B (ethyl bromide).

6. **Final Arrangement**:
- The decreasing order of the rate of β-elimination reaction with alcoholic KOH is:
- A > C > B.

### Conclusion:
The correct order is:
1. A: `CH3-CH(CH3)-CH2Br` (1-bromo-2-methylpropane)
2. C: `CH3-CH2-CH2Br` (1-bromopropane)
3. B: `CH3-CH2Br` (ethyl bromide)