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Arrange the following alkyl halides in d...

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`

A

`A gt B gt C`

B

`C gt B gt A`

C

`B gt C gt A`

D

`A gt C gt B`

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The correct Answer is:
To arrange the given alkyl halides in decreasing order of the rate of beta-elimination reaction with alcoholic KOH, we need to analyze the stability of the alkenes formed from each alkyl halide. The more substituted the alkene, the more stable it is, and thus, the faster the rate of the elimination reaction. ### Step-by-Step Solution: 1. **Identify the Alkyl Halides:** - A: `CH₃C(CH₃)(H)(C)CH₂Br` - B: `CH₃CH₂Br` - C: `CH₃CH₂CH₂Br` 2. **Draw the Structures:** - For A: The structure is a branched alkyl halide with two methyl groups attached to the second carbon. - For B: The structure is a straight-chain alkyl halide (ethyl bromide). - For C: The structure is a straight-chain alkyl halide (propyl bromide). 3. **Determine the Beta Hydrogens:** - In beta-elimination, the hydrogen atom from the beta position (the carbon adjacent to the carbon bearing the halide) is removed. - For A, the beta position has 6 alpha hydrogens (2 from each methyl group). - For B, the beta position has 0 alpha hydrogens (no substituents). - For C, the beta position has 3 alpha hydrogens (from the propyl group). 4. **Assess the Stability of the Alkenes Formed:** - A forms a di-substituted alkene (more stable due to having more substituents). - B forms a non-substituted alkene (least stable). - C forms a mono-substituted alkene (more stable than B but less than A). 5. **Rank the Stability:** - A (most stable, di-substituted) > C (mono-substituted) > B (least stable, non-substituted). 6. **Final Order of Rate of Beta-Elimination:** - The order in decreasing rate of beta-elimination with alcoholic KOH is: - A > C > B ### Conclusion: The final arrangement of the alkyl halides in decreasing order of the rate of beta-elimination reaction with alcoholic KOH is: - **A > C > B**
To arrange the given alkyl halides in decreasing order of the rate of beta-elimination reaction with alcoholic KOH, we need to analyze the stability of the alkenes formed from each alkyl halide. The more substituted the alkene, the more stable it is, and thus, the faster the rate of the elimination reaction. ### Step-by-Step Solution: 1. **Identify the Alkyl Halides:** - A: `CH₃C(CH₃)(H)(C)CH₂Br` - B: `CH₃CH₂Br` - C: `CH₃CH₂CH₂Br` ...
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