Which of the following alkyl halides is the most reactive substrate in an $E_2$ reaction?

• A. ${\mathrm{(CH}}_3{)}^3\mathrm{CBr}$
• B. ${\mathrm{CH}}_3\mathrm{CHBr}{\mathrm{CH}}_3$
• C. ${\mathrm{CH}}_3\mathrm{CH}{\mathrm{CH}}_2\mathrm{Br}$
• D. ${\mathrm{CH}}_3\mathrm{Br}$

Answer: (A)

Verified step by step guidance

1. Step 1: Understand the nature of the E2 reaction mechanism. E2 is a bimolecular elimination that requires a strong base and a good leaving group, and it proceeds via a single concerted step where the base removes a proton while the leaving group departs.
2. Step 2: Recall that the reactivity of alkyl halides in E2 depends on the degree of substitution of the carbon bearing the leaving group. Tertiary alkyl halides (3°) are generally more reactive in E2 than secondary (2°) or primary (1°) because the transition state is stabilized by alkyl substituents through hyperconjugation and inductive effects.
3. Step 3: Identify the degree of substitution of each alkyl halide given: (CH\_3)\_3CBr is tertiary, CH\_3CHBrCH\_3 is secondary, CH\_3CH\_2CH\_2Br is primary, and CH\_3Br is primary.
4. Step 4: Consider steric hindrance: although tertiary alkyl halides are hindered, E2 reactions tolerate steric bulk better than SN2, so tertiary alkyl halides still react well in E2.
5. Step 5: Conclude that the most reactive substrate in an E2 reaction is the tertiary alkyl halide (CH\_3)\_3CBr because it forms the most stable transition state and the elimination is favored over substitution.