Which of the following is the least reactive substrate in an $E_2$ reaction?

• A. ${\mathrm{CH}}_3{\mathrm{CH}}_2{\mathrm{CH}}_2\mathrm{Br}$ (1-bromobutane)
• B. ${\mathrm{CH}}_3{\mathrm{CH}}_2\mathrm{Br}$ (1-bromopropane)
• C. ${\mathrm{CH}}_3\mathrm{Br}$ (methyl bromide)
• D. ${\mathrm{(CH}}_3{)}_3\mathrm{CBr}$ (tert-butyl bromide)

Answer: (A)

Verified step by step guidance

1. Step 1: Understand that an E2 reaction is a bimolecular elimination where a base removes a proton (β-hydrogen) from the substrate while the leaving group departs simultaneously, forming a double bond.
2. Step 2: Recall that the reactivity of substrates in E2 depends largely on the degree of the carbon attached to the leaving group (primary, secondary, tertiary) and the availability of β-hydrogens for elimination.
3. Step 3: Analyze each substrate: methyl bromide has no β-hydrogens, so it cannot undergo E2; primary alkyl bromides (1-bromopropane and 1-bromobutane) have β-hydrogens but are less reactive than tertiary; tert-butyl bromide is tertiary and highly reactive in E2 due to the stability of the transition state and the ease of β-hydrogen removal.
4. Step 4: Compare the primary substrates (1-bromopropane and 1-bromobutane) by considering steric hindrance and chain length; longer chains can sometimes slightly affect reactivity but generally both are primary and less reactive than tertiary.
5. Step 5: Conclude that methyl bromide is unreactive in E2 due to lack of β-hydrogens, tert-butyl bromide is most reactive, and among the primary substrates, the least reactive is the one with the least favorable conditions for elimination, which is 1-bromobutane.