Which of the following molecules is most likely to undergo an $E1$ reaction when stirred in water?

• A. ${\mathrm{CH}}_3\mathrm{CH}=\mathrm{CHBr}$ (allyl bromide)
• B. ${\mathrm{(CH}}_3{)}_3\mathrm{CBr}$ (tert-butyl bromide)
• C. ${\mathrm{CH}}_3\mathrm{CH}-\mathrm{Cl}$ (ethyl chloride)
• D. ${\mathrm{CH}}_3\mathrm{CH}-\mathrm{Br}$ (ethyl bromide)

Answer: (B)

Verified step by step guidance

1. Step 1: Understand the mechanism of an E1 reaction. E1 (unimolecular elimination) involves two main steps: first, the leaving group departs to form a carbocation intermediate; second, a base removes a proton from a β-carbon, resulting in the formation of a double bond.
2. Step 2: Identify the stability of the carbocation intermediate formed after the leaving group leaves. More stable carbocations favor the E1 mechanism because the rate-determining step is carbocation formation. Carbocation stability order is tertiary > secondary > primary > methyl.
3. Step 3: Analyze each molecule's structure to determine the type of carbocation that would form upon loss of the leaving group (Br or Cl). For example, tert-butyl bromide forms a tertiary carbocation, allyl bromide forms an allylic carbocation (which is resonance stabilized), ethyl chloride and ethyl bromide form primary carbocations (less stable).
4. Step 4: Consider the solvent and conditions. Water is a polar protic solvent that stabilizes carbocations and can promote E1 reactions, especially with stable carbocations.
5. Step 5: Conclude which molecule is most likely to undergo E1 based on carbocation stability and reaction conditions. The molecule that forms the most stable carbocation intermediate (typically tertiary or resonance-stabilized) will favor the E1 mechanism.