Which of the following best describes the intermediate and major organic product formed in an E1 reaction of $C_2{H}_5{\mathrm{Br}}_1$ with methanol as solvent?

• A. The intermediate is a carbanion and the major product is $C_2{H}_6$ (ethane).
• B. The intermediate is a carbocation and the major product is $C_2{H}_4$ (ethylene).
• C. The intermediate is a bromonium ion and the major product is $C_2{H}_5\mathrm{OH}$ (ethanol).
• D. The intermediate is a free radical and the major product is $C_2{H}_5\mathrm{Br}$ (ethyl bromide).

Answer: (B)

Verified step by step guidance

1. Identify the substrate and reaction conditions: The substrate is ethyl bromide (C\_2H\_5Br), and the solvent is methanol, which is a polar protic solvent. This environment favors an E1 mechanism.
2. Recall the E1 reaction mechanism: E1 reactions proceed via a two-step process where the first step is the formation of a carbocation intermediate by loss of the leaving group (Br\^-).
3. Determine the intermediate: Since the leaving group (Br) departs first, the intermediate formed is a carbocation, specifically an ethyl carbocation (C\_2H\_5\^+).
4. Consider the major product formation: After carbocation formation, a proton is lost from a beta-carbon to form a double bond, resulting in an alkene. For ethyl carbocation, this leads to ethylene (C\_2H\_4) as the major product.
5. Exclude other options: The intermediate is not a carbanion, bromonium ion, or free radical in an E1 reaction. Also, the major product is not ethane, ethanol, or ethyl bromide under these conditions.