Given $2-\mathrm{bromo}-2-\mathrm{methylbutane}$, what alkene is the major product formed in an $E1$ reaction?

• A. $C_{2}H{5}_2-C=\mathrm{CH}{2}_1$ (2-methyl-2-butene)
• B. $\mathrm{CH}{2}_1=\mathrm{CH}-\mathrm{CH}{2}_1-\mathrm{CH}{3}_1$ (1-butene)
• C. $\mathrm{CH}{3}_1-\mathrm{CH}{2}_1-\mathrm{CH}=\mathrm{CH}{2}_1$ (1-butene)
• D. $\mathrm{CH}{3}_1-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}{3}_1$ (2-butene)

Answer: (A)

Verified step by step guidance

1. Identify the structure of 2-bromo-2-methylbutane. It is a tertiary alkyl halide with the bromine attached to the second carbon, which also has a methyl substituent.
2. Recall that an E1 reaction proceeds via a two-step mechanism: first, the leaving group (Br) departs, forming a carbocation intermediate; second, a base removes a proton from a β-carbon to form the alkene.
3. Determine the carbocation intermediate formed after the bromide leaves. Since the bromine is on a tertiary carbon, the carbocation formed is tertiary and relatively stable.
4. Identify all possible β-hydrogens adjacent to the carbocation center that can be removed to form alkenes. Consider the formation of more substituted (more stable) alkenes according to Zaitsev's rule.
5. Draw the possible alkenes formed by elimination of β-hydrogens and compare their substitution patterns. The major product will be the most substituted, most stable alkene, which in this case is 2-methyl-2-butene.