When $3$-bromo-$3$-methylheptane is treated with a strong base, how many different alkenes (constitutional isomers) can be formed as major products via E$2$ elimination?

• A. $2$
• B. $4$
• C. $3$
• D. $5$

Answer: (B)

Verified step by step guidance

1. Identify the structure of 3-bromo-3-methylheptane. It is a seven-carbon chain (heptane) with a bromine substituent and a methyl group both attached to the third carbon.
2. Determine the possible β-hydrogens available for E2 elimination. Since the bromine is on carbon 3, look at carbons 2 and 4 (the carbons adjacent to carbon 3) for hydrogens that can be abstracted by the strong base.
3. For each β-carbon (C2 and C4), consider all unique hydrogens that can be removed to form a double bond. Remember that elimination forms an alkene between the carbon bearing the leaving group (C3) and the β-carbon where the hydrogen is removed.
4. Draw all possible alkenes formed by elimination at C2 and C4. Take into account that the methyl substituent at C3 can influence the formation of different alkene isomers due to possible branching and position of the double bond.
5. Count the number of unique constitutional alkene isomers formed. Each distinct position or branching pattern of the double bond counts as a different alkene. This total gives the number of major E2 elimination products.