Which of the following alkenes is the major (Zaitsev) product formed in the E2 dehydrohalogenation of $C_3{H}_7\mathrm{Br}$ (2-bromopropane) with a strong base?

• A. $C_4{H}_8$ (butene)
• B. $C_2{H}_4$ (ethylene)
• C. $C_3{H}_6$ (propene)
• D. $C_3{H}_8$ (propane)

Answer: (C)

Verified step by step guidance

1. Identify the substrate and the reaction type: The substrate is 2-bromopropane (C\_3H\_7Br), and the reaction is an E2 dehydrohalogenation with a strong base, which typically removes a proton adjacent to the carbon bearing the leaving group (bromine) to form an alkene.
2. Determine the possible β-hydrogens: In 2-bromopropane, the bromine is attached to the second carbon. The β-carbons are the first and third carbons, each having hydrogens that can be abstracted by the base to form a double bond.
3. Apply Zaitsev's rule: The major product in an elimination reaction is usually the more substituted alkene, which is more stable. This means the double bond will form where it results in the greatest number of alkyl substituents attached to the alkene carbons.
4. Draw the possible alkenes formed by removing a β-hydrogen: Removing a hydrogen from either the first or third carbon will form propene (C\_3H\_6), which is the only possible alkene product here since the molecule has only three carbons.
5. Conclude that the major product is propene (C\_3H\_6), as it is the more substituted and stable alkene formed by elimination from 2-bromopropane following Zaitsev's rule.