The haloalkane, 2 -iodo-2-methylpentane, can be prepared by treating an alkene (A) with hydroiodic acid. (A) can be (a) 2 -methylpent-2-ene (b) 2 -methylpent-1-ene (c) 2 -methyl-but-2-ene (d) both (a) and (b)

When a hydrogen halide reacts with an alkene, regioselectivity plays a crucial role in determining the major product. According to Markovnikov's rule, when an unsymmetrical alkene reacts with a hydrogen halide, the hydrogen atom gets added to the carbon atom that has more hydrogen atoms, while the halogen atom gets added to the carbon atom with fewer hydrogen atoms. This rule helps predict the major product formed during the reaction.

Apply Markovnikov's rule to each given alkene to determine which one would produce 2-iodo-2-methylpentane when treated with hydroiodic acid (HI). (a) 2-methylpent-2-ene: \[ CH_{3} - \textbf{C} = \textbf{C} - CH(CH_{3}) - CH_{2} - CH_{3} \] Upon reacting with HI, Iodine will attach to the first carbon and hydrogen will attach to the second carbon, resulting in 2-iodo-2-methylpentane. (b) 2-methylpent-1-ene: \[ CH_{2} = \textbf{C} - CH(CH_{3}) - CH_{2} - CH_{2} - CH_{3} \] Upon reacting with HI, Iodine will attach to the second carbon and hydrogen will attach to the first carbon, also resulting in 2-iodo-2-methylpentane. (c) 2-methyl-but-2-ene: \[ CH_{3} - \textbf{C} = \textbf{C} - CH(CH_{3}) \] Upon reacting with HI, Iodine will attach to the first carbon and hydrogen will attach to the second carbon, producing an incorrect product (2-iodo-2-methylbutane).

Since both (a) 2-methylpent-2-ene and (b) 2-methylpent-1-ene produce the desired product (2-iodo-2-methylpentane) upon reacting with hydroiodic acid, the correct answer is (d) both (a) and (b).