Account for the regioselectivity and stereoselectivity observed when 1-methylcyclopentene is treated with each reagent. (a) \(\mathrm{BH}_{3}\) (b) \(\mathrm{Br}_{2}\) in \(\mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{Hg}(\mathrm{OAc})_{2}\) in \(\mathrm{H}_{2} \mathrm{O}\)

When 1-methylcyclopentene is treated with borane (\(\mathrm{BH}_{3}\)), hydroboration-oxidation occurs. This reaction is known for its anti-Markovnikov selectivity, meaning that the hydrogen adds to the more substituted carbon of the alkene, while the boron atom adds to the less substituted carbon. This regioselectivity can be explained by the fact that boron is less electronegative than carbon, leading to the partial positive charge on the more substituted carbon. The reaction is also syn-selective meaning that the hydrogen and boron atoms add to the same face of the alkene. Once the hydroboration step is complete, the product is treated with hydrogen peroxide and a base to replace the boron with an OH group and obtain an alcohol as the final product.

When 1-methylcyclopentene reacts with bromine (\(\mathrm{Br}_{2}\)) in \(\mathrm{H}_{2}\mathrm{O}\), a halohydrin forms. The reaction proceeds through a bromonium ion intermediate, which is formed due to the electrophilic nature of bromine. Water comes in as a nucleophile and attacks the carbon atom with the greater positive charge, which is the more substituted carbon. This results in Markovnikov addition of OH and Br groups to the alkene. The stereochemistry of the reaction is anti, meaning that the OH and Br groups will add on opposite faces of the alkene, leading to a trans product.

In the presence of mercury(II) acetate, \(\mathrm{Hg}(\mathrm{OAc})_{2}\), 1-methylcyclopentene undergoes an oxymercuration-demercuration reaction. This reaction is also known for its Markovnikov selectivity, and it proceeds through a mercurinium ion intermediate, which is similar to the bromonium ion in the previous reaction. The nucleophile, in this case water, attacks the more substituted carbon resulting in the formation of an alcohol. After the addition of water, a reducing agent, such as sodium borohydride (\(\mathrm{NaBH}_{4}\)), is used to replace the mercury atom with hydrogen. The addition of the two groups is syn, with the OH and H groups adding to the same face of the alkene.