(Z)-2-Butene reacts with Br_2/H_2O. The resul | vedclass

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(D) $1$. The reaction of $(Z)-2$-butene with $Br_2/H_2O$ proceeds via the formation of a cyclic bromonium ion intermediate, followed by anti-attack of

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Equal amounts of $I$ and $II$

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(D) $1$. The reaction of $(Z)-2$-butene with $Br_2/H_2O$ proceeds via the formation of a cyclic bromonium ion intermediate, followed by anti-attack of water, resulting in a racemic mixture of $(2R, 3R)-3$-bromo$-2$-butanol and $(2S, 3S)-3$-bromo$-2$-butanol.$2$. When these bromohydrins are treated with a base like methoxide $(CH_3O^-)$, the hydroxyl group is deprotonated to form an alkoxide ion.$3$. This alkoxide ion then performs an intramolecular $S_N2$ attack on the carbon bearing the bromine atom, displacing the bromide ion.$4$. Since the $S_N2$ reaction involves inversion of configuration at the carbon atom undergoing the attack, the stereochemistry of the starting bromohydrins leads to the formation of $(cis)-2,3$-epoxybutane (also known as $cis-2,3$-dimethyloxirane).$5$. The structures $I$ and $II$ represent the two enantiomers of $cis-2,3$-epoxybutane, which are formed in equal amounts as a racemic mixture.

$(Z)-2$-Butene reacts with $Br_2/H_2O$. The resulting bromohydrin when treated with methoxide in methanol undergoes an intramolecular $S_N2$ reaction. Taking into consideration the stereochemical consequences of the reaction mechanism involved, choose the final product$(s)$ of these transformations.

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