(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.

List-$I$	List-$II$
$(A)$ Benzene	$1$. Phosgene
$(B)$ Ethylene	$2$. Silver mirror
$(C)$ Acetaldehyde	$3$. Mustard gas
$(D)$ Chloroform	$4$. $(4n + 2) \pi$ electrons
	$5$. Carbylamine

$(Z)-2$-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.

Similar Questions

Which of the following reactions is $NOT$ correctly represented?

Consider the above sequence of reactions. $151 \ g$ of $2-$bromopentane is made to react. The yield of the major product $P$ is $80 \%$ whereas the yield of $Q$ is $100 \%$. The mass of product $Q$ obtained is $........ \ g$. (Given molar mass in $g \ mol^{-1}$: $H: 1, C: 12, Br: 80$)

Match the following lists:
List-$I$ List-$II$
$(A)$ Benzene $1$. Phosgene
$(B)$ Ethylene $2$. Silver mirror
$(C)$ Acetaldehyde $3$. Mustard gas
$(D)$ Chloroform $4$. $(4n + 2) \pi$ electrons
$5$. Carbylamine

What will be the final product in the given reaction?

When $but-3-en-2-ol$ reacts with aqueous $HBr$,the product formed is:

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