For the reaction of trans-but-2-ene with IBr, | vedclass

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Question

(A) The reaction of $trans-but-2-ene$ with $IBr$ involves the electrophilic addition of $Br^+$ to the alkene,forming a cyclic bromonium ion intermedia

Vedclass · Accepted Answer

Two products are formed in this reaction.

Vedclass · Answer

(A) The reaction of $trans-but-2-ene$ with $IBr$ involves the electrophilic addition of $Br^+$ to the alkene,forming a cyclic bromonium ion intermediate.Since $IBr$ is a polar interhalogen compound,$Br$ is more electronegative than $I$,making $Br$ the electrophile $(Br^{\delta+})$ and $I$ the nucleophile $(I^{\delta-})$.The attack of the nucleophile $(I^-)$ on the cyclic bromonium ion occurs from the side opposite to the bromonium bridge (anti-addition).For $trans-but-2-ene$,the anti-addition of $IBr$ results in the formation of a pair of enantiomers.Statement $A$ is incorrect because the reaction produces a racemic mixture of enantiomers,which are two distinct stereoisomers,but the question asks for the wrong statement regarding the mechanism or product nature. Actually,looking at the options,$A$ is correct (two products are formed),$B$ is correct (it is anti-addition),$C$ is correct (cyclic bromonium ion is formed),and $D$ is correct (the products are enantiomers). However,in the context of $IBr$ addition to an unsymmetrical alkene or specific stereochemistry,if the reaction were to produce meso compounds or diastereomers,that would be the error. Given the standard interpretation,all statements are technically true for this specific reaction. If forced to choose,$A$ might be considered 'wrong' if one considers the racemic mixture as a single product,but usually,it is two enantiomers. Let's re-evaluate: $trans-but-2-ene + IBr \rightarrow$ racemic mixture of $2-bromo-3-iodobutane$. This consists of two enantiomers. Thus,all statements are correct. Assuming a typo in the question,we identify the most likely intended answer.

For the reaction of $trans-but-2-ene$ with $IBr$,which statement is incorrect?

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