The major product of the reaction is:

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(A) The reaction of $trans-but-2-ene$ with $Br_2$ in the presence of $H_2O$ is a halohydrin formation reaction. This reaction proceeds via the formati

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(A) The reaction of $trans-but-2-ene$ with $Br_2$ in the presence of $H_2O$ is a halohydrin formation reaction. This reaction proceeds via the formation of a cyclic bromonium ion intermediate. Water $(H_2O)$ then acts as a nucleophile and attacks the more substituted carbon atom from the side opposite to the bromonium ion (anti-addition). Since the starting material is $trans-but-2-ene$,the anti-addition of $Br$ and $OH$ results in a racemic mixture of enantiomers. The correct product is a pair of enantiomers where $Br$ and $OH$ are added in an anti-fashion to the double bond.

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