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(D) The reaction of an alkene with $Br_2$ in the presence of a nucleophilic solvent like $CH_3OH$ proceeds via the formation of a cyclic bromonium ion

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(D) The reaction of an alkene with $Br_2$ in the presence of a nucleophilic solvent like $CH_3OH$ proceeds via the formation of a cyclic bromonium ion intermediate. $1$. The double bond attacks $Br_2$ to form a cyclic bromonium ion. $2$. The nucleophile $(CH_3OH)$ then attacks the more substituted carbon atom of the bromonium ion from the side opposite to the bromine atom (anti-addition). $3$. Based on the stereochemistry of the starting material,the $CH_3O$ group will be added to the more substituted carbon,and the $Br$ atom will be on the adjacent carbon,resulting in the anti-addition product as shown in the solution image.

What is a likely product of the reaction shown?

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