In the given reaction,what is A?4-methylcyclo | vedclass

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(B) The reaction is an $S_N2$ mechanism.$S_N2$ reactions proceed with inversion of configuration at the chiral center where the leaving group is attac

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(B) The reaction is an $S_N2$ mechanism.$S_N2$ reactions proceed with inversion of configuration at the chiral center where the leaving group is attached.In the starting material,the $Br$ atom is on one side of the cyclohexane ring (let's assume it is 'up').During the $S_N2$ attack by $OH^-$,the nucleophile attacks from the side opposite to the leaving group $(Br^-)$.This results in the $OH$ group being on the opposite side of the ring relative to the original position of the $Br$ atom.Since the $CH_3$ group is at the $4$-position,the relative stereochemistry between the $CH_3$ group and the $OH$ group will be inverted compared to the $CH_3$ and $Br$ groups.Therefore,the product $A$ is the trans-isomer of $4$-methylcyclohexanol.

In the given reaction,what is $A$?
$4-methylcyclohexyl bromide + OH^- \xrightarrow{S_N2} A$

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In the given reaction,what is $A$?$4-methylcyclohexyl bromide + OH^- \xrightarrow{S_N2} A$