Identify product A in the following reaction: | vedclass

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(B) The reaction proceeds via an $S_N2$ mechanism. In an $S_N2$ reaction,the nucleophile $(OH^-)$ attacks the electrophilic carbon from the side oppos

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$4-$methylcyclohexanol (trans-isomer)

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(B) The reaction proceeds via an $S_N2$ mechanism. In an $S_N2$ reaction,the nucleophile $(OH^-)$ attacks the electrophilic carbon from the side opposite to the leaving group $(Br^-)$,resulting in an inversion of configuration (Walden inversion). Starting with the given $4-methylcyclohexyl$ bromide,the $Br$ atom is on one side of the ring. The $S_N2$ attack by $OH^-$ will occur from the opposite side,leading to the formation of the trans-isomer of $4-methylcyclohexanol$.

Identify product $A$ in the following reaction:
$4-methylcyclohexyl bromide + OH^- \xrightarrow{S_N2} A$

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Identify product $A$ in the following reaction: $4-methylcyclohexyl bromide + OH^- \xrightarrow{S_N2} A$