The product in the reaction (figure) + HBr is | vedclass

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(C) The reaction involves the acid-catalyzed ring opening of an epoxide. In the presence of $H^+$,the oxygen atom of the epoxide is protonated,making

Vedclass · Accepted Answer

$1-$bromo$-1-$methylcyclohexan$-2-$ol

Vedclass · Answer

(C) The reaction involves the acid-catalyzed ring opening of an epoxide. In the presence of $H^+$,the oxygen atom of the epoxide is protonated,making it a better leaving group. $HBr$ provides the nucleophile $Br^-$. For an unsymmetrical epoxide,the nucleophile attacks the more substituted carbon atom because it can better stabilize the partial positive charge in the transition state. In this case,the tertiary carbon (the one with the $CH_3$ group) is more substituted than the secondary carbon. Therefore,$Br^-$ attacks the tertiary carbon,and the $C-O$ bond breaks to form a hydroxyl group at the secondary carbon. The final product is $1-bromo-1-methylcyclohexan-2-ol$.

The product in the reaction (figure) $+ HBr$ is

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