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(A) The reaction of $Methylcyclohexane$ with $Br_2$ in the presence of light $(h
u)$ is a free radical substitution reaction.The stability of the int

Vedclass · Accepted Answer

$1-$Bromo$-1-$methylcyclohexane

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(A) The reaction of $Methylcyclohexane$ with $Br_2$ in the presence of light $(h
u)$ is a free radical substitution reaction.The stability of the intermediate free radical determines the major product.In $Methylcyclohexane$,the tertiary $(3^\circ)$ carbon at the $1$-position is the most substituted carbon.Abstraction of the hydrogen atom from this $3^\circ$ carbon generates a $3^\circ$ free radical,which is the most stable radical among the possible options.Therefore,the bromine atom preferentially attacks this position to form $1-Bromo-1-methylcyclohexane$ as the major product.

Predict the major product for the following reaction:
$Methylcyclohexane \xrightarrow{Br_2, h\nu} \text{Major Product}$

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Predict the major product for the following reaction:$Methylcyclohexane \xrightarrow{Br_2, h\nu} \text{Major Product}$