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(B) The reaction proceeds via an $S_N1$ mechanism because the substrate is a secondary alkyl halide and the solvent is polar protic $(H_2O)$.$1$. The

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$1-$propylcyclohexanol

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(B) The reaction proceeds via an $S_N1$ mechanism because the substrate is a secondary alkyl halide and the solvent is polar protic $(H_2O)$.$1$. The first step is the formation of a secondary carbocation by the loss of the $Cl^-$ ion.$2$. This secondary carbocation undergoes a $1,2-H^-$ shift to form a more stable tertiary carbocation.$3$. The nucleophile ($OH^-$ from $NaOH$) then attacks the tertiary carbocation to form the major product,$1-propylcyclohexanol$.

Consider the above chemical reaction. Product $A$ is:

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