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For example, let us say we have 2.2-dimethycyclohexanol reacting with $\ce{HBr}$. We know that the oxygen gets protonated and then leaves as water, but the question does it leave on its own or does a backside attack happen? If its $\mathrm{S_N1}$ we have 1-bromo-1,2-dimethylcyclohexane as the major product. If its $\mathrm{S_N2}$, we have 1-bromo-2,2-dimethylcyclohexane as the major product. This is assuming we are at 0℃.

I believe that its $\mathrm{S_N1}$ because bromide is a weak base.

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Actually, secondary alcohols undergoes $\mathrm{S_N1}$ or $\mathrm{S_N2}$ depending upon the substrate you are using. For e.g. In your case(as provided above) the substrate has steric hindrance nearby alcoholic group and the carbocation will be stable due to the migratory aptitude of the methyl group. Hence, here should be $\mathrm{S_N1}$ and you have to see the structure of substrate to decide $\mathrm{S_N1}$ or $\mathrm{S_N2}$ for secondary alcohols. Also, alcohols are not good leaving groups also! But after protonation the water is a very good leaving group.

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  • $\begingroup$ Yes, here it'll be $\mathrm{S_N1}$ preferably. Why not $\mathrm{S_N2}$?, one cause is steric crowding in the adjacent carbon and for the attacking nucleophile and leaving group nature comparison see here, chemistry.stackexchange.com/questions/75356/… $\endgroup$ – chail10 Jun 8 '17 at 9:54

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