$\ce{R-OH + HBr -> R-Br + H2O}$

An $\ce{R+}$ carbocation is formed during this reaction.

If $\ce{R}$ is any alkyl group instead of methyl or ethyl, can the beta hydrogen to the hydroxy group be removed to form a double bond, similar to a dehydration??


It can happen.

It practice, you avoid it happening by keeping the temperature low. $\ce{Br-}$ is a better nucleophile than $\ce{H2O}$ is a base, so at lower temperatures, the substitution is faster.

If the elimination happens, alkenes react with $\ce{HBr}$ to form alkyl bromides. As long as acid is present, the carbocation can reform. For example, with isopropyl alcohol

Direct SN1 Substitution $$\ce{(CH3)2CH-OH + HBr -> (CH3)2CH-OH2+ + Br-}$$ $$\ce{(CH3)2CH-OH2+ -> (CH3)2CH+ + H2O}$$ $$\ce{(CH3)2CH+ + Br- -> (CH3)2CH-Br}$$

Substitution by elimination-addition $$\ce{(CH3)2CH-OH + HBr -> (CH3)2CH-OH2+ + Br-}$$ $$\ce{(CH3)2CH-OH2+ -> (CH3)2CH+ + H2O}$$ $$\ce{(CH3)2CH+ + H2O <=> CH3-CH=CH2 + H3O+}$$ $$\ce{(CH3)2CH+ + Br- -> (CH3)2CH-Br}$$

In some cases, two processes may appear to generate different alkyl halides, but this would be no different than if a carbocation rearrangement occurred.

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