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$$\ce{R-OH + NaBr + H2SO4 -> R-Br + NaHSO4 + H2O}$$

Due to sulphuric acid, the $\ce{-OH}$ group will get protonated and convert into $\ce{-H2O+}$ which is a great leaving group in comparison to the hydroxyl group. Therefore I assumed it would readily form a carbocation and then $\ce{Br-}$ could attack to form the alkyl bromide.

That is, it would follow the SN1 pathway (since the reaction is happening in 2 steps, first the formation of carbocation and then the attack of $\ce{Br-}$) but I also read somewhere that we might need to have a look on the substrate, if it is primary then it would follow SN2.

I am a bit confused after so many conditions and situations being introduced so I was hoping if someone could give me a brief and crisp explanation about this, and is there any particular factor that needs to be focused on before judging this mechanism?

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  • $\begingroup$ Both reaction happen but the extent of each reaction depends on what the R is and to some extent on the solvent. $\endgroup$ May 6 at 13:21
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    $\begingroup$ Alright, so is it safe to assume that it would majorly follow SN2 if it is a primary alcohol and SN1 for a tertiary alcohol? Also, what should be our approach for a secondary one? Do we simply conclude that SN2 and SN1 both happen in a considerable amount? I'm not sure about the solvent part because my book didn't mention anything else but the reaction that I mentioned. $\endgroup$ May 6 at 14:45
  • $\begingroup$ What you are asking is too broad and would be difficult to cover in one answer. Read this (and the linked pages) and see if it answers your question. $\endgroup$
    – S R Maiti
    May 6 at 16:07
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yeah so basically no reaction is purely SN1 or purely Sn2 its always a mixture of both and the overall order which we see is nothing but what that majority of reaction follows so now going by ur question, we do have to check about the substrate to inspect SN2 mechanism but in this case, since the reaction conditions are made in such a way that the overall reaction follows SN1 (as it is acid-catalyzed) hence we would compare the rate of reaction in terms of RDS (FORMATION OF CARBOCATION) and hence the more stable the carbonation is the faster the reaction proceeds.

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