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In polar protic solvents, $\ce{Br-}$ and $\ce{I-}$ are considered to be very good nucleophiles. Now, here the solvent is alcohol which is polar protic. So, why don't secondary alcohols react with $\ce{HBr/HI}$ via $\mathrm{S_N}$2 and go by $\mathrm{S_N}$1 instead?

Notably, primary alcohols do undergo $\mathrm{S_N}$2, and secondary carbons can definitely undergo $\mathrm{S_N}$2 as well, so something must be special about water as a leaving group that makes us say that secondary alcohols primarily undergo $\mathrm{S_N}$1.

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  • $\begingroup$ The hydroxyl group is not a good leaving group. Thus, the reaction is not simple substitution reaction. It is acid catalyzed and go through a cabocation intermediate based on the alcohol syructure (e.g., a primary alcohol does not go through carbocation intermediate). $\endgroup$
    – Mathew Mahindaratne
    May 28, 2022 at 17:21
  • $\begingroup$ @MathewMahindaratne Yes, but both SN1 and SN2 can occur after protonation of alcohol( as we know because SN2 seen in primary and SN1 in tertiary)...now, in secondary alcohol SN2 should occur because Br- and I- are strong nucleophiles? $\endgroup$
    – user121185
    May 28, 2022 at 17:32
  • $\begingroup$ You said it is in alcohol medium so alcohol is also a nucleophile. Weak but has plenty. So product is a mixture. Nonrtheless, the secondery alcohol reaction is whether SN1 or SN2 depends also on solvent. If it is polar protic SN1 is predominate due to stabilization of carbocation intermediate. If it is polar aprotic, SN2 predominate. $\endgroup$
    – Mathew Mahindaratne
    May 28, 2022 at 17:44
  • $\begingroup$ @MathewMahindaratne But aren't Br- and I- strong nucleophiles in polar protic? So does that mean all nucleophiles(no matter how strong) will show SN1 in polar protic? $\endgroup$
    – user121185
    May 28, 2022 at 17:49

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