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My teacher told that the mechanism for reaction of alcohol with $\ce{NaBr + H2SO4}$ follows SN1 when alcohol is 2°/3° and SN2 when it is primary or methanol. He told that the reactivity of alcohol also follows the order 3° > 2° > 1° which is that of SN1.

But I couldn't understand why is it the case. The nucleophile is $\ce{Br-},$ which is anionic nucleophile—so it should always follow SN2.

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    $\begingroup$ @Aniruddha Deb , it's not just NaBr , it's NaBr + H2SO4 , which would protonate the -OH group into a good leaving group. $\endgroup$ – Abhinav Sep 20 at 15:09
  • $\begingroup$ Primary alcohols will form primary carbocation which is not very stable , also the nucleophile here is a strong anionic nucleophile which would favour SN2 . $\endgroup$ – Abhinav Sep 20 at 15:12
  • $\begingroup$ @AniruddhaDeb Read carefully the post you linked. OP's teacher is right. $\endgroup$ – Mithoron Sep 20 at 17:00
  • $\begingroup$ OH is a poor leaving group. After protonation, it gets converted to one of the best leaving groups which leave the substrate immediately. This results in a carbocation $\endgroup$ – Soumyadwip Chanda Sep 20 at 17:06
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    $\begingroup$ Can people please stop proposing primary and methyl cations in their mechanisms? $\endgroup$ – Zhe Sep 20 at 17:11

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