# Mechanism of reaction of alcohol with NaBr + H2SO4

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.

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