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.

  • 1
    $\begingroup$ @Aniruddha Deb , it's not just NaBr , it's NaBr + H2SO4 , which would protonate the -OH group into a good leaving group. $\endgroup$
    – Abhinav
    Commented Sep 20, 2020 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
    Commented Sep 20, 2020 at 15:12
  • $\begingroup$ @AniruddhaDeb Read carefully the post you linked. OP's teacher is right. $\endgroup$
    – Mithoron
    Commented Sep 20, 2020 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 $\endgroup$ Commented Sep 20, 2020 at 17:06
  • 3
    $\begingroup$ Can people please stop proposing primary and methyl cations in their mechanisms? $\endgroup$
    – Zhe
    Commented Sep 20, 2020 at 17:11

1 Answer 1


A highly important fact is that any sodium halide + acid produces hydrogen halide (gaseous) which dissolves in water to form hydrohalic acid.

$$\ce{2Na\color{green}{X}_{(s)} + H2SO4_{(aq)} -> Na2SO4_{(aq)} + 2H\color{green}{X}_{(aq)}}$$

This hydrohalic acid will lead to $\mathrm{S_N1}$ substitution with carbocation formation (important to note as carbocations carry a possibility for ionic migration to attain stability) and gives the final product.

enter image description here

An exception, though, is $\ce{NaI}$, for which $\ce{H2SO4}$ behaves as a good oxidising agent and oxidises the $\ce{HI}$ produced to $\ce{I2}$. In order to prevent this, we use $\ce{H3PO4}$ for the reaction.

  • $\begingroup$ Well, I'm tempted to put another downvote. OP had some stupid misconception about nucleophile, which should have gotten it closed. You didn't touch this bit, but messed up depiction of how salts and acids behave. It really has noting to do with what happens if you react solid salt with concentrated acid. $\endgroup$
    – Mithoron
    Commented May 13 at 16:39
  • $\begingroup$ @Mithoron is it "The nucleophile is Br−, which is anionic nucleophile—so it should always follow SN2." $\endgroup$ Commented May 14 at 3:35
  • $\begingroup$ It looks like this. I'm mostly concerned that you make some misleading impressions, though. "any sodium halide + acid"? Not any halide - not iodide, or halides of plenty of metals or non-metals. And only with some concentrated acids. $\endgroup$
    – Mithoron
    Commented May 14 at 12:15
  • $\begingroup$ That general statement did not work - but for NaX (X=F, Cl, Br) and KX (the highly ionic solids) it works right? $\endgroup$ Commented May 15 at 0:53

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