My textbook introduced $\ce{PBr3}$ by stating that it leads to an inversion in stereochemistry since it always goes through an SN2 pathway, unlike using $\ce{HBr}$.

My question is that since $\ce{Br-}$ is a reasonably good nucleophile, so I assume it would most likely proceed through SN2 in a substrate with a secondary carbon anyway. Both pathways seem to me to be extremely similar with the exact same attacking species eventually ($\ce{Br-}$) and an excellent modified leaving group attached ($\ce{H3O+}$ and $\ce{OHPBr2}$).

What exactly is the difference and the advantage (other than being non-acidic) that make halogenation by $\ce{PBr3}$ a better method?

  • $\begingroup$ HBr can give rise to great amounts of alkene through elimination due to the SN1 pathway with secondary alsohols. $\endgroup$
    – Waylander
    Nov 4, 2020 at 7:19
  • $\begingroup$ How would this be avoided when using PBr3? $\endgroup$
    – Tsz
    Nov 4, 2020 at 14:56
  • 1
    $\begingroup$ There is no SN1 pathway with PBr3 $\endgroup$
    – Waylander
    Nov 4, 2020 at 19:31

1 Answer 1


The problem with $\ce{HBr}$ is that it can almost always cause an SN1 reaction, even if the alcohol is primary.

Consider the reaction of propanol with $\ce{HBr}$. Even though the resulting leaving group would make a primary carbocation if it left on its own, this can be avoided via a concerted rearrangement, thus making it possible for an SN1 reaction to happen. This doesn't happen with $\ce{PBr3}$ of its difference in mechanism.

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  • $\begingroup$ If, as your mechanism shows, the reaction is concerted, then there is no primary carbocation formed. Is this true? $\endgroup$
    – user55119
    Nov 28, 2022 at 19:26
  • $\begingroup$ Yes, which is what makes the SN1 reaction viable even for primary carbocations. I'm not exactly sure why (since the LG is good in the PBr3 reaction), but Solomons & Fryhle 10th edition mentions "The reaction of an alcohol with PBr3 does not involve formation of a carbocation and usually occurs without rearrangement of the carbon skeleton" in chapter 11.9, which means that it wouldn't have issues of potential concerted rearrangements. $\endgroup$
    – chemN00b
    Nov 28, 2022 at 19:37
  • $\begingroup$ Your diagram shows HBr, not $\ce{PBr3}$ . I'd be surprised if a primary alcohol didn't give a primary bromide with $\ce{PBr3}$, i.e., an $\ce{SN2}$ reaction. But with HBr, you show the rearrangement as a concerted hydride shift, indicating no intermediary primary carbocation $\endgroup$
    – user55119
    Nov 29, 2022 at 1:18
  • $\begingroup$ I believe I was unclear with my wording. I am trying to convey that the water group leaving on its own does not involve the concerted rearrangement (and would thus result in a primary carbocation), but because the concerted rearrangement happens, the formation of a primary carbocation is avoided entirely. Hopefully my edit better reflects this. $\endgroup$
    – chemN00b
    Nov 29, 2022 at 3:18

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