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

My question is that since 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 (Br-) and an excellent modified leaving group attached (H2O+ and OHPBr2).

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

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  • $\begingroup$ HBr can give rise to great amounts of alkene through elimination due to the SN1 pathway with secondary alsohols. $\endgroup$ – Waylander Nov 4 '20 at 7:19
  • $\begingroup$ How would this be avoided when using PBr3? $\endgroup$ – Tsz Nov 4 '20 at 14:56
  • $\begingroup$ There is no SN1 pathway with PBr3 $\endgroup$ – Waylander Nov 4 '20 at 19:31

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