The following reaction has been troubling me for a while now.

enter image description here

Since peroxides won't give the anti-Markovnikov product when used with HCl, I cannot think of any reagents that would result in the product shown in the image above. There doesn't seem to be any Wagner–Meerwein rearrangement either. This left me wondering, what reagent(s) should be used to form the product shown above.


2 Answers 2


One way to do it would be do first do a hydroboration-oxidation reaction, which leads to the anti-Markovnikov addition of $\ce{H2O}$. So you will get an $\ce{-OH}$ group instead of $\ce{-Cl}$. Then, we can replace the $\ce{-OH}$ group with $\ce{-Cl}$ by treating it with $\ce{PCl3}$ or $\ce{PCl5}$.

  • 2
    $\begingroup$ MsCl/Et3N will also accomplish this and is easier to handle than PCl3 $\endgroup$
    – Waylander
    Commented Apr 15, 2020 at 13:44

1.Using 1%alkaline KMnO4 (Bayers reagent) 2.By pinnacole pincolone rearrangement we get aldehyde at terminal position 3.using suitable reducing agent reduce( Aldehyde)it to OH 4.Use PCl3 or PCl5 to replace OH by Cl


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