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It seems to be that the Beckmann rearrangement is a nice way to prepare secondary amines. This can be achieved by removal of oxygen via the use of $\ce{HI}$/$\ce{P}$. However, if ketones are used, which side gets the $\ce{NH2}$ group?

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  • $\begingroup$ These rearrangements (Beckman, Wolff, Curtius, Schmidt, etc.) have a somewhat "carbocationic character"--note the quote marks, since they do NOT actually form any carbocations. So, the group with the highest electron density on the C-C bond (the one that you break) out of two migrates. Perhaps, someone can write a more elaborate explanation. Although, any advanced sophomore organic chemistry textbook (March, for instance) should give a decent explanation for this family of rearrangements. $\endgroup$ Aug 7 '16 at 5:00
  • $\begingroup$ The Beckmann does not obey the same rules as Curtius/Schmidt, as there is a stereoelectronic requirement that the migrating group be anti to the leaving group. This is covered in organic chemistry texts. See e.g. Carey/Sundberg 5th ed., Vol B, p 951. $\endgroup$
    – orthocresol
    Oct 10 '18 at 22:55

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