chloechem - 22-5-2017 at 20:32
Hi, I'm using the Hunsdiecker reaction to convert [2-(Aminomethyl)phenyl]acetic acid (red) to 1-[2-(Bromomethyl)phenyl]methanamine (green). How do I
stop, or at least reduce the formation of 1-[4,6-Dibromo-2-(bromomethyl)phenyl]methanamine (blue) (or more than one bromine), and then convert the
bromine to a methyl group (1-(2-Ethylphenyl)methanamine (black)). Thank you.
[Edited on 23-5-2017 by chloechem]
[Edited on 23-5-2017 by chloechem]
clearly_not_atara - 22-5-2017 at 23:18
This is not a wise choice of intermediate. The product will likely convert to 2H-indoline, and the formation of this side-product cannot be prevented
under the conditions of the Hunsdiecker reaction.
However, under normal Hunsdiecker conditions, ring-bromination will not be significant, and can be ignored.
But instead consider e.g. reducing the acid to an alcohol with LiAlH4
CuReUS - 23-5-2017 at 07:51
Why don't you directly reduce the COOH to CH3 ? http://www.sciencemadness.org/talk/viewthread.php?tid=73024#...