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Reduction of quaternary ammonium salt

  1. Jan 4, 2015 #1
    I am confused with the reduction of quaternary ammonium salt.
    ##Ph-CH_2-CH_2-N^+Me_3##
    on reduction with Zn-Hg in HCl will not work. The -NMe3 is not disturbed(given in my text). But why?
    Also the text says On reduction with LiAlH4 or NaBH4, -NMe3 would be disturbed. what would be the product? In which way will the reaction proceed?
    Wolf-kishner reduction will disturb the group. But what would the final product be?
     
  2. jcsd
  3. Jan 9, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Jan 10, 2015 #3

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    Are you certain you haven't confused reduction of the anion of a quaternary ammonium compound with the cation, and omitted the identity of the anion being reduced?
     
  5. Jan 10, 2015 #4
    Wolf kishner reduction means you have OH- from potassium hydroxide and It will react with the N+ cation. Zn-Hg in HCl gives H+ ions which have no effect on the cation. Both liAlH4 and NaBH4 give H- ions which will react with N+ ions. I think I have got it. Can you ask someone to verify this?
     
  6. Jan 10, 2015 #5

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    Are you talking about an "acid-base" neutralization reaction? There is no N+ cation in aqueous solution.
     
  7. Jan 11, 2015 #6
    I may be wrong, but I don't remember quaternary ammonium compounds as particularly good substrates for reductions.
    The Me3N+ group looks like a leaving group to me, and that Ph in 2 may allow elimination pathways.
    So, treating the compound in the OP with bases may give styrene; nucleophiles (Nu-) may give Ph-CH2-CH2-Nu (so perhaps hydride donors would give ethylbenzene? Not sure though).
    In any case, I don't see why one would try Wolff-Kishner or Clemmensen conditions (normally used to reduce phenyl ketones to alkylbenzenes) on a compound like that.
    Or maybe the 'text' is trying to show us that you'd have side reactions if you tried to reduce a carbonyl group on a molecule that also contains a quaternary ammonium group.
    Bizarre though. :O/
     
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