Reduction of quaternary ammonium salt

[AdityaDev]

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?

 

[Bystander]

reduction of quaternary ammonium salt.

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?

 

[Bystander]

Are you talking about an "acid-base" neutralization reaction? There is no N⁺ cation in aqueous solution.

 

[lavoisier]

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/