Reduction of quaternary ammonium salt

In summary, the reduction of a quaternary ammonium salt is a complex process and can result in different products depending on the conditions used. Reduction with Zn-Hg in HCl will not disturb the -NMe3 group, while reduction with LiAlH4 or NaBH4 will. The final product and reaction pathway will also vary depending on the reagents used. Additionally, using Wolff-Kishner or Clemmensen conditions on a compound containing a quaternary ammonium group may result in side reactions. Overall, the reduction of quaternary ammonium salts is not a straightforward process and requires careful consideration of the reaction conditions and potential side reactions.
  • #1
AdityaDev
527
33
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?
 
  • #3
AdityaDev said:
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?
 
  • #5
Are you talking about an "acid-base" neutralization reaction? There is no N+ cation in aqueous solution.
 
  • #5


The reduction of quaternary ammonium salts can be a bit confusing because it depends on the specific reducing agent used. In the case of Zn-Hg in HCl, this reducing agent is not strong enough to break the carbon-nitrogen bond of the -NMe3 group. This is because the Zn-Hg in HCl only has a moderate reducing power and is not able to reduce the -NMe3 group. Therefore, the -NMe3 group remains unchanged in the final product.

On the other hand, when using stronger reducing agents like LiAlH4 or NaBH4, the carbon-nitrogen bond of the -NMe3 group can be broken, resulting in the formation of the corresponding amine. This is because these reducing agents have a higher reducing power and are able to reduce the -NMe3 group.

The exact product of the reduction reaction will depend on the specific quaternary ammonium salt used and the reaction conditions. In general, the reaction will proceed by the reduction of the quaternary ammonium salt to form a tertiary amine, followed by further reduction to a secondary amine, and finally to a primary amine. However, other products may also be formed depending on the specific reagents and reaction conditions.

In the case of the Wolf-Kishner reduction, the -NMe3 group will be disturbed and the final product will be the corresponding primary amine. This is because the reaction conditions involve high temperatures and strong reducing agents that are able to break the carbon-nitrogen bond of the -NMe3 group.

In summary, the reduction of quaternary ammonium salts can be complex and the specific product will depend on the reducing agent and reaction conditions used. It is important to carefully consider the properties of the reducing agent and the structure of the quaternary ammonium salt to predict the outcome of the reaction.
 
  • #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/
 

1. What is a quaternary ammonium salt?

A quaternary ammonium salt is a compound that contains a positively charged nitrogen atom surrounded by four organic groups. It is commonly used as a disinfectant or surfactant due to its ability to disrupt cell membranes and kill bacteria.

2. How is quaternary ammonium salt reduced?

Quaternary ammonium salt can be reduced by using a reducing agent, such as sodium borohydride or lithium aluminum hydride. This process involves the transfer of electrons from the reducing agent to the positively charged nitrogen, resulting in a reduction of the molecule.

3. Why is the reduction of quaternary ammonium salt important?

The reduction of quaternary ammonium salt is important because it can lead to the formation of a new compound with different properties. This can be useful in the development of new drugs, pesticides, or other chemicals with specific functions.

4. What are the potential hazards of working with quaternary ammonium salt?

Quaternary ammonium salt can be toxic if ingested or inhaled, and can cause skin and eye irritation. It is important to handle this compound with caution and use appropriate personal protective equipment.

5. How is the reduction of quaternary ammonium salt monitored?

The reduction of quaternary ammonium salt can be monitored using various analytical techniques such as nuclear magnetic resonance (NMR) spectroscopy, mass spectrometry, or high performance liquid chromatography (HPLC). These techniques can help determine the success and purity of the reduction reaction.

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