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Topic: Zn/HCl Reductive Coupling Mechanism  (Read 9042 times)

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o alquimista

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Zn/HCl Reductive Coupling Mechanism
« on: March 26, 2006, 07:26:17 AM »
R.G. Amiet, The Preparation of Lucigenin, Journal of Chemical Education, Volume 59 no. 2, February 1982

"To a refluxing solution of N-methylacridone (1g) in ethanolic HCl (50 ml ethanol, 10 ml conc. HCl) is added zinc dust (3.2 g) portionwise over 15 min. The mixture is refluxed for a further 30 min, then added to water (100 ml) to give a green precipate of the crude bis-acridinium compound."

Does anyone know the mechanism of this reductive coupling reaction? Attached is my guess.




Offline Albert

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Re:Zn/HCl Reductive Coupling Mechanism
« Reply #1 on: March 26, 2006, 07:31:33 AM »
Do you mean Clemmensen reduction?

o alquimista

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Re:Zn/HCl Reductive Coupling Mechanism
« Reply #2 on: March 26, 2006, 09:11:13 AM »
Due to the reagents I already figured out that it is a Clemmensen reduction. I was wondering how the coupling of the two aromatic ketons takes place (mechanism). As in the above N-methylacridone --> bis-acridinium example.

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Re:Zn/HCl Reductive Coupling Mechanism
« Reply #3 on: March 26, 2006, 07:26:50 PM »
For a homoogeneous mechanism (the one Albert posted is a heterogeneous mechanism) of the Clemmenson reduction you form a benzyllic radical (in this case doubley benzyllic) so you could have a radical-radical coupling to make the first carbon-carbon bond.  Alternatively, you could draw a mechanism with an doubley benzyllic carbanion and an extended iminium ion in order to make the carbon-carbon bond.  I think I favor the radical coupling, myself.

o alquimista

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Re:Zn/HCl Reductive Coupling Mechanism
« Reply #4 on: March 27, 2006, 08:02:32 AM »
In this example the product has one extra double bond, namely the coupling bond. If it's a reaction between two radicals, there is one double bond too little.

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Re:Zn/HCl Reductive Coupling Mechanism
« Reply #5 on: March 27, 2006, 02:06:37 PM »
You would couple two ketyl type radicals and then eliminate water.

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