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Topic: friedel-crafts acylation of benzene  (Read 16733 times)

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Offline yanks6621

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friedel-crafts acylation of benzene
« on: January 29, 2007, 08:53:13 PM »
Give the structure of the product obtained when benzene is acylated with succinic anhydride in the presence of AlCl3

Does the lone pair on the O not doubled bonded attack the Al in AlCl3? Can this result in an attack by the benzene double bond on the O?

 ???

Offline lavoisier

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Re: friedel-crafts acylation of benzene
« Reply #1 on: January 30, 2007, 02:15:57 PM »
No, benzene will attack the carbon as usual.

However, your observation is important: actually when you use anhydrides in FC acylations, you need a stoichiometric amount of AlCl3 (as opposed to a catalytic amount in FC alkylations with alkyl halides) because the metal coordinates the carbonyl groups of the anhydride.

Offline wainblatrobert

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Re: friedel-crafts acylation of benzene
« Reply #2 on: March 02, 2007, 04:16:07 PM »
The same product is obtained through acylation with an ahydride as with a acid chloride...you get acetophenone or methyl-phenyl-ketone in your case:

benzene-C(=O)-CH3 and acetic acid as a byproduct...the AlCl3 is not used up in the reaction

Offline eris

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Re: friedel-crafts acylation of benzene
« Reply #3 on: March 02, 2007, 05:24:57 PM »
The key factor here is that the succinic anhydride is cyclic, and when the benzene attacks one carbonyl, the leaving group that completes the reaction is the other carboxylate, which will open the ring.  wainblatrobert, your answer would apply for acetic anhydride, but not for succinic (a different structure).

Offline wilson

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Re: friedel-crafts acylation of benzene
« Reply #4 on: March 04, 2007, 04:14:08 AM »
Is this mechanism slightly different since succinic anhydride does not have a halogen?

I can't figure out the exact way this reaction goes. Nevertheless I came out with a product. Is it correct?

Offline desperad0oo7

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Re: friedel-crafts acylation of benzene
« Reply #5 on: March 05, 2007, 09:40:28 PM »
Basically the lone pair on the --o-- of the anhydride will attack the lewis acid AlCl3 creating a pisitive charge on the oxygen.

Then one of the c-o bonds on the Oxygen will break heterolytically and go to the oxygen leaving you with an acyl group and AlCl3--Junk.
The acyl group will then attack the benzene as usual.

Hope this is helpful.

Offline vietknight

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Re: friedel-crafts acylation of benzene
« Reply #6 on: April 15, 2007, 01:20:12 AM »
Can someone draw the reaction mechanism for this reaction?

I don't understand how AlCl3 contributes to the breaking of the ring...what is the purpose of AlCl3 in this reaction...The response made by lavoisier is not very clear. Without AlCl3 what would happen...there is no Halide in the reaction to do alkylations.

What does it mean when the metal "coordinates" the carbonyl groups to the anhydride?

I've drawn my possible answer to the product but I'm not sure if its correct..by just ignoring that AlCl3 is even there...
« Last Edit: April 15, 2007, 01:31:52 AM by vietknight »

Offline Carcul

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Re: friedel-crafts acylation of benzene
« Reply #7 on: April 15, 2007, 07:02:02 AM »
That's right: the structure posted by Vietknight is the one obtained after treatment of the structure posted by Wilson with water.

Now an exercise: what's the product expected by treatment of Vietknight's struture with some acid as catalyst?

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