Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Rutherford on January 21, 2013, 12:30:56 PM
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How do anhydrides activate AlCl3? I see this first time and I need to finish this scheme.
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This is just a normal Friedel-Crafts acylation.
Look on this page for the Haworth reaction
http://en.wikipedia.org/wiki/Friedel–Crafts_reaction (http://en.wikipedia.org/wiki/Friedel–Crafts_reaction)
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From looking at the scheme I tried to work out the mechanism, is it okay?
In the mechanism I wrote that the second C atom got the acyl group attached to it, but it could be any of the C atoms. How should I know which one it is?
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I don't really like the acylcarbocation, but I suppose it may be stabilised by the Al.
Alkyl groups on the benzene ring are ortho/para directors, so I guess you will get some para substitution product.
Don't forget that when the pi electrons of the aromatic ring attack the acylcation you generate a cation in the benzene ring which is resonance stabilised.
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"Don't forget that when the pi electrons of the aromatic ring attack the acylcation you generate a cation in the benzene ring which is resonance stabilised."
Knew that but didn't know how to draw it. Thanks very much for the help. I got one last question. In the Haworth reaction on the wikkipedia link you posted, the product is reduced and then H+ is added so the R-COOH group cyclizated to form a cyclic ketone. How is this possible?
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Similar mechanism protonate on of the acid and cyclisation via the pi electrons as usual.