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Topic: synthesis 1,3,5, tri-t-butylbenzene  (Read 7333 times)

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synthesis 1,3,5, tri-t-butylbenzene
« on: February 02, 2005, 07:43:55 PM »
I have been struggling with this all week.   :'(
what is the machanism to synthesize 1,3,5, tri-t-butylbenzene from 1,4, di-t-butylbenzene and t-butyl chloride and aluminumchloride (Freidel-crafts rxn)?  I've tried rearrangement to produce a tertiary carbacation and that doesn't seem to work.  Is it simply that from the 1,4, di isomer t-butylbenzene is formed and then the 1,3,5 isomer is synthesized after 2 F-C substitutions?

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Re:synthesis 1,3,5, tri-t-butylbenzene
« Reply #1 on: February 02, 2005, 08:33:32 PM »
I'm not sure if I recall correctly, but I think that the 1,4-di-t-butylbenzene alkylates to form the Meisenheimer complex, as you would expect for a typical Freidel-Crafts alkylation, but the adjacent t-butyl group is too large for direct elimination of the hydrogen to re-aromatize, so there is a 1,2-hydride shift to give the tetrahedral carbon at the 4 position of the ring.  The t-butyl group on that carbon can then fall off to regenerate a tertiary carbocation and leave 1,3-di-t-butylbenzene.  Eventually, that t-butyl carbocation will find the 5 position of the ring and do a normal Friedel-Crafts to get to your product.

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Re:synthesis 1,3,5, tri-t-butylbenzene
« Reply #2 on: February 03, 2005, 11:32:31 AM »
Thanks :)

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