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Topic: Friedel Craft reaction question  (Read 2823 times)

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Offline pigeonsATW

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Friedel Craft reaction question
« on: February 17, 2013, 02:05:39 PM »


I'm not sure which ring the CH3CO would attach to. Also, how would I know whetherCOC1=CC=CC=C1 or[O-][N+](=O)C1=CC=CC=C1 has a stronger influence on the regiochemistry of the reaction?
Thanks

Mod edit: IMG code. Dan
« Last Edit: February 17, 2013, 05:26:09 PM by Dan »

Offline orgopete

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Re: Friedel Craft reaction question
« Reply #1 on: February 17, 2013, 03:10:43 PM »
A Friedel-Crafts is an electrophilic reaction. Which ring do you think would be a better electron donor?
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Offline pigeonsATW

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Re: Friedel Craft reaction question
« Reply #2 on: February 17, 2013, 04:31:33 PM »
A Friedel-Crafts is an electrophilic reaction. Which ring do you think would be a better electron donor?
OK I might be completely off here, but this is what I think... The resonance structures for anisole gives a negative charge on the ortho and para Carbons, making these atoms electron donors.
 C\[O+]=C1/[CH-]C=CC=C1 :resonance:C[O+]=C1C=C[CH-]C=C1 :resonance:C\[O+]=C1\[CH-]C=CC=C1
Similarly resonance structures for nitrobenzene give a positive charge on the ortho and para Carbons. (poor electron donors).
[O-][N+]([O-])=C1[CH+]C=CC=C1 :resonance:[O-][N+]([O-])=C1C=C[CH+]C=C1 :resonance:[O-][N+]([O-])=C1[CH+]C=CC=C1
And as nitrobenzene is attached to the meta position of anisole it has no electrons that it can donate to nitrobenzene.
This eliminates A, B and D.
Of C and E, I'm assuming steric hindrance plays a part in making E more favorable.
Please confirm whether my long winded explanation makes sense / is correct. ???
Thanks!

Offline Dan

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Re: Friedel Craft reaction question
« Reply #3 on: February 17, 2013, 05:28:56 PM »
Please confirm whether my long winded explanation makes sense / is correct. ???

Sounds sensible to me.
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