December 01, 2021, 01:23:26 AM
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Topic: Nucleophilic aromatic substitution when EWG and LG are ortho...  (Read 1083 times)

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Offline kekie

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So I was drawing out the mech for Nucleophilic aromatic substitution when the EWG (electron withdrawing group) and LG (leaving group) are ortho to each other, but when there is no pi bond between the two carbons, and I ran into this;

Why not instead of doing this;

Does it just do this?;


Does it do that? If not, why?
Is this question wholly meaningless because of the way pi electrons are distributed in benzene rings?

Offline Nescafe

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Re: Nucleophilic aromatic substitution when EWG and LG are ortho...
« Reply #1 on: May 18, 2014, 04:24:39 PM »
Since everything is in resonance it does not matter how you draw it. Typically, the (-) charge is placed on the neighboring carbon which then rapidly kicks back into the ring and kicks of the leaving group in order to reestablish aromaticity of the ring. The first step (nuc) coming in is the RDS.

see this (http://highered.mcgraw-hill.com/sites/dl/free/0073375624/825564/Nucleophilic_Aromatic_Substitution.pdf)

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