Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: kekie on May 18, 2014, 03:22:34 PM
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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;
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fs27.postimg.org%2Fwef11oxrn%2Fnormal_Mech.png&hash=97e225d61bffe0385768b2b7fe79a2943ce99c6d)
Does it just do this?;
(https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fs27.postimg.org%2Fq1zvruupf%2Fpotential_Mech.png&hash=6d3275515711eba5a3b26b055cb56c60631265b5)
Does it do that? If not, why?
Is this question wholly meaningless because of the way pi electrons are distributed in benzene rings?
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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)