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Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: retrospect3 on May 21, 2021, 05:28:55 AM

Title: Lithation of 2-chloropyridine with LDA
Post by: retrospect3 on May 21, 2021, 05:28:55 AM

So I was completing the mechanism with the lithation of 2-chloropyridine with LDA and through research I found that the proton at the C-3 position is removed to allow for a nucleophillic attack at the C-3 position. But I'm confused on why the proton is removed rather than Chlorine since chlorine is extremely electronegative.

Essentially, just asking why is the proton at C-3 removed rather than the chlorine

Title: Re: Lithation of 2-chloropyridine with LDA
Post by: Orcio_87 on May 21, 2021, 01:43:16 PM

Quote

I found that the proton at the C-3 position is removed to allow for a nucleophillic attack at the C-3 position

Removing of C-3 H⁺ will disable nucleophilic substition at this place (Nu^- will not attach to C^-).

LiN[CH(CH₃)₂]₂ is strong base so substition follows rather elimination-addition mechanism (product will be mixture of 2- and 3-substituted pyridine).