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I was told by my teacher that the following reaction proceeds through SNi:

enter image description here

Moist $\ce{Ag2O}$ is basically $\ce{AgOH}$, what role does it play in this reaction's mechanism? I think this is what happens, but I am not sure:

  1. $\ce{AgOH}$ attacks $\ce{Cl-}$ and during this process the $\ce{Ag-O}$ bond breaks and an intimate ion pair is formed between the carbocation and $\ce{Cl-}$.
  2. Since a partial bond still exists between $\ce{Ag}$ and $\ce{O}$, $\ce{O}$ is forced to attack from the same side from where $\ce{Cl-}$ left and the reaction proceeds to completion without Walden inversion.

Is my reasoning correct? Or is there a different mechanism?

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    $\begingroup$ As for the reaction mechanism, I would rationalize it as follows: we know that AgX bond is very strong. A transient bond between Ag and Cl readily forms. A "curved arrow" goes from C-Cl to Cl-Ag (charge is transfered from C-Cl to Ag-Cl). At the same time, a curved arrow goes from the Ag-O(H) bond to a new C-O(H) bond, in a sort of internal sn2, being diatomic, fashion with a 4 elements reaction intermediate (C-Cl-Ag-O). The four members cyclic intermediate probably could favor a syn attack, with no stereochemical inversion. $\endgroup$
    – user32223
    Apr 22, 2018 at 17:17
  • $\begingroup$ @The_Vinz Thanks, it looks very similar to my proposed one. $\endgroup$
    – Archer
    Apr 22, 2018 at 17:20
  • $\begingroup$ This would be reasonable if there were actually reacting AgOH molecules. In such molecule coordination number of Ag is too low. $\endgroup$
    – Mithoron
    Apr 22, 2018 at 21:22

1 Answer 1

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If you suspect that there is something different going on here, and I believe you are right.

My take, in the presence of lab light, $\ce{Ag2O}$ and $\ce{H2O}$ actually results, in the formation of the hydroxyl radical ($\ce{.OH}$) and the hydrogen atom radical ($\ce{.H}$) via the creation of a powerful photocatalyst couple $\ce{Ag2O-Ag}$ as outlined in this article: Ag2O as a New Visible‐Light Photocatalyst: Self‐Stability and High Photocatalytic Activity.

The photoelectric introduced electron hole $\ce{h+}$ and electron $\ce{e-}$ can act as follows in the present of water:

$\ce{H2O <=> H+ + OH-}$

$\ce{H+ + e- -> .H}$

$\ce{OH- + h+ -> .OH}$

The hydrogen atom radical is likely responsible for the extraction of the chlorine atom from the organic forming a product of $\ce{HCl}$.

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