# What is the SNi mechanism of the reaction of moist silver oxide with a halide?

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

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?

• 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.
– user32223
Apr 22, 2018 at 17:17
• @The_Vinz Thanks, it looks very similar to my proposed one. Apr 22, 2018 at 17:20
• This would be reasonable if there were actually reacting AgOH molecules. In such molecule coordination number of Ag is too low. Apr 22, 2018 at 21:22

## 1 Answer

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 $$\ce{HCl}$$.