Upon treatment with Tollens' reagent (ammoniacal silver(I) nitrate), aldehydes are oxidised to carboxylic acid, and silver(I) is reduced to silver metal.
I am trying to find a mechanism for the this reaction online, but the only thing I can find is the balanced equation. Can someone propose or help me find the mechanism?
Here are the two half reactions:
$$\begin{align} \ce{[Ag(NH3)2]+ + e- &-> Ag^0 + 2NH3} \\ \ce{RCHO + 3OH- &-> RCO2- + 2H2O + 2e-} \end{align}$$
which together yield the overall reaction
$$\ce{2[Ag(NH3)2]+ + RCHO + 3OH- -> 2Ag^0 + RCO2- + 4NH3 + 2H2O}$$
Here is a diagram of the reaction mechanism. The carbonyl group is oxidized in the process and the $\ce{Ag^+}$ is reduced. The resultant oxidized aldehyde (now a radical cation) reacts with hydroxide to form a tetrahedral intermediate. A gem-diol like intermediate is formed via a hydrogen shift, which then continues on to the final carboxylate anion.
A Google search for "Tollens' test mechanism" gave me a link to this article: J. Chem. Res. 2011, 35 (12), 675–677. A quick glance at the text of the article leads to the conclusion that the currently proposed mechanism is as follows:
$$\begin{align} \ce{R-CHO + H2O &-> R-CH(OH)2} \\ \ce{R-CH(OH)2 + Ag+ &-> R-C^.(OH)2 + H+ + Ag^0} \\ \ce{R-C^.(OH)2 + Ag+ &-> R-COOH + H+ + Ag^0} \end{align}$$
In strongly alkaline solution (pH > 10) the mechanism changes to the following:
$$\begin{align} \ce{R-CHO + OH &-> R-CH(OH)O-} \\ \ce{R-CH(OH)O- + Ag+ + OH- &-> R-C^.(OH)O- + H2O + Ag^0} \\ \ce{R-C^.(OH)O- + Ag+ + OH- &-> R-COO- + H2O + Ag^0} \end{align}$$
However, it seems that there is no direct evidence for these exact mechanisms, though they seem believable. That is normal, though; proving any mechanism is a long and tedious task that can take decades of dedicated studies, so we have to live with it.
The balanced equation can tell you a lot. This is the balanced equation:
$\ce{R-CHO + 2[Ag(NH3)2]+ + 3HO- -> R-COO- + 2Ag + 2H2O + 4NH3}$
Let's check out what's happening:
1) First, the aldehyde is being oxidized. Specifically, the carbon in the aldehyde is being oxidized; it is losing electrons to silver.
Oxidation half reaction: $\ce{R-CHO -> R-COO- + 2e^- + H+}$
Oxidation states of C: $~~~~~~+1~~~~~~~~~~~~~+3$
2) Mass isn't balanced; we need a source of $\ce{O^2-}$ (negative 2 oxidation state oxygen atoms). Given that this reaction is happening with a solution of silver ammonia, it makes thermodynamic sense that your best source of $\ce{O^2-}$ oxygen is the hydroxide anion: $\ce{HO^-}$.
Water is a possible source but it's pricier to heterolytically cleave two $\ce{H-O}$ bonds as opposed to just one $\ce{H-O}$ bond.
Thermodynamically favorable: $\ce{H-O^- -> H^+ +O^2-}$
Not so thermodynamically favorable: $\ce{H-O-H -> 2H^+ +O^2-}$
3) The freeing of one $\ce{O^2-}$ yields the hydrogen proton, $\ce{H+}$. This is an acid; it will react immediately with the strongest base in the system, which would be the hydroxide anion.
Also, note that a hydrogen proton is freed from the aldehyde. The hydrogen atom is freed heterolytically as to yield 2 electrons, which go to the silver ammonia complex.
$\ce{R-CHO -> R-COO- + H+ + 2e^-}$
Two protons require two additional hydroxide anions to react with; this is why we use a total of three hydroxide anions and this yields two water molecules.
$\ce{2H+ + 2HO- -> 2H2O}$
4) The above explains the hard stuff. The reduction of silver is easy:
$\ce{[Ag(NH3)_2]+ + e^- -> Ag + 2NH3}$
The silver in the silver ammonia complex ion has an oxidation state of +1; gaining an electron allows solid silver to precipitate out, giving a positive Tollens test for aldehydes. The ammonia remains unchanged.
