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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?

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3 Answers 3

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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.

reaction mechanism

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    $\begingroup$ How would we explain from this mechanism whether primary alcohols give a positive Tollens' test or not? $\endgroup$ Commented Mar 16, 2018 at 5:33
  • $\begingroup$ @GaurangTandon Your question is quite different from the original question so you might want to post it as a new question. My view would be that since a typical alcohol does not have a readily available electron (e.g. high-lying HOMO as a carbonyl has), an alcohol would not undergo 1 electron oxidation and react with Tollens reagent in a normal fashion to produce the characteristic silver mirror. That said, if you heat things up, or have oxidizing impurities in the sample, then all bets are off and anything could happen. $\endgroup$
    – ron
    Commented Mar 16, 2018 at 21:26
  • $\begingroup$ Oh, that's interesting. Thanks! Though do have a look here, as you said, gentle heating is causing silver ppt to form, as bon says $\endgroup$ Commented Mar 17, 2018 at 2:01
  • $\begingroup$ @GaurangTandon No, he says it produces a "fine black precipitate" with the alcohol. $\endgroup$
    – ron
    Commented Mar 17, 2018 at 3:21
  • $\begingroup$ Do you mind posting a new question about this? I don't know how to phrase it whilst avoiding being called a duplicate :/ Or please tell me how to? $\endgroup$ Commented Mar 17, 2018 at 3:25
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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.

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  • $\begingroup$ I couldn't see any mechanism or text describing the mechanism on the link you provided, you could? Also, your mechanism suggests that silver(I) will oxidize saturated carbons containing a hydrogen, for example, dimethoxymethane. Is this reported in the literature? $\endgroup$
    – ron
    Commented Jun 25, 2014 at 21:03
  • $\begingroup$ 1) You have to download and read the article for it. 2.a) It was reported that aldehydes forming stable hydrates, like $\ce{CCl3CHO}$, reacts much faster. 2.b) Tollen reagent does give false positive, but I am too lazy to find full list. $\endgroup$
    – permeakra
    Commented Jun 25, 2014 at 21:10
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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.

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  • $\begingroup$ The aldehyde is not losing electrons to oxygen. It's losing electrons to silver. Note that silver goes from the +1 oxidation state to the 0 oxidation state. $\endgroup$
    – jerepierre
    Commented Jun 25, 2014 at 20:05
  • $\begingroup$ I see what you're saying; I was referring to the cause of the +1 to +3 oxidation state change of the carbon from reactant to product. $\endgroup$
    – Dissenter
    Commented Jun 25, 2014 at 20:06
  • $\begingroup$ You will never form oxygen dianion in an aqueous solution. It is far too basic. In this case, deprotonating hydroxide is equally thermodynamically disfavorable to doubly deprotonating water. $\endgroup$
    – jerepierre
    Commented Jun 26, 2014 at 15:08
  • $\begingroup$ What's your source? Do you have a citation? $\endgroup$
    – Dissenter
    Commented Jun 26, 2014 at 15:10
  • $\begingroup$ I don't know what the pKa of hydroxide is. I've never seen it in a table, presumably because it's not acidic at all. Here's a discussion though: chemicalforums.com/index.php?topic=62198.0 According to your mechanism, hydroxide is the destination of the proton that comes off the original hydroxide. So we can estimate the equilibrium constant for that reaction. The pKa of hydroxide is 50 (estimate) and the pKa of water is 16. The equilibrium constant for the deprotonation of hydroxide by hydroxide then is 10^(16-50), which is a very small number. $\endgroup$
    – jerepierre
    Commented Jun 26, 2014 at 22:31

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