$$\ce{HCOOH ->[\text{Fehling's solution}] CO2 + H2O}$$

The points discussed in a previous question do not account for the fact that formic acid also gives positive Fehling's test since it is not enolisable. Then what is the reason/mechanism for the reaction?

  • $\begingroup$ I thought that the exact mechanism is not known in case of both fehlings and tollens. $\endgroup$
    – user57048
    Aug 23 '18 at 6:15
  • $\begingroup$ Formic acid is something like an aldehyde, it has a hydrogen atom attached to a carbonyl group. $\endgroup$ Aug 23 '18 at 9:47
  • $\begingroup$ Because Fehling's solution oxidizes enolizable aldehydes to carboxylic acids, it does not preclude all other oxidations. Draw cupric formate and move formate electrons toward copper. $\endgroup$
    – user55119
    Aug 23 '18 at 23:17

Here is a short simple available answer provided online by Prachi Sawant in 2016 to quote:

Formic acid has both aldehydic (-CHO) and carboxylic (-COOH) functional groups. Hence, it gives both Tollen's and Fehling's test positive.

Reaction of formic acid with Tollen's reagent:

$\ce{HCOOH + 2[Ag(NH3)2]+ +2OH- → 2Ag + CO2 + 2H2O + 4NH3 }$

Reaction of formic acid with Fehling's reagent:

$\ce{HCOOH + 4OH- + 2Cu^{2+}(in complex) → CO2 + 3H2O + Cu2O }$

Now, here is my attempt on the possible underlying mechanic, which is likely complex. I start with the above net reaction, which in reality, is probably not completely accurate, but nevertheless, may provide some insight into actual half-cell reactions, beginning with the reduction of the organic cupric complex:

$\ce{Cu^{2+} + e^- -> Cu^+ }$

where the insoluble Cu2O (the test indicator) is formed. The corresponding other half-cell is less clear accept for the liberation of an electron:

$\ce{COOH^- -> .HOCO + e^- }$

which may speculatively may be followed by:

$\ce{.HOCO + .HOCO -> H2O + CO + CO2 }$ (speculation)

And/or, this paper favoring non-conventional ionized species in the context of formic acid as possessing more stability as demonstrated, for example, by:

$\ce{HCOOH -> [.HCOOH]^+ + e^- }$

With its source cited below, followed by:

$\ce{[.HCOOH]^+ -> H2O .CO^+ }$ (a distonic ion, per 2nd cited source)

$\ce{OH^- + H2O .CO^+ -> H2O + COOH^- }$ (speculation)

To quote a cited source above :

Rate constants for the radical-induced hydrogen abstraction from formic acid, HCOOH, are presented here. Only those reactions leading to the formation of $\ce{.HOCO}$ were investigated.

Per a source: 'The gas phase chemistry of the formic acid radical cation $\ce{.[HCOOH]^+}$ Mechanism for exchange of the hydrogen atoms: a quantum chemical investigation', to quote:

Earlier labelling experiments had indicated that prior to the loss of H the hydrogen atoms in ionized formic acid, $\ce{.HCOOH^+}$, can become positionally equivalent via an unknown pathway. Using ab initio molecular orbital calculations we have located a minimum energy reaction pathway for hydrogen exchange reactions which takes place close to the dissociation level and proceeds via flexible ion/dipole complexes. Along this path three equilibrium structures of different atom connectivity are found: $\ce{.HCOOH^+}$, 1, ionized formic acid; two forms of the surprisingly stable ion/dipole complex $\ce{.HOHCO^+}$, 2, 3; and the distonic ion $\ce{H2O CO.^+}$, 4. Ion 1 is separated from ion 2 by a barrier of 142 kJ mol−1 and this barrier corresponds to a 2A′ → 2A″ surface crossing. The barrier for the proton transfer 3 → 4 is only 25 kJ mol−1 relative to 3; the transformation $\ce{.HCOOH^+ -> H2O .CO^+}$, formally a 1,2-hydrogen shift, takes place via the ion/dipole complex $\ce{HO· HCO^+}$, $\ce{H2O .C^+}$ is the reacting configuration for decarbonylation and this reaction is calculated to take place at the dissociation limit in full agreement with experiment.

Where ‘decarbonylation’ can also be effected by several transition metal complexes, (source), as in the current context with an organic copper complex.


According to some people:

If you look carefully, you will find both the aldehyde radical and the carboxyl radical in formic acid:

formic acid 1

formic acid 2

The red line shows the aldehyde functional group character and the blue line indicates the carboxyl functional group character.

As a result, it has mainly characteristics of carboxyl compound, but sometimes shows carbonyl character as well which happens in case of tollen's reagent and fehling's solution reacting with it.

You can identify it by doing the 2:4-DNPH test to which the formic acid doesn't respond.

But to me and most other people:

This reaction certainly is given by aldehydes, it is not specific for this class of compound since many other acids, such as tartaric or citric, similarly reduce silver nitrate. The reduction of mercuric chloride to mercurous chloride by formates likewise offers no support for the aldehyde theory, since aldehydes in general, including formaldehyde, cannot effect this change. Actually, formates give none of the reactions which characterize an aldehyde; thus they fail to restore the colour to Schiff's reagent, produce no reaction with 2.4 dinitrophenylhydrazine hydrochloride, and above all do not reduce Fehling's solution, which is probably the most specific of all aldehyde reagents.

  • 1
    $\begingroup$ They’re not radicals; at best, they are functional groups. But formic acid really doesn’t undergo any typical aldehyde reactions, so calling the carbonyl group an aldehyde is missing the point … $\endgroup$
    – Jan
    Nov 1 '19 at 11:32

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