Oxidation of methanoic acid (formic acid) to carbon dioxide and water

Question

Out of the carboxylic acids, methanoic acid $(\ce{HCOOH})$ is the only one that can be further oxidised by potassium permanganate because the top half of the molecular has an aldehyde functional group and hence exhibits most of the properties of aldehydes. I am trying the understand the mechanism for the oxidation of methanoic acid. In the oxidation of aldehydes we get a carboxylic acid, so why does the oxidation here give carbon dioxide and water? I have tried searching extensively for the mechanism, but all sources are the same; they just state that methanoic acid oxidises to carbon dioxide. This is not at all obvious or intuitive to me.

Answer 1

Indeed, methanoic acid $(\ce{HCOOH})$ is the only carboxylic acid that can be further oxidized by oxidizing agents such as potassium permanganate because it contains exchangeable proton attached to carbonyl moiety. However, this oxidation is not simple as it seems (Ref.1). Accordingly, the stoichiometric reaction is: $$\ce{2MnO4- + 3HCOOH + 2H+ -> 2MnO2 + 3CO2 + 4H2O}$$ According to Ref.1, the reaction follows a complex mechanism as stated by the abstract:

The kinetics of the permanganate oxidation of formic acid in aqueous perchloric acid were examined in the temperature range 15 to 35°. The reaction, $\ce{2MnO4- + 3HCOOH + 2H+ -> 2MnO2 + 3CO2 + 4H2O}$, appears to proceed through two independent paths in which the rate-determining steps are bimolecular reactions of permanganate with formic acid and formate ion, respectively. The kinetics are thus of the form, $—d[\ce{MnO4-}]/dZ = [\ce{MnO4-}][\ce{HCOOH}](k_1 + k_2K_i[\ce{H+}])$, where $k_1= \pu{1.1 \times 10^9 exp[ —16400/RT] M-1 sec-1}$ and $k_2 = \pu{7.8 \times 10^9 exp[ -13000/RT].M-1 sec-1}$. The formate ion reaction $(k_2)$ exhibits a large deuterium isotope effect $(k_\ce{HCOO-}/(k_\ce{DCOO-} = 7)$ which suggests cleavage of the $\ce{C-H}$ bond in the rate-determining step. The absence of a corresponding isotope effect in the formic acid reaction $(k_1)$ suggests that it proceeds by a different mechanism. A difference in mechanism for the two paths is also indicated by the observation of a considerable solvent deuterium isotope effect for $k_2$ $(k_2^\ce{H2O}/k_2^\ce{D2O} = 0.38)$ but not for $k_1$. The reaction is susceptible to pronounced catalysis by $\ce{Fe^{+++}}$ (but not by $\ce{Na+, Ag+, Cu^{++}}$ or $\ce{Co^{++}}$). The kinetics suggest that the catalytic path involves reaction of $\ce{HCOOH}$ or $\ce{HCOO-}$ with a $\ce{FeMnO4^{++}}$ complex. It seems likely that the initial reduction products of $\ce{MnO4-}$,in the uncatalyzed and catalyzed reactions, are $\ce{Mn(V)}$ and $\ce{Mn(VI)}$, respectively.

Accordingly, both formic acid $(\ce{HCOOH})$ and formate ion $(\ce{HCOO-})$ subjected to oxidized by permanganate ion, but seemingly involves different reaction paths (and possibly different mechanisms). This conclusion is based on faster reaction on $\ce{HCOO-}$ ion compared to $\ce{HCOOH}$ (The rate of reaction of $\ce{HCOO-}$ with $\ce{MnO4-}$ exceeds that of $\ce{HCOOH}$ by a factor of more than $10^3$) and large isotope effect difference between two reactions (see above abstract).

Earlier workers (e.g., Ref.2) have proposed that the rate-determining step in the permanganate oxidation of formate ion involves a two-equivalent reduction of permanganate leading to the formation of $\ce{Mn(V)}$ as an intermediate (this applies to oxidation of formic acid as well). There are several possible types of mechanisms by which this may be accomplished. Followings represent some of the more obvious limiting cases (Ref.1):

• Hydride ion transfer from formate to permanganate: $$\ce{HCO2- + MnO4- -> CO2 + HMnO4^2-} \tag A$$
• Transfer of two electrons from formate to permanganate (possibly coupled with proton transfer to a water molecule): $$\ce{HCO2- + MnO4- -> HCO2+ + MnO4^3-} \tag B$$ Note: Here, $\ce{HCO2+}$ can be consider as $\ce{H+ + CO2}$.
• Transfer of an oxygen atom from permanganate to formate: $$\ce{HCO2- + MnO4- -> HCO3- + MnO3^-} \tag C$$ Ref.2 has shown detailed study of $\ce{^{18}O -}$ exchange during the reaction, which suggests the mechanism may consists of all three possibilities (A, B, & C). Ref.2 suggest some possible complex mechanisms. For instance, following is one:

Recently, this oxidation mechanism has been studied in organic solvents using organic soluble permanganate ion and suggested following mechanism for formic acid oxidation (Ref.3):

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References:

1. Sandra M. Taylor and J. Halpern, "Kinetics of the Permanganate Oxidation of Formic Acid and Formate Ion in Aqueous Solution," J. Am. Chem. Soc. 1959, 81(12), 2933–2937 (DOI: https://doi.org/10.1021/ja01521a004).
2. Kenneth B. Wiberg and Ross Stewart, "The Mechanisms of Permanganate Oxidation. II. The Oxidation of Formate Ion," J. Am. Chem. Soc. 1956, 78(6), 1214–1216 (DOI: https://doi.org/10.1021/ja01587a034).
3. Donald G. Lee and Joaquin F. Perez-Benito, "Kinetics and mechanism of the oxidation of formic acid by methyltributylammonium permanganate in methylene chloride solutions," J. Org. Chem. 1988, 53(24), 5725–5728 (DOI: https://doi.org/10.1021/jo00259a022).

Answer 2

Methanoic acid (or formic acid) is $\ce{HCOOH}$. It can be oxidized into carbonic acid $\ce{H2CO3}$. But this carbonic acid $\ce{H2CO3}$ is not stable. It is spontaneously decomposed into $\ce{CO2}$ and $\ce{H2O}$, whatever the origin of this $\ce{H2CO3}$. So it's allowed to state that methanoic acid is oxidized directly into $\ce{CO2}$ and $\ce{H2O}$, if the existence of the intermediate $\ce{H2CO3}$ is not mentioned.