# Product of oxidation of alkanes

I'm trying to find the product of the oxidation of iso-butane using KMnO4, i know that alkanes undergo step wise oxidation with strong oxidising agents to finally give carboxylic acids.. With straight chain carbons, its very straightforward - ethane becomes acetic acid etc. But with branched chain alkanes I don't know how to proceed. Can someone tell me if there's a general rule for oxidation of alkanes- to get the product?

• Are you sure of the premise? KMnO4/H+ is a test for alkenes vs. alkanes. Just the benzyl position undergoes ox. Isn't so? And in general,anyway, if an alkane react than is a multisite reaction. Like in bromination. – Alchimista Feb 11 '19 at 14:34
• Actually, the reactants in the reaction are given as KMnO4/ [O] , so I think oxidation is implied. – karun mathews Feb 11 '19 at 15:27
• Oh ya I remember now, yes KMnO4/H+ is a test for unsaturated hydrocarbons, but KMnO4 is also used as an oxidant according to my textbook, but the only examples given are oxidation of methane and ethane.. – karun mathews Feb 11 '19 at 15:30
• You might want to look at the Kuhn-Roth oxidation although it is run with chromic acid. – user55119 Feb 11 '19 at 19:34

Potassium permanganate, $$\ce{KMnO4}$$, is the most common and probably the most applicable oxidizing agent discussed in organic chemistry teaching classes. $$\ce{KMnO4}$$ is a very strong oxidizing agent, and is capable of oxidizing a wide range of organic molecules. Usually, it will oxidize organic compounds to appropriate carboxylic acids, but, the final products can vary on the conditions used. If you googled enough to find reduction potentials of $$\ce{KMnO4}$$, you may find following conditions, which are the most used reduction conditions of $$\ce{KMnO4}$$ (Wikipedia): \begin{align} \ce{MnO4- + H+ + e- &-> HMnO4^-} & \mathrm{E^o} &= \pu{0.90V}\tag{1}\label{1}\\ \ce{MnO4- + 8H+ + 5e- &-> Mn^2+ + 4H2O} & \mathrm{E^o} &= \pu{1.51V}\tag{2}\label{2}\\ \ce{MnO4- + 2H2O + 3e- &-> MnO2(s) + 4OH-} & \mathrm{E^o} &= \pu{1.69V}\tag{3}\label{3}\\ \ce{MnO4- + 4H+ + 3e- &-> MnO2(s) + 2H2O} & \mathrm{E^o} &=\pu{1.7V}\tag{4}\label{4} \end{align} Other manganese containing compounds other than $$\ce{KMnO4}$$, which can also be used as oxidizing reagents are listed below: \begin{align} \ce{MnO2(s) + 4H+ + e- &-> Mn^3+ + 2H2O} & \mathrm{E^o} &= \pu{0.95V}\tag{5}\label{5}\\ \ce{MnO2(s) + 4H+ + 2e- &-> Mn^2+ + 2H2O} & \mathrm{E^o} &= \pu{1.23V}\tag{6}\label{6}\\ \ce{HMnO4- + 3H+ + 2e- &-> MnO2(s) + 2H2O} & \mathrm{E^o} &= \pu{2.09V}\tag{7}\label{7}\\ \end{align}

Yet, as most of the viewers already commented, I also believe $$\ce{KMnO4}$$ does not oxidize alkanes, except in following situations:

$$\ce{KMnO4}$$ oxidizes alkane carbon atoms, which contain sufficiently weak $$\ce{C-H}$$ bonds such as those in benzylic positions ($$\ce{C-H}$$ bonds in the $$\alpha$$-positions of substituted aromatic rings). These oxidizable alkyl benzene substituents can be -$$\ce{CH3}$$, or any $$1^\mathrm{o}$$-alkyl groups such as -$$\ce{CH2CH3}$$, or $$2^\mathrm{o}$$-alkyl groups such as -$$\ce{CH(CH3)2}$$. All of them will be oxidized to carboxylic acid group, -$$\ce{CO2H}$$ (See Scheme 1 below):

It is also important to note that when the benzylic carbon is a $$3^\mathrm{o}$$-$$\ce{C}$$ (e.g., corresponding alkyl groups such as $$-\ce{C(CH3)3}$$ group in tert-butylbenzene, see the example given in RHS of Scheme 1), the oxidation reaction does not happen.

In addition to benzylic carbons, $$\ce{KMnO4}$$ also oxidizes weak $$\ce{C-C}$$ bonds such as $$\pi$$-bonds in alkenes and alkynes (Scheme 2):

$$\ce{KMnO4}$$ also oxidizes $$\sigma~\ce{C-C}$$ bonds such as that in glycol-moiety (e.g., $$\ce{HO-C-C-OH}$$) and $$\ce{C^{\mathrm{sp^2}}-O}$$ bonds such as those in phenolic compounds (e.g., it oxidizes phenol to 1,4-benzoquinone, commonly known as para-quinone).

At last, as mentioned above, conditions of the reaction can be adjusted such a way that desired compound can be achied in high yields. For example, see following example (Ref.1):

Note that no benzylic carbon or -$$\ce{OCH2}$$- was oxidized here.

Although tertiary $$\ce{C-H}$$ bond in isobutane (preferred IUPAC name: 2-methylpropane; $$\ce{CH3CH(CH3)CH3}$$) is relatively weaker than othe alkane $$\ce{C-H}$$ bonds, I’m highly skeptical of the statement of “Oxidizing agents such as potassium permanganate readily oxidize a tertiary hydrogen atom to a hydroxyl group, e.g., isobutane is oxidized to t-butanol.” (vide infra). To my knowledge, the literature evidence for that type of oxidation reactions does not exist. Nevertheless, 2-methylpropane (isobutane) can be oxidized to 2-methylprop-2-ol (t-butanol) by reacting supercritical-phase isobutane with air in the presence of a catalyst such as $$\ce{Pd/C}$$ at high temperature, Instead of using $$\ce{KMnO4}$$ (Ref.2):

Abstract: tert-2-Butyl alcohol can be synthesized efficiently on $$\ce{SiO2 –TiO2}$$ or $$\ce{Pd–C}$$ catalyst if air is directly introduced to the supercritical-phase isobutane.

tert-Butanol (TBA) is the initial material in methyl tert-butyl ether (MTBE) production and methyl methacrylate (MMA) synthesis. MTBE is high-octane-number gasoline additive and MMA is a resin material for plastic synthesis (e.g., PMMA or poly-methyl methacrylate). Thus, one would have wondered why scientists did not attempted to synthesize TBA in a such an easy way like using $$\ce{KMnO4}$$.

However, in a different note, almost exclusive $$\ce{C-H}$$ bond activation has been observed in the reaction of linear and branched alkanes with $$\ce{CeO2+}$$ (Ref.3)

Reference:

1. A. Abiko, J. C Roberts, T. Takemasa, S. Masamune, “$$\ce{KMnO4}$$ revisited: Oxidation of aldehydes to carboxylic acids in the tert-butyl alcohol - aqueous $$\ce{NaH2PO4}$$ system,” Tetrahedron Letters 1986, 27(38), 4537-4540 (https://doi.org/10.1016/S0040-4039(00)84997-7)
2. L. Fan, Y. Nakayama, K. Fujimoto, “Air oxidation of supercritical phase isobutane to tert-butyl alcohol,” Chem. Commun. 1997, (13), 1179-1180 (DOI:10.1039/A702308A).
3. A. Gunay, K. H. Theopold, “$$\ce{C-H}$$ Bond Activations by Metal Oxo Compounds,” Chem. Rev. 2010, 110(2), 1060–1081 (DOI: 10.1021/cr900269x).
• Thanks a lot for your detailed answer. But as you said that alkanes don't generally get oxidised , is the example given by @Anubhab Days right? I haven't a reference of il finars volume 1, so I'm not sure. – karun mathews Feb 14 '19 at 5:49
• @Karun mathews: please see my late addition to the original answer for reasoning. – Mathew Mahindaratne Feb 14 '19 at 17:43
• Thanks for the very researched answer. If I find anything else I'll post it here then.. – karun mathews Feb 15 '19 at 4:12
• @MathewMahindaratne Indeed, I had also not heard of oxidation of tertiary hydrogens by KMnO4 until I came across that statement in my answer. Although it is unlikely, Finar may be wrong, as in, it doesn't contain any reference to a real paper to support his claim. – Anubhab Das Feb 15 '19 at 9:05
• @Anubhab Das: Please post any new finding on this subject. I'm really interested to them. – Mathew Mahindaratne Feb 15 '19 at 18:26

Quoted directly from IL Finar's Organic Chemistry Volume 1 -

"Oxidising agents such as potassium permanganate readily oxidise a tertiary hydrogen atom to a hydroxyl group, e.g., isobutane is oxidised to t-butanol."

• Ah ok thanks. So could I assume a general rule that 3° H>2° H>1° H for more likelihood of oxidation? – karun mathews Feb 12 '19 at 10:02