Can you suggest most favorable mechanism for toluene oxidation to carboxylic acid using aq Hypo solution, K2CO3 and aq. KOH?
first step is to abstract an H on toluene.
Ar-CH2-H is possibly the most easily abstractable.
Emmanuel you are correct but I don't think that the reagent K2CO3+KOH have any ability to dehydrogenate a toluene molecule. Some type of oxidizing agent such as molecular oxygen is necessary for such H-abstraction.
What is an oxidant in your system? Toluene is not soluble in water.
What seems to be certain is that in basic conditions the OCl- anion, specifically the oxygen in it, is the oxidant. On how this anion excerts its oxidation activity you can draw several mechanisms. Perhaps, the most simple and plausible is the following reaction in aqueos basic medium: OCl- + H2O = Cl- + 2 .OH. The formed hydroxyl radicals can go to the toluene phase and abstract all benzylic(methyl) protons of your toluene giving you benzoic acid (sodium salt) at the end. Note, that the reaction will be facilitated by the formation of benzyl alcohol and once benzaldehyde is formed that can also give you half equivalent benzoic acid salt and half equivalent of benzylalcohol in the presence of KOH via the Cannizzaro reaction.
Dear Imre,
Hypochlorite has a very long shelf-life (stable). if hypothetically the reaction OCl- + H2O = Cl- + 2 .OH is possible, then it should be very slow. In addition , hydroxyl radical will quickly react with Cl- giving Cl(.) atom. The possible reaction mechanism Cl2 + H2O ↔ HClO + Cl− + H+, Cl2 + ambient light -> 2 Cl(.). Cl(.) is a strong hydrogen atom acceptor.
Dear Yurii,
Perhaps you think about the oxidation reaction of HOCl in acidic conditions, where chlorine is the main oxidant. Please check the reactions of NaOCl, the behaviour of OCl- anion (especially its pH dependence!) and its use in oxidation reactions below pH 10-11 including toluene to benzoic acid. Under these conditions, oxygen is the oxidant, regardless whether it goes the way I've shown above or by less likely mechanisms such as 2 OCl- → 2.O + 2Cl- →O2 + 2 Cl- (Cl- + ClO2- is not really expected at this pH) or by an attack of the O side of OCl- on carbon. Anyway, when .OH radicals are formed in the absence of suitable CH-s under these conditions they will self-destruct by the brutto reaction of 2 .OH → H2O2 or 4 .OH → 2 H2O + O2, which will regenerate the initial state by NaCl + H2O2 → NaOCl +H2O or by 2NaCl + O2 → 2NaOCl, respectively, i. e. nothing happens for outside. According to the literature if a .Cl radical, more exactly a .Cl2- (through Cl- + .Cl → .Cl2- ) was formed that would take electron from OH- (or H2O) to yield .OH radical and not the opposite.
Regards,
Hypochlorite is prepared through the reaction
Cl2 + 2 NaOH =NaCl + NaClO + H2O
This reaction is reversible. Commercial solutions always contain significant amounts of sodium chloride. If HO(.) is formed, the only possible reaction is HO(.) + Cl- -> HO- + Cl(.). In the absence of C-H bonds, 2 Cl(.) -> Cl2, As far as I know the decomposition of hypochlorite never produces O2. If O2 is formed, this reaction can be formally defined as water oxidation by hypochlorite.
What really matters is that a basic aqueous solution of NaOCl will give you oxygenated (and not chlorinated!) products in reaction with organic compounds (containing suitable H-s to abstract or with many three valent P-compounds, phosphines, phosphites, etc.
Hypochlorite can also be prepared as follows:
2NaCl + O2 → 2NaOCl
The oxidation of organic molecules by NaOCl or its decomposition to NaCl and O2 is catalyzed by a number of transition metal compounds, such as Co, Cu, Fe, Ni, Ru etc. and by some radical initiators such as TEMPO.
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