Can someone help me identify the steps for the following reaction? I'd like to keep track of everything that happens. I can't figure out what happens with oxidation in this context, and thus can't get started.
It's 2-methylpropan-1-ol oxidized with potassium dichromate(VI) in aqueous sulfuric acid to yield 2-methylpropanoic acid (also known as isobutyric acid).
Answer
A dichromate salt and aqueous sulfuric acid are used to prepare chromic acid in situ, which is among the most powerful oxidizing agents commonly available. A primary alcohol is first oxidized to an aldehyde (via a chromate ester intermediate). Under acidic conditions, the aldehyde's oxygen acts as a base and abstracts a proton, the carbonyl carbon is attacked by water (whose oxygen acts as a nucleophile) and simultaneously the carbonyl π-bond breaks, and then finally the resulting molecule is deprotonated by another water molecule, yielding a diol. That diol is then subject to further oxidation by the remaining chromic acid, which occurs by the same mechanism as the initial oxidation to an aldehyde. The ultimate result is a carboxylic acid.
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