# Oxidizing 2-Methylpropan-2-ol using potassium dichromate. Why is heating required?

In Chemistry class, we had to combine $\ce{K2Cr2O7}$ with 2-Methylpropan-2-ol and heat it for the oxidation to occur.

But why do we have to heat the mixture when, for example, butanol, can be oxidized via use of the same reagent but without heating it?

Tertiary alcohols (such as 2-methyl-2-propanol, in your specific case) cannot be directly oxidized by $\ce{Cr(VI)}$ salts like $\ce{K2Cr2O7}$ because they have no $\alpha$-hydrogen. On the other hand, a primary alcohol (such as 1-butanol) will be readily oxidized, even under mild conditions and with reagents much gentler than $\ce{K2Cr2O7}$.
With heating and sufficiently acidic conditions, a tertiary alcohol can be eliminated to form an alkene, which can then be oxidatively cleaved to yield two molecules with carbonyl groups. In the case of 2-methyl-2-propanol, one would presumably obtain acetone and either formaldehyde or formic acid, depending on how far the oxidation is permitted to go. (Though, under conditions vigorous enough to effect the oxidative cleavage of the alkene, it's highly unlikely the aldehyde could be isolated. I would also note that carbon dioxide is a product of oxidative cleavage of terminal alkenes when $\ce{KMnO4}$ is the oxidizer, but whether this occurs with $\ce{Cr(VI)}$ salts is unknown to me.) This is really the only reaction I can imagine to have plausibly occurred, if any.
To the best of my knowledge, $\ce{K2Cr2O7}$ is not very efficient for this purpose, and the procedure itself strikes me as pretty odd if the oxidative cleavage described above is the intended reaction. It may be that no reaction was intended to happen, since oxidation reactions of this type can be used as chemical tests to differentiate alcohols by degree of substitution.