Chiron - 3-7-2020 at 11:26
This is a trick question asked in the practice questions for the unit I'm studying.
CH3C(CH3)2OH + KMnO4/OH- + heat -> ?
This seems like a trick question because my understanding is that tertiary alcohols are not readily oxidized under this format without the aid of an
acidic solution. Heated KMnO4 shouldn't do anything in a basic solution. Whereas cold KMnO4 in a basic solution would yield a vicinal diol.
My reflex is to write the formation of alkene CH3-C(CH3)=CH2 but this doesn't seem right. This is org chem 101 so they're not asking for anything
fancy. My answer right now is "no reaction". It's a trick because they are combining different conditions hoping you'll mess up.
Or am I wrong?
DraconicAcid - 3-7-2020 at 11:44
It's not a trick question - "No reaction" is a perfectly reasonable answer for many such questions.