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Author: Subject: KMnO4 oidation of tertiary alcohol

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[*] posted on 3-7-2020 at 11:26
KMnO4 oidation of tertiary alcohol

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?
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[*] posted on 3-7-2020 at 11:44

It's not a trick question - "No reaction" is a perfectly reasonable answer for many such questions.

Please remember: "Filtrate" is not a verb.
Write up your lab reports the way your instructor wants them, not the way your ex-instructor wants them.
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