Orgo: ethylbenzene reacting with K2Cr2O7

[imn0tokei]

Hi guys. I'm memorizing the road maps from destroyer and I am confused when I encountered the reaction of ethylbenzene with K2Cr2O7 and H+. Is the product supposed to be the one on the left or the one on the right? I thought it would be the one on the right because there are 2 carbons coming of, thus, you will still have 2 carbons coming off benzene after the oxidation. OR does K2Cr2O7 acts similarly to KMnO4 where it reacts with the allylic hydrogen to make carboxylic acid?

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[Predental1995]

It would be the one on the left because similar to KMnO4, it oxidizes an alkyl group on a benzene ring into benzoic acid

 

[imn0tokei]

Thank you so much!

 

[Incis0r]

The only thing you need for that is a free proton on the benzylic position. It always oxidizes at the benzylic position. Even if there was a straight chain alkane 45 carbons long, only the benzylic would be oxidized and the rest would be "chopped off."

 

[DAT Destroyer]

Hi guys. I'm memorizing the road maps from destroyer and I am confused when I encountered the reaction of ethylbenzene with K2Cr2O7 and H+. Is the product supposed to be the one on the left or the one on the right? I thought it would be the one on the right because there are 2 carbons coming of, thus, you will still have 2 carbons coming off benzene after the oxidation. OR does K2Cr2O7 acts similarly to KMnO4 where it reacts with the allylic hydrogen to make carboxylic acid?

Oxidation of ethyl benzene with either dichromate and acid...or permanganate and acid work-up both five benzoic acid.

Hope this helps.

Dr. Romano