Cr2O3 DAT Destroyer vs. McMurry

[oral09]

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Mcmurry said Cr2O3 will oxidize alcohol to carbocylic acid but DAT destroyer said it work similar with PCC, which is not strong enough to produce carboxylic acid.

Which one should be follow if there the question like this on the DAT?

 

[DentalDeac]

I was reading in Kaplan today, and PCC is only strong enough to oxidize a primary alcohol to an aldehyde or a secondary alcohol to a ketone.

Na2Cr2O7 can oxidize a primary alcohol all the way to a carboxylic acid

 

[Simiam]

Cr2O3 is AKA Jones reagent which will oxidize any primary alcohol to a carboxylic acid.

 

[DRHOYA]

Cr2O3 is AKA Jones reagent which will oxidize any primary alcohol to a carboxylic acid.

👍

 

[jdent]

Which question in the destroyer are you refering to?
PCC is a mild oxidizing reagent, while K2Cr2O7 and jones catalyst will oxidize to a carboxylic acid.

 

[War Eagle]

As a general rule of thumb, I would always go with a textbook over a study guide.

 

[gentile1225]

I agree, Cr2O3 will oxidize primary alcohols to carboxylic acids. PCC will only oxidize to aldehydes in the case of primary alcohols, and ketones in the case of secondary alcohols. Tertiary alcohols can not be oxidized by either reagent.

 

[pistolpete007]

i think he is refering to the road map

 

[oral09]

Which question in the destroyer are you refering to?
PCC is a mild oxidizing reagent, while K2Cr2O7 and jones catalyst will oxidize to a carboxylic acid.

look at the road map...

 

[mddang]

Is it CrO3 with pyridine????

If it is, then it is similar to PCC and oxidizes primary alcohols to aldehydes.

 

[jdent]

I think that you are referring to road map number 1. From the start molecule treatment with PCC and K2Cr2O7 both result in the same thing. Yes your right it seems like they are both accomplishing the same thing but thats only here because its in a cyclic molecule. They are both being oxidized. If you look at other examples it is more clear. But your right, PCC is mild and product is aldehyde, while Cr2O7 will oxidize.