K2Cr2O7 oxidize double bond ??

[atlanta213]

Advertisement - Members don't see this ad

I just wonder K2Cr2O7 oxidize double bond to OH like KMnO4..

Also Does all strong oxidizing agent like KMNO4 oxidize aldehyde, alcohol, double bond( all three to COOH) and not ketone? right?

Thanks

 

[doc3232]

hmmmm, good question.
I will go with yes.
I think it will fully oxidize to ketones (for 2nd) and carbox acids for primary carbons.
It is like CrO3 and water and acid and also like HCrO4
All those chromates do the same thing

Someone should correct me if I am wrong. Not very sure.

 

[IDoWhatIWant]

I just wonder K2Cr2O7 oxidize double bond to OH like KMnO4..

Also Does all strong oxidizing agent like KMNO4 oxidize aldehyde, alcohol, double bond( all three to COOH) and not ketone? right?

Thanks

You mean FROM OH? Yeah.

 

[condoleezarice]

You should know KMnO4 can do a lot of different things to an alkene. Regular KMnO4 will just make an alkene an alcohol. Hot KMnO4 will cleave an alkene into a ketone OR carboxylic acid, depending what is coming off the carbonyl group(1 alkyl group->carboxylic acid, 2 alkyl groups->ketone). If there is only 1 alkyl group off the carbonyl carbon, it will first go to aldehyde, then oxidize further to the carboxylic acid.

I believe ketones can be oxidized with KMnO4, heat, and acid. It's tougher to do than oxidizing the rest, hence the heat.

 

[doc3232]

Summation of oxidizers
Chromates: oxidize fully.
Ozone followed by Zn and water will form ketones and aldehyes from double bonds
cold dilute KMnO4 turns double bonds to diols, one OH for each carbon in the double bond
Hot KMnO4 in base will oxidize fully
Can't remember the other oxidizer but I think all the rest oxidize fully.
Hope this helps

 

[atlanta213]

Thanks. I got everything you guys said except this..

so.. K2Cr2O7 oxidizes double bond to OH like KMnO4?

and K2Cr2O7 and KMnO4 oxidize aldehyde and ketone to COOH?

 

[doc3232]

It depends what kind of KMnO4 you use.
Whether dilute and cold
or hot and in basic conditions.