You are right about the alpha carbon relating to a carbonyl. Some textbooks will also use it to describe the carbon bearing the hydroxyl group as I did in the video.
Great video! I just have one question about the oxidation of an aldehyde, If oxidation mechanism consists of losing an alpha hydrogen and gaining an oxygen bond why don't you lose the hydrogen on the aldehyde when we oxidise it ? Thanks
After abstracting the hydrogen atom from the oxygen atom, your alpha carbon would've 5 bonds and that's not possible. So you've to abstract an alpha hydrogen that the alpha carbon obeys the octet rule.
Sir everything is good but in my material they gave that oxidation takes place with tertiary also and forms alkene by using copper.i need explanation about this
Someone correct me if I am wrong: When you put chromate anion together an alcohol, the chromate anion gets reduced and the alcohol gets oxidized, so the "original" chromate ion is consumed as a reactant, and "new" chromate anion is produced. This process will keep going util all or close to all of the "original" chromate ions are used up in most conditions.
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I also think this step will be less likely to occur, because the pka of the chromate ester should be much higher than the pka of water. Maybe it's an intramolecular proton shift.
No it's +4, part of the mechanism in the video is missing a curly arrow, HCrO3 has only one chromium-oxygen double bond so it's 1 x Cr=O, 1 x Cr-OH and 1 x Cr-O(-). The chromium (IV) species is itself capable of carrying out further oxidation or 1) disproportionating to Cr(V) and Cr(III) i.e. 2 Cr(V) --> Cr (III) + Cr (V), where Cr(V) can then also serve as an oxidant or 2) itself be oxidized by Cr(VI) i.e. Cr(VI) + Cr(IV) --> 2 Cr(V) where Cr(V) can....you get the picture :).
7:20 for Mechanism !
You are right about the alpha carbon relating to a carbonyl. Some textbooks will also use it to describe the carbon bearing the hydroxyl group as I did in the video.
Great video! I just have one question about the oxidation of an aldehyde, If oxidation mechanism consists of losing an alpha hydrogen and gaining an oxygen bond why don't you lose the hydrogen on the aldehyde when we oxidise it ? Thanks
After abstracting the hydrogen atom from the oxygen atom, your alpha carbon would've 5 bonds and that's not possible. So you've to abstract an alpha hydrogen that the alpha carbon obeys the octet rule.
Did u finish school yet?
Thank you for this. It was very clear and exactly what I needed!
Great Vid! What program did you use to write?!
Thanks so much for posting,it was really helpful..
i love this! love your way of thinking
Good explanation, Thank you sir.
How do the hydrogens give two electrons to the carbons if they only have one electron to give? Have I misinterpreted something?
Sir everything is good but in my material they gave that oxidation takes place with tertiary also and forms alkene by using copper.i need explanation about this
Someone correct me if I am wrong:
When you put chromate anion together an alcohol, the chromate anion gets reduced and the alcohol gets oxidized, so the "original" chromate ion is consumed as a reactant, and "new" chromate anion is produced. This process will keep going util all or close to all of the "original" chromate ions are used up in most conditions.
+Au Ail cromate is a kation with Oxidation state of 6+, in CrO4^2- ? if i remember right?
It also get reduced yes, into Cr3+.
now I understand... thanks
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How can water (weak base ) deprotonate the chromate ether? whats the pka pf the chromate ester?
I also think this step will be less likely to occur, because the pka of the chromate ester should be much higher than the pka of water. Maybe it's an intramolecular proton shift.
why is the water took the alpha hydrogen?
isnt the oxidation number of the chromium in HCrO3 5+ (2x O=-4 and 1x OH=-1)?
No it's +4, part of the mechanism in the video is missing a curly arrow, HCrO3 has only one chromium-oxygen double bond so it's 1 x Cr=O, 1 x Cr-OH and 1 x Cr-O(-). The chromium (IV) species is itself capable of carrying out further oxidation or 1) disproportionating to Cr(V) and Cr(III) i.e. 2 Cr(V) --> Cr (III) + Cr (V), where Cr(V) can then also serve as an oxidant or 2) itself be oxidized by Cr(VI) i.e. Cr(VI) + Cr(IV) --> 2 Cr(V) where Cr(V) can....you get the picture :).
Thank you, sir. Great help
Thnks
Thank you so so much
Wow thanks 🌹💙
thank youu sir
Excellent
What happened to the H bonded oxygen pls tell me
Too good
probably 'paint' with a digital pen thingie
you can learn anything #
Thank you sir