Man you are the coolest chemistry teacher I have seen in my 4 years of bachelor degree

Outstanding explanation

Thank you so much for your content! Due to Corona virus there were no lectures in organic chemistry at my university and you're helping so much to learn the basics! Thank you!

Thank you professor !

Truly incredible

Excellent video.

I'm in graduation and watching his channel I've been watching him for years now

thank you!

thanks🙏🙏

You are awesome. I'm from India and your language is so clear..one more reason to love Chemistry ☺

@Alchemist_171

4 жыл бұрын

Lol

@Alchemist_171

4 жыл бұрын

go home

A brilliant explanation. Thank you so much 😊,

highly appreciated

@sniperammow4865

5 жыл бұрын

Safwan Hassan lol

amazing

Lol we are doing this is class today, I am ahead boi!

Good

Hi! What would happen if we formed a double enolate? Which site would react?

Why is the thermodynamic enolate more stable? What are the effects of the methyl group, what are they actually stabilizing?

@ProfessorDaveExplains

4 жыл бұрын

more substituted alkenes are more stable in general, hyperconjugation

@edonSG

4 жыл бұрын

@@ProfessorDaveExplains thanks for your quick answer. I still dont get it, as far as i know hyperconjugation means the alkyl Groups are able to push electrones to a carbon that has electrone deficite, but in this case the negative oxygen is already pushing negative charge to the double bound that has a high electrone density. So why should that be stable? If it was a carbocation then it would make sense, propably im not getting where a carbocation is being formed here?

@ProfessorDaveExplains

4 жыл бұрын

No carbocation, the same principle applies to any sp2 carbon.

Sir in case for thermodynamically controlled reaction the two possible enolates have approximately the same free energy of activation as you said , so if the free energy of activation is same then both have equal rates . And if both have equal rates then there is no way that one is major and the other is minor.

@dang.5258

2 жыл бұрын

Theoretically yes, but you are missing the Alkene stability. Hydroxide can de-protonate forming an Alkene that has 4 groups and the other 3. Therefore the one with four groups will form in higher percent.

What if NaOEt

@ProfessorDaveExplains

4 жыл бұрын

ethoxide is still fairly sterically unhindered, once you introduce branching you see steric effects

Hi

Hii

I want to marry Professor Dave. T-T