Formation of imines and enamines | Aldehydes and ketones | Organic chemistry | Khan Academy

[@ujjwalrawat5105]

Couldn't find a better teacher!

[@shoaib456]

The person that makes the video is really good explaining concepts, I have been watching a lot your videos for my Ochem class and they are really helpful. Thank you so much.

[@brianm.4368]

How can you justify a direct nucleophilic attack on the carbonyl carbon without protonating the oxygen first? From what I can see, this reaction shouldn't meaningfully progress once the pH in solution eclipses 6-7 (optimum pH = 4-6).

[@rachaelmwaura1464]

Thank you sooo much ...❤🥺🥺🥺

[@almiraalmira1451]

Makes a lot of sense, thank you so much!

[@alexmartinez3683]

wow, thank you

[@RockbandDrummer321]

Beyond helpful videos

[@zeyrania]

You saved my ass in organic chemistry, thanks a lot

[@miriamramos8873]

Thank you for all your help.

[@arianacisneros1477]

holy stepwise mechanism, batman!

[@afc90501]

thank you so much!

[@ariadinjones481]

Hey, was wondering why u kept using the lone electrons on your lsat 2 examples instead of HSO4- Wouldn't the O on the SO4- molecule be a better place to take the H+

[@tahani91kh80]

What's affecting on reductive elimination

[@edis9493]

How come oxygen cannot just take the proton from the first amine that attacked rather than taking one from an amine floating in solution... steric reasons? 9:15

[@user-gx7em6dd8n]

At 5:31, why does acid protonate the OH rather than the N of the carbinolamine? My thinking is that NH3 is a stronger base than OH, so NH3 would be protonated over OH. Can anyone help me reason where I'm going wrong please?

[@ujjwalrawat5105]

Amanda Wong it's just being done to explain the product, giving mechanisms is nothing but an estimated guess as to what really the compounds would be undergoing in order to form the final product. In case acid protonates the N of carbinolamine, the final product would be different. It will happen though in nature, but this fact can't be used to explain the product formation here. Hope you get it.

[@gmcenroe]

Remember the principle of microscopic reversibility. All of these steps are in equilibrium, so yes you could reprotonate the N but then you would be back to the previous intermediate rather then going forward to the desired product. That is why in practice you want to remove water from the reaction mixture for instance by azeotropic distillation which drives the reaction to completion.