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Pls make a series pertaining to JEE Advanced Organic Chemistry, as I absolutely adore the way that you make concepts in organic chem so simple...

@prakhyatpandey5341

Ай бұрын

You would get tons of support! Love from India!

This was an incredible explanation; thank you so much for taking the time to make it, and then for putting it up on KZread. I studied the Michael addition and the Robinson annulation before, but I've certainly forgot some of the finer points, like the effect of basicity on the type of reaction. For a lot of reasons, I had a great time watching your video!

Thank you for the videos you make! They are always super helpful. I always understand whatever I came for when I click on your videos.

so loving, every point well elaborated into detail.

You are an incrediblly the best teacher and best explainer ever

Great explanation, thanks!

Your videos always help so much!!! Thank youu 🙏❤️

@millerzion6863

2 жыл бұрын

Instablaster.

@HasanKhan-qz5uq

2 жыл бұрын

Insta I'd ?

Because of this video I was able to understand this concept and apply it to other problems thank you!!

15:33 The minor product is the one shown, the major product would be deprotonation of the alpha-carbon to the right of the ketone on the top left, as this would result in the most highly substituted and therefore stable double bond.

@zacharyelfallah1070

3 жыл бұрын

I don't think because that would form a 4 member ring which is less stable than a 6 member ring.

@PunmasterSTP

3 жыл бұрын

@@zacharyelfallah1070 I think Famhe is referencing the formation of the final alpha-beta unsaturated ketone, and I think his reasoning is correct. I'm not sure how much steric factors would come into play (i.e. the strain produced by having the double bond in the side where the two rings are fused vs. having it between an alpha carbon and a carbon shared by both rings) but I think the argument about substitution is valid.

@kendalldoer5466

2 жыл бұрын

Yeah @samter steric factors should not matter because we're using KOH which is not bulky, so I think Famhe is right, the major product would be the other double bond because it's more stable to deprotonate the top alpha proton

your videos are really helpful

THANK YOUUU! I finally understand it

Thank you again!

Thanks sir it's very helpful

which is the alpha H in the keto in your first step

Hey please if the C alpha between C=O and C-OH is not disponible to losean H which one can WE use

thank you

Thanks man!

Could the last OH- have taken the alpha H next to the other carbonyl group, and formed a double bond between the two rings instead?

@ateata7854

4 жыл бұрын

Unfavorable due to geometry

@infernape716

4 жыл бұрын

@Jm Cresencio That's right. The dehydration of an aldol product forms a double bond in conjugation with the original carbonyl.

3:07 why does it not undergo aldol reaction in presence of OH- and 2 ketones ?

Does it have to form a 6 member ring?

love you again

Thanks

what is different between intramolecular aldol and robinson annulation? this reaction reversibel or irreversibel?

@bruno0_u

2 жыл бұрын

Not sure about reversability but the intramolecular Aldol is just the second step of the Robinson annulation. Robinson Annulation is just 1) Michael Addition (α, β unsaturated ketone) followed by 2) Intermolecular Aldol (1,2 direct)

Is he saying micro addition or Michael addition? The caption keeps saying micro

@tahj420

4 жыл бұрын

michael

Awesome

But isnt OH group not a good leaving group?

@RhysGreenable

2 жыл бұрын

that's why it needs heat to overcome the activation energy

@kendalldoer5466

2 жыл бұрын

It's a good enough leaving group if the rxn in being done in base (which it is). If it is being done in acidic conditions than you need C-OH2+

I swear all of these carbonyl reactions are so similar

yo michael; sheeetttyy