This was an incredible explanation; thank you so much for taking the time to make it, and then for putting it up on RUclips. 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!
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 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.
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
Thanks so much for your help! One thing, around 11:40 you said that stronger bases prefer to attack at the carbonyl carbon than the beta carbon. Isn't it that stronger nucleophiles prefer to attack at the carbonyl carbon than the beta carbon? Not the same thing, right, since weaker bases are stronger nucleophiles, and stronger bases are weaker nucleophiles? Thanks!
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)
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You are an incrediblly the best teacher and best explainer ever
This was an incredible explanation; thank you so much for taking the time to make it, and then for putting it up on RUclips. 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!
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...
You would get tons of support! Love from India!
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.
I don't think because that would form a 4 member ring which is less stable than a 6 member ring.
@@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.
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 always help so much!!! Thank youu 🙏❤️
Instablaster.
Insta I'd ?
so loving, every point well elaborated into detail.
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.
Thanks so much for your help! One thing, around 11:40 you said that stronger bases prefer to attack at the carbonyl carbon than the beta carbon. Isn't it that stronger nucleophiles prefer to attack at the carbonyl carbon than the beta carbon? Not the same thing, right, since weaker bases are stronger nucleophiles, and stronger bases are weaker nucleophiles? Thanks!
your videos are really helpful
Great explanation, thanks!
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?
Unfavorable due to geometry
@Jm Cresencio That's right. The dehydration of an aldol product forms a double bond in conjugation with the original carbonyl.
Thanks sir it's very helpful
what is different between intramolecular aldol and robinson annulation? this reaction reversibel or irreversibel?
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)
Thank you again!
Hey please if the C alpha between C=O and C-OH is not disponible to losean H which one can WE use
3:07 why does it not undergo aldol reaction in presence of OH- and 2 ketones ?
THANK YOUUU! I finally understand it
which is the alpha H in the keto in your first step
Is he saying micro addition or Michael addition? The caption keeps saying micro
michael
Thanks
Tyyy❤
Thanks man!
thank you
love you again
Does it have to form a 6 member ring?
Awesome
But isnt OH group not a good leaving group?
that's why it needs heat to overcome the activation energy
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