Robinson Annulation
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- čas přidán 3. 01. 2015
- The Robinson Annulation is really neat, you guys. It's a Michael Addition followed by an intramolecular Aldol Condensation, and it makes a ring! Contain your excitement, you don't want a public disturbance violation.
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Studying the day before my mcat, and this was the easiest and quickest explanations I could find and understand. Thank you SO much!
How was the exam????
@@andy0695 killed the chemistry/physics section! can't speak so much for the rest LOL
This is wonderful. Thank you so much for your time!
I am a great fan of this reaction mechanism, I do believe the OH is eliminated after electrons are pushed to the carbonyl oxygen in that last step. Like an E1cB
David Wells yes indeed, the last part of an aldol condensation is an elimination, as shown here!
One of the best on CZcams. Thanks.
Professor Dave!!!!! I love your videos. They are extremely helpful, thank you for making them!!
+Mary Bass my pleasure! spread the word!
OMG thank you so much ! This video is very useful, and it's short, effecient. Thank you Prof
We got more information with less span of time....thank you very much prof.dave 🤗
You're honestly the greatest! Thank you :)
Thanks for making it so simplistic.
awesome teaching ...thanku so much professor u have helped me alot for my xams.
Thank you for that simple explanation !
Awesome tutorial! It is a very cool reaction indeed.
Thanks! This was super helpful
Brilliant! Thanks!
Thank you very much for your effort
Thank you very much for the good and clear effort :)
Robinson annulation? More like "This is some great knowledge and information!" Thanks again so much for making and sharing such amazing videos.
I have never left a comment on a video before but you are literally the best ever teacher on youtube and I wish you were my teacher-which you kind of are! Thank you so much for this!
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I have been studying through your amazing videos since 3 weeks!
My exam is on Monday and I hope to do well!
Wish me luckkk🤞🏻
I know it's been awhile, but how did your exam go?
Easy and good explanation
thank you sir.
you're a solid guy, Dave.
Thanks from Brasil. Great job, Professor.
Thank you Pro.
you're so amazing!
Really so helpful
Thank you professor 🤓
Just watched two of your videos. They are very good and well explained. I was looking for some explanation for some name reactions during studying for my organic chemistry II examn. The script was very bad but just spending five minutes on your video gave me all explanation i needed thanks!!
Just giving the links of your vids to the students would be better than reading the scrpit ;)
Thanks from Germany!
I know it's been a long time, but how did your exam go and how have your studies gone since then?
The exam went well, I got a pretty good mark in the oral exam.
But I havent got much contact with OC since then because I actually studied biochemistry, OC was just a subject we had to do
@@simonbedenbender3087 I'm glad it went well! How did the rest of biochem go, and where did you end up career or academic-wise?
2020 coronavirus pandemic-sees professor dave cough-NO
What is the overall reaction? I mean what do we write as the reagents for this rxn?
WHERE were your videos when I was in first year?? Your videos are so effective and your style of simplifying complicated concepts is beyond brilliant. Would it be possible for you to make some videos on retrosynthesis? and maybe the organic chemistry of palladium?
Thank you
momijidoll13 I will absolutely add retrosynthesis to the list of next topics! Thanks for watching!
For the step after the tautomerism step, would the base used not have to be an LDA or stronger base to give the kinetic/less substituted product?
T-5hr until my orgo exam and Dave out here saving my ass once again... not the hero I deserved, but the hero I needed.
JazaqAllah Sir
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How do you know which carbon to deprotonate (which enolate to form) during the intramolecular aldol condensation step?
Also, at the end, do we really have hydroxide as a leaving group? Not water?
+Darlina Liu good questions! first, it's absolutely true that you could enolize on either side of the carbonyl for the second step, but a four-membered ring is much less likely to form, so enolization on the inside of the carbonyl might just be reversed to reform the carbonyl until enolization occurs on the other side, at which point the ring might snap shut to get the six-membered ring. and yes hydroxide is a reasonable leaving group in strong basic conditions, if there's a bunch of hydroxide swimming around in solution one more won't hurt.
+Professor Dave Explains Thanks for the thoughtful replies, Professor Dave!! Your videos are so so helpful, wish I discovered them earlier!
Thanks sir
Very helpfull tnx, can u pls next time film tje whole rection when u are done so we can see it
I like your way in explanation Is there any video in this account about (syn) and (anti) or (Si/ Ri).
nice video sir
So this is kind of similar to a Dieckmann condensation being an intramolecular version of Claisen condensation?
Hi, sir. In regards to your mechanism at around 3:51 , I was wondering if there should be an arrow showing the movement of the bonding pair of electrons to the oxygen that will later have the negative charge?
+Alyaa Hakim ah yes, you're right, it seems i've omitted an arrow!
fascinating
In the first step the enolization happens on the more substituted carbon. Why isn't the reaction controlled by a kinetik mechanism in basic conditions? (why not an attac on less substituted carbon?)
Like I say in the video it's just because that proton is the most acidic! The conjugate base has resonance across two carbonyls.
J'adore merci beaucoup
4:35, isn't OH- typically a bad leaving group?
Sir, it's very useful learns organic chemistry mechanism sir. I have one question about the Robinson annulation reaction sir. That is what is the stereo chemistry behind in the last water molecule elimination step sir
hmm, no stereochemistry that i can think of, it's just an E2
beast.
nice sir
merci beaucoup
sir how do we know whether the product will form ring or not. why can't it remain linear sir.
sir I mean that how do we know for given two reagents, Robinson annulation will take place.
how come you use hydroxide as a base? i've been taught to always use sodium ethoxide for these mechanisms and michael addition itself
+its Skudzy hydroxide is the classic, that's what i typically see. methoxide or ethoxide are fine too. kind of arbitrary i think.
okay thank you for the answer!
Will you get a racemic micture? The methyl group can be a wedge or a dash correct?
oh good question, i suppose so, i would probably have to draw it out with chairs to know for sure
@@ProfessorDaveExplains Alright, I was thinking the enolate can attack the Michael acceptor from any face (re so si) so you'd get a racemic mixture.
well at that stage the molecule would be achiral because its meso, but the product is chiral, so i'm actually not positive if there is anything that would favor one isomer over another, there may be information in the chairs
@@ProfessorDaveExplains Yeah exactly, thanks!
Nice
Hey Prof. Dave. Me again. Pestering you by now probably. But, will be happy to donate by next month hopefully as I do deeply appreciate you answering my ridiculous questions. On the tauteromerization part (probably spelled it wrong) why does the OH grab that hydrogen on Carbon 4 shortly before the ring is made? I believe it's Carbon 4, it appears to be primary with 3 hydrogens. Anyway why that Carbon and not any on the other Carbons? Not surprised that curiosity hasn't killed me yet !
so it's the pi bond that grabs the hydroxyl proton, whereas the OH sigma bond because a CO pi bond. check out my tutorial on michael addition, i more thoroughly discuss tautomerization!
Thanks for the clear explanation! I am curious what the significance of your tattoo is?
it's the diagram from which the parsec is derived!
sei un mago
can you do a robinson annulation without a di-carbonyl reagent?
steven schulster I wouldn't think so, as you need one carbonyl to enolize for the first step, and the other to act as the electrophile in the second step.
Isnt OH a bad leaving group though?
in strongly basic conditions it's no problem
sir i love chemistry and also you
Why doesn't the Robinson Annulation always happen because the product of a michael addition creates two carbonyl groups that could react as such?
Eva C well it depends on the structure of the electrophile, it needs to be able to form a six-membered ring in the transition state of the second step, so if there is a phenyl or tert-butyl group alpha to the carbonyl it won't work. there could also be issues due to sterics. but fundamentally you are correct that many michael additions could result in subsequent robinson annulation.
If I pass my organic chemistry exam on Monday, I'll have you to thank!
I know it's been a while, but did you pass?
@@PunmasterSTP to be honest, I can't remember if it came up on whatever exam it was for, but I sit here waiting to start a PhD in polymer chemistry in April so I can say the video is at least good enough not to have you fail
@@dancingdodo2768 That's amazing, and that sounds like an awesome field to do research in! Is your ultimate goal to go into industry, stay in academia, or do something else?
@@PunmasterSTP the PhD is sponsored by the company I work for but afterwards, I don't really have a preference. I just want to do research. I haven't got the patience for the stupidity of commercialisation
@@dancingdodo2768 I gotcha. Please forgive my ignorance, but what type (or subtypes) of polymer chemistry are there? Which one(s) would you hope to conduct?
i think it doesnt stop there
hi, i really like your videos and i am having a hard time understanding Hofmann exhaustive methylation . Could you please upload a video explaining it's mechanism and everything.
+Kaustav Das nitrogen lone pair does SN2 on methyl iodide until no more methyls can be accommodated!
+Professor Dave Explains if the nitrogen lone pair is in resonance with the ring then what happens?
+Kaustav Das it can still be methylated, it will just probably be less kinetically favorable and also bear a formal positive charge in the product.
I gotta tell my university about you...
+Nickolai indeed you must! spread the good word.
Any chance you could post organic chem videos on phenols, amines to amide conversion and Mass spect? :)
Oh and you are absolutely wonderful. The text book is way to thick in its terminology to understand
+Nickolai i will add to the list! i definitely plan to do all forms of spectroscopy/spectrometry.
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Moves a bit fast