Hi Frank, These videos are good! As for the reason, in the UK we believe there is no well understood mechanism. We use a simple model assuming a perpendicular approach. The Oxaphosphatane is key. If we imagine a stickman with the head being the phosphorous of the ylid and the legs being the electronegative group of the ylid + Hydrogen. The left arm being the oxygen of the carbonyl and the right being a bulky R group and Hydrogen. The Electronegative group will be on the right leg during a stabilised reaction and away from the oxygen making the electronegative group above the plane during rotation to form the oxaphosphatane as the bulky R group will be point down away from the phosphorous, the electornegative group will be pointing up. However when it is unstabalised (no resonance) the R group on the ylid prefers to originate away from the bulky R group on the carbonyl thus it will be on the left leg and pointing down in the oxaphosphatane as the phosphorous always rotates to be next to the oxygen that it binds to in the transition state. O -|- /\ Here's your reason my friend :)
Hey Frank, just watched your video and it was really useful. I think E is produced with a stabilised ylide because the intermediate of the Z is reversible and E intermediate is not. So all the Z reverses back into the original reagents and reforms until only the E product is left.
The unstabilized ylide is more reactive due to a lack of resonance stabilization. Therefore, it will tend to undergo the fastest reaction possible (under kinetic control); this leads to the less stable (Z) stereoisomer. Contrarily, the stabilized ylide is less reactive due to resonance stabilization and can "hold out" longer for the slower reaction, leading to a more stable product (E stereoisomer) under thermodynamic control.
Hi Frank, I have a suggestion for your video. Whenever you finished writing mechanisms or notes on the board, you may want to leave like 3 seconds for viewers to copy.
I like the way u teaching . it's very easy to understand compare to others ppl. erm. do u understand the thermodynamics chapter of physic ? because I can't find any video of thermodynamics that are easy to understand about. @.@
Thanks Alice! :] I appreciate the nice feedback. I'll try to keep my videos nice and simple then haha. And as for Physics, perhaps this site can help you out a little. socratic.org/physics
I don’t think this is correct. Yes the resonance stabilization helps but it would still be a mixture of E and Z alkenes. To form >90% E alkenes you would need a EWG that makes the Ylide stable. With aromatic systems you would your Semi stabilize the Ylide. For Z it’s clear.
Dude you should work on drawing your structures and the words you use. You have drawn arrows from a low pair on carbon??? What lone pair???? I would advice students to steer clear of these lectures.
+Harry Shirley www.masterorganicchemistry.com/tips/the-wittig-reaction/ It is perfectly okay to draw arrows from a lone piar on carbon. That's how MOC does it, along with pretty much all the resources I've seen online as well. Unless you're saying you prefer to see a Phosphorus - Carbon double bond attack which I agree is totally fine as well since that's the resonance structure.
this is technically incorrect. the arrow should never show movement from the charge, it comes from the lone pair that provides the atom with the negative charge
Thank you so much. Nobody on youtube talks about Cis Trans for Wittig. Only you! Thanks!
Hi Frank,
These videos are good! As for the reason, in the UK we believe there is no well understood mechanism. We use a simple model assuming a perpendicular approach.
The Oxaphosphatane is key.
If we imagine a stickman with the head being the phosphorous of the ylid and the legs being the electronegative group of the ylid + Hydrogen. The left arm being the oxygen of the carbonyl and the right being a bulky R group and Hydrogen.
The Electronegative group will be on the right leg during a stabilised reaction and away from the oxygen making the electronegative group above the plane during rotation to form the oxaphosphatane as the bulky R group will be point down away from the phosphorous, the electornegative group will be pointing up.
However when it is unstabalised (no resonance) the R group on the ylid prefers to originate away from the bulky R group on the carbonyl thus it will be on the left leg and pointing down in the oxaphosphatane as the phosphorous always rotates to be next to the oxygen that it binds to in the transition state.
O
-|-
/\
Here's your reason my friend :)
Hey Frank, just watched your video and it was really useful.
I think E is produced with a stabilised ylide because the intermediate of the Z is reversible and E intermediate is not. So all the Z reverses back into the original reagents and reforms until only the E product is left.
luke hick Ahhhh that does sound like a reasonable cause behind it. Thanks for sharing Luke!
Love your energy and passion while explaining in your videos! Thank you so much!
The unstabilized ylide is more reactive due to a lack of resonance stabilization. Therefore, it will tend to undergo the fastest reaction possible (under kinetic control); this leads to the less stable (Z) stereoisomer. Contrarily, the stabilized ylide is less reactive due to resonance stabilization and can "hold out" longer for the slower reaction, leading to a more stable product (E stereoisomer) under thermodynamic control.
Sir you are amazing.I'll recommend your channel to all my friends prepararing for competitive exams.
no one ever could explain it better, thank you!
You’re very welcome! Happy Friday 🥳!
You are the real MVP MANN !!!
Thank you so so much!!! this is the best orgo video on youtube
Hi Frank, I have a suggestion for your video. Whenever you finished writing mechanisms or notes on the board, you may want to leave like 3 seconds for viewers to copy.
Your vids are exceptional! Thanks Frank!
Crystal clear! Thank you so much!!!
would the E alkenes be the major product in all cases because of sterics? what would explain the Z alkene being the major product?
hey!!! what about OMe group as in Pph3=CHOMe? will it be unstable through mesomeric and form Z alkene or stable by inductive to form E?
My professor dead told us the easiest way to remember this is Z = Zame Zide lmao its lame but you will never forget the difference xD
My prof says Z is for Zis (cis-)
Awesome
Real cool man, an awesome study aid. your the man, respect :)
Thanks! Haha really appreciate the positive feedback. Keep up the hard work with Orgo and I'm sure you'll do good.
F
Thank you!
You’re welcome! The Wittig Rxn is actually pretty easy peasy right? 😄 Where are you taking Orgo rn?
Super helpful thank you!!
I like the way u teaching . it's very easy to understand compare to others ppl. erm. do u understand the thermodynamics chapter of physic ? because I can't find any video of thermodynamics that are easy to understand about. @.@
Thanks Alice! :] I appreciate the nice feedback. I'll try to keep my videos nice and simple then haha. And as for Physics, perhaps this site can help you out a little. socratic.org/physics
Welcome^^ .seriously I always watch to ur video during exam week:) it's really very very very help me a lot>< thank u sooo muchhhhhhhhhhh
Mate, try to find it out from Clayden's 2nd edition
Sweet, thanks! *****
Nice
I don’t think this is correct. Yes the resonance stabilization helps but it would still be a mixture of E and Z alkenes. To form >90% E alkenes you would need a EWG that makes the Ylide stable. With aromatic systems you would your Semi stabilize the Ylide. For Z it’s clear.
Dude you should work on drawing your structures and the words you use. You have drawn arrows from a low pair on carbon??? What lone pair????
I would advice students to steer clear of these lectures.
+Harry Shirley www.masterorganicchemistry.com/tips/the-wittig-reaction/ It is perfectly okay to draw arrows from a lone piar on carbon. That's how MOC does it, along with pretty much all the resources I've seen online as well. Unless you're saying you prefer to see a Phosphorus - Carbon double bond attack which I agree is totally fine as well since that's the resonance structure.
this is technically incorrect. the arrow should never show movement from the charge, it comes from the lone pair that provides the atom with the negative charge