Wittig Reaction Mechanism

[@TheOrganicChemistryTutor]

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I never thought about the four-membered ring transition state in terms of electronegativity, but that makes perfect sense! Thanks again for sharing all of these wonderful explanations. You are doing ochem students everywhere a great service!

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[@OrganiceseOrgChem]

The Wittig is Z-selective. The Horner-Wadsworth-Emmons and other stabilized ylids with similar mechanisms give E-stereochemistry.

[@mariahthompson5789]

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[@ItisMe-n2u]

3:02 shouldn't E and Z be invalid here.... I guess we need to use cis and trans over here as there is CH3 present on both the carbons... so ig the left one is cis and right one is trans and not E and Z... correct me if i'm wrong

[@Rhopalocera1]

Question: What compounds are used to go from ph3P to ph3P=R? If they're alkyl halides then why do compounds with Li work? Please explain.

[@Tahazif_TheCool22]

Yes, alkyl halides are used to attach the alkyl group to Pph3 in presence of BuLi. The mechanism is as follows: You have Pph3 and RX with BuLi. The alkyl halide has polarity between it's bond and the phosphorus in Pph3 has lone pair which gets attracted with the partial positive polarity of alkyl in alkyl halide Pph3 + RX ---> (ph3)-P(+)-R The Phosphorus has made a bond with alkyl and it has positive charge in it. Now comes BuLi which deprotonates the alkyl and make a complex with Li. (ph3)-P(+)-R + BuLi -----> [(ph3)-P(+)-R(-)]Li The alkyl now has a negative charge. Remember the the positive and negative don't make any pie bond due to presence of Li. Now it is ready to attack alkyl ketone. This is what I learnt in my class. Hope you might find something new in it. It's hard to write structure here, you can draw them accordingly to understand better. Hope it helped!

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[@miriamcalvoortiz7140]

5:01 for the mechanism

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[@Shadow12aven]

Question, @7:45 shouldn't the resonance form of Ph3P have a positive charge under phosphorus since it has four bonds?

[@PunmasterSTP]

I think the phosphorus atom has five bonds, counting the sigma and pi bonds to the carbon atom separately. So the formal charge is #valence electrons - #bonds - 2*#lone pairs = 5 - 5 - 0 = 0.

[@ronmasse]

This mechanism is like a dance. I love it.

[@payalshukla983]

I am in class 11 th from India it is in our syllabus thank you

[@casualcasual1234]

Hello, at 10:00, when forming the cyclo compound, when do I know the electrons are transferred from 1 bond to another bond or from 1 bond to an atom e.g. Oxygen / Carbon? Is there ant general concept I should be known about? Thanks

[@tomatrix7525]

Tony Lee There are many different scenarios which control this, no single rule really, here though in your question, the two carbons are attracted because of their opposing electronegativoty and positivity. You also need to know carbon never has 5 bonds, so when this 5th bond is made, it must drop another to go back to 4, it chooses to break one of the bonds on oxygen because it tries to break a bond without loosing any atoms if possible, so if it broke any other bonds atoms would leave. Know this oxygen withdraws the electrons after the bond breakage, as it is more electronegative than carbon. This negative oxygen readily combines with the positive pph3, forming that structure.

[@himanshuangane2948]

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[@alizzyb]

cis product is preferred, per my organic chemistry instructor and the textbook.

[@aswathym2048]

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[@osamudiamenogbonmwan5201]

I notice you have p-c double bonds; why?

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Could you do a video of the baeyer williger reaction? 🥺

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[@aradhanakaushik413]

Hey sir I have a query let's suppose we have 2 unsymmetrical carbonyl gp. which will gonna react then?

[@gargitripathi6331]

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[@donnadong5837]

can you talk about the stabilized and non stabilized ylides?

[@hlongshlongs8729]

If you haven't found the help, search the video, Knowbe, "witting reactions". @11-12:30. It helped me. :)

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Why can’t the ylide carbon take away α-hydrogen from acetaldehyde?

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6:48

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[@maxleibrecht7484]

It’s not correct!!! Yes you get a mixture but in the example of using a alkyl-Ylide you will get >= 90% Z alkene because it’s a labile Ylide. 50/50 would be only if you use a Aromatic ring or system as the funktional group of the Ylide. For E alkene you have to use very stable Ylides with EWG funktional groups on the Ylide!!!! 10:59

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