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I’ll have a think about them but I usually focus on an underlying teaching topic rather than a specific molecule. On a glance it looks like Buchwald-Hartwig couplings would be useful for these.

🙂 You’re welcome - glad you enjoyed

Thanks 🙂 yes - agreed in hindsight I should’ve at least said locally statically similar or something. I will bear this in mind for future videos

🙂 Thanks. A fiddly one to work out what I wanted to do with this topic.

🙂 Got more time back to pick up where I left off finally

🙂 Thanks. Am glad to have the time again to make more content

🙂 Thanks. It’s an area I’m hoping I can expand into for sure and hopefully spark some curiosity beyond the usual topics

🥳 Thanks. Too many work deadlines this year - feels good to be back and play with making these videos

Thanks 🙂 glad it was helpful

I think it’s correct. Note the it’s flipped over from the example in this paper with the aldehyde in the product on the right rather than the left - a rotation not a mirror image. macmillan.princeton.edu/wp-content/uploads/aldehyde-aldol.pdf

@@CasualChemistry oh, now I see that! Thank you🫶🏻

The main idea is there has to be a difference between the two sides of the carbonyl do differentiate between. The NEt3 is bulky enough but not too bulky to be able to get the H off the ethyl group but not the isopropyl group just on sterics. This is the dominant thing going on. I’m going to touch on this in my next video: another factor is the isopropyl group’s bulk favours the boron Lewis acid sitting on the ethyl group side. There’s also a Stereoelectronic effect that helps make the ethyl group’s protons a bit more acidic. These reactions are done at low temperatures when you need to be careful to select so small differences in activation energies are enough to get selectivity.

You’re welcome 🙂

It has its place this one, though I’ll admit I’ve only ever done one of these and it was out of desperation. If going via an alkene and hydrogenation, there will be a problem if there’s an alpha stereocentre that you want to keep pure.

Sulfur chemistry gets everywhere (!) 🙂

Sure - I’ll try and remember in future. It’s not so conventional with this sort of system but I get your point.

Thanks 🙂

Not really so high - the pKa of H2 is about 35, and you can go up to 50 with alkanes. H- has its 2 electrons in a quite low energy 1s orbital with a decent effective nuclear charge. NH3 is about 33. So the hydride and the sodamide anions are similar stability. Probably some reversibility in this specific reaction but then the final deprotonation is definitely irreversible and pushes the equilibrium to the product as the H2 gas released goes into the surroundings. I’ve heard people say the Na+ cation helps, but I don’t buy it.

The Neber reaction needs a stronger base to form an anion stepwise and also the specific oxime C=N geometry locked in (eg TsO away from a big R group). To be honest I don’t think you see the Neber reaction very often even if you form the oxime anion as the mechanism has a high activation energy. There are other pathways likely to compete with it, such as other types of nucleophilic attack. The Beckmann doesn’t require the anion formation

(Sorry for the slow reply - I’ve been doing my dissertation. Back on the videos now 🙂)

Well harsh as in not tolerant of many functional groups but there isn’t much going on in this molecule so should be fine. The DielsAlder with butadiene is problematic as the dienophile does not have an electron withdrawing group attached. So both needs a higher temp than usual, but worse there’s no way to stop the butadiene reacting further with the product alkenes, and then further again into a mixture. There’s also a risk that the specific alkyne starting material here would isomerise, and also the product into conjugation.

@@CasualChemistry ahhh I see, yo so how did you get into this? I'm in undergrad still and I want to work in org synthesis, any tips?

Sorry for slow reply - I was busy finishing my dissertation. Back on to the videos now 🙂 I’m afraid there’s not an easy answer here other than a lot of practice on my part. I find it easiest to try and get as much of the molecule as possible in the plane of the paper - and definitely keep all the reaction arrows in the plane too else the 3D gets messy. Playing with a tetrahedron model can be a good idea too so you can visualise that important shape in several ways. Eg both like a pyramid but also like a triangle with a pointy out bit from one perspective. And from another perspective like two lines joining two atoms, one behind the other, at right angles.

Will do - they’re in my plan. I’m back on the video making now as my dissertation is finished 🙂

Hi. So sorry for the slow reply - I caught caught up finishing my dissertation. Well one of my main reasons for making these videos is because of a lack of resources on this. The Baran med chem lecture notes on his website are pretty good for this sort of thing

@@CasualChemistry oh yes Baran is great. Good luck with the dissertation! I’m also writing and it is a big weight! Thanks ☺️

Thanks 🙂 If I’m totally honest, practice is by far the most effective way of getting good and try not to just memorise things. It’s more like learning maths I reckon - focus on methods and patterns between reactions. If you haven’t done so already, a solid understanding of qualitative molecular orbital theory is a really good way to unpack problems logically

🙂 Thanks

Yep - that’s what’s in the mechanism

Yes - so the order of steps needs to be considered carefully. Use of the nitrile avoids this issue in the forward synthesis

😊 Thanks. They’re from a lot of practice though

That’s the idea, yes

:) Glad it was helpful

That’s true but that just means that can happen faster. Here everything is reversible, and only when the epoxide can anything happen. When it does fragment, it can’t go backwards either then

@@CasualChemistry he said that,he had given some rules ,including this rule,if all these rules were not followed in exams he would not give marks.

I’m afraid that sound like a teacher who might not know the subject so well. Also, the epoxide is a strained system - the epoxide lone pairs are not pure sp3 anyway

If you mean 8 membered ring for the delocalised part replacing benzene, that probably won’t work. That ring would be anti aromatic and not planar. If you mean 8 membered ring on the right hand side, it would be much trickier to form as the angles do not work well for cyclising like that. You might just form a dimer with another molecule instead.

@@CasualChemistry to which extent,it works

I don’t think the 8 membered ring would work very well

At the start it’s just reversible up until the water is removed by the drying agent. Also, the nitrogen line pair is higher energy than the oxygen as it’s a less electronegative atom (lower effective nuclear charge)

@@CasualChemistry thank you

Thanks 🙂

Sure- there are other ways of getting the same product, but usually via fully pre-functionalising the pyridine at the nitrogen. Common methods included making the pyridinium cation by tosylation, or using a O-substituted pyridine N-oxide

Glad they’re helpful 🙂 I realised that there wasn’t much useful stuff out there freely available on these topics. There’s a big reactivity window so should be find. Using dilute (mild) acid with the enamine hydrolysis at room temperature. The N is much more basic than an O lone pair, and the C=N pi bond is less strong than a carbonyl bond. The ester hydrolysis would require a more aggressive set of conditions and probably at high temperature.

@@CasualChemistry thank you so much!

Hmm - well more of a Wurtz coupling. But you’d struggle to get just cross-coupling of two different aromatics that way if you want a generalisable reaction.

Glad you enjoyed the video! It was very deliberately to present this topic more coherently than you normally see in textbooks etc

True and that's what depicted in the content!@@CasualChemistry

🙂 Spoilers for quite a way after high school

🙂 Thanks. I’ve been working to trying to cover some range with my retrosynthesis videos as there weren’t many similar resources out there - hopefully you can find a useful mixture of intro to advanced level videos

😎 You’re welcome - glad it was useful

Yes - that’s exactly correct. The video hasn’t included the work up procedure involving acid as it’s not so conventional to do so. The choice of ethoxide throughout is also to ensure an equilibrium too, but I agree that product will be largely deprotonated until acid work up.

@@CasualChemistry I understand, if the product will be deprotonate until we add the acid at the end and we have the reaction between the ester and the OH- before that wouldn't be hard that the OH- attacks an already negatively charged molecule, I hope this makes sense, I am asking because we did the exact steps you outlined and we got a very yellowish-orange solution, I still don't have the yield but I was thinking that maybe we got side reactions because the product is described as white and i thought that maybe by neutralising the solution before adding the OH- could limit the side reactions. I have also considered that maybe I am over complicating it and it's just impurities from the reagents most likely the α,β unsaturated ketone. I would be curious to hear your opinion.

So it’s a very long time ago since I did something close to this but one thing I would say is with this sort of chemistry (and enolates almost in general) yellow/orange solutions are quite common. A small amount of impurity colour goes a very long way - probably something involving a conjugated system here such as a small amount of an enol tautomer. As in you still have the product but some fraction of it is sitting in a very coloured isomeric form - almost certainly cleanable by flushing through a plug of silica (best case) or by column chromatography, or with dimedone specifically, try a recrystallisation. There will inevitably be some by products to with this largely mix and heat sort of process too - different enolates can form in lesser extents through out, eg deprotonating the alpha beta unsaturated ketone and it doing a different aldol. But I think my gut instinct without any further data is to stick with it - you’ve probably got the right thing in good yield even if yellow - you probably need another purification step.

@@CasualChemistry Thank you very much! That makes a lot of sense. I will do that :)