Love your videos. I have one concern about the Benzyl protecting, usually the secondary amine will be benzylated as well, therefore i would use TBDMS or other silyl PGs
Thanks for the feedback 🙂 more videos on the way when my in-person teaching load drops down a bit. This is a fair comment and definitely a workable solution in practice. Given how early in the synthesis this benzylation step is I’d definitely give my proposal a go to see the results, with NaH in excess and then setting the BnBr as the limiting reagent (added slowly too) to maximise yield of alkoxide substitution.
Agreed. It's a useful little fragment to have in your back pocket for control and not getting yourself stuck with C=C isomers that don't head to your intended product.
@@CasualChemistry Yeah! It's funny, I'm actually in the process of transposing an isopropylidene malonate to a prenyl type nucleophile, so cool to see related chem.
:) Definitely will be keeping this going but it's as a hobby alongside my main job. I've got quite a lot of experience in organic chemistry laboratories and also I've been teaching undergraduate/postgraduate organic chemistry for a long time now.
planning a medchem viva soon, anyone know any good resources for example questions for retrosyntheses (+ w answers ina ddiiton to these amazing videos ;)) many thanks!!!
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
Can you explain how the reaction for a e1cb elimination of a methoxy works, as in step 2 to 3 in the synthesis of minoxidil? How is a methoxy a favorable leaving group? Am I just missing something like heat to drive the reaction? en.wikipedia.org/wiki/File:Minoxidil_synthesis.png
I think this is all about the context of the reaction. I'd agree that methoxide is never observed as a leaving group in SN1, SN2, E1 or E2, and this is mainly due to the thermodynamic/kinetic driving forces for each of those reactions. It is a leaving group in E1CB and addition-elimination (e.g. saponification of an ester). For E1CB, the rate-determining step is the one with the leaving group next to the enolate, or equivalent, which importantly here has a negative charge itself. The stability of the charge is actually better on the methoxide than when it's on the enolate - you can see this from the pKaH values (15ish for methoxide, 20ish for enolates). This helps contribute nicely to the enthalpy driving force, delta H for just this alone would be negative. Delta S for that step is also favourable (positive) purely from the fact that the step is one species going to two. It's sort of intramolecular in this respect which also lowers the activation energy barrier for the process as a kinetic effect (not present in SN2 for example). There is another major favourable enthalpy change (negative) coming from the formation of the very strong C=O bond as part of the step, and more than that it forms a delocalised system too in the product. So basically there are a load of extra driving forces that make the delta G favourable and also contribute to there being a lower activation energy for the E1CB mechanism. Another factor is that, say in the example you gave in minoxidil, the methoxide leaving group, after it's left it can't add back in by conjugate addition so easily - usually explained by it being too hard a nucleophile for a soft electrophile. The amine nucleophile a hard/soft borderline nucleophile and so in competition with the methoxide, it will always do conjugate addition in preference.
@@CasualChemistry I don't know why my reply keeps getting deleted, but thanks for the explanation, helps a bunch. Do you happen to have a full reaction mechanism for this, as with step 2-3 with minoxidil? Only paper I found with a quick search still left me empty handed. Advanced Synthesis & CatalysisVolume 361, Issue 23 p. 5284-5304 Review "Cyclisation Reactions Involving Alkyl Enol Ethers"
@@olefiend The nitrile is behaving in the same way as a carbonyl. So this system is an alpha,beta-unsaturated system which is electrophilic (soft) at the beta carbon - the one that starts with the OMe attached. So send in the amine nucleophile doing a Michael addition first and delocalising electrons up into the nitrile N. The charge then kicks back down doing an E1CB. Next the hydroxylamine acts as a nucleophile - the N lone pair is the HOMO as it's higher in energy than the oxygen. This is doing a direct (and probably reversible) addition to the nitrile carbon.
@@CasualChemistry So simple! Thanks. Looking at it doesn't make sense at first, but thinking of the nitrile as an electron reservoir like with carbonyls... duh! Thanks again! Edit: Ran into the same reaction with Pyrimethamine. This is no longer bugging me. en.wikipedia.org/wiki/File:Pyrimethamine_traditional_synthesis.png
This channel is underrated!
Thanks for the feedback 🙂 please do share some links about if you know people who might be interested in my content
Please continue to make videos! Loving the casual chemistry ⚗
Many thanks 😃 will do!
Amazing work
😀 Thanks for the feedback. I’m glad you enjoyed the video
Love your videos.
I have one concern about the Benzyl protecting, usually the secondary amine will be benzylated as well, therefore i would use TBDMS or other silyl PGs
Thanks for the feedback 🙂 more videos on the way when my in-person teaching load drops down a bit. This is a fair comment and definitely a workable solution in practice. Given how early in the synthesis this benzylation step is I’d definitely give my proposal a go to see the results, with NaH in excess and then setting the BnBr as the limiting reagent (added slowly too) to maximise yield of alkoxide substitution.
Becaude alcoxyde is best nucleophylic than amine, truth? Cause his negative charge
Malonate alkylidene conj. addition is a neat workaround. The vinyligous amide should be deactivated enough to avoid double addition.
Agreed. It's a useful little fragment to have in your back pocket for control and not getting yourself stuck with C=C isomers that don't head to your intended product.
@@CasualChemistry Yeah! It's funny, I'm actually in the process of transposing an isopropylidene malonate to a prenyl type nucleophile, so cool to see related chem.
:)
I have a Question how do u understand this you are best prof continue
:) Definitely will be keeping this going but it's as a hobby alongside my main job. I've got quite a lot of experience in organic chemistry laboratories and also I've been teaching undergraduate/postgraduate organic chemistry for a long time now.
planning a medchem viva soon, anyone know any good resources for example questions for retrosyntheses (+ w answers ina ddiiton to these amazing videos ;)) many thanks!!!
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 ☺️
Do you believe this pharmaceutical agent was effective for ...what's been in the news a lot since 2020?
I don't feel qualified to comment on the results of clinical trials as I'm not a biologist/medic
Can you explain how the reaction for a e1cb elimination of a methoxy works, as in step 2 to 3 in the synthesis of minoxidil? How is a methoxy a favorable leaving group? Am I just missing something like heat to drive the reaction?
en.wikipedia.org/wiki/File:Minoxidil_synthesis.png
I think this is all about the context of the reaction. I'd agree that methoxide is never observed as a leaving group in SN1, SN2, E1 or E2, and this is mainly due to the thermodynamic/kinetic driving forces for each of those reactions. It is a leaving group in E1CB and addition-elimination (e.g. saponification of an ester).
For E1CB, the rate-determining step is the one with the leaving group next to the enolate, or equivalent, which importantly here has a negative charge itself. The stability of the charge is actually better on the methoxide than when it's on the enolate - you can see this from the pKaH values (15ish for methoxide, 20ish for enolates). This helps contribute nicely to the enthalpy driving force, delta H for just this alone would be negative. Delta S for that step is also favourable (positive) purely from the fact that the step is one species going to two. It's sort of intramolecular in this respect which also lowers the activation energy barrier for the process as a kinetic effect (not present in SN2 for example).
There is another major favourable enthalpy change (negative) coming from the formation of the very strong C=O bond as part of the step, and more than that it forms a delocalised system too in the product.
So basically there are a load of extra driving forces that make the delta G favourable and also contribute to there being a lower activation energy for the E1CB mechanism. Another factor is that, say in the example you gave in minoxidil, the methoxide leaving group, after it's left it can't add back in by conjugate addition so easily - usually explained by it being too hard a nucleophile for a soft electrophile. The amine nucleophile a hard/soft borderline nucleophile and so in competition with the methoxide, it will always do conjugate addition in preference.
@@CasualChemistry I don't know why my reply keeps getting deleted, but thanks for the explanation, helps a bunch. Do you happen to have a full reaction mechanism for this, as with step 2-3 with minoxidil? Only paper I found with a quick search still left me empty handed.
Advanced Synthesis & CatalysisVolume 361, Issue 23 p. 5284-5304
Review
"Cyclisation Reactions Involving Alkyl Enol Ethers"
@@olefiend The nitrile is behaving in the same way as a carbonyl. So this system is an alpha,beta-unsaturated system which is electrophilic (soft) at the beta carbon - the one that starts with the OMe attached. So send in the amine nucleophile doing a Michael addition first and delocalising electrons up into the nitrile N. The charge then kicks back down doing an E1CB.
Next the hydroxylamine acts as a nucleophile - the N lone pair is the HOMO as it's higher in energy than the oxygen. This is doing a direct (and probably reversible) addition to the nitrile carbon.
@@CasualChemistry So simple! Thanks. Looking at it doesn't make sense at first, but thinking of the nitrile as an electron reservoir like with carbonyls... duh! Thanks again!
Edit: Ran into the same reaction with Pyrimethamine. This is no longer bugging me.
en.wikipedia.org/wiki/File:Pyrimethamine_traditional_synthesis.png