Also please do share the channel around to others that might be interested. My aim is to try and put some different style of videos on Chemistry out into the education space
🙂 Only a small number of undergrads get that with my current job. Hoping this channel expands my teaching range - in fact, a big reason why I’m making these videos
Another possible synthesis (albeit not for teaching purposes) is: start from commercial 4,6-dichloropyridimine, add the phenethylamine by SNAr, add hydrazine by SNAr, then Fischer Indole synthesis with 4-OH-acetophenone
Very true - there are definitely alternative routes to this molecule depending on starting materials and general preferences. You'd have to be careful with your equivalents in the hydrazine step (and also be careful from a chemical hazard point of view using it (!)) to prevent forming an N-N linked dimer of your 6-rmembered ring heterocycles.
Great content! But I don't think that you will end up with the methoxy-pyrimidine after condensation with formamide (on your second last step) ; instead I assume you will have the hydroxy-pyrimidine / pyrimidone. As I said, I love this kind of educational content and this comment is just nitpicking ;-) Keep it up!
Agreed. I’m making a bit of a compromise there on what would be “steps” I guess but I thought would probably make things a bit more clear to people newer to heterocyclic chem. Thanks for the feedback though - it’s massively appreciated knowing that people enjoy the videos 🙂 more on the way
There are definitely alternative routes to the product here. The nitrile is probably more common in practice on an industrial scale for minimising purification steps later.
I think it’s safe. The SNAr is possible on the chlorine on the nitrogen heterocycle because the intermediate negative charge ends up on a N, which is stable because it’s electronegative. On the benzene, there’s no place to delocalise an intermediate negative charge too that would be stable, so the SNAr reaction can’t occur at on that ring as it is.
If I may ask a question as an undergrad, in the last substitutions on the aromatic structure, why does the OH/Cl substitution only occur on the nitrogen ring, and not on the phenol as well?
An SNAr reaction only works if you can delocalise the intermediate’s negative charge to a good place. Here that wouldn’t be possible on the all-carbon ring - normally you’d need a nitro substituent (or equivalent) to help forming a “Meisenhiemer intermediate”. Heterocycles with the N atom embedded can do this substitution chemistry really easily in contrast as it’s electronegative.
I’m not sure you would get selective pyrrole formation over other competing reactions in that step. For example, it is an enamine after all, so it’s nucleophilic at the beta carbon, not necessarily at the nitrogen lone pairs. But even assuming you found conditions to promote that reactivity, why not intramolecular attack of the amine into the methyl ester? Seems a bit messy
We’re definitely well into the land of thermodynamic control with heterocyclic chemistry like this. There are likely side mechanisms that just reverse because the path that wins out will head to the major thermodynamic sink of the aromatic product. Enamines are nucleophile as both at N and C - the HOMO’s electron density is mainly at the ends of the 3-centre pi system. In fact the coefficients will be larger at the N as it’s the more electronegative atom. So you can see reactivity at both ends provided they’re not already fully substituted. You often see enamines in chemistry that are fully saturated at N so they deliberately only go at carbon - say an enamine that is made from a secondary amine such as pyrrolidine or piperidine. In this video that’s not the case so reaction out of either end is ok. In the ring closure, the reaction out of the carbon isn’t so favourable as it would be 3-enolendo-exo-trig by Baldwin’s rules. If the 3-membered ring did form it would likely just reverse anyway to relieve strain. Granted there are other heterocycles that might form in much lower yield but the one drawn with the attack of the protonated nitrile is probably preferred kinetically, and also thermodynamically as the product will have a favourable N lone pair in the 2position of the purple set up for further delocalisation on top of the aromatic system.
I’m assuming we’re talking about the attack at the P atom: I’ve used what are commonly used conventional curly arrows for this that make a direct analogy to the carbonyl. However using arrows on 3rd row elements, particularly here with the +5 oxidation state, as the bonding in the POCl3 itself isn’t massively well represented by the also conventional way I’ve drawn its structure. I believe there is experimental evidence that the mechanism (in solution phase) isn’t really a “pure” SN2 or A-E, but as ever with chemistry there are shades of grey between the strict mechanism classes that work well for top row elements like carbon.
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The additional info on stereochemistry is helpful. Nice series, thanks!
You’re welcome 🙂
Just discovered this channel. Cannot believe this is free content :)
Glad you enjoy the videos 🙂 It’s a hobby for me and, as a teacher, I’m glad people find them useful
Also please do share the channel around to others that might be interested. My aim is to try and put some different style of videos on Chemistry out into the education space
“Thank you for your help, Mario! But your princess is in another heterocycle!”
lol 🤣 There's always more heterocycles - they get everywhere!
Absolutely love all your retrosynth videos!
Thanks! More in the pipeline. I'm trying to make sure there's some new topic in each of them.
If you want any tips/input on your videos, feel free to reach out to me on Discord :)
That would be cool for some feedback. I’ll contact you on Discord 🙂
Finally new content here :) very helpful retrosynthesis!
Glad you enjoyed the new video. Life stuff got in the way earlier in the year so back on to this now
More one dream in mind: to be ur student:)
🙂 Only a small number of undergrads get that with my current job. Hoping this channel expands my teaching range - in fact, a big reason why I’m making these videos
Another possible synthesis (albeit not for teaching purposes) is: start from commercial 4,6-dichloropyridimine, add the phenethylamine by SNAr, add hydrazine by SNAr, then Fischer Indole synthesis with 4-OH-acetophenone
Very true - there are definitely alternative routes to this molecule depending on starting materials and general preferences. You'd have to be careful with your equivalents in the hydrazine step (and also be careful from a chemical hazard point of view using it (!)) to prevent forming an N-N linked dimer of your 6-rmembered ring heterocycles.
Thank you so much! Great video (as usual).
You’re welcome 🙂 Glad you enjoyed the video
Thank You so much 😍
Stay Blessed 🤩
Thanks 🙂 Hope the video was helpful
@@CasualChemistry Too helpful ❤️
Love your content. Liked and subscribed.
Keep it up!
Thanks for the feedback - much appreciated :) Glad you enjoy the videos
0:34 chiral resolution go brrrr
🤣 definitely the best option here. Check out the prices in Sigma Aldrich
Actually in SnAr methoxy groups are good leaving groups when treated with amines (like aminolysis of esters)
Definitely alternative options available. I figure you’d pick based on the scale and other factors like cost if doing it for real.
Great content!
But I don't think that you will end up with the methoxy-pyrimidine after condensation with formamide (on your second last step) ; instead I assume you will have the hydroxy-pyrimidine / pyrimidone. As I said, I love this kind of educational content and this comment is just nitpicking ;-) Keep it up!
Agreed. I’m making a bit of a compromise there on what would be “steps” I guess but I thought would probably make things a bit more clear to people newer to heterocyclic chem. Thanks for the feedback though - it’s massively appreciated knowing that people enjoy the videos 🙂 more on the way
Why not directly transesterify the Iminoester to the amidine?
There are definitely alternative routes to the product here. The nitrile is probably more common in practice on an industrial scale for minimising purification steps later.
Might be being a bit thick, but on your chlorination step at the end, wouldn’t you effectively get both methoxy groups being substituted?
I think it’s safe. The SNAr is possible on the chlorine on the nitrogen heterocycle because the intermediate negative charge ends up on a N, which is stable because it’s electronegative. On the benzene, there’s no place to delocalise an intermediate negative charge too that would be stable, so the SNAr reaction can’t occur at on that ring as it is.
If I may ask a question as an undergrad, in the last substitutions on the aromatic structure, why does the OH/Cl substitution only occur on the nitrogen ring, and not on the phenol as well?
An SNAr reaction only works if you can delocalise the intermediate’s negative charge to a good place. Here that wouldn’t be possible on the all-carbon ring - normally you’d need a nitro substituent (or equivalent) to help forming a “Meisenhiemer intermediate”. Heterocycles with the N atom embedded can do this substitution chemistry really easily in contrast as it’s electronegative.
Thanks for the swift response, great video :)
You’re welcome 🙂
I’m not sure you would get selective pyrrole formation over other competing reactions in that step. For example, it is an enamine after all, so it’s nucleophilic at the beta carbon, not necessarily at the nitrogen lone pairs. But even assuming you found conditions to promote that reactivity, why not intramolecular attack of the amine into the methyl ester? Seems a bit messy
We’re definitely well into the land of thermodynamic control with heterocyclic chemistry like this. There are likely side mechanisms that just reverse because the path that wins out will head to the major thermodynamic sink of the aromatic product.
Enamines are nucleophile as both at N and C - the HOMO’s electron density is mainly at the ends of the 3-centre pi system. In fact the coefficients will be larger at the N as it’s the more electronegative atom. So you can see reactivity at both ends provided they’re not already fully substituted. You often see enamines in chemistry that are fully saturated at N so they deliberately only go at carbon - say an enamine that is made from a secondary amine such as pyrrolidine or piperidine. In this video that’s not the case so reaction out of either end is ok.
In the ring closure, the reaction out of the carbon isn’t so favourable as it would be 3-enolendo-exo-trig by Baldwin’s rules. If the 3-membered ring did form it would likely just reverse anyway to relieve strain.
Granted there are other heterocycles that might form in much lower yield but the one drawn with the attack of the protonated nitrile is probably preferred kinetically, and also thermodynamically as the product will have a favourable N lone pair in the 2position of the purple set up for further delocalisation on top of the aromatic system.
4:55 i think the mechanism here is sn2 not addition-elimination
I’m assuming we’re talking about the attack at the P atom: I’ve used what are commonly used conventional curly arrows for this that make a direct analogy to the carbonyl. However using arrows on 3rd row elements, particularly here with the +5 oxidation state, as the bonding in the POCl3 itself isn’t massively well represented by the also conventional way I’ve drawn its structure. I believe there is experimental evidence that the mechanism (in solution phase) isn’t really a “pure” SN2 or A-E, but as ever with chemistry there are shades of grey between the strict mechanism classes that work well for top row elements like carbon.