Retrosynthesis 12, PKI-166 - Organic Chemistry

Retrosynthetic analysis of PKI-166 - a EGFR tyrosine kinase inhibitor that inhibits pancreatic cancers. The retrosynthesis uses a cheap chiral pool starting material and involves the construction of the bicyclic nitrogen heterocycle from scratch.
#orgo #organicchemistry #ochem #synthesis #science
The retrosynthesis begins by separation of the aromatic component of the molecule from the single stereocentre at the C-N bond. This disconnection is sensible as an amine can be attached to an aromatic ring when it acts as the nucleophile in an SNAr substitution reaction. The amine nucleophile is actually cheap and readily available from the chiral pool as 1-phenylethylamine. 1-Phenylethylamine is obtained as a single enantiomer by chiral resolution with L-malic acid, which occurs naturally as part of the citric acid cycle and the Calvin cycle in biochemistry. A racemic mixture of 1-phenylethylamine is first synthesised from the reductive amination of acetophenone with ammonia. When this racemic mixture is treated with the single enantiomer of malic acid, two diastereomeric salts form. The salt formed with D-1-phenylethylamine crystallises out of solution, whereas the salt formed with L-1-phenylethylamine stays in solution allowing for easy physical separation. The malic acid salts can then be treated with a base to return the free amine and the malic acid resolving agent washed away. Single enantiomers of 1-phenylethylamine are themselves used frequently in organic synthesis as chiral resolving agents (for other chiral resolutions).
Turning to the nitrogen heterocycle, we can envisage creating each half - the pyrrole-like half and the pyridone-like half - separately by ring closing reactions. Firstly, the aryl chloride required for the SNAr substitution with 1-phenylethylamine must be disconnected first as it is reactive. These 2-chloropyridine type structures can be easily constructed from the parent pyridone using a deoxygenating reagent, such as phosphoryl chloride (POCl3). The 6-membered ring part of this nitrogen heterocycle can be disconnected between the two nitrogens and extracting the carbon which is of the same oxidation level of an amide. The bonding pattern can therefore be constructed by condensing a reagent such as formamide (HCONH2) or methyl formate (HCOOMe) between two nitrogen centres. This disconnection leaves behind a trisubstituted pyrrole (2,4,5-substituted). This type of pyrrole ring can be synthesised from a variety of methods exploiting either the inherent reactivity of the pyrrole heterocycle or by de novo construction of the aromatic heterocycle from a linear precursor, which is proposed in this video.
When the pyrrole ring is disconnected between the 1- (N) and 2-positions, the linear precursor for cyclisation will be a 1,4-diX (1,4-difunctionalised) carbonyl species, which will form the 5-membered heterocycle. 1,4-diX compounds are often synthesised by using Umpolung chemistry (reversed polarity). One possible option would be to use an alpha-halocarbonyl reagent that incorporates the C2 and C3 from the target pyrrole, and target it’s soft electrophilic centre with a soft nucleophile. An appropriate soft nucleophile would be based on a 1,3-dicarbonyl species (based on a malonate) which would also bring in the correct oxidation levels at both C5 and the branched position coming off C4. An anion of a 1,3-dicarbonyl species would form the soft nucleophile at the position that would become the C4 position in the target pyrrole.
The alpha-halo carbonyl required for the Umpolung chemistry above can be simply constructed by monobromination, for example, of the 4’-methoxy-acetophenone (4-acetylanisole). This is compound is pretty cheap in itself, but could be synthesised by Friedel-Crafts acylation of anisole. The standard nucleophilic reactivity of a benzene ring bearing a pi electron donating group would ensure that the major product of such a Friedel-Crafts acylation would be substitution in the para position.

[@eddie9350]

Just discovered this channel. Cannot believe this is free content :)

[@CasualChemistry]

Glad you enjoy the videos 🙂 It’s a hobby for me and, as a teacher, I’m glad people find them useful

[@CasualChemistry]

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

[@katiefrisk980]

“Thank you for your help, Mario! But your princess is in another heterocycle!”

[@CasualChemistry]

lol 🤣 There's always more heterocycles - they get everywhere!

[@theonearney205]

Absolutely love all your retrosynth videos!

[@CasualChemistry]

Thanks! More in the pipeline. I'm trying to make sure there's some new topic in each of them.

[@eastofthegreenline3324]

The additional info on stereochemistry is helpful. Nice series, thanks!

[@CasualChemistry]

You’re welcome 🙂

[@joshg.6536]

Finally new content here :) very helpful retrosynthesis!

[@CasualChemistry]

Glad you enjoyed the new video. Life stuff got in the way earlier in the year so back on to this now

[@grgvv]

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

[@CasualChemistry]

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.

[@crabcrab2024]

Thank you so much! Great video (as usual).

[@CasualChemistry]

You’re welcome 🙂 Glad you enjoyed the video

[@user-salim77]

More one dream in mind: to be ur student:)

[@CasualChemistry]

🙂 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

[@giovannipelissero1886]

Love your content. Liked and subscribed. Keep it up!

[@CasualChemistry]

Thanks for the feedback - much appreciated :) Glad you enjoy the videos

[@That_Chemist]

If you want any tips/input on your videos, feel free to reach out to me on Discord :)

[@CasualChemistry]

That would be cool for some feedback. I’ll contact you on Discord 🙂

[@ammarhaider5453]

Thank You so much 😍 Stay Blessed 🤩

[@CasualChemistry]

Thanks 🙂 Hope the video was helpful

[@ammarhaider5453]

@@CasualChemistry Too helpful ❤️

[@andrewjin6618]

0:34 chiral resolution go brrrr

[@CasualChemistry]

🤣 definitely the best option here. Check out the prices in Sigma Aldrich

[@David-fh6rr]

Why not directly transesterify the Iminoester to the amidine?

[@CasualChemistry]

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.

[@Shiro-bp6hb]

4:55 i think the mechanism here is sn2 not addition-elimination

[@CasualChemistry]

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.

[@konstantinkonstantinov7078]

Actually in SnAr methoxy groups are good leaving groups when treated with amines (like aminolysis of esters)

[@CasualChemistry]

Definitely alternative options available. I figure you’d pick based on the scale and other factors like cost if doing it for real.

[@fritzcat1174]

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!

[@CasualChemistry]

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

[@niyudi]

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?

[@CasualChemistry]

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.

[@niyudi]

Thanks for the swift response, great video :)

[@CasualChemistry]

You’re welcome 🙂

[@bradduff3279]

Might be being a bit thick, but on your chlorination step at the end, wouldn’t you effectively get both methoxy groups being substituted?

[@CasualChemistry]

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.

[@anthonytam1803]

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

[@CasualChemistry]

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.