Why does substitution occur 2 carbons away from nitrogen? Remove if not allowed

[aced126]

This might be considered a synthesis question so remove if not allowed, but I'm trying to understand why a group substituted at the described position.

https://www.erowid.org/archive/rhodium/pdf/nichols/nichols-psilocin.pdf.

That link is Nichols' improvement on psilocybin. What I'm trying to understand is why, in the first step, the oxalyl chloride substitutes in the 2 position of the indole ring. There's already an oxygen group on carbon 7, which should increase reactivity at carbon-4 and to a lesser extent carbon-6 due to hindrance from the phenyl group. Also the phenyl group itself should undergo substitution a bit I would've thought. Yet the reaction proceeds at 77% at the carbon 3 position. Wikipedia says the carbon 3 position is 10^13 times more reactive than a benzene carbon. Why is this so?

Once again, please remove/edit if this is against the rules.

 

[adder]

This is due to the pyrrole-like nitrogen atom. Heteroarenes like furan or pyrrole are much much more reactive than benzene, so when you fuse them into a benzene ring, the heteroatom being part of the aromatic system will have the dominating directing effect. Position 3 in indole is the most reactive position towards electrophilic substitution, if you draw resonance structures, you will see that electrophilic substitution at C3 is extremely favoured (the resonance structure with formal negative charge at C3 has the other ring in tact). As you can see in the paper, the substituted indole reacts with oxalyl chloride without a Lewis acid catalyst while substituted benzenes with electron-donating groups still need catalysis for acylation.

 

[aced126]

Thanks. What makes the heteroarenes more reactive than benzene? Having some trouble trying to draw out the resonance structures as well. When I draw an electrophile attached to the 2 carbon and a carbocation on the 3 carbon, I find that a tertiary carbocation forms at carbon 8, along with carbocations at carbon 6 and 4. This seems like 2 substitution has pretty stable resonance forms. When an electrophile is attached to the 3 carbon and a carbocation on the 2 carbon, the only resonance structure I can draw is the aromatic nitrogen electrons forming a double bond and the nitrogen gaining a formal charge of +1. Is this sole resonance form more stable than all the other carbon resonance forms? If this is so then it makes sense...

 

[adder]

Heteroarenes can be more or less reactive than benzene, it depends whether the heteroatom is electron-donating or electron-withdrawing and these properties arise from the difference in electronegativity between heteroatoms and carbon atoms. In pyridine for instance the nitrogen atom is electron-withdrawing and withdraws electron density from the ring thus making it less reactive towards electrophilic substitution. On the other hand in 5-membered heteroarenes like furan, thiophene, and pyrrole the heteroatom is donating electron density to the ring and thus makes it more reactive towards electrophilic substitution. It all can be easily explained with resonance structures so you should definitely look into resonance and induction as they're very useful, well, actually necessary to understand a lot of stuff. Generally speaking, heteroarenes containing nitrogen atoms can have pyridine-like or pyrrole-like nitrogen atoms, in indole just like in pyrrole the nitrogen atom is electron-donating.

Pyridine - Organic Chemistry by Clayden

Five-membered heterocycles - Organic Chemistry by Clayden

Benzo-fused heterocycles - Organic Chemistry by Clayden

As you can see in the last link, electrophilic attack at C2 disrupts the other aromatic ring while attack at C3 has the other ring left in tact, so it is favoured. The positive charge will mostly lie on the nitrogen atom as you've already figured out yourself.

Reactivity of various aromatic compounds can be also explained by the resonance energy that must be lost during electrophilic aromatic substitution. A good example is naphthalene compared to benzene, the former has a resonance energy of 61 kcal/mol, and the latter only 36 kcal/mol. When benzene undergoes electrophilic substitution, aromaticity is lost when sigma complex is formed and resonance energy of 36 kcal/mol is lost. However, when naphthalene undergoes electrophilic substitution, it only loses 61-36 kcal/mol, i.e. 25 kcal/mol (the value differs in different textbooks, but the point is it's lower for naphthalene than it is for benzene), the aromaticity of only one ring of naphthalene is lost as sigma complex is formed, the other ring is in tact, so the resonance energy of 36 kcal/mol is retained. Naphthalene is thus more reactive towards electrophilic aromatic substitution than benzene because it loses less resonance energy during sigma complex formation. Of course different factors come into play with more complex compounds and hence you can't really use resonance energy to fully explain differences in reactivity, pyridine for instance has a lower resonance energy than benzene, but it is strongly deactivated and will often react with electrophiles at the nitrogen atom.

BTW, Organic Chemistry by Clayden is a perfect textbook to get into more advanced organic chemistry. It contains all the basics you need to get into organic chemistry and then builds on them to expand your knowledge. I definitely recommend it to you and if you're really just starting, some more basic textbooks by Wade or McMurry may be useful for you as well. I hope that helps.

 

[aced126]

Thanks a lot. I'll order the book. I did read a while ago a bit about aromaticity and how the lone pair on the nitrogen in pyridine is localised on the nitrogen, but the lone pair in pyrrole or indole needs to participate in aromaticity to satisy Huckel's law(?). It's annoying because I'm trying to understand all of this fully but it seems that I'd have to take a course in quantum physics to truly do so!

 

[adder]

You don't need to go deeply into quantum physics. It's simply the way valence electrons of the nitrogen atoms are arranged in pyrrole and pyridine. The Huckel's rule of 4n+2 pi electrons is related to the stability difference between the closed-ring and open-ring counterparts (e.g. benzene and hexane-1,3,5-triene, cyclobutadiene and buta-1,3-diene, and so on), if closed-ring counterpart is the more stable one, then it is aromatic, if it is less stable, then it is antiaromatic, if there is no difference, then it's simply non-aromatic. And Huckel noticed that in pairs with closed-ring counterparts being more stable, those counterparts followed the rule of 4n+2 pi electrons, while in pairs with closed-ring counterparts being less table, the rule was 4n pi electrons (another requirement for aromaticity/antiaromaticity is planarity so there are exceptions, basically all the other requirements should be checked before checking if the Huckel's rule applies). For one-ring aromatic compounds n=1, so there must be 6 pi electrons in the pi system.

Both in pyrrole and pyridine the nitrogen atom is sp2-hybridized (which is actually kind of an exception), so there are three hybrid orbitals and one unhybridized p orbital. In pyrrole hybrids are used to bind with 2 carbon atoms and 1 hydrogen atom, these bonds are parallel to the plane of the ring, the remaining one p orbital with 2 electrons is part of the pi system of the ring (6 pi electrons in total, 4 from 4 carbon atoms, 2 from the lone pair of the nitrogen atom; the pi system is always made of electrons in unhybridized orbitals). In pyridine 2 hybrids take part in bonds with neighbouring carbon atoms, 1 hybrid is the lone pair parallel to the plane of the ring, and 1 unhybridized p orbital with 1 electron is part of the pi system (6 pi electrons in total, 5 from 5 carbon atoms, 1 from nitrogen atom). Basically it's good to first draw an aromatic compound with sigma bonds only and then think what is going on with the rest of electrons, Kekule formulas aren't really the best depictions of what happens in reality, but nonetheless they are very useful for drawing mechanisms.

 

[Dresden]

It's because carbon #3 of indole is adjacent to the electron withdrawing benzene ring.

 

[blowpipe]

An aryl group is not electron-withdrawing. Following that logic and considering that what happens is electrophilic aromatic substitution, indole being the nucleophile, you'd have to say that C-3 is the most nucleophilic carbon, meaning that the benzene ring is rather electron-donating. I think it's already been said that the reason substitution occurs at C-3 is because of the resonance structures that favor that carbon the most (read: the least disfavorable when thinking about aromaticity of the intermediate).

 

[Dresden]

Oh yeah, sorry, I may have got the electron withdrawing versus electron donating thing backwards, but it is really just a moot point, as the reason substitution preferentially occurs at C #3 of indole is still because that carbon is immediately adjacent to the benzene ring, like I said before.

 

[atara]

http://en.wikipedia.org/wiki/Vinylogy

 

[Hammilton]

This thread isn't about old ways to listen to music, stay on topic Atara.