3
$\begingroup$

I am studying SNAr mechanism from this webpage. I am confused as to why this convoluted mechanism occurs. It seems as though we simply create an anion when the hydroxide attacks in order to quench it directly after by pushing out the chloride. Why does the hydroxide not push out the chloride in a typical substitution reaction? Also, what evidence do we have for the Meisenheimer complex actually forming?

Meisenheimer complex

$\endgroup$
2
$\begingroup$

Basic level explanation: SN2 occurs only at sp3 carbons, where you can get backside attack.

Intermediate: Addition to the double bond generates a highly stabilized anion. This is a lower energy transition state than the concerted SN2.

Advanced: The reaction occurs at the site of maximum frontier molecular orbital overlap. The HOMO (highest occupied molecular orbital, the hydroxide nucleophile) adds to the LUMO (lowest unoccupied molecular orbital, π* of the arene). To do SN2, one would need to interact with σ*, which is not the LUMO, and has an unreachable trajectory (points towards the middle of the ring).

$\endgroup$
0
$\begingroup$

Well, you do not typically find substitution reactions occurring at olefins. Think about it.

You would come up with a sort of situation where the 'leaving group' would be a lone pair of electrons (C-C $\pi$ electrons). This has been done with organometallic reagents like lithium organocuprates, involving oxidative addition and reductive elimination sequences. But that is typically not the nucleophilic substitution that organic chemists like to think about.

So, in this case, 'pushing out' the ipso chloride in an SN2 fashion is not an option. In fact, the ipso position will often have a fluoride to stabilize the Meisenheimer complex.

And because this is an equilibrium process, you often find that the reaction is being driven forward by some means, because a lot of the intermediates are pushed back to the starting material. In this particular case, I think deprotonation of the final product (p-nitrophenol) would be a suitable way.

As for the evidence of existence of the Meisenheimer complex, I'll point you to the first compound/intermediate here. The steric factors of the ethoxy groups prevent the collapse of the adduct.

$\endgroup$
  • $\begingroup$ Oh I see. So is it because the Sn2 transition state is sterically hindered by the benzene ring that it cannot proceed? $\endgroup$ – bGe Mar 19 '18 at 17:53
  • $\begingroup$ No it's got nothing to do with sterics. As I said, normal Sn2 on an olefin is simply not done. It's orbitally unfavorable (no $\sigma^*$ and $\sigma$ overlap between the electrophile and the incoming nucleophile). As for the isolated intermediate, the ring does not even remain planar due to the interactions between the o-nitro and ethoxy groups. $\endgroup$ – Sagnik Mar 20 '18 at 0:59
0
$\begingroup$

I've just been reading an article in the September 2018 issue of Chemistry World "Textbook aromatic substitution mechanism overthrown". Apparently there's a recent paper (I haven't read it yet, but I intend to) claiming that in most cases the reaction is concerted, without formation of a discrete Meisenheimer intermediate. (Ref E E Kwan et al, NAt. Chem. 2018, 10, 917 (DOI: 10.1038/s41557-018-0079-7)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.