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So, I came across this reaction as an example of E2 elimination.

enter image description here

Our teacher told us that this is what we call E2' reaction, as in, the $\ce{OH^-}$ takes the terminal $\ce{H^+}$, followed by resonance and then removal of $\ce{Cl^-}$.

This didn't really make sense to me at first by how he has taught us the mechanism for E2, which is through concerted single step $\beta$-elimination, while this one seems like $1,4$-elimination, and at least three steps by the path he told us.

I tried to make it make sense by the following pathway:

enter image description here

Question: Is this correct? What exactly is happening here? This doesn't seem like resonance to me as that doesn't involve partial charges, I think. Does this happen for even longer chains, like $1,6$ or $1,8$ eliminations?

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    $\begingroup$ This is essentially correct. Remember that there is a big energy gain from aromatisation - you are pushing rocks downhill $\endgroup$
    – Waylander
    Feb 25 at 11:37
  • $\begingroup$ @Waylander Thanks for the comment. I do understand the compound would gain a lot of stability, relatively, due to aromatic ring. Are there any issues with the mechanism I've proposed? $\endgroup$ Feb 25 at 13:33
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    $\begingroup$ Mechanism looks OK $\endgroup$
    – Waylander
    Feb 25 at 15:55
  • $\begingroup$ @waylander Thanks! $\endgroup$ Feb 25 at 18:45

2 Answers 2

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Yes the mechanism is correct. When ever a nucleophile is already present in a compound then its is favourable that intramolecular attack would occur as they are faster than intermolecular attack. In this case the pie electron attacks and removes chlorine and as you have taken a strong base simultaneously protonation will occur in one step. So it is like E2. If you had taken a weak base then the reaction would occur through carbocation path that is 1)SNGP(Neighbouring group participation) then 2)E1

And about longer chains is that longer the chain the intramolecular attack can be slow. Or It is possible that ring formation takes place like 6,5,3 membered ring according to the substrate.

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I would assume that the molecule is deprotonated and then loses the chloride ion to give the aromatic product.

Either caroon atom 3 (with the chlorine) or carbon atom 6 (with only a pair of hydrogens attached off the ring) could be reversibly deprotonated; either way a conjugated pentadienide carbanion is formed with the negative charge formally distributed over three atoms. But to go on to the aromatic product there must be a good anionic leaving group from the other one of the two carbons mentioned above. Deprotonating C-3 means a hydride ion has to leave from C-6, which is not favored so the deprotonation ends up getting reversed. Deprotonating C-6 allows C-3 to release its more favorable chloride ion and thus enables formation of the aromatic product.

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