I was asked to find the product of potassium ethoxide reaction with 1-[2-(chloromethyl)phenyl]propan-2-one (1).

I came up with two different paths, each leading to the same product.

1-[2-(chloromethyl)phenyl]propan-2-one (1) on reaction with potassium ethoxide gives a carbanion 2.

This carbanion 2, in conjugation with $\ce{C=O}$ gives enolate 3. This enolate can follow path 1, an $\mathrm{S_N}{2}$ reaction (Williamson ether synthesis) to give 4.

The same carbanion 2 in conjugation with benzene ring gives 6 via allylic rearrangement(Path 2).6 could undergo a possible Diels-Alder reaction giving again 4.

My question is, will the reaction follow Path 1, a Williamson ether synthesis route or will it follow Path 2, a Diels-Alder reaction.

Suggested Path 1 & 2

  • $\begingroup$ Path 1 looks the more probable as path 2 breaks the aromaticity of the system $\endgroup$ – Waylander Jun 16 at 6:38
  • $\begingroup$ @ Waylander , got it.However , is path 2 at all possible , a diels alder route.That would be a different route $\endgroup$ – Chakravarthy Kalyan Jun 16 at 7:03
  • 3
    $\begingroup$ Unrelated to the actual issue of which path is more likely, but I don’t think path 2 can be called a Diels-Alder reaction. It resembles more an 6-pi Electrocyclic Ring Closing reaction. $\endgroup$ – IanC Jun 16 at 8:00
  • $\begingroup$ Wouldn't you get a different product (2-oxa~ instead of 1-oxa~) via Path 2? The O and a C seem to have changed places in the last step, between 6 and 4 $\endgroup$ – szentsas Jun 17 at 16:05

Your Answer

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

Browse other questions tagged or ask your own question.