In this reaction, KOtBu is supposed to be used to form an enolate first and then ethyliodide is added. I just can't figure out the reaction mechanism of this reaction, can someone help?
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$\begingroup$ Looks like the enolate forms in the gamma position i.e. in conjugation with the double bond and then alkylates through the alpha position leaving the double in beta/gamma. Why does it not go the other side? Not sure, possibly due to the conjugation $\endgroup$– WaylanderCommented Apr 10, 2023 at 11:16
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1$\begingroup$ Similar question here: chemistry.stackexchange.com/q/132890/16683 In my answer to that question, there's one additional step in that the double bond migrates back into conjugation with the carbonyl group. Of course, that's not possible here because the alpha carbon is quaternary. So it stops there. There are two interesting aspects: (1) why does it deprotonate there — it's probably because deprotonating at the gamma carbon leads to the more thermodynamically stable enolate. (2) why does the extended enolate react at alpha instead of gamma — not sure! $\endgroup$– orthocresolCommented Apr 10, 2023 at 12:51
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1$\begingroup$ @orthocresol what we're not seeing is the detailed breakdown of reaction products. There may be some gamma alkylation in the product mixture. $\endgroup$– WaylanderCommented Apr 10, 2023 at 14:00
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$\begingroup$ @orthocresol Thank you for your answer, I hope I understood correctly since my English isn't perfect. I edited the question to show the answer I ended up with, does it seem correct? $\endgroup$– roosaveeraCommented Apr 11, 2023 at 13:35
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1$\begingroup$ @roosaveera Yes, except that the first proton you're removing is from the wrong carbon. Also, you can submit that as an answer to your own question (and in fact that is encouraged!) :) $\endgroup$– orthocresolCommented Apr 11, 2023 at 13:37
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