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What is the mechanism for this reaction? The tertiary amine is converted into an amide and the ring gains a double bond. I can't figure out how the C-O double bond would be formed.

enter image description here

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  • $\begingroup$ Oh, that's the easy part, but can you figure out what happens earlier? That's more interesting part. $\endgroup$ – Mithoron Feb 15 at 1:32
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    $\begingroup$ Find the nucleophile. Find the electrophile. Combine. $\endgroup$ – Zhe Feb 15 at 2:24
  • $\begingroup$ Hint. The amide is a formyl amide. Where does the H attached to the carbonyl come from? $\endgroup$ – Zhe Feb 15 at 2:25
  • $\begingroup$ Can chloroform react with sodium hydroxide? $\endgroup$ – Waylander Feb 15 at 7:53
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reaction mechanism

This reaction is very similar to Reimer-Tiemann reaction which also consists of the reagent $CHCl_{3}$ + $NaOH$. The mechanism of that reaction is believed to have dichlorocarbene intermediate too. A good point to remember is that the dichlorocarbene intermediate is often used to form cyclopropane or carbonyl groups.

reimer tiemann reaction

This is the mechanism for the reimer tiemann reaction. We can see the similarity in the this reaction and the one you have mentioned previously.

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  • $\begingroup$ Mm, that's better. Although the arrows still suggest an intramolecular proton transfer which I don't like... $\endgroup$ – orthocresol Feb 24 at 11:29
  • $\begingroup$ @orthocresol I am not sure if I should discuss this here, but I will be honest, I am a 12th grade student and hence couldn't understand the terms you used earlier. I don't understand what correction there should be here. Can you please elaborate in simple terms, or can we discuss it in chat? $\endgroup$ – napstablook Feb 24 at 15:39
  • $\begingroup$ To deprotonate a X–H bond, the base (let's say B) needs to come from the opposite side of the hydrogen, i.e. X–H···B. That's essentially like an SN2 reaction, and the reason is the same: you need to get to the σ* orbital. This makes an intramolecular deprotonation essentially impossible, because that lone pair that is supposedly doing the deprotonation can't reach the σ* orbital. More likely that the negative charge picks up a proton from somewhere else, and the proton on the ring is lost some other way. $\endgroup$ – orthocresol Feb 24 at 15:59
  • $\begingroup$ Apart from that nitpick, your mechanism is fine, though; your edit fixed the other things I pointed out. But I'd still ask you to type out your text, instead of attaching a picture of written text. It's easier for everybody else to read, and can be searched for. $\endgroup$ – orthocresol Feb 24 at 16:02
  • $\begingroup$ I will grant that that the σ* orbital seems very difficult align. I just thought that the aromaticity would drive it to be essentially SN1 type acid base reaction. I guess the arrows seem to direct people into thinking this is SN2( in which case should I just get rid of them and add equilibrium arrows?) I will update this answer to text format once my major exams get over (because I don't mainly know how to do TeX and draw custom images.) That being said thanks for your inputs. $\endgroup$ – napstablook Feb 24 at 16:23

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