I was going through a problem in the book Advanced problems in organic chemistry by Chouhan:

problem figure

I figured out that since di-isopropyl amide will act as a hindered base it will take out the following hydrogens (in the form of $\ce{H+}):$

  1. the one where the carbanion formed is in conjugation with benzene ring
  2. the one where carbanion formed is in conjugation with nitrogen.

After that these carbanion(s) will attack respectively on their adjacent $\ce{C-C}$ σ-bonds and the nitrogen(s) will depart in the form of $\ce{N2},$ this gives me option (a) as the answer, but the solution manual proposes a different reaction mechanism:


I am not able to understand the formation of the three-membered ring, please help. Is there a mistake in my thinking or are there some other concepts involved?

I am an undergraduate student of organic chemistry.

  • 4
    $\begingroup$ see en.wikipedia.org/wiki/Shapiro_reaction and the section on "catalytic Shapiro reaction" - that suggests that the intended starting material is the cyclopropane, which means that that your book probably has an error. Mind you, this book is also abbreviating isopropyl as "i-Ph", so I'm inclined to suggest that you deserve a better book. [edit: I meant aziridine; silly me.] $\endgroup$ – orthocresol Aug 12 '20 at 15:45
  • 1
    $\begingroup$ The Catalytic Shapiro Reaction: DOI:10.1021/JA951422P Answer 46(a) is correct, the aziridine ring is missing in the reactant of the question. $\endgroup$ – user55119 Aug 12 '20 at 15:56
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    $\begingroup$ So, I attempted to provide a mechanism before actually reading the answer or comments. I immediately got stuck because of conservation of mass. The starting material has 30 hydrogen atoms but the products only have 28 hydrogens. There's no oxidant here, so something immediately had to be wrong. $\endgroup$ – Zhe Aug 12 '20 at 16:24
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    $\begingroup$ user55119 already told you that that the given answer is correct. I gave you a link to a Wikipedia page that says the same thing. Both of us told you that your starting material in your question is wrong. I don't think there's much else to be said, to be honest. Someone else could write it up into an answer, though. $\endgroup$ – orthocresol Aug 12 '20 at 16:24
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    $\begingroup$ @Zhe: You are correct. I immediately realized there are two double bonds in the products while SM has only one. So, as as orthocresol said, aziridine has to be there to begin with. Also note that in the solution, it starts with exyra double bond (terminal one). That confirms the sloppiness of authors (poor proofreading). $\endgroup$ – Mathew Mahindaratne Aug 13 '20 at 6:53

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