In the Organic Syntheses preparation of azulene (http://www.orgsyn.org/demo.aspx?prep=CV7P0015) there are two reactions. First is a clevage of the C-N bond in pyridine and the second one is the annelation of a seven-memebered ring:

Reaction scheme

I suppose that the first step of the first reaction is a nucleophilic substitution by pyridine at the 1-chloro-2,4-dinitrobenzene, but I have no idea how the mechanism could continue. Could some one please help me? Thank you.

Also, for the 2nd reaction, is it an electrophilic substitution at the cyclopentadieneyl anion, or does it have an entirely different mechanism?


1 Answer 1


The @orthocresol link to the Zincke reaction is a good start to the mechanism. You are correct about the first step. Dimethylamine adds to the electrophilic pyridinium salt 1 as illustrated in 1 --> 5. The reaction is driven by loss of the poor nucleophile (decent leaving group), DNP-NH2. The (Z)-isomer 5 is in equilibrium with the (E)-isomer 6 but the reactions leading to azulene 12 require a cisoid conformation to allow for cyclization.

Cyclopentadienide anion, which is delocalized but for convenience is shown as a localized anion, displaces dimethylamine which is protonated by the methanol in the reaction mixture (see your link to Organic Synthesis). Fulvene 8 must generate the geometry shown to complete the remaining steps to azulene (9 --> 12).

ADDENDUM: An added thought. The transformation of 8 ---> 10 may be a 4n+2 disrotatory thermal electrocyclization that forms the new sigma bond of 10 with cis hydrogens.

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    $\begingroup$ Nice answer. The only issue I see is that 8 and 9 are resonance structures instead of separate species. $\endgroup$
    – Zhe
    Feb 24, 2019 at 4:14
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    $\begingroup$ @Zhe: So were 5 and 6. $\endgroup$ Feb 24, 2019 at 6:34
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    $\begingroup$ True enough for 8 and 9. I broke 8/9 down into discreet steps for easier understanding. I'll bracket them and use resonance arrows. 5 and 6 involve more than resonance. Most references show an equilibrium. Actually, one has to get to 8/9 whether it is thru 5 or 6. $\endgroup$
    – user55119
    Feb 24, 2019 at 13:44
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    $\begingroup$ @MathewMahindaratne Actually, the double bond geometry prevents that from being the case. $\endgroup$
    – Zhe
    Feb 24, 2019 at 14:53
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    $\begingroup$ @Zhe: Thanks for pointing it out. I agree with you. Isomerization makes the difference. :-) $\endgroup$ Feb 24, 2019 at 19:02

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