What is the reaction mechanism here?
I guess the $\ce{SnCl2}$ reduces $\ce{NO2}$ to $\ce{NH2}$ but then what happens?
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2$\begingroup$ I'm thinking there is a retro-Diels-Alder in there somewhere $\endgroup$– WaylanderCommented Aug 4, 2020 at 11:05
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$\begingroup$ Can you please show me how exactly, I tried didn’t work $\endgroup$– lidyaCommented Aug 4, 2020 at 15:36
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$\begingroup$ If I knew exactly I would post an answer. I do not have database access to find the paper where this is reported. $\endgroup$– WaylanderCommented Aug 4, 2020 at 15:39
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$\begingroup$ @Waylander A side-note about the speculated occurrence of a retro-DA, I'm sceptical; because in the starting material, I count 12, but in the product only 11 carbon atoms. $\endgroup$– ButtonwoodCommented Aug 4, 2020 at 21:11
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$\begingroup$ @Buttonwood Count again! There are 13 in the SM $\endgroup$– WaylanderCommented Aug 4, 2020 at 22:15
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1 Answer
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This reaction was conducted by Ho and Liao1. Nitro compound 1 is reduced to the nitroso compound 2 which undergoes a [3+3] electrocyclization to form 3 (4). Tautomerization of 4 to 5 is followed by a retro-"Diels-Alder" reaction to open to 6. Conjugate addition of the nitrogen to the unsaturated aldehyde moiety of 6 to afford dihydro pyridine 7. Tautomerization of 7 to 8 is followed by the loss of acetaldehyde. [Structures are from the paper; my numbering.]
ADDENDUM: The mechanism provided in the paper does not address a number of stereochemical issues. Nitro compound 1 is obviously formed as an endo-Diels-Alder product between (E)-(2-nitrovinyl)benzene and 1,3-cyclopentadiene. The bicyclic compound 3 (4) must have a cis-fusion given the mechanism of the [3+3]-electrocyclization (2 → 3) (vide infra). However, the stereochemistry arising from the retro-Diels-Alder reaction would not be expected to kinetically form the (Z,Z)-diene 6 but rather the (E,Z)-isomer 6a. Clearly, 6a is incapable of cyclization as prescribed. But 6a is a masked, vinylogous malonaldehyde that is susceptible to enolization and isomerization to 6b or 6c. While 6c can cyclize by the reported procedure of conjugate addition, there is also the probability that ring closure occurs by electrocyclization of 6b.
- Tse-Lok Ho and Po-Yau Liao, Tetrahedron Lett., 1994, 35, 2211. DOI:10.1016/S0040-4039(00)76799-2
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$\begingroup$ I already upvoted your answer. But I'm curious about the elimination of acetaldehyde. What is the driving force? Is that mean $\beta$-amino acids are prone to leave $\ce{CO2}$? $\endgroup$ Commented Aug 9, 2020 at 17:00
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1$\begingroup$ @Mathew: Wouldn't be surprised if the zwitterion of beta-aminopropionic acid gives ethylene under some set of conditions. In the present case, becoming aromatic is certainly worth something. If you mix 9 and 10 under some set of conditions, I'd be surprised to see any 8. $\endgroup$ Commented Aug 9, 2020 at 17:25
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$\begingroup$ @Mathew: I've thinking about your question again. There are stannic salts generated in the reduction. Perhaps they facilitate loss of acetaldehyde. So the question is, will Sn(IV) catalyze the addition of acetaldehyde to 3-phenylpyridine (9) to form 8? $\endgroup$ Commented Aug 9, 2020 at 20:29
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$\begingroup$ Might not. Otherwise, the last elimination step would have been reversible. I think gaining aromatically is a very good point for the elimination. In my experience, I didn't find any decomposition of $\beta$-amino acids during their synthesis. $\endgroup$ Commented Aug 9, 2020 at 21:33
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1$\begingroup$ Mostly a quibble, but I was wondering if the last step could be conveniently drawn as such. A bit reminiscent of beta-ketoacid decarboxylation. i.sstatic.net/Byn29.jpg (cc @MathewMahindaratne) $\endgroup$ Commented Aug 12, 2020 at 7:01
