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I came across a chemistry problem posted on Discord. The objective is to propose a reagent that will enable the following reaction to occur:

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One proposed answer for the mysterious Reagent D is sodium methanethiolate.The claim is made on the basis of the fact that patent US6777575B2 describes this exact sort of synthesis and calls for using MeSNa as a reagent in this situation. A counterargument is that the nitro group is stabilized through conjugation, and that methoxy or trifluoromethyl would be a better leaving group in this situation. Which claim is correct? In the reactant, can nitro act as a leaving group and allow for substitution with methanethiolate? Why or why not?

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  • $\begingroup$ It sure is usual leaving group in plenty of SNAr reactions. CF3 isn't better and exchanging methoxy wouldn't be thermodynamically favorable afaict. $\endgroup$
    – Mithoron
    Oct 29, 2021 at 16:46

1 Answer 1

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The nitro group (or nitrite anion) can indeed act as a leaving group in nucleophilic aromatic substitution, especially if the ring is electron-deficient (as it is in this problem). Wikipedia offers another example, also using an electron-deficient ring:

In the compound methyl 3-nitropyridine-4-carboxylate, the meta nitro group is actually displaced by fluorine with cesium fluoride in DMSO at 120 °C.[1]

Nucleophilic aromatic substitution with nitrite as leaving group

What enables displacement of the nitro group is its electronegativity, combined with the fact that it actually does not conjugate all that well with the ring. Conjugation of the nitro group to the ring would require increased formal charge separation with a positive charge being imparted to the ring, which becomes more unfavorable if the ring is electron-deficient.

Cited Reference

  1. Tjosaas F, Fiksdahl A (February 2006). "A simple synthetic route to methyl 3-fluoropyridine-4-carboxylate by nucleophilic aromatic substitution". Molecules (Basel, Switzerland). 11 (2): 130–3. https://doi.org/10.3390/11020130. PMC 6148553. PMID 17962783.
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