Why benzenediazonium bromide and iodide are of little interest?

Question

When we think of aryldiazonium compounds, the first compound that comes to mind is benzenediazonium chloride. Sure thing, the compound is versatile and has been used to make a vast array of organic compounds and thus serves its purpose as a useful reagent. But what about other halides like benzendiazonium fluoride, bromide and iodide? There is little to no literatureon these compounds suggesting they are of little interest. What makes them so un-noteworthy? Is it because of stability? Is it because of unfavourable reaction conditions? Or is it because BDC is so popular that other halide compound hasn't been tested or experimented with?

Answer 1

There is indeed very little to be found in the literature.

A recent Russian paper (Ref.1) mentions benzenediazonium bromide as a reactant. This thesis from 1975 (Ref.2) contains a preparation of benzenediazonium bromide and an attempted preparation of benzenediazonium iodide; it notes that the iodide is unstable as does this reference here. The discussion section of this Organic Syntheses' preparation of fluorobenzene (Ref.3) mentions an alternative preparation by formation of benzenediazonium fluoride and its decomposition referencing work from the 1900s.

My overall thought is that they are probably less stable than the well-known chlorides and undergo reaction in lower yields so they are of little interest.

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References:

1. Yu. V. Ostapiuk, V. S. Matiichuk, N. I. Pidlypnyi, N. D. Obushak, "Convenient synthesis of $\alpha$-bromo ketones by the Meerwein reaction," Russian Journal of Organic Chemistry 2012, 48, 519–522 (https://doi.org/10.1134/S1070428012040094).
2. Geoffrey D. Picard, "Sulfonium Salts, Aliphatic and Aromatic Diazonium Salts, Formation and Isolation," Honors Theses 1975, 2119 (https://digitalworks.union.edu/theses/2119)(PDF).
3. D. T. Flood (Checked by W. W. Hartman and J. R. Byers), "Fluorobenzene," Org. Synth. 1933, 13, 46 (DOI: 10.15227/orgsyn.013.0046) and Organic Syntheses, Coll. Vol. 1943, 2, 295 (PDF).

Answer 2

Mostly it's a matter of economy. In most diazonium salt synthesis the counterion is not involved in the organic reactions; it's just a spectator. So we choose chloride because hydrochloric acid is a cheap and effective chemical to acidify the aromatic amine+nitrite mixture and make the diazonium ion in the first place. Ergo an aryldiazonium chloride.

The paper (Ref.1) mentioned by Waylander in the comments and reference cited in his answer involves bromide as the counterion because, exceptionally, this ion does serve as a reactant. In that paper the ketone reacts with the diazonium ion to give a benzylic cation, which then combines with the bromide counterion from the original diazonium salt to give the bromine-bearing product.

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The case of benzenediazonium fluoride is special. In addition to the economic aspect in applications where the halide is a spectator, we cannot effectively make the fluoride salt in cases where the fluorine would serve as a reactant. Hydrogen fluoride is a strong acid as such, but in dilute aqueous solution it turns into a tightly bound ion pair, $\ce{H3O^+F^-}$, which is then too weak to effect the diazonium ion formation (see here for a discussion of this property). Instead, in the Balz-Schiemann reaction, tetrafluoroboric acid is used to form the diazonium tetrafluoroborate, whose anion then acts as a fluoride ion source to form fluorine-bearing products. Fluorobenzene and 4-fluorobenzoic acid are among the products reached via this method.

Benzenediazonium tetrafluoroborate has a safety advantage over the chloride. The latter can be explosive if isolated, apparently due to internal reaction vetween the cation and anion of the salt. The tetrafluoroborare, with a less reactive anion, is safer to isolate and store.

References:

1. Yu. V. Ostapiuk, V. S. Matiichuk, N. I. Pidlypnyi, N. D. Obushak, "Convenient synthesis of $\alpha$-bromo ketones by the Meerwein reaction," Russian Journal of Organic Chemistry 2012, 48, 519–522 (https://doi.org/10.1134/S1070428012040094).

Answer 3

Good answer by @Waylander and @OscarLanzi. They provided some literature evidence of the synthesis procedure of Benzenediazonium bromide and iodide and discussed its usage and worth and explained why they are not that important as compared to benzenediazonium chloride. My answer just revolves on the actual synthesis procedure(slightly abridged in my answer) of benzenediazonium bromide and attempted synthesis of benzenediazonium iodide sincerely taken from 1975 thesis(Ref. 2 of @Waylander's answer). Thus fully completing the discussion.

Benzenediazonium Bromide

Take 250 ml copper vessel. Add 20g ice and 20g conc. nitric acid. Maintain a temperature of -10 °C with constant stirring of the mixture. To this, add 5g of aniline. The reaction is exothermic so keep mixture under -5 °C by pouring liquid nitrogen outside the vessel. A white to tan colored suspension is formed. To this, add 3g of finely powdered sodium nitrite in small amounts for 30 seconds till the tan colored disappears thus the reaction is complete.

3 ml of bromine is added to 100 ml water saturate with sodium bromide. Cooled at -5 °C and then added to diazonium mixture. A yellow orange crystal is formed immediately. Maintain constant stirring for 15 minutes. Then the solid is filtered and cooled to 0 °C. 5 g of wet solid is obtained(dec. at 78 °C) which was confirmed to be benzenediazonium tribromide. It was added to 100 ml acetone. 1.5 g(18% yield) of benzenediazonium bromide(dec. 108 °C) were obtained and bromobenzene as side product.

Benzenediazonium Iodide

Repeat the above procedure for making the diazonium mixture. Then add 4ml bromine to water saturated with potassium iodide. Cooled at -5 °C and then added to diazonium mixture. Reddish brown crystal were formed which decomposed immediately at -10 °C. The product has a boiling point of 188 °C and was identified as iodobenzene, not benzenediazonium iodide, in the thesis.