Why benzenediazonium bromide and iodide are not important?

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? Chemists and scientist do not take them noteworthy as there is little to no literature evidence of their synthesis. Are the compounds so unstable that chemists neglect them? Nobody has attempted to make the compounds in the way benzenediazonium chloride is made. Why is it so? Is it because of unfavourable reaction conditions? Can't we use hydrogen bromide or hydrogen iodide to make benzenediazonium bromide/iodide respectively? Like the following reaction:

$$\ce{C6H5NH2 + HNO2 + HX -> C6H5N2+X- + 2H2O ~~~~~~~~ X = F, Br, I}$$

What are the reasons for the non-existance of/not preparing other benzenediazonium halides? Is it just because of their instability or feasibility or any other reasons? Or is there is any (other) way to (theoretically) make these compounds (if they exist)?

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.

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 with that iodobenzene was formed as side product.