# Why is nitrate anion substituted but bromide anion is not in Von Richter reaction?

In Von Richter reaction which is a nucleophilic aromatic substitution. As according to this source, there is only a 37% yield of the product.I want to know what are the competing side reactions which results in such a low yield of the desired product. I searched for it on the internet but couldn't find anything related to it.

Secondly, I want to know the reason behind the substitution of nitrate anion as the major product although bromide anion is a better leaving group according to this table taken from Wikipedia

wikipedia

• please mention the reason of downvote so I can edit my post accordingly and not commit the same mistake in future posts Jun 19 '21 at 11:05
• I would expect that cyanide displacement of Br is one of the major byproducts Jun 19 '21 at 12:39
• I expected the same thing but it would be better if someone can give a reference for it Jun 19 '21 at 13:12
• Well, I think you see yourself how bad capitalisation, punctuation and spelling you've got there. Jun 19 '21 at 13:45
• Worth reading this paper pubs.acs.org/doi/abs/10.1021/jo01149a007 Jun 19 '21 at 15:06

The Von-Richter reaction's accepted mechanism is given by-

As you can see, this is not a simple case of say, removal of $$\ce{NO2-}$$ ion, instead it involves removal of $$\ce{N2}$$ which provides thermodynamic stability.

the tables you use have been calculated on the basis of $$\ce{pK_a}$$ values which don't give the right answers in many cases due to solvent effects and the intermediates involved in specific reactions. Wikipedia itself states

It is important to note that the list given above is qualitative and describes trends. The ability of a group to leave is contextual. For example, in $$S_nAr$$ reactions, the rate is generally increased when the leaving group is fluoride relative to the other halogens. This effect is due to the fact that the highest energy transition state for this two step addition-elimination process occurs in the first step, where fluoride's greater electron withdrawing capability relative to the other halides stabilizes the developing negative charge on the aromatic ring. The departure of the leaving group takes place quickly from this high energy Meisenheimer complex, and since the departure is not involved in the rate limiting step, it does not affect the overall rate of the reaction

This order is a general trend and as the mechanism in this reaction suggests, is not the end-all be-all determining factor.

Coming to why this reaction gives low yield, is mainly due to the hydrolysis of $$\ce{CN-}$$ ion by water/alcohol to $$\ce{HCOO-}$$/$$\ce{CH3COO-}$$. This is was suggested because according to this paper$$^1$$

the recovery of much unreacted starting material suggest that hydrolysis (or alcoholysis) competes with the nitro compound for cyanide ion to the extent that hydrolytic destruction of cyanide is the yield-limiting factor; it is known that cyanide ion is rapidly hydrolyzed in water solution at elevated temperatures

There is one thing missing from wikipedia that von-richter reactions were conducted in elevated temperatures and in sealed tubes $$(\ce{150^oC-165^oC})$$ both originally by von richter himself and in subsequent studies for mechanism*.

There may be other competitive side reactions but were not documented for this reaction (because it is of low yield anyway, there was not much work on it.)

*The paper by J.F.Bunnett,J.F.Cormack and Frank C.Mckay$$^1$$ (linked by @waylander) cites-

von Richter’s reactions were run in “alcohol” solution in sealed tubes at $$\ce{180^oC-200^oC}$$ or higher.

and the letter by Prof. Myron Rosenblum $$^2$$ while explaining the mechanism with evidence from isotopic labelling cites-

9.5 mmoles of p-chloronitrobenzene, when treated with 20 mmoles of potassium cyanide, and 6.2 mmoles of ammonium nitrate in a sealed tube at $$\ce{160^oC}$$ for 1.75 hours, gave nitrogen gas containing 0.75% $$\ce{N_2^29}$$

It is normal for nucleophilic $$S_nAr$$ to be conducted around this temperature** so this is not surprising thing to be missed by wiki, it also means that $$S_nAr$$ of cyanide on bromine can probably be competitive but experiments predict a majority of unreacted products according to this table in most solvent cases:

interestingly both this table and the letter by Prof. Rosenblum (46.5% yield for the same reaction but with chlorine) suggest different yields to what wikipedia suggests.

TL;DR : Leaving group order doesn't matter here because of different mechanism involved converting $$\ce{-NO2}$$ to $$\ce{-N2-}$$ which is a very good leaving group and yield is low because of hydrolysis of $$\ce{CN-}$$ ion at the preferred temperature.

**Ref :. Morrison & Boyd sec 26.7 (Nucleophilic aromatic substitution: bimolecular displacement)

cited material:

$$1$$ : J.F.Bunnett, J.F.Cormack and Frank C.Mckay ,"Mechanism and reactivity in aromatic nucleophilic substitution reactions"J. Am. Chem. Soc. 1958 5(3), 481-490 doi 10.1021/jo01149a007

$$2$$ : Rosenblum, M. The Mechanism of the von Richter Reaction. J. Am. Chem. Soc. 1960, 82, 3796–3798; doi 10.1021/ja01499a090

• Kindly see this about citing papers. Thanks! Jul 11 '21 at 10:57
• @napstablook While aiming to bring the literature references into the style of J. Am. Chem. Soc. , the style recommended to use on chemistry.se, I noticed that reference #1 links to a publication of different year, journal and publisher than the one it (currently) states. So, which one is reference #1? Jul 11 '21 at 11:26
• @Buttonwood Thanks for pointing out! Apparently wikipedia had a different source cited (both by J.F.Bunnett) I didn't think to cross-check. I will change the link Jul 11 '21 at 11:29
• does the removal of bromine by SnAE can be a side reaction ? Jul 12 '21 at 0:55
• @Samardeepsingh I have mentioned that while it could have been, experimental results suggested that the reactant is largely unreacted (see the table) which means that it is mostly hydrolysis. Now for only few cases of glycol the yield of both product and the amount of unreacted reactant is low. I suspect this is because of the SnAE side reaction, however there is not much experimental observation about this reaction, since it is not very useful.. Jul 12 '21 at 0:58