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Recently I was reading about identifying N-terminal amino acid residues using Sanger's reagent (1-fluoro-2,4-dinitrobenzene). The following image showing the reaction is taken fro Wikimedia Commons:

Sanger's method for protein N-terminus sequencing

The free N-terminus amino group performs a nucleophilic aromatic substitution to get the first product; after that we hydrolyse the adduct. Why doesn't the Ar–N bond in the final product also undergo hydrolysis to give 2,4-dinitroaniline, though? There are many examples (e.g. isocyanides, cyanides) where heating with strong acid causes cleavage of the C–N bond. Is it because resonance with the aromatic ring causes the nitrogen lone pair to be not basic?

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    $\begingroup$ Please note my edit to your question: (1) The chemical names are wrong; substituents (fluoro, dinitro) are joined to the parent chain (benzene) by hyphens, not separated by spaces. (2) The $\mathrm{S_NAr}$ mechanism is not an "ary $\mathrm{S_N2}$". (3) No need to put three or four consecutive question marks; one is enough. $\endgroup$
    – orthocresol
    Dec 2 '21 at 15:00
  • $\begingroup$ @orthocresol brother can you please check my understanding in the comments that i made on your answer? $\endgroup$ Dec 3 '21 at 12:39
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You raise the examples of isonitriles and nitriles undergoing C–N bond cleavage via hydrolysis. That is true, but it is worth thinking about how this happens. Both isonitriles and nitriles are converted to amides first when reacted with aqueous acid. Now, amides will certainly be hydrolysed at the last stage of the peptide sequencing: you can see that in the image itself, as the amide bonds holding the polypeptide together are broken (so that the labelled residue can be detected).

However, amides behave quite differently from amines. The C–N bond you're asking about ($\ce{R-NH-Ar}$, Ar = 2,4-dinitrophenyl) is an amine, not an amide. The comparison to isonitriles / nitriles is therefore not relevant. Amines simply don't undergo typical $\mathrm{S_N2}$ reactions (quaternary ammonium salts do, but that's a different matter entirely), so the C–N bond doesn't get broken via hydrolysis. It is true that there is resonance with the electron-withdrawing dinitrophenyl group, which reduces the availability of the nitrogen lone pair, but this is actually irrelevant: even if that aryl group were just a methyl group, the conclusion would not be affected.

Note that this has nothing to do with the C–N bond strength, either. The C–N bond in an amide has partial double bond character, and is stronger than the C–N bond in an amine (which is a plain single bond). The question is not which is stronger, it is which has an available pathway for cleavage that doesn't involve too high an activation energy. Amides can be cleaved via a nucleophilic acyl substitution mechanism: although this is not easy to accomplish (you need strong acid and lots of heat), it is still possible. Amines, on the other hand, could conceivably be cleaved via a $\mathrm{S_N2}$ or $\mathrm{S_N1}$ mechanism. Neither is very favourable in this case (amines are very poor leaving groups, and the carbocation above is not stable at all, with that electron-withdrawing carbonyl group next to it).

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    $\begingroup$ I think what I got confused on was thinking that R-NH-R' If the Nitrogen in this formed a coordinate bond with a proton then the quaternary amine so formed will be a good leaving group but it turns out even then it doesn't. amines form salts with acids and this doesn't change the fact that they are bad leaving groups. Here the only possibility was nucleophilic acyl substitution. That is infact not related to Rnh2 being a good leaving group or not. It is just a property of acid derivatives to show this reaction whenever surrounding medium is deficient of the group that is attached to carbonyl $\endgroup$ Dec 3 '21 at 7:47
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    $\begingroup$ Even if RNHR' form salts with some acid(ie donates a lp to $H^+$) even then we wont expect it to break the original C-N bonds that are present in the salt. kindly confirm these thoughts of mine. $\endgroup$ Dec 3 '21 at 7:50
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    $\begingroup$ from the last line of my first comment sir I wanted to say that that the equilibrium of acyl sn2 will be drawn to the products only if there is a deficiency of the group being displaced in the surrounding med( to prevent reverse reaction)(le charterlier principal) $\endgroup$ Dec 3 '21 at 8:00

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