I've several doubts about this ICHO synthesis problem, which was a part of the preparatory problem package of 2018 (so yes, it's not a part of an ongoing contest)

enter image description here

Before sharing the answer given, I'd like to share my thoughts and efforts towards the problem -

  • The first step is clearly an oxidation step, performed in presence of the peroxy acid above the arrow; which is subsequently reduced to the corresponding carboxylic acid. The oxidation product should be an amine oxide, but the question is, which one? (We have three nitrogens to play with)

  • For the sake of continuation, let's assume A is what they expect it to be (check the image below). Now, I'd expect more than just one hydrogen of the ring to be replaced by deuterium in the base catalysed tautomerism step. What's the next step (acid catalysed) for? Also, in basic medium (1) the initial ester would undergo saponification to acid.

  • B to C is a reducing step. Pd/C in presence of formic acid is known to be a reducing agent, but how do we know it reduces the amine oxide back to amine? (It'd be great if some sources could be provided for this, I couldn't find any)

    • C to the final product is pretty direct, the carboxylic acid group converts to acyl halide which undergoes nucleophilic addition by the ammonia derviative.

enter image description here

Thanks a lot.

P.S. Only a sound explanation for the A to B conversion remains - rest has been taken care of by the first posted answer. Please make additions/include relevant stuff for other conversions if you find any.

  • 1
    $\begingroup$ I would love to see a mechanism for the acid catalysed deuteriation of the N-Oxide adjacent carbon. I feel that the base is for saponification only and it doesnot catalyse deuteriation. $\endgroup$ Jan 13, 2019 at 6:50
  • $\begingroup$ That's basically an enolate homologue. If you're reading this question, you probably know everything needed to understand that already. @EshaManideep $\endgroup$
    – Zhe
    Jan 13, 2019 at 13:42
  • $\begingroup$ There's no reason why you can't form the N-oxide in both positions. The Pd hydrogenation should reduce both of them to the amine at the end. $\endgroup$
    – Zhe
    Jan 13, 2019 at 13:43
  • $\begingroup$ Then, wouldn't deuterium substitution happen at the other ring too ? I guess n N-Oxide formation occurs at essentially one place @Zhe $\endgroup$ Jan 13, 2019 at 13:47
  • $\begingroup$ @EshaManideep That's a good point. In light of that, the better answer is that the pyridine nitrogen is less sterically hindered than the pyrazole has an aryl substituent close to the N. This is somewhat easy to see if we draw out the hydrogens on the aryl rings explicitly. $\endgroup$
    – Zhe
    Jan 13, 2019 at 23:44

1 Answer 1


You have asked quite a number of questions in your post so let's tackle it one by one.

Why does the oxidation take place at that particular nitrogen atom?

To understand this, we first need to acknowledge that peroxyacids $\ce {RCO3H}$ are electrophilic oxidising agents. Specifically, the oxygen atom attached to the $\ce {H}$ is the electrophilic atom. From the frontier MO perspective, the reactive LUMO of the peroxyacid is the $\ce {O-O \sigma^*}$. This implies that the most nucleophilic nitrogen atom is likely to be favourably oxidised.

In fact, the oxidation of pyridine, which is a chemical species containing a nucleophilic $\ce {N}$ atom, to form pyridine $\ce {N}$-oxide using peroxyacids is well-known. It is documented by Clayden, Warren & Greeves (2012) on p. 730 as a method of increasing the reactivity of pyridine towards both electrophilic and nucleophilic aromatic substitution reactions. Notice that the bottom $\ce {N}$-containing ring actually resembles a pyridine fused with another benzene ring. However, note that the lone pair on $\ce {N}$ is not delocalised into the ring(s). Thus, there is no significant decrease in nucleophilicity of the bottom $\ce {N}$ in this compound relative to pyridine's $\ce {N}$ atom, suggesting that oxidation of the peroxyacid can very likely take place there.

Of course, we must be thorough in our analysis and also take a look at the relative nucleophilicities of the other two $\ce {N}$ atoms. Immediately, we will notice that for the $\ce {N}$ atom that appears to be making 3 single bonds has a lone pair that is heavily delocalised, in both the benzene ring below and the 5-membered ring it is in. Thus, we can eliminate the possibility of the the oxidation taking place at that nitrogen atom.

For the final top-most $\ce {N}$ atom, the lone pair may appear to be of the same nature of that of pyridine, i.e. held in an $\ce {sp^2}$ hybrid orbital. However, note that the presence of the adjacent $\ce {N}$ atom may result in inductive withdrawal of electron density from that $\ce {N}$ atom, decreasing its nucleophilicity.

Ultimately, I believe this first part is trying to test students if they are able to relate the structure and reactivity of pyridine towards oxidation by peroxyacids to the bottom pyridine-like ring. If students can make the connection, then they would instinctively know that the oxidation has taken place at that $\ce {N}$ atom.

From A to B what happened?

Well... There are a number of things which happened here. The most obvious being the first step: Basic hydrolysis occurs, with the leaving of $\ce {EtO^-}$. The next step is an electrophilic aromatic substitution, with the acidic deuterium ions as the electrophile. For pyridine $\ce {N}$-oxide, the 2- and 4-positions are activated towards EAS reactions. This can be shown simply by drawing the resonance structures of the compound with an $\ce {N=O}$ double bond. Similarly, the EAS takes place at the 2-position, relative to the $\ce {N}$-oxide group. It is difficult to explain why it doesn't take place at the 4-position as well. In general, for this type of structure elucidation questions, we would need to predict structures by looking at the resultant structures as well, especially if they are given. In this case, we see that only the 2-position has deuterium substitution taking place and we should simply leave it as that. Of course, if anybody has a good explanation for why that is the case, please propose it. Also, the other positions on the other rings may not be activated enough for the substitution to take place. Thus, we do not see substitutions taking place at those positions. Again, I am not able to provide a rigorous explanation for why that is the case.

Finally, the third step with $\ce {H2O}$ effectively makes the pH of the medium neutral again and also changes the carboxylic acid group from $\ce {COOD}$ to $\ce {COOH}$. This is simply due to a concentration effect. The high concentration of $\ce {H2O}$ simply pushes the equilibrium towards species containing $\ce {COOH}$.

What about from B to C?

From Wikipedia, it seems that $\ce {B}$ to $\ce {C}$ is simply a reduction step. The ammonium formate decomposes to give $\ce {NH3}$, $\ce {H2}$ and $\ce {CO2}$ in the presence of $\ce {Pd/C}$. The hydrogen gas evolved then gets adsorbed onto the palladium catalyst, where it can reduce the $\ce {N}$-oxide to give back the original $\ce {N}$-heterocycle.

And the last step involving conversion of C to the final product is as you have proposed.


Clayden, J.; Greeves, N.; Warren, S. Organic Chemistry (2nd ed.). Oxford University Press: New York, 2012.

  • $\begingroup$ Is there any source for B to C, which clearly says that amine oxides are reduced back to amines? Of course, the logic is sound but we can never really be sure about reduction reactions without a credible source. $\endgroup$ Jan 13, 2019 at 8:10
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    $\begingroup$ Ammonium formate + Pd/C is a standard reaction for catalytic transfer hydrogenation. I've used it many times. If a reduction has been recorded using hydrogenation with H2 gas then catalytic transfer hydrogenation will do it. $\endgroup$
    – Waylander
    Jan 13, 2019 at 8:35
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    $\begingroup$ This paper reports the deoxygenation of pyridine N-oxides by transfer hydrogenation with palladium catalysis risweb.st-andrews.ac.uk/portal/en/researchoutput/… $\endgroup$
    – Waylander
    Jan 13, 2019 at 8:49
  • $\begingroup$ "For the final top-most N atom, the lone pair may appear to be of the same nature of that of pyridine, i.e. held in an sp2 hybrid orbital. However, note that the presence of the adjacent N atom may result in inductive withdrawal of electron density from that N atom, decreasing its nucleophilicity." Please correct me if I'm wrong but - I think that, since electron density is lost by the top most N atom, shouldn't its nucleophilicity increase? Since it has less electron density, its a better place to attack for the electron-rich oxygen atom of m-CPBA, right? $\endgroup$
    – McSuperbX1
    Mar 11, 2020 at 15:32
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    $\begingroup$ @McSuperbX1 You are mistaken. Firstly, m-CPBA is an electrophilic oxidising agent. See here for greater clarity: en.m.wikipedia.org/wiki/Prilezhaev_reaction The nitrogen atom donates a lone pair towards the electron-poor oxygen atom of m-CPBA. Secondly, the decrease in electron density of an atom would decrease its nucleophilicity in general. For example, consider ammonia and pyridine. Due to the lower electron density at N in pyridine due to delocalisation, pyridine is significantly less nucleophilic compared to ammonia. $\endgroup$ Mar 12, 2020 at 22:55

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