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imine formation

Why does the third group react in this example and not the first?

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closed as off-topic by Mithoron, user55119, Tyberius, A.K., Jon Custer Nov 28 '18 at 3:41

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  • $\begingroup$ Perhaps, you can also elaborate on your thoughts as well? $\endgroup$ – Tan Yong Boon Nov 26 '18 at 7:42
  • $\begingroup$ Well the 1 group is an amide and the 2-3 group is a hydrazide. $\endgroup$ – A.K. Nov 27 '18 at 3:49
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Imine formation essentially involves an nucleophilic amine and an electrophilic carbonyl group. To decide which is the nucleophilic $\ce {N}$ atom which performs the nucleophilic attack on the electron-deficient $\ce {C}$, we would have to first consider each $\ce {N}$ atoms nucleophilicity, which we can interpret here to be the ability of the $\ce {N}$ atom to donate its lone pair of electrons.

For 1 and 2, the lone pair is delocalised into the carbonyl group and they can be seen as behaving as amides. This delocalisation significantly reduces their nucleophilicity. As discussed by Ron here, the delocalisation is so strong that restricted rotation about the $\ce {C-N}$ bond is observed. For 3, there is no such delocalisation present, thus the lone pair of the $\ce {N}$ is readily donated to electrophiles, making it the likely nucleophilic atom for the imine formation.

Another possible effect that raises the nucleophilicity of that $\ce {N}$ atom is the alpha effect, also discussed here. The following diagram (p. 513) from Clayden, Warren, & Greeves (2012) illustrates the molecular orbital theory basis of the effect. Although it is used to illustrate the effect for peroxide, the same concept can be extended to our case of nitrogen atoms. The adjacent $\ce {N}$ atom with the lone pair can increase the nucleophilicity of the $\ce {N}$ atom. However, the alpha effect may be less prominent for our case since the lone pair of the adjacent $\ce {N}$ is also being strongly delocalised into the carbonyl group. Nonetheless, it is an interesting consideration, further justifying why 3 should be most nucleophilic among the three.

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Reference

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

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This is semicarbazide semicarbazone reaction which is a nucleophilic addition reaction with the elimination of water. This addition reaction, like any other, involves a nucleophile (lone pair on N here, which one - we will see) on the nucleophilic center (carbonyl group). Thus the reaction will go more forward if the reacting nucleophile is more electron rich and thus more nucleophilic.

Now, 1st and 2nd groups are participating in conjugation (I presume you know about resonance, etc.) with -C=O group and hence their lone pair electrons (cause for nucleophilicity for them)are delocalised and thus less available making them weak nucleophiles. Thus, the 3rd group is the most nucleophilic and proceeds with the reaction. Hope this helps!

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