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I know this that when a primary amine is reacted with a carbonyl compound, an imine is formed:

Imine formation

I'm wondering why the position of this equilibrium lies to the product side / right-hand side. Why is the formation of the C=N double bond favoured?

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  • $\begingroup$ similarly to chemistry.stackexchange.com/questions/10722/… $\endgroup$ – Mithoron Jan 16 '18 at 15:50
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    $\begingroup$ No, i didn't get it! $\endgroup$ – Selena Jan 16 '18 at 16:04
  • $\begingroup$ If we simply compare the relative reactivities of the product and reactant sides, we can see that the attack of the more nucleophilic lone pair of the nitrogen would attack faster than that of the oxygen. The ratio of the rate constants gives the equilibrium constant. Voila $\endgroup$ – Eashaan Godbole Jan 30 '18 at 7:32
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In practice, a drying agent is sometimes added to the reaction: I've used molecular sieves before, but a cursory database search brings up a few other options, e.g. magnesium sulfate, silica gel, or Montmorillonite (this list is probably not exhaustive). These remove water from the system and pull the equilibrium position over to the right-hand side.

That's one possibility, but I don't think that's all there is to it - I think a lot of imine formations proceed perfectly ok even without dehydrating agents being added to them. However, you'll have to hear about these from somebody else, as I don't know what would drive those reactions. It could well be that simply adding an excess of either amine or carbonyl drives the reaction to completion.

Interestingly, a 2009 paper by Saggiomo and Lüning1 describes their investigations into supposed imine formation in water. Apparently, both the carbonyl compound (benzaldehyde or salicylaldehyde) and amine (aniline) do not dissolve in water - not particularly surprising. The imine did not form appreciably in water and only started to form when water was removed during workup and purification.

P/S This is just the top hit I found on Google, it's hardly a complete literature search.


  1. Saggiomo, V.; Lüning, U. On the formation of imines in water—a comparison. Tetrahedron Lett. 2009, 50 (32), 4663–4665. DOI: 10.1016/j.tetlet.2009.05.117.
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See,We compare any reaction thermodanimaclly and kinetically Let me put the thermodynamical aspect first which is that the formation of water is the very good sign of thermal stability for a reaction and a energy of 572 KJ/mol is realesed and hence helps to meet with the enrgy demands for formation of carbon and nitrogen bond And suprisingly carbon and nitrogen bond is stable due to presence of lone pair on nitrogen and high density of electrons around it. Now coming back to thermodynamics the carbon nitrogen energy is comparatively very high 614KJ/mol for this carbon and nitrogen bond. Also important thing note oxygen is more electronegative but yet as wr see oxygen is ready to leave with hydrogen to form carbon nitrogen bond double bond.

Now Kinetics The product hence formed is by first order Kinetic nad hence also it is easy for hydrogen to migrate to alchol group and hence give water and with it leave it as to provide a negative charge to nitrogen and henceforth the water is left to make the way for carbon and nitrogen bond to form.Also notice there will be a positive charge on oxygen so would leave easilyand hence there will a negative charge on "NZ" leaving formation of the double bond. See this for data https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Bond_Energies and http://witcombe.sbc.edu/water/chemistryelectrolysis.html

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  • $\begingroup$ I strongly recommend not using all caps, this is considered rude. I also suggest using some paragraphs. When you're typing your answer, you need to leave one blank line between paragraphs. $\endgroup$ – orthocresol Jan 29 '18 at 19:14
  • $\begingroup$ Should i edit it $\endgroup$ – Rishi Kakkar Jan 29 '18 at 19:17
  • $\begingroup$ Of course. $\phantom{}$ $\endgroup$ – orthocresol Jan 29 '18 at 19:17
  • $\begingroup$ Its ok now or something else $\endgroup$ – Rishi Kakkar Jan 29 '18 at 19:22

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