Condensation reactions are those that involve removal of a water molecule. When you analyze the reaction between an aldehyde and hydroxyl amine, I don't understand why it's not possible for the acidic hydrogen of the aldehyde to be abstracted by the $\ce{OH-}$ radical of the amine, thus leaving behind an amide ($\ce{R-CO-NH2}$) rather than an oxime being formed?

  • 6
    $\begingroup$ The aldehyde hydrogen (the H in -CHO) is not acidic, so it can't leave as H+. Furthermore the hydride ion is an exceedingly poor leaving group, so it can't leave as H-. Point is, there's no way of getting that hydrogen away from the carbon. $\endgroup$ Oct 3, 2015 at 12:25

2 Answers 2


I don't understand why it's not possible for the acidic hydrogen of the aldehyde to be abstracted by the $\ce{OH-}$ radical of the amine

There is a lot wrong in this sentence and that’s the key to the answer. For one, there is no hydroxide radical (neither in hydroxyl amines nor anywhere else); if anything it is a hydroxyl radical $\ce{HO^{.}}$ — but it is not present in hydroxyl amine. If you check out the mechanism of the condensation, you will realise that no basic hydroxide species occurs:

hydroxyl creation mechanism

If any basic species ever exists in here, it is the oxygen anion in the first step. Note especially that the hydroxyl group on nitrogen never loses its proton and thus never becomes basic in this mechanism.

I can hear you asking ‘so why doesn’t the oxyanion in step two abstract the neighbouring hydrogen?’ And this is really the key:

That hydrogen is not acidic!

Never, never, never ever deprotonate this hydrogen on an aldehyde. The only way to make it moderately acidic (but you would still need very strong bases to do so) is something along the lines of the Corey-Seebach umpolung. It generates a cyclic S,S-acetal out of your aldehyde which is stable. Its $\mathrm{p}K_\mathrm{a}$ is low enough that it is actually accessable. That of an aldehyde is not.

Also, to finally generate an amide, you would need some kind of migration of the hydroxyl oxygen to the aldehyde carbon after hypothetic deprotonation. I’m not saying it’s impossible, but rather unlikely (but that is difficult to judge due to the unlikelyness of the previous step). Note that an amide’s carbon has an oxidation state of $\mathrm{+ III}$ while the oxime’s carbon has $\mathrm{+I}$. You need a redox mechanism to get there.


It is not the $\ce{-OH}$ which would attack the $\ce{-C=O}$ bond i.e., the carbonyl bond of aldehyde, but its the lone pair of electrons on $N$ which will attack it.

To make it simple, first of all you need to know whats happenings here, the carbonyl bond $\ce{-C=O}$ is not electrons rich because of the EN $O$ bonded to it, so the best neucleophile will be the one to attack the $\ce{-C=O}$ bond.

General important method to remember

Any carbonyl group, aldehyde or ketone when reacts with (excess of) $\ce{NH2-Z}$ will give $\ce{H2O}$ and $\ce{C=NZ}$, where Z can be any thing $H$ or $-OH$, etc.In an acidic Medium.

Basically what you need to remember is $\ce{H2}$ of $\ce{C=NZ}$ reacts with $=O$ of $\ce{-C=O}$ to form water. this is an easy way to remember but this is not really happenings, not directly, there are few steps occuring in between.


$\ce{R-HC=O}$ + $\ce{NH2-OH ->R-HC=N-OH + H2O}$

$\ce{R-HC=N-OH}$ will oxidise to $\ce{R-HC=N=O}$

  • 1
    $\begingroup$ The last statement (R−HC=N−OH will tautomerise to R−HC=N=O) is incorrect. As written, that's an oxidation, not a tautomerization. $\endgroup$
    – jerepierre
    Dec 15, 2015 at 20:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.