- Formation of hydrates
- Formation of hemiacetals and hemiketals
- Acid and base catalyzed formation of hydrates and hemiacetals
- Formation of acetals
- Acetals as protecting groups and thioacetals
- Formation of imines and enamines
- Formation of oximes and hydrazones
- Addition of carbon nucleophiles to aldehydes and ketones
- Formation of alcohols using hydride reducing agents
- Oxidation of aldehydes using Tollens' reagent
Formation of oximes and hydrazones
How aldehydes and ketones can react with hydroxylamine to form oximes or hydrazine to form hydrazones. Created by Jay.
Want to join the conversation?
- In these reactions, why is the nitrogen acting as the nucleophile instead of the oxygen? The oxygen has more electron density, so why wouldn't it react with the partially positive carbon instead?(5 votes)
- The O does act as a nucleophile in competition with N, but it's a dead-end process.
Reaction with O gives the reversible formation of a hemiketal.
Reaction with N forms the oxime in an essentially irreversible process as the adduct dehydrates.
As the ketone gets used up, the hemiketal decomposes in an attempt to maintain the concentration of the ketone (Le Châtelier's Principle)(11 votes)
- In an intermediate step of forming an oxime: why doesn't the OH on the hydroxylamine form water and make a leaving group (as it already does with the double bonded O on the reactant)? approx @3:00(6 votes)
- I think the N and the O are very similar in electronegativity, so the O can't pull out the electrons from the N. Hence, they stay stuck together.(3 votes)
- at7:10why doesn't the other nitrogen of the hydrozine react with another molecule of cyclohexanone resulting in R=N-N=R (where R=cyclohexane)?(3 votes)
- The lone pair on the NH₂ group in the hydrazone R₂C=N-NH₂ is delocalized by resonance with the C=N bond, so it isn't as nucleophilic as the lone pair in H₂N-NH₂ itself.(3 votes)
- Does the hydrazone have sterioisomers like the oxime?(1 vote)
- Yes, hydrazones have E and Z stereoisomers, but the energy barrier for interconversion is so low that it is difficult to isolate the isomers.(3 votes)
- What would be the subclassification of the stereoisomer formed at6:10?
Thanks :^)(1 vote)
- The one on the left is E, and the one on the right is Z.
I think the IUPAC names would be:
(1E)-1-(1-methylethyl)-N-hydroxyethanimine and (1Z)-1-(1-methylethyl)-N-hydroxyethanimine, but I find the nomenclature for these compounds counterintuitive, so no guarantees.
The following seems somewhat helpful:
http://www.adichemistry.com/organic/basics/isomerism/geometrical/geometrical-isomerism.html (search for: oxime)(2 votes)
- Jay said toward the beginning, that the acid was protonated amine. However, toward the end, he just said acid catalyzed. For all these reactions, can the acid catalysis be from the protonated amine? Or can you also add another acid? Or would a strong vs. weak acid proceed via another reaction? I'm curious in case the MCAT puts H2SO4 in there or something as a trick to totally undo what I've learned.(1 vote)
- At6:00he says that these are stereoisomers of oximes.Are they enantiomers or other kind of stereoisomers?(1 vote)
- No, they ain't enantiomers because there is not a chiral Carbon.(1 vote)
- with the last example given, is there any minor product formed of the cyclohexanone reacting with the secondary amine in the other molecule forming an enamine product?(1 vote)
- When comparing hydrazone and the oxime, which one is more stable? Is it oxime, since oxygen is more electronegative than nitrogen?(1 vote)
- It depends on the number of resonance structures formed by the compounds!(1 vote)
- whether cyclopentanone will react with cyclohexanone with hydroxylamine in the same way leading to beckmann arrangement?(1 vote)
Voiceover: In the last video we saw the mechanism to make imines. And to make an imine, we started with an aldehyde or ketone, added an amine, used an acid catalyst, and we formed our imine. And if this Y here is equal to a hydrogen, or an alkyl group, we called it an "imine." If that Y is equal to an OH, we would call it an "oxime." So let me go ahead and write that, so this would be an oxime as your product. And so the word, "oxime" is kind of like a combination of "oxygen" and "imine." And then, another derivative would be if the Y group was equal to NH two, or if the Y group were equal to NH, and then have another alkyl group in there, so R double prime, we would call it a "hydrazone," so this would be a hydrazone derivative, and so, the mechanism would be the same as what we talked about before. And if we formed any of these, an imine, an oxime, or a hydrazone, and we wanted to go from those products, back to our original starting aldehyde or ketone, we could just increase the concentration of water. So, an excess of water, again, an acid-catalyzed reaction, could push your equilibrium back this way, and give you back your amine, and your aldehyde or ketone. So, let's take a look at an example of that. So, here we have our imine, and to it we're adding aqueous acid, so an excess of water here, and so, we're gonna get back our original aldehyde or ketone. And if we think about it, if we look at our product over here, we have a nitrogen double-bonded to a carbon, and then an R group, and an R group, so nitrogen double-bonded to a carbon, R group, and R group; so, it's R and R prime, so going backwards, we'd be starting with a ketone here, and so we would get cyclohexanone. So let's go ahead and draw cyclohexanone in, as one of our products. So, for our hydrolysis of an imine, we would get cyclohexanone, and then, our other product, we could see right in here, this portion, we would get back our amine, and so we would get back our nitrogen, and then, think about the mechanism, we lost two protons in that mechanism, so we get back our primary amine, like that. And so, this reaction works, of course, for oximes or hydrazones as well. But let's look into the formation of oximes and hydrazones here. So, let's look at another reaction. Once again, we're starting with cyclohexanone, but this time, we're dealing with hydroxyl amine, so this guy right here, once again, with an acid catalyst. And so, here our OH is our Y group, so let's go back up here, and identify that again, let me use a different color. So our OH is our Y, so right here; it's this portion, so we're gonna put an OH on the nitrogen that's double-bonded to our carbon. So let's go ahead and draw the products, for our oxime here. So, we're going to start with our ring. So we have our ring here, and then we're gonna have our carbon double-bonded to a nitrogen, and this time, instead of it having a hydrogen or an alkyl group, we're gonna have an OH, so we have an OH right here, and let's go ahead and put in some lone pairs of electrons, and so this would be formation of an oxime, so this is our oxime product. And, oximes are more stable than imines, and the reason for that has to do with the fact that we have this oxygen here, with a lone pair of electrons, so we could think about moving in this lone pair of electrons into here, and pushing these electrons off, onto this carbon. So, if we were to draw a resonance structure, we could go ahead and show, once again, here's our ring, and then here we have our carbon bonded to our nitrogen, and then, now there would be a double bond between this nitrogen and this oxygen, and this oxygen would still have a lone pair of electrons on it, giving it a plus one formal charge; this nitrogen would have a lone pair of electrons on it, and then we move some electrons out onto this carbon, so this carbon right here gets a negative one formal charge. So let's show some of those electrons. So, these electrons right here, on the oxygen, moved into here, and then we could show these electrons, kicking off onto this carbon, to form a carbanion here, for our resonance structure. So we can de-localize some of those electrons, and you could think about this resonance structure over here, with this little bit more negative charge, helping to stabilize this carbon, which we know, is partially positive. So it's partially positive, over here on the left, and so that electron density from that resonance structure helps to stabilize it a little bit, and so this is one way to look at why an oxime is more stable than an imine. so let's look at this ketone, so, so far, we talked about using a symmetrical keytone, with R groups that are the same on both sides, but this time, we're dealing with an un-symmetrical ketone, and so, when we add our hydroxylamine, we're gonna get an oxime product, but we get two possible products here. So let me go ahead and draw them out, so because we're dealing with an un-symmetrical ketone to start with, we could show lone pair of electrons and nitro on one side, and the OH on the other side, so that's one of our possible products. And then we could show a stereoisomer to that, we could show the lone pair on the left side this time, and the OH on the right side. And so, these are stereoisomers, so if we examine them a little bit more in detail, the OH is opposite side from this, if you're thinking about the double bond, and the double bond here, you can think about the OH being on the same side as this. So, stereoisomers are possible, if you're not starting with a symmetrical ketone, here. And the same thing happens with imines, but oximes are more stable than imines, and so it's easier to isolate the oxime isomers, once they are formed. And so, that's something to look out for, on reactions. Alright, let's do another reaction: and this one's a little bit different than before. Alright, so instead of having an OH here, we have an NH two as our Y component here, so let's go back up, and look at our generic reaction again. So, an NH two is now our Y right here, and so, when the Y is equal to an NH two, we're dealing with a hydrazone, as our product. So, once again, same mechanism, but let's think about what the product would look like here: So, reaction of hydrazine, so this guy right here is called, "hydrazine," with cyclohexanone, is going to give us our ring, so a double bond to this nitrogen here, and then, this time, our Y is going to be NH two, so, we go ahead and put "NH two" coming off of here like that. And so, this would be a hydrazone, so put in my lone pairs of electrons right here. And so, once again, hydrazones are also more stable than imines, and, once again, this had to do with the fact that this nitrogen here has a lone pair of electrons, and so, we can do the same thing that we did before. So, this is a hydrazone, hydrazone product, so, "hydrazone," like that. Alright, let's do another example of a formation of a hydrazone: So let's look at this guy, little bit more complicated looking, but you can see that we have here our NH two, and then we have all of this; we have all of this up here as well, and so we can think about this part reacting with our carbon EEL, so acid-catalyzed reaction again. So it looks a little bit intimidating, let's get a little bit more space here, but really, we're just gonna form hydrazone, so you can just think about all this as being an R double prime group, if you want to. So let's get a little bit more space, and let's draw the product: So, we're going to have our ring, like that, and then we're going to have it double bonded our nitrogen, so you think about this nitrogen, right here, and then, we can just go ahead and draw the rest of the molecule, here. So, since its symmetrical, it doesn't really matter which side I put it, so I'll just put it on the left here. So nitrogen, bonded to a hydrogen, and then we have our benzene ring here, so go ahead and put that in, like that, and then we have these nitro groups, coming off of our benzene ring, and so, this is a famous, historical reaction. So this compound that we are reacting our cyclohexanone with, is two four DNP, so "2, 4 DNP," or "2,4-Dinitrophenyl-hydrazine." So this would be two nitro, four nitro, dinitrophenyl-hydrazine, so you can see this derivative here, of hydrazine. And the reason why this is useful, historically, is this is a diagnostic test for an aldehyde or a ketone, because it reacts with aldehydes or ketones, usually to give an orange or red precipitate, and that solid usually has a good melting point, so you can characterize aldehydes and ketones. And so, before the advent of things like NMR, this helped with structure determination, and there are all kinds of tables, listing these hydrazone derivatives, and so this is more of a historical reaction, than anything else. But it is kind of interesting, to look at it here.