- Aldehydes and ketones questions
- Nomenclature of aldehydes and ketones
- Physical properties of aldehydes and ketones
- Reactivity of aldehydes and ketones
- 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
- Cyclic hemiacetals and hemiketals
How addition of acid or base can catalyze the formation of hydrates and hemiacetals. Created by Jay.
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- How do you differentiate which carbon the base OH- attacks? In the enolate ion formation in the 'alpha carbon' section it deprotonated alpha carbon exclusively, but in this one it attacks the carbonyl carbon exlusively. How do you know when it will do what? (i.e. deprotonate alpha carbon so that it can form enolate vs. attack carbonyl carbon to form either gem diols or hemiacetal)(13 votes)
- what's the point of acid catalyzation, if water can attack carbonyl carbon even if carbonyl oxygen isn't protonated?(3 votes)
- when you catalyze it with an acid the equilibrium is shifted towards the product more than the reactants .(6 votes)
- For acid-catalyzed reactions, why wouldn't nucleophilic attack occur before protonation of the carbonyl oxygen? Or does the order of these steps not matter?(1 vote)
- Protonation of the carbonyl oxygen comes first, because it makes the carbonyl carbon even more susceptible to nucleophilic attack.(6 votes)
- you said this forms a hemiacetal but you used H+ as well as an alcohol -- so wouldnt that ALWAYS form an acetal. is there any way to stop at the hemiacetal(3 votes)
- Hemiacetals form quite nicely in an equilibrium reaction whether or not H⁺ is present.
RCHO +R'OH ⇌RCH(OH)OR'
Acetals need the H⁺ to protonate the OH group and facilitate its removal, but the reaction is still an equilibrium.
RCH(OH)OR' + R'OH ⇌ RCH(OR')₂ + H₂O
To isolate the acetal, you have to remove the water as it forms (with a Dean-Stark apparatus), otherwise the acetal will hydrolyze back to the hemiacetal.(3 votes)
- could the pi bond move to the next carbon in the R group instead of to the O?(1 vote)
- How does alkoxide make a better nucleophile that hydroxide?(2 votes)
- An alkoxide is a species where we have a anionic oxygen with a ionically associated (bonded) with a cationic alkali metal. Whereas a hydroxide is essentially an alcohol group. Note that once in aqueous solutions, the cationic alkali metal will dissociate. Of course if we dissolve an alcohol the OH group remains intact. Thus it's a comparison of whether an O- or an OH group is more nucleophilic in solution. And of course, the O- is certainly more nucleophilic given that the rest of the molecule is identical. Hope this helps.(2 votes)
- Which would occur faster, the acid catalyzed or base catalyzed?
Which ones are the rate determining steps out here?(2 votes)
- The speed to the reactions may not be so easy to figure out, as to the rate determining steps. But you can easily put together that, in base-catalyzed reaction, the rate determining step is when the pair of eletrons in OH attacks the carbonyl carbon, as in the acid-catalyzed reaction is a much more stable pair of eletrons that performs the attack over the eletrophile :)(1 vote)
- Which is more nucleophilic OR or OH?(2 votes)
- OR is essentially more nucleophilic as the carboxyl group has one sigma bond and one pi bond ,if you look closer after ionizaton the carbon gets nucleophilic in nature but whereas OH after ionizaton disassociates into H+ and O- so you cant really comment whether OH molecule is nucleophilic or electrophilic (and even after dissociation of OH ,O and H exist as independent ions ).So relatively speaking OR is the only molecule which has both electrophile and nuclophile hence OR is a nucleophilic molecule .(1 vote)
- Why would the alcohol be deprotonated by hydroxide in the base-catalyzed hemiacetal formation? The pKa of water is 15.7, whereas the pKa of alcohols are on average around 17. Is there a limitation on which alcohols can be used in hemiacetal formation?(1 vote)
- Ahh, but the hydrate is not an ordinary alcohol. It is a gem-diol. The electron withdrawing effect of one OH group enhances the acidity of the other. The pKas of gem-diols are usually in the range of 13 to 14. So a gem-diol is from 1000 to 10 000 times as acidic as an ordinary alcohol.(2 votes)
Voiceover: We've already seen how to form hydrates and hemiacetals in un-catalyzed reactions; in this video, we're going to see how to form hydrates and hemiacetals, in both acid and base-catalyzed versions. And so, we'll start with the acid-catalyzed: So here we have an aldehyde, or a ketone, and let's do hydration first. So, we know that in a normal hydration reaction, you just have to add water, but in an acid-catalyzed version, you would have to add a proton source, so H plus, and so you'd form hydronium, or H three O plus. And so, in an acid-catalyzed reaction, the first thing that's gonna happen is protonation of your carbonyl oxygen. So, lone pair of electrons on your oxygen here, are gonna pick up a proton from hydronium, leaving these electrons behind, here. So let's go ahead and show what happens: So we're going to protonate the carbonyl oxygen here, so we're gonna have a hydrogen attached, and give this a plus one formal charge, on our oxygen now. Our carbon is still bonded to an R group, and a hydrogen over here, and so, we could draw a resonance structure for this; we could show these pi electrons here, moving off, onto the oxygen, so let's go ahead and do that. So now, this top oxygen here would have two lone pairs of electrons around it, and we took a bond away from this carbon, so if we took a bond away from this carbon, we get a plus one formal charge. So let's go ahead, and put resonance brackets in here, and then, let's follow those electrons. So these pi electrons in here, move out onto that top oxygen, taking a bond away from your carbonyl carbon right here; that's gonna give it a full positive charge in this resonance structure, so plus one formal charge. And so, this makes your carbonyl carbon more electrophilic, which means a nucleophile can attack it better; a nucleophile's gonna be more attracted to it, and that makes your carbonyl more reactive, so acid catalysts work, by making your carbonyl carbon more electrophilic, which makes it more reactive, in a nucleophilic addition reaction. And so, another water molecule is gonna come along here. Let's go ahead and make that yellow again. So, a water molecule is gonna come along, and function as our nucleophile, next in our mechanism. So, lone pair of electrons on the oxygen are going to attack our carbonyl carbon, and so let's go ahead and show that. So, the oxygen adds on to our carbon here, and this oxygen still had two hydrogens attached to it, and it had used up one lone pair of electrons; it still has one more, which gives it a plus one formal charge. This carbon is still bonded to this oxygen, this oxygen also has two lone pairs of electrons on it, like that. And so, we can show our intermediate, so we have a hydrogen over here, like that. So let's show those electrons. So, lone pair of electrons on the oxygen here, are going to attack our carbonyl carbon, and form a bond, so those are these electrons in here. And then, once again, those pi electrons in here, in our carbonyl, had kicked off onto our oxygen, like that. So, that gives us our intermediate, and we're almost to the formation of a hydrate, the last step would be to deprotonate: So a molecule of water comes along, this time, it's gonna function as a base, and take this proton away, leaving these two electrons behind, on this oxygen. So let's go ahead, and show our product, formation of a hydrate, so we have a carbon bonded to two OH groups now, and so we form our hydrate, or our gem-diol. Let's follow those electrons now. So, let's say that these electrons right in here, moved off onto this top oxygen, doesn't really matter which lone pair you say they are, and that's formation of our hydrate. We could have followed through as a ketone, so instead of an aldehyde, we could have started off here as a ketone, and so, this would've been our prime, and then, this would've been our prime, and then, this would've been our prime, and then, finally, this would've been our prime; so, that's still formation of a hydrate. If we wanted to form a hemiacetal, we wouldn't have started with water; we would have started with an alcohol, so let's go ahead and show that. So let's make this R double prime OH, and, of course, an acid-catalyzed version, so R double prime OH and H plus, instead of forming a hydrate, this is gonna form a hemiacetal, so instead of H three O plus, which is what we had in our acid-catalyzed hydration, we're gonna protonate the alcohol, so let me go ahead and change this, this would no longer gonna be an "H," here, this would be "R double prime," like that, and so, when we protonate our carbonyl oxygen here, and make our carbon more electrophilic, this wouldn't be water as our nucleophile; this would be alcohol. So our alcohol molecule would be a nucleophile, and attack our carbonyl carbon, and so we could follow through that R double prime. So this wouldn't be an "H," this would be "R double prime," and therefore, in our final product, this wouldn't be an "H," this would be "R double prime," and hopefully you can see, that this is now a hemiacetal. It's actually difficult to isolate hemiacetals most of the time, and if you're doing a reaction in acid, usually, the hemiacetal is going to move on to form a full acetal. So let's go ahead, and show the structure of a full acetal. And, this isn't just one step, and I won't go through the steps in this video, that'll be in the next video, but I just wanted to show you the structure of a full acetal here. So we have R double prime, and lone pairs of electrons on our oxygen; and we have, over here, R double prime and lone pairs of electrons on our oxygen, so this would be an acetal, and let's, once again, compare structures. So, with an acetal, we have this OR double prime group, and another OR double prime group on our acetal. For our hemiacetal, we had only one of those OR double prime groups, and so, once again, we'll see much more details about acetals, in the next video. Alright, we can also form hydrates and hemiacetals, using base to catalyze the reaction, so, let's look at the mechanism for that. And so, let's first start with hydration again, and, over here on the right, is our aldehyde, or our ketone, in the un-catalyzed version, once again, we used water as our nucleophile. We know that our carbonyl carbon, right here, is partially positive, our oxygen is partially negative, and so, our water molecule functioned as our nucleophile, and attacked our carbonyl carbon. But in a base-catalyzed version, a base can take off this proton here, on water, and make it into a stronger nucleophile, so, it would form the hydroxide anion, which we know is a better nucleophile than water. So full, negative one formal charge on our oxygen, is gonna make our hydroxide anion a better nucleophile, and this is what's going to attack in a base-catalyzed version, so, nucleophilic attack, our nucleophile attacks our electrophile, pushes these electrons off, onto our oxygen, so let's go ahead, and show the results of that. So, now we would have, our OH is now going to be bonded to our carbon, so let's go ahead, and put that in there, so now we have only two lone pairs of electrons on our oxygen, so let's follow some electrons once again. So, these electrons right here, are gonna form a bond between that oxygen and that carbon, so let's say it's this right here, and, also bonded to that carbon, we would have an oxygen, which had two lone pairs of electrons: it got another one, so now it has a negative one formal charge, so let's follow those electrons too, so make those red. So right in here, these electrons go off onto that oxygen, once again, it doesn't really matter which lone pair you say they are: let's just say it's those. And also bonded to our carbon, we would have our R group, and our hydrogen, so here is our intermediate, and, in the next step, to form a hydrate, we would have to protonate. So, we would have a water molecule right here, could now function as an acid, so let's go ahead and show that. So this is gonna function as a base, so let's say these electrons, in red, pick up a proton from water, leaving these electrons behind, on the oxygen, and that will form our hydrate product, 'cause now we have this oxygen, over here on the left, and now we have this oxygen over here, on the right, has now been protonated, so let's go ahead and show those electrons in red, grabbed this proton, and so it formed this bond, right here, and we have formed our hydrate, so let me go ahead and put in this R, and this H. Once again, we could've carried this through, starting with a ketone, so let's do that really fast. So, instead of a hydrogen here, we could've had our R prime, so it would've been an R prime here, would've been an R prime here, and that's, of course, a hydrate as well. If we want to form a hemiacetal, once again, we would not start with water; we would start with an alcohol, so let's make this R double prime. And, if we use a base, we would take off the proton on our alcohol, to form an alkoxide, so this would be R double prime here, and an alkoxide anion is more nucleophilic than alcohol just by itself. And so the alkoxide attacks, and so, instead of this being a hydrogen, this would now be R double prime, and we could've shown this step, this wouldn't be water; this would be our alcohol, so I could make this R double prime here, and then, finally, for our hemiacetal product, this wouldn't be hydrogen; this would be R double prime. And so we can see, we have now formed a hemiacetal as our product. Now, in base, a hemiacetal will not react to form an acetal, so, that's kind of an important point, when you're using acetals as protecting groups, they are stable in base. So you can't form acetals in base, and so, therefore, once you have an acetal, it's stable to base. And so, we'll see much more about that in the next few videos.