- 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 aldehydes (or ketones) react with alcohols to form hemiacetals (or hemiketals). Created by Jay.
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- It would be nice if there was an explanation to what a hemiacetal is as compared with an acetal(19 votes)
- The difference between hemiacetals and acetals is as follows: A hemiacetal has a ROROH group, while an acetal has a ROROR group.(20 votes)
- Could the first intermediate (2:50) react intramolecularly? The proton would come from the positive oxygen. The negative oxygen would attack the proton and the formerly-positive oxygen would now be neutral. Is that possible?(3 votes)
- It's possible but unlikely, because the bond angles would have to be 90 ° in the transition state.
Another possible process would be intermolecular transfer between two of the intermediates, but there are so few of them that it is unlikely they would collide with each other.
The most likely process is deprotonation by a solvent molecule, because there are many solvent molecules available to collide with the intermediate.(10 votes)
- cant we use water instead of the alcohol in the reaction?(3 votes)
- No. If you use water, you will get a 1,1-diol, which is too unstable to be isolated, because the equilibrium shifts back to starting materials.
RCHO + H₂O ⇌ RCH(OH)₂
Most aldehydes form equally unstable hemiacetals.
RCHO + R'OH ⇌ RCH(OH)OR'
The most common exceptions are mono- and disaccharides (sugars).(5 votes)
- Okay, so in the very first step, we have a nucleophillic attack of oxygen. my question is that wouldn't it face steric hindrance or electrostatic repulsion from the oxygen in the aldehyde group? and if there is more electropositive alkane chain in position of 'R' , wouldn't the oxygen go there?(2 votes)
- To answer your second question: the carbonyl carbon is extremely likely to be the most electropositive carbon in the molecule, we know this because a resonance structure can be drawn where the carbonyl carbon has a full positive charge and the oxygen has a full negative charge. As for your first question, this makes aldehydes and ketones very electrophilic and subject to nucleophilic attack, regardless of any steric hindrance from the R group(s).(2 votes)
- At around3:40in the video, when the protination of the negatively charged Oxygen is done, we form an alkyl oxide, can this cause the molecule (the hemiacetal/ketal) to react further? like a substitution reaction?(2 votes)
- How exactly did those two chair conformations formed? I didn't get it Sal.(2 votes)
- The sigma bonds in the intermediates can always rotate. In order for the intramolecular reaction to occur, the lone pairs on the OH want to be as close to the carbonyl of ketone, or aldehyde, as possible, and to do so they will adopt the conformers. The chair conformers are conformers of cyclic product that forms as a result of the intramolecular attack.(1 vote)
- At2:35, can the oxygen that has the negative charge grab the hydrogen, that attaches to the positive oxygen, instead of using another R-OH group to deprotonate the hydrogen? I hope you understand my question. Please help thx(1 vote)
- It's quite probable that the same R-OH molecule facilitates the transfer of the proton from the positively charged oxygen to the negatively charged oxygen. Therefore, while Jay shows the deprotonation and the protonation as two distinct steps, they probably merge into one which amounts to the transfer of the proton from one oxygen to the other oxygen, as you suggest.(2 votes)
- where is the name "hemiacetal" derived from?(1 vote)
- Probably because the first reaction was carried out with acetaldehyde.
CH₃CHO + CH₃OH → CH₃CHOHCH₃ → CH₃CH(OCH₃)₂
acetaldehyde → hemiacetal → acetal
hemi means half, and the hemiacetal was half-way from acetaldehyde to the final product.(2 votes)
Voiceover: In the previous video, we saw how to make hydrates from aldehydes and ketones, and this video will show how to form hemiacetals from either an aldehyde or a ketone. And so the difference is, instead of adding water, this time we're adding an alcohol and so this reaction is at equilibrium. And then over here on the right, our product this time is a hemiacetal, so this is a hemiacetal right here. So in terms of the mechanism for the formation of a hemiacetal, it's completely analagous to the formation of a hydrate, right? We have our carbonyl situation over here on the left, for our aldehyde or ketone, with the oxygen being more electronegative and withdrawing some electron density away from our carbonyl carbon. So we have a partially negative oxygen and a partially positive carbonyl. So this carbonyl carbon right here is partially positive, so it is electrophilic. And so alcohol can act as a nucleophile. So our alcohol molecule, I'll go ahead and draw it out here, is going to function as our nucloephile in this reaction. So lone pair of electrons on our oxygen is going to attack our carbonyl carbon right here, and push these pi electrons in here off onto our oxygen. So our oxygen already had two lone pairs of electrons like that. So let's go ahead and show the result of that nucloephilic attack. So the oxygen now is bonded to our carbon. So here's our oxygen, it's formed a bond to our carbon. And there was a hydrogen attached to our oxygen, and also an R double prime group, like that. And there was still one lone pair of electrons on our oxygen, giving our oxygen a plus one formal charge. This carbon was bonded to another oxygen, which used to have two lone pairs of electrons, it just picked up another lone pair, so it now has a negative one formal charge like that. And then we still had an R group and then a hydrogen. So let's follow some electrons. All right, so these elctrons right here in magenta on our alcohol formed a bond between the oxygen and the carbon, so these electrons right in there. And then our pi electrons in our carbonyl here, moved out onto our oxygen, so it doesn't really matter which lone pair we say it is. Let's just say it's that one right there. And so this is our intermediate. All right, next step would be to deprotonate this intermediate here, so we could have another molecule of alcohol, so let's go ahead and draw that in here, so another molecule of an alcohol, which in this case can function as a base. All right, so we can think about lone pair of electrons on this alcohol taking this proton, and leaving these electrons behind on this oxygen. So let's go ahead and draw the results of that acid base reaction. So we deprotonate, and now we have our oxygen bonded to our R double prime group, and then we still have this oxygen over here on the right, with a negative one formal charge. We still have an R and a hydrogen. So on this oxygen right there, is now another lone pair of electrons which came from the deprotonation step. Let's go ahead and show that. All right, these elctrons in here, between the oxygen and the hydrogen are now on the oxygen, so that takes away the plus one formal charge from the oxygen, and we have this. So we're obviously extremely close to our product, we only need one more acid-base reaction. And alcohols are amphoteric like water, so they can function as acids and donate protons as well. So let's go ahead and show that. So we have another molecule of alcohol come along, and a lone pair of electrons takes this proton, leaves these electrons behind. And then we protonate, and we can form our hemiacetal. Once again, we could have carried through an R prime group, right, like that if we had started with a ketone, and we would get this right here instead of the hydrogen for our hemiacetal product. So both are hemiacetals. And so this is the general mechanism to form a hemiacetal. Now this is not acid or base catalyzed, so I will cover that in the next video. So the formation of hemiacetals, usually the equilibrium is actually favors the formation of your aldehyde or ketones, so it's usually back here to the left. However, for formation of five or six numbered rings in an intra-molecular hemiacetal formation, the equilibrium is actually to the right. And so this is a very important reaction. So let's take a look at it right here. So here we have over here an aldehyde and an alcohol in the same molecule, so this is going to be an intra-molecular hemiacetal reaction. So we're going to push the [equilibrium] to the right, so it's actually to the right from these, we form a cyclic hemiacetal over here on the right. Let's go ahead and number our carbon so we can try to figure out what happened here. This carbon right here will be number one, two, three, four, and five like this. And we know what happens in the mechanism, we know that the alcohol functions as a nucleophile, and attacks the carbonyl carbon. So this oxygen right here must be able to swing around and attack this carbonyl carbon and push these electrons off onto this oxygen. So let's go ahead and follow our carbons over here on the right. So this carbon turns out to be number one. This carbon's number two, three, four and five. And so this oxygen right here on our ring was this oxygen, and then let's go ahead and label the other oxygen as well. So this oxygen becomes this oxygen. And so formation of cyclic hemiacetals is extremely important in biochemistry, and carbohydrate chemistry. And so let's go ahead and go through this in a little bit more detail here. And so here I have the exact same molecule, except I've shown it in a different conformation. I've shown the sigma bonds here rotated a little bit differently. So all these sigma bonds in here have free rotation. So they're going to rotate to allow our necleophile to attack our electrophile a little bit easier. So I can think about lone pairs of electrons on our oxygen, and now it's a little bit easier to see the nucleophilic attack. It's going to attack right here at this carbon, and push these electrons in here off onto this oxygen. So let's go ahead and show the result of that. So now I have my oxygen, so it is bonded to this carbon now, and I've formed my ring. So let's follow those electrons, these electrons right here in magenta now form this bond right here. And I'm going to say that this oxygen right here in red is actually going to go up in the plane, so we'll come back to that in a minute. So this oxygen right here is going to go up. It had two lone pairs of electrons around it. It picked up another one, so that's where our negative one formal charge is. The oxygen in our ring still has the hydrogen bonded to it, and a lone pair of electrons that gets a plus one formal charge like that. And then I'm saying this hydrogen here is down, so this is our intermediate. And without showing all of the acid-base steps here, let's just think about what happens. We know that we next deprotonate, so a base comes along and takes this proton, and then these electrons in here move off onto our oxygen. And then we know after that we protonate our negative charge right here. So let's go ahead and draw one of the possible products. So we have our oxygen, that's part of our ring here, drawn in the chair conformation. We now have an OH right here, and then, so we have an OH equitorial and then we have a hydrogen axial here as one of our possible products. And then two lone pairs of electrons on this oxygen. Hopefully you can see how this is one of the possible products and that here we drew it with like a flat plane, and here we've drawn it in more of a chair conformation. So the OH equatorial for this one. Let's go back to our original situation over here on the left, where the nucleophile attacked the carbonyl carbon. We know that the geometry at our carbonyl carbon is trigonal planar, so it's possible the nucelophile could attack from the opposite side and if that happened, then the oxygen would go down relative to the plane. So let's go ahead and show that now. We have another possibility here. So let's go ahead and draw it out. So we have our oxygen is now going to be part of our ring, so we've formed our ring. This time instead of putting the oxygen going equatorial, we're going to put the oxygen going down. We're going to put it going axial, and so it has three lone pairs of electrons around it, so negative one formal charge. And the hydrogen goes equatorial like that, and then we still have a lone pair of electrons on this oxygen, and then hydrogen and a plus one formal charge. So this is another possibility. So once again, thinking about the mechanism, we know a base deprotonates, so these electrons kick off onto here, and then we know that next we protonate our negative one charge like that, and that forms our hemiacetal. So let's go ahead and draw it over here. So here's another product. So let's go ahead and this time we have our hydrogen equatorial, and our OH is now axial. And so these are our two possibilities. When I think about this carbon right here, I think about stereo chemistry of this carbon. This is a chiral center. It's chiral. So we have these two as our possible products. Now this is extremely important when you get into carbohydrate chemistry, so these acetals differ at carbon one. So this would be the carbon one position, and when you do carbohydrate chemsitry, this is called the anomeric carbon. And if the OH is up or down, those are different anomers. This is extremely important, once again for something like glucose. And so we'll definitely talk about that in a later video. So the formation of cyclic hemiacetals is an extremely important reaction to understand.