- 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 acetals are used as protecting groups. Thioacetals are similar to acetals, but form from reaction of an aldehyde with a thiol (not an alcohol). Created by Jay.
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- at4:00why doesn't the carboxylic acid get turned into an acetal?(15 votes)
- Only aldehydes form acetals. Ketones, as in this case, form ketals. Carboxylic acids form neither.
In fact, under the mild conditions of ketal formation, they don't even form esters with the ethylene glycol.
That's why you can protect the ketone while you do a reaction on the carboxylic acid.(11 votes)
- Why would I use this seemingly more complex route, if I could just use BH3/THF as shown in the video "Reduction of carboxylic acids"? There you say that BH3/THF reduces the carboxy-group without touching the carbonyl-group.(7 votes)
- This reaction protects the carbonyl group for more than just this example. In fact, this will work with an ester group attached as well. Not to mention, simply using LiAlH4 will reduce the ketone as well. Sometimes this process is needed in lab or practice, which is why it is important to know as an option.(5 votes)
- Doesn't the 'ethylene' imply the presence of a double bond when in fact only single bonds (associated with -anes) exist?(4 votes)
- Yes, strictly speaking you are right. Its proper name is ethane-1,2-diol but it is very commonly referred to as ethylene glycol. This is because it is manufactured from ethylene. Likewise, there is, for example, propylene glycol which is propane-1,2-diol and which also has no double bonds. It's a case of remembering these non-systematic names because you will see them used a lot by chemists.(5 votes)
- It is necessary to use 1-2-ethanediol for this reaction? Could we have used a simpler acetal such as CH3OH?(2 votes)
- Not necessarily, any alcohol works, but 1,2-ethandiol's reaction with carbonyl groups to form acetals is much more favored by equilibrium than single alcohols due to entropy.
To protect a ketone using a straight chain alcohol, like methanol, you react two molecules of alcohol and a ketone and produce one molecule of water and an acetal. To protect a ketone using a diol, you react one molecule of diol and a ketone and produce one molecule of water and an acetal. In the first example, you reacted 3 molecules to form 2, which decreases entropy (since entropy favors having more molecules bouncing around in solution), whereas in the second example you reacted two molecules to form 2, which is much more favorable from an entropy standpoint.
This means you don't need to use as much alcohol or acid or heat to completely react the ketone with the alcohol if you're using a diol as opposed to a single alcohol. In terms of protecting on a test or homework assignment, it doesn't matter, because an acetal from methanol will protect a carbonyl just like an acetal from 1,2-ethandiol would.(3 votes)
- At8:10when talking about reducing the Cyclo-dithiol acetal and adding the two hydrogen groups, is this all done by Raney Nickel?(1 vote)
- Yes, Raney Nickel is the only reagent needed to react with the cylic thioketal to produce an alkane. Raney Nickel is also good for the hydrogenation of Alkenes and Alkynes.(2 votes)
- around4:00, can the acid protonate the hydroxyl portion of the carboxylic group, thereby forming a water molecule which may become a leaving group?(1 vote)
- Yes, it can, and it will form an oxonium ion which is a good leaving group. But it won’t leave, because the carbocation that would form is highly unstable.(2 votes)
Voiceover: We've already seen how to form acetals. If we start with an aldehyde or a ketone, and we add an excess of alcohol in an acidic environment we can form our acetal. We talked about ways to increase the formation of your acetal by removing, in this case, water from your reaction to drive the equilibrium to the right. But let's say you increase the concentration of water, so let's say you increase the concentration of this. You would push the equilibrium to the left, and you would hydrolyze your acetal and convert it back into your original aldehyde or ketone and your alcohol. So this can actually be a useful reaction if you want to use an acetal as a protecting group. And so let's look at an example of hydrolyzing an acetal. So, if we have this acetal over here on the left, which is the same one that we made in the previous video, let's see how to analyze the products that we would make by hydrolizing it. So we add HCl as our acid catalyst, and we add an excess of water. And so we're going to get back our original aldehyde and alcohol here. So let's analyze the structure of this. So this carbon right here is the important one so let me go ahead and change color here. So this carbon right here in magenta, if we go back up to here that's this carbon we're talking about here. And so we know that we have a carbon attached to that so that would be this R group right here. And then we also have a hydrogen coming off of that carbon, so that would be this hydrogen right here. And so immediately we can work backwards and see we have an R group we have a hydrogen, we're dealing with an aldehyde right here. Specifically a two-carbon aldehyde, right? So there's two carbons on this, so acetaldehyde is one of our products. So let's go ahead and draw acetaldehyde here, a two-carbon aldehyde. And then our other product, we know is going to be an alcohol. So if we analyze the structure of our acetal, we should be able to figure out what kind of alcohol it is. So if we look at that, that's like this portion right here, which we know came from our alcohol. So all we have to do is add on a hydrogen and then we have the structure of our alcohol product here. So it would be a four-carbon alcohol, so one, two, three, and four, and the exact same thing down here. So our other product would be butanol. And we have two equivalents of it. So that'd be one, two, three, four carbons. So, by looking at our acetal and thinking about hydrolyzing it and thinking about where those portions came from, we can easily come up with the products of this hydrolysis of this acetal. So let's see how we could use this reaction in, if we were trying to synthesize this molecule over here on the right. And so, if we're trying to synthesize this molecule from this molecule, first you might be able to think you could figure it out by looking at the functional groups. Alright, so we have a ketone over here on the left and we want a ketone over here for our product. We also have a carboxylic acid over here to the left, and then we have an alcohol over here on the right. So, if you're thinking about how to complete this kind of transformation, you might think to reduce the carboxylic acid to form your alcohol. And so, you could do that with something like lithium aluminum hydride. So you might think, and you could just add some lithium aluminum hydride here, in the first step. And the second step, add a source of protons to protonate your alkoxide anion to form your alcohol as your product. The only problem with trying to do this in one step is lithium aluminum hydride is also going to reduce your ketone here, to form a secondary alcohol. And so, this will not work. The first thing you have to do is protect the ketone, and then you can use your lithium aluminum hydride. So, we're going to use, we're going to protect our ketone using an acetal, of course. And we're going to react it with ethylene glycol. So if we react our starting compound here with ethylene glycol, and we use an acid catalyst, we're going to form an acetal. Alright, so the ethylene glycol is going to react with the ketone portion of the molecule to form an acetal, specifically a cyclic acetal. Alright, so over here on the left we still have our carboxylic acid. And over here on the right, now we're going to have an oxygen bonded to this carbon, another oxygen bonded to this carbon, and they are, of course, connected. And so, that's our cyclic acetal, which came from this oxygen, this carbon, this carbon, and this oxygen. Right, so that's this oxygen, this carbon, this carbon, and this oxygen. So acetals are stable in basic conditions. And so, now we can add our lithium aluminum hydride, so we add our lithium aluminum hydride here in the first step. And that is going to reduce our carboxylic acid. And so in the second step we can add a source of protons, and we can add some excess water. And so we can protonate the alkoxide anions to form our alcohol product. And also we can add an excess of water in an acidic environment it's going to hydrolize our acetal. So we're also going to hydrolize our cyclic acetal here, and get back our original ketone. And so we've now been able to make our desired product. And so that's one way to use an acetal as a protecting group. Alright, let's look at another type of reaction here. So this one's formation of a thioacetal. So this is completely analogous to the formation of an acetal. So you start off with an aldehyde or a ketone. And this time, instead of using an alcohol you're going to use a thiol, so instead of having an oxygen here, you have a sulfur. So you need acidic conditions, and you can even use something like boron trifluoride here, which can function as a Lewis acid catalyst for this reaction. And you form a thioacetal as your product. So once again you have your R double prime group coming from your thiol, and then you have a sulfur here instead of an oxygen. And then you have two of these to form your thioacetal. And so, one reason you might want to form a thioacetal instead of an acetal, is thioacetals have an additional reaction that they undergo, and we can use it in this transformation. So let's say our goal was to go from the product on the left to the product on the right. And so you can see that we have reduced our ketone here to form this cyclohexene portion. So reducing your ketone. So the first thing we could do, to do this transformation, is to form a thioacetal. So it's going to be, again, analogous to the formation of an acetal. This time we're going to use, instead of a diol, we're going to use something with two SH groups instead here, so a dithiol here. So we can use an acidic environment here. And we're going to form a cyclic thioacetal. So very similar to what we did before. The ring is here, the carboxylic acid is going to be untouched. And instead of an oxygen directly bonded to our ring, we're going to have a sulfur. And the same thing on this side, a sulfur. And then our two carbons. So once again, let's follow those atoms here. So, a sulfur, two carbons, two carbons, and a sulfur. So here's your sulfur, two carbons, two carbons and then another sulfur, so a cyclic thioacetal And then there's actually a reaction for converting this compound into our target compound. And this involves a special kind of nickel, so if we use Raney nickel here. It's a special kind of finely divided nickel that has already adsorbed some hydrogens. And what it's going to do is add some hydrogens to this carbon right here. So we're going to add two hydrogens to this carbon. And you can see that we have reduced the thioacetal and we have formed our desired product. So this is another way to reduce a ketone here, to this alkane portion of the molecule. And so there are other ways to do it, and this just gives you another tool that you can use for synthesis problems.