Concept: Concept: Protecting Groups3m
Now let's talk about a synthetic application of acetals and that's to use acetals as protecting groups. Let's look at these two molecules above me. What we’ll find is that these molecules are in equilibrium. We have learned word that if you expose a carbonyl to alcohol and acid, you can basically make an acetal. In this case since this is a cyclic acetal, I would expect to be using a cyclic diol in combination with acid. I’ll just put H+ to give me my cyclic acetal.
What's important about the difference between these is that there's a huge difference in reactivity between my original carbonyl and my acetal. Notice that my carbonyl has an extremely reactive partial positive pi bond. Whereas an acetal does have dipole so it does have dipole similar to the carbonyl. But notice that they're located on sigma bonds not on a pi bond. All of these bonds are extremely difficult to break. In fact, an acetal is about as reactive as an ether. Just remember how reactive ethers were. Pretty much unreactive. Ethers barely do anything. All that you can do is combust them. Sure, they blow up. But then others in combustion they don't react and that’s because these single bonds are extremely difficult to break.
If I ask you which one is the safer version, if I was to run a reaction somewhere else on the molecule, which one is safer to have around, a carbonyl or an acetal? The answer is an acetal because an acetal really isn't going to react with almost anything. Whereas a carbonyl is so reactive, we’ve learned they can even react with water, so we want to protect it. That's the whole idea behind a protecting group. Acetals are used to protect sensitive aldehydes and ketones from reaction with other reagents since they’re reversible. The idea being that you can turn the carbonyl in to an acetal, do your reaction somewhere else and when you're done, you can go ahead and hydrolyze it back to the original carbonyl, sparing it from reaction with your second reagent.
I want you to look at this example. I want you guys to devise a synthesis for this. I want you to look at these two molecules and figure out what would be the best way to make that first molecule into the second one. I’m going to give you a hint. This is not a one-step reaction. If you try to do it in one step, you’re going to fail. Try to think about what reactions you could and what sequence to make this transformation happen and then I'll show you.
Concept: Example: Acetal Protecting Group5m
Alright, so the first thing you might have noticed is that actually switched the molecule on you guys by a little bit. I decided to put a nitrogen here instead of an O H. It's not really going to change the reaction at all, it's just that I did this for clarification because there were some unwanted side reactions with the O H that I'm trying to avoid. So anyway let's just look at this transformation, how could we make this happen? Well the transformation that I'm trying to produce is going from an amide to an amine and in case you don't remember this reaction or maybe you haven't even learned it yet the way to transform an amide to an amine is just to use a reducing agent, there's tons of reducing agents that will do this but I guess the most common is lithium aluminum hydride so we know that lithium aluminum hydride is a very powerful reducing agent and it's strong enough to get rid of that carbonyl and turn it into an amine.
So that's great to know, awesome, is that the answer for this question? No, I told you it's not going to be one step it's not that easy it's not you just put LAH and you're done but guys the reason is because LAH is a very powerful reducing agent and it's going to react with my ketone as well, I have a ketone, but look in the end product I need to also retain that ketone I can't just get rid of it and turn it into an alcohol so how can I reduce part of the molecule while keeping the ketone intact? You guessed it, we're going to use an acetal protecting group. So that means that before we do anything we're going to have to protect this ketone so let's go ahead and start. My first reagent is going to be, let's go ahead and use some kind of alcohol and acid to perform an acetal reaction. The most common protecting groups are usually the cyclic acetals, it doesn't have to be cyclic but that just seems to be the one that is kind of like the go to in the lab. So let's just use 1, 2 ethanediol which would be this guy. So it's a two carbon chain with two alcohols and acid, H plus. So what that's going to do is transform my ketone into an acetal so let's go ahead and draw that everything here is the same by the way, acetals don't react with amides so that's why nothing's going to happen down there but over here I'm now going to have O, O and two carbons in between. So that is my cyclic acetal protecting group, this is my protecting group. Cool, awesome guys, so now I've got my protecting group in place. What's awesome about these protecting groups is that, is that a very reactive molecule now? No, this is basically a diether and we know that ethers don't react much at all so now I can go ahead and reduce this molecule without fearing that I'm going to lose my ketone. So now let's go ahead to the next step, this is where I use my lithium aluminum hydride and that's just going to blast away at this amide. So what I'm now going to get is it's going to turn my amide into in an amine but I'm still going to have my protecting group in place.
Now one of the advantages of my acetal protecting group is that we know it's reversible so that means that what can I use to reverse it and take this off and make it back into the ketone? I can just use dilute acid, dilute H plus. There's lots of different ways you could write this, it could just H 3 O plus that's fine. So we know that instead of this being a one step reaction this is actually a three step transformation where one I placed my protecting group, two I used my reducing agent and then three I just used H 3 O plus to hydrolyse that acetal back off and regenerate the original carbonyl. Alright guys, so that is the way that we use acetal protecting groups. Let's go ahead and move on to the next page.
Concept: Practice 1: Selective Reduction5m
Hey guys, let's take a look at the following practice question. So here it says provide the chemical steps necessary for the following synthesis. So if we're looking here, we're going to say it looks like we start out with an aldehyde and we end with an aldehyde. Then what occurred? Here we have a ketone and it looks like the carbonyl group, this group here, got reduced to a methylene group, a C H 2 group. So our ketone became an alkyl group. Now we know that there's two ways at least that we know of that can get rid of a carbonyl entirely and change it into a methylene. We can use Wolff-Kishner reduction, we can use Clemmensen reduction and also we've seen that we can use ethyl acetal groups followed by raney nickel and H 2 to change a carbonyl into a C H 2 group. Now what's the issue with that? The issue is that if I try to do this it would reduce both my aldehyde carbonyl and my ketone carbonyl so all we have to do here is we have to somehow protect the carbonyl of the aldehyde first then reduce the ketone carbonyl and then reattain the carbonyl for the aldehyde.
Now remember how do we protect carbonyl know groups of aldehydes and ketones? We protect them by creating acetal groups. So what I'm going to do here first is here's my aldehyde, here's my ketone, here I can use if I want a diol with our acid-catalyst on the bottom and remember this is going to protect the carbonyl group of my aldehyde and remember it protects my aldehyde before my ketone because aldehydes are more reactive than ketone carbonyls. So our cap would look like this. Remember diols help to create caps. If I had used two moles of an alcohol then that would give us an acetal group, it just wouldn't be in cap form. Now the carbonyl group of my aldehyde is safe and secure so now I can use, let's see, we could use Wolff-Kishner here which happens through basic means and here that would reduce the carbonyl group of my ketone. Now here the reason I chose to do Wolff-Kishner instead of Clemmensen is remember that acetal groups are unreactive when you use base and Wolff-Kishner happens under basic means however if I had used Clemmensen reduction, Clemmensen reduction uses zinc mercury amalgam over acid and remember your acetal groups are susceptible to attack by acids we don't want to use acids. Acids would eat into my acetal group and convert it back into a carbonyl which could then possibly be reduced by Clemmensen reduction so here the best approach would be in order to make sure that cap stays where it is, undergo basic conditions in forms of reduction.
So now we've reduced that carbonyl group from the ketone now I can finally take off the cap so I throw in some acidified water, H 3 O plus and that gives me back my carbonyl of my aldehyde plus this ethyl chain right here. So remember we can use acetal groups to protect our carbonyls and in order of protecting aldehydes are more reactive so they would react first, ketones are less reactive so they are less likely to react first. Then remember we want to make sure that the acetal group is not affected so you don't want to undergo any type of acidic type of reaction so avoid using Clemmensen, Wolff-Kishner which happens under basic conditions is a much safer bet. It would reduce the carbonyl carbon from the ketone and not affect the acetal cap that we have for the aldehyde carbonyl.
Concept: Practice 2: Selective Reduction5m
Hey guys, let's take a look at the following question. So here it says provide the chemical steps necessary for the following synthesis. So if we notice here what's going on? We have an aldehyde here and it's still an aldehyde here, then we have here a nitrile group and somehow that nitrile carbon gets converted into a ketone. So now recall, how do we change a nitrile group into a ketone group? The answer is we use grignard's reagent. Now here's the thing, if I use grignard's reagent at this point, aldehyde carbonyls are very active so if I use grignard at this point it has a good chance of attacking this carbonyl carbon here and changing that aldehyde into a secondary alcohol so what I need to do here is I need to protect that carbonyl carbon of the aldehyde. Now how do we protect carbonyl groups of aldehydes and ketones? We protect them by using alcohols to make an acetal group. So here I choose to use two moles of an alcohol and don't forget the acid-catalyst on the bottom to help this all get started.
So what's going to happen here is that's going to protect that aldehyde carbonyl. So right there we have an acetal group there so it's protected. Now I can bring in grignard's but what does my grignards look like? Well here this nitrile carbon here is this same carbonyl carbon here which would mean that this C H 3 group, where did that piece come from? That piece came from the grignards reagent itself. So here we have C H 3 and remember grignards is ionic so it's not really together. The R group will be negative, the M G X part will be positive. Remember nitrogen is more electro negative so it's partially negative, the carbon here is partially positive. Since that carbon is partially positive it's going to attract this negative carbon here which attacks that carbon causing this bond to break and come over here. So what are we going to get? We're going to get a nitrogen making two bonds and having two lone pairs so it's going to be negatively charged. Here goes a C H 3 group I just added, the M G B R part is still positive so it will be attracted to that negative carbon and remember we're still protecting that carbonyl group of the aldehyde so that acetal group is still there. Now finally what do we do? We bring in some H 3 O plus. Now the H 3 O plus will do two things. First of all, we say that this here, a C double bonded to an N, is equivalent to an imine and when we use water or N H 3 positive on an imine, it's going to basically wash away that nitrogen and leave an oxygen behind. At the same time that H 3 O plus will also wash off the acetal group and give me back the carbonyl group of my aldehyde.
So here at the end I'll get my aldehyde restored and this carbon here becomes this carbon here and remember because of the H 3 O plus it becomes a carbonyl which is still connected to this C H 3 that originally came from the grignards reagent. So here this is how we would solve a question like this and remember the whole point of making acetal groups is to protect the carbonyl group of an aldehyde or ketone. Here we have to first protect the aldehyde before we could use grignard's because if we didn't, the grignard would most likely come in and attack the carbonyl carbon of the aldehyde and give us a secondary alcohol which is not what we wanted to form in the first place.
What product is expected from this reaction?
What is the expected product from this reaction sequence?
Predict the major product for the following reaction, paying attention to the regio- and stereochemistry where appropriate.
Provide the product for the following reaction