Now let's say we have an aldehyde and we choose to react it with an alcohol. What type of product will we get in this reaction? Let's say it's catalyzed by an acid. The product will be a hemiacetal.
So we're going to get an OR group. plus an OH group. So that's a hemiacetal. Now it doesn't have to stop here.
The hemiacetal can react with another methanol molecule under acidic conditions. And so we're going to get two OR groups. So in this case, This product is called an acetal. Now if we take the acetal and react with H3O+, under acidic conditions, well it has to be acidic, this is already acidic, we can go back to the aldehyde compound. And so this whole process is reversible.
So let's say we have a cyclic ketone. Go ahead and predict the major product if it reacts with methanol under acidic conditions. So this can go all the way to an acetal. Let's say if we use excess methanol. So we can get two OR groups.
And then, if we react with H3O+, we can regenerate the original ketone. Now let's go over the mechanism for the formation of an acetyl group. So because the reaction occurs under acidic conditions, the first step is protonation. So once we add the hydrogen to the oxygen atom, this group becomes more electrophilic, more susceptible to nucleophilic attack by the methanol molecule.
And so the oxygen atom of methanol is attracted to the partially positive carbonyl carbon, and so the methanol attacks there. Now the double bond is going to break, so we're going to get an OH group. And here we have an oxygen attached to a hydrogen and a methyl group. And now that oxygen has a positive charge.
Now, what we need to do at this point is we need to get rid of this hydrogen atom. Because right now, this group is a good leaving group. Anytime oxygen has three bonds, it's a good leaving group. And so this OH group, what it could do is take a lone pair, form a double bond, and just kick this group out, giving us this compound.
So that's going to take us a step back, which we don't want to go in that direction. And so what we need to do is get rid of the hydrogen. So let's use another methanol molecule to act as a weak base. And so it's going to take away this hydrogen for us, which is what we wanted to do.
And so now we have an OR group and an OH group. So now the methanol that took the hydrogen away It's now protonated, and so what I'm going to do is take a hydrogen from that protonated alcohol and transfer it to the OH group. We want the OH group to be a good leaving group because we need to get rid of it. And so let's take a lone pair from oxygen from the OCH3 group, form a pi bond, and let's kick out water. So this arrow should be pointing towards the oxygen atom.
It didn't look like it was doing that before. And so this is what we now have. So what do you think the next thing... is going to be.
What's our next step here? So right now we have this group which is highly electrophilic because if we draw the resonance structure we can put a positive charge on the carbonyl carbon. So what we're going to do is take another methanol molecule and react it with the carbonyl carbon. So it's going to attack here, breaking this pi bond. So this is what we now have.
In addition to the OCH3 group, we have another OCH3 group but with a hydrogen atom attached to it. And so the last step in this mechanism is deprotonation. So let's use another methanol molecule to get rid of that hydrogen.
And so we're going to have two OCH3 groups. And so this is the mechanism for the formation of the acetyl functional group. Now let's go over the next reaction. So let's say if we have cyclohexanone and we wish to react it with ethylene glycol.
What's going to happen? Now in the previous examples, we saw that when mixing an aldehyde or a ketone with one alcohol molecule, we got a hemiacetal. But when mixing it with two alcohol molecules, we got an acetal.
Acetylene glycol is one molecule but with two alcohol functional groups. So what we're going to get is the acetyl but a cyclic acetyl. And it's going to look like this.
So we have two OR groups connected to each other. And this reaction is useful as a protecting group. Now we can remove the protecting group by reacting it with H3O+.
And so that is going to regenerate the ketone. So here is an example that will illustrate the use of the protecting group. So here we have a molecule that contains a ketone and an ester functional group.
How can we use a protecting group in a way that we can reduce the ester to an alcohol, but not reduce the ketone to an alcohol? Now, if we use, for instance, sodium borohydride. Sodium borohydride can reduce the ketone to an alcohol, but it can't reduce the ester.
And so we're going to get this product, which is what we don't want. Now, if we use lithium aluminum hydride, it will reduce everything. it can reduce the ester into an alcohol but it will also reduce the ketone to an alcohol now we don't want to use sodium borohydride because we need to reduce the ester to an alcohol and sodium borohydride can't do that so we can eliminate that option now we want to use lithium aluminum hydride because it can reduce the ester to an alcohol however we don't want it to reduce the ketone so what should we do It's important to understand that ketones are more reactive than esters.
And so what you need to do is use a protecting group. So that's where ethylene glycol comes into play. This is going to react with the ketone because it's more reactive than esters.
And so here is the protecting group. And now we can use lithium aluminum hydride to reduce the ester while keeping the ketone safe. So once we use lithium aluminum hydride, we're going to get this product. The OCH3 group is going to leave as methanol, and so we're not concerned with that side product. Now, once we reduce the ester to an alcohol, then we can add H3O+, to remove the protecting group, converting it back into a ketone.
And so that's one way in which you can use a cyclic diol in order to form... a patechnigroup, which is a cyclic acetone.