In this video, we're going to focus on reactions associated with ethers and epoxides. So first, let's talk about how we can synthesize an ether. One of the most efficient ways to do this is to use the Williamson ether synthesis reaction.
And so what happens is, the first thing you need to do is you need to use a strong base, such as sodium hydride, or... sodium amide. But let's use sodium hydride, NaH. The hydride ion, with its negative charge, is attracted to the partially positive hydrogen atom, which is bonded to the partially negative oxygen atom.
And so this hydride ion grabs the hydrogen, and then these electrons go back to the oxygen. And so what we're going to get is an alkoxide ion, and also hydrogen gas. as a side product. Now hydrogen gas will leave the solution, it's just going to go up there.
And then in the second step, we can react this with an alkyl halide like ethyl bromide. And so this step is an SN2 reaction, which means we need to use methyl and primary halides. Secondary halides could favor an E2 reaction, so you don't want to use a secondary alkyl halide.
And this will give us an ether. And so that's the basic idea behind the Williamson ether synthesis reaction. Now let's work on another example. Go ahead and try this problem.
Let's say if we have phenol, and we're going to react it with sodium hydroxide in the first step, and in the second step we're going to react it with methyl bromide. Go ahead and predict the major product. Now for the last example, we use 1-butanol.
And the pKa of most alcohols is around 16-18. And so for this, you need to use a strong base like sodium hydride or sodium amide. Sodium hydroxide, if you use it, the reaction will be in equilibrium. It can take off the hydrogen, but not in good yield. The pKa for phenol is 10, and so it's low enough that we can use sodium hydroxide to get the alkoxide ion in good yield.
So let's write up a mechanism. So in the first step, the hydroxide ion will take off the hydrogen. And the reason why phenol is more acidic than a typical alcohol is because the conjugate base is stabilized by resonance. The negative charge on the oxygen can delocalize into the ring, thus stabilizing it.
In the next step, we can react phenoxide with methyl bromide. And so we're going to get an ether. We're going to get methyl phenyl ether.
And so that's another example of the Williamson ether synthesis reaction. Now let's go over another example. So consider this molecule. What's going to happen if we react it with tert butoxide? Go ahead and predict the major product of this reaction.
There's two things that tert butoxide can do. It can behave as a nucleophile, attacking the carbon, kicking out the chlorine atom. Or, it can behave as a base, abstracting the hydrogen atom. And in fact, both of these things can happen.
However, because tert butoxide is a bulky base, hysterically hindered, due to all of these methyl groups, it prefers to behave as a base rather than a nucleophile. So by making the base more sterically hindered, you can increase its specificity. You can make it behave more as a base, less of a nucleophile.
And so we're going to focus on this reaction where it takes off the hydrogen. But just keep in mind that both things could happen. So you can get a mixture of products.
in this example. However, we can increase the yield of this reaction by making this base more sterically hindered. There are other bases out there that are so sterically hindered that they don't act as nucleophiles, they just only act as bases.
Now once we take off this hydrogen and we get the alkoxide ion, this oxygen is going to attack itself. not itself but it's gonna the molecule attacks itself and so you're gonna have an intramolecular reaction and so it's going to close and form a ring let's call this one two three four five six so we're gonna get a six membered ring So let's say this is 6. This is going to be 1, 2, 3, 4, and 5. Attached to carbon 4, we have a methyl group. And this is basically it.
So this is the final product of this reaction. So we have basically an ether inside of a ring. Now, there are other ways in which we can make an ether. particularly if we start with an alkene.
So if we react the alkene with methanol under acidic conditions, notice what can happen. The alkene will react with the hydrogen and the hydrogen will go on a primary carbon so that we can get a more stable secondary carbocation intermediate. And then the methanol molecule will attack the carbocation and so we're going to get this species. And then in the final step, another methanol molecule will take away this hydrogen.
And so the end result is that we have an ether. Another way in which we can make an ether is the oxymercuration-demercuration, actually the alkoxymercuration-demercuration reaction of alkenes. So we're going to use mercury acetate with, let's use ethanol. CH3CH2OH followed by sodium borohydride. The end result is that we're going to add the ether here on the secondary carbon.
This reaction proceeds with moccormick-carvey geochemistry. And so all we need to do is take off the hydrogen and just add this group to the secondary carbon. So that's another way in which you can make ethers, using the alkoxymercuration-demercuration reaction.
Now let's talk about the acid-catalyzed cleavage of ethers. Ethers don't react with bases for the most part. They're unreactive towards them.
But to acids, they tend to cleave. So let's react to this ether with hydroiodic acid. What do you think is going to happen?
Now you need to determine if the reaction will proceed by an SN1 mechanism or an SN2 mechanism. So notice that this carbon is tertiary, which means that we can get a fairly stable tertiary carbocatine intermediate. So therefore, this reaction is going to go via the SN1 mechanism. The first step is protonation.
The purpose of this is to make the oxygen a good leaving group. And then in the next step, the oxygen is going to leave. And so we're going to get a tertiary carbocadon intermediate.
And the first side product that we have is methanol. And then in the second step, iodide could come in, giving us an alkyl halide. And so these are the two products that we get in this reaction.
If we add one equivalent of HI. Now it's important to understand that if we add excess HI, it can react with methanol. So if we add another HI molecule, then this oxygen will be protonated in the first step.
By the way, this will occur by an SN2 mechanism. So we're going to get this intermediate, and then in the second step, iodide is going to attack from the back in an SN2 reaction, kicking out the water molecule, given methyl iodide. So let's summarize what's going on here. So let's say if we add HI in a one-to-one reaction with this ether, or basically if we add one equivalent of it.
then we're going to get these two products, tert-butyl iodide and methanol. So the iodide ion will go on the tertiary carbon. It's going to go on the more substituted carbon of the ether. Now, if we add to the same ether, excess HI, this reaction will go to completion.
We're still going to get tert-butyl iodide, but it's going to react with methanol, giving us methyl iodide. So keep that in mind. So if you add an acid like HI or HBR in excess, you should get no alcohol as your final product.
Now there are some reagents where you can't convert the alcohol into an alcohol halide, such as phenol, for example. you can't convert that into phenyl iodide. It's just not going to happen. But for other regular alcohols, if you use H-I in excess, all of them will be converted to an alcohol halide.