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One way to make ethers is to use the Williamson ether synthesis, which is where you start with an alcohol, and you add a strong base to deprotonate the alcohol. Once you deprotonate the alcohol, you add an alkyl halide, and primary alkyl halides work the best. We'll talk about why in a minute. And what happens is you end up putting the R prime group from your alkyl halide on to what used to be your alcohol to form your ether like that. So let's look at the mechanism for the Williamson ether synthesis, where you start with your alcohol. We know that alcohols can function as weak acids. So if you react an alcohol with a strong base, something like sodium hydride, we know that the hydride portion of the molecule is going to function as a strong base. This lone pair of electrons is going to take that proton, which is going to kick these electrons off onto the oxygen. So if we're drawing the product of that acid-based reaction, we now have an oxygen with three lone pairs of electrons around it, giving it a negative 1 formal charge. And we call that an alkoxide anion, which would interact with the positively charged sodium ion floating around. So there's some electrostatic or ionic interaction between those opposite charges. And here's where you introduce your alkyl halide. So if we draw our alkyl halide, it would look like this. And we know that there's an electronegativity difference between our halogen and our carbon, where our halogen is going to be partially negative, and our carbon is going to be partially positive. Partially positive carbon means that that carbon wants electrons. It's going to function as an electrophile in the next step of the mechanism. And a lone pair of electrons in the oxygen is going to function as a nucleophile. So opposite charges attract. A lone pair of electrons on our nucleophile are going to attack our electrophile, our carbon. At the same time, the electrons in the bond between the carbon the halogen are going to kick off onto the halogen like that. So this is an SN2-type mechanism, which is why a primary alkyl halide will work the best, because that has the decreased steric hindrance compared to other alkyl halides. So what will happen is, after nucleophilic attack, we're going to attach our oxygen to our carbon like that, and we form our ether. So if we wanted to, we could just rewrite our ether like this to show it as we added on an R prime group like that. Let's look at an example of the Williamson ether synthesis. So if I start with a molecule over here on the left, and it's kind of an interesting-looking molecule. It's called beta-naphthol. And so beta-naphthol has two rings together like this, and then there's an OH coming off of one of the rings, like that. So that's beta-naphthol. And in the first part, we're going to add potassium hydroxide as our base. Now, potassium hydroxide is not as strong of a base as sodium hydride is, but in this case, it's OK to use a little bit weaker base. So the lone pair of electrons on the hydroxide are going to take that proton, leaving these electrons behind on the oxygen. So when we draw the conjugate base to beta-naphthol-- and we can go ahead and show that-- we're going to take off that proton, which is going to leave that oxygen there with three lone pairs of electrons, giving it a negative 1 formal charge. So this is our alkoxide anion. And this alkoxide anion is resonance stabilized. So a resonance-stabilized conjugate base stabilizes the conjugate base, which makes beta-naphthol a little bit better acid than other alcohols that we will talk about. So since beta-naphthol is a little bit more acidic, that's why it's OK for us to use a weaker base for this example. So potassium hydroxide is strong enough to take away the acidic proton in beta-naphthol because the conjugate base to beta-naphthol is resonance stabilized. So in the second step, once we have formed our alkoxide anion, this is where we add our alkyl halide. So if I add my alkyl halide in my second step-- let's see if we can have enough room here-- I'm going to use methyl iodide as our alkyl halide. So methyl iodide looks like that. And once again, we know this carbon is going to be the electrophilic carbon, so nucleophile, electrophile. So a lone pair of electrons on the oxygen attacks the carbon, kicks these electrons off onto the iodide, and we form our product. So let's go ahead and draw the ether product that will result. So these rings are going to stay the same like that. And we now are going to have our oxygen attached to a methyl group, which came from the methyl iodide like that. So we formed our product. This product is called nerolin, which is a fixative used in perfume. So this has an interesting smell to it. So if you ever get a chance to do this Williamson ether synthesis, it's just interesting to see what nerolin smells like, what it looks like, and to think about it as being a component of some perfumes. Let's think about synthesizing an ether. So if you were given a problem where the question said something like, OK, here is the ether that you want to synthesize. What would you need in order to do so? So you need to think about, OK, there's my ether, and I'm going to make it from some other things over here. And if I analyze the alkyl groups attached to my ether, and I have a methyl group over here, and this would be like a cyclohexyl group over here. And one of those two groups I'm going to use for my alkyl halide. You want to use the group that's the least sterically hindered since it's an SN2-type mechanism. So you want to go with the methyl group. So in your second step, you would need to add something like methyl iodide. That's the least sterically hindered, so that's going to improve your yield on this reaction. So that's the second step. And in the first step, you'd have to add a strong base, so we'll use sodium hydride here. And your alcohol, therefore, must come from this. So this must be where your alcohol comes from. So if I'm going to show my starting alcohol, it would have to look like this. So if I add that alcohol in the first step, sodium hydride, I take off that proton, form an alkoxide, that alkoxide nucleophilic attacks the methyl iodide to add the methyl group on, and to form the ether on the right. So that's how to think about using the Williamson ether synthesis. So think about retrosynthesis and think about which alkyl group is the best one to use for your alkyl halide.