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Let's look at how to prepare sulfides from thiols. So over here on the left, I have my thiol. And to that thiol, I'm going to add sodium hydroxide. And the sodium hydroxide is going to deprotonated the thiol, which is then going to react with this alkyl halide in the second step of the reaction, to produce my sulfide as my product. So here's my sulfide right here. This reaction is the analog of the Williamson ether synthesis, which we've seen in earlier videos. So in that video, we started off with an alcohol, and we reacted our alcohol with a strong base in the first step and an alkyl halide in the second step, and we formed an ether as our product. So we can go ahead and draw our ether in here like that. So the thiol is the sulfur analog into an alcohol, and a sulfide is the sulfur analog to an ether. Let's look at the mechanism to make sulfide. So if I start with my thiol right here-- so I have carbon bonded to sulfur bonded to a hydrogen. And then two lone pairs of electrons on that sulfur. And if I think about the difference in electronegativity between carbon and sulfur, there's actually not much of a difference in terms of numbers. So this is not a very polar bond actually. That's different from what we see with an alcohol. So up here, if we look at the alcohol, I know that oxygen is much more electronegative than carbon. So this oxygen here would get a partial negative. And this carbon on the left that it's bonded to would get a partial positive. So there's much more of an electronegativity difference in alcohols. In thiols, there's not really that much of a difference. But thiols can still function as nucleophiles because these lone pairs of electrons are located further away from the nucleus than the lone pair of electrons in oxygen, because sulfur is a larger atom. So those electrons are more polarizable, and so thiols are actually excellent nucleophiles. So if we go back here to our mechanism, we're now going to add sodium hydroxide, which is a base. So we can go ahead and put OH minus over here. So the hydroxide anion is going to function as a base. And a lone pair of electrons are going to take this proton and leave these electrons behind on the sulfur. So let's go ahead and draw the conjugate base to the thiol. So we now have carbon bonded to sulfur, and this sulfur now has three lone pairs of electrons, giving it a negative one formal charge. So this is called a thiolate anion. So let me just go ahead and write that. And thiolate anions are very stable. That negative charge on the sulfur-- since sulfur is a large atom, you can spread out that negative charge over a very large area. So the thiolate anion is relatively stable, and that makes thiols more acidic than alcohols. OK? So alcohols don't have the same type of stabilization since alcohols are smaller. So thiols are actually very acidic, and that's why we can use sodium hydroxide here to deprotonated our thiol to form the thiolate anion. In the second step, we add our alkyl halide. And so here's my alkyl halide. And the alkyl halide does have a polarized bond, right? The difference in electronegativity between a halogen a carbon atom is fairly large. So this halogen here is going to get a partial negative charge. And this carbon is going to get a partial positive charge. So thiols are good nucleophiles. Thiolate anions are even better nucleophiles. And so the thiolate anion is going to function as the nucleophile. The partially positive carbon is going to function is the electrophile, and we're going to get an SN2 type mechanism, where our strong nucleophile attacks our electrophile and kicks these electrons off here onto the halogen. And we can go ahead and form our product. So this is an SN2-type mechanism. And we end up with the sulfur now bonded to two R groups. And they could, obviously, be the same R groups. Or they could be different R groups. And so we've formed our sulfide like that. Let's do an example of the preparation of sulfide. So we're going to start with-- let's see. Let's start with this one right here. All right. So we start with this molecule. And to that thiol, we're going to add sodium hydroxide in the first step. So let's go ahead and write sodium hydroxide here. A Na plus and then OH minus like that. And then, in the second step, we're going to add an alkyl halide. So let's add this as our alkyl halide-- so ethyl bromine. So when I think about the mechanism, I know the first step is an acid base reaction. The electrons on the hydroxide anion-- so one of these electron pairs here-- are going to take this proton-- that's the acidic proton on my thiol-- leaving these electrons behind on the sulfur. So I can go ahead and draw the resulting thiolate anion. So I can go ahead and draw that. Let me see what we have here. We have our ring, and then we have our sulfur. And our sulfur now has three lone pairs of electrons around it like that. Thiolate anions are excellence nucleophiles. And when I look at my alkyl halide, once again, I know the electronegativity difference between bromine and this carbon here. I'm going to give bromine a partial negative charge. This carbon is going to be partially positive. So the thiolate anion's going to act as a nucleophile. And an SN2-type mechanism and a lone pair of electrons, here, are going to attack my electrophile. And we can go ahead and form our product. So now we have our ring here, which is connected to our sulfur. And our sulfur, now, just picked up 2 more carbons, right? Because these electrons in here are going to kick off onto the bromine, and we end up putting an ethyl group onto that sulfur. So there are 2 carbons now on that sulfur like that. And the sulfur has 2 lone pairs of electrons. And so we formed our sulfide. Now, if I were to name this sulfide, it's a lot like naming ethers. So I could use the common way of naming this and treat those as alkyl groups. And if I look on the right, this would be an ethyl group. So I could go ahead and start naming it. I could say it's ethyl-- and if I look at the alkyl group on the left side-- this is a phenyl group, right here. So it's ethyl. And then phenyl since I'm calling the alphabet rule here-- e before p. And then finally, I know it's a sulfide. So I can just go ahead and finish the nomenclature by saying sulfide here. So ethyl phenyl sulfide is the sulfide produced in this analog of the Williamson ether synthesis.