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Studying for a test? Prepare with these 8 lessons on Alcohols, ethers, epoxides, sulfides.
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Video transcript
In this video we're going to see how to prepare alkyl halides from alcohols. And so if we start with this alcohol over here on the left, and we add SOCl2, which is called thionyl chloride, and pyridine to it. We're going to substitute a chlorine atom for the OH group. And this mechanism occurs via an SN2 type mechanism, which means that it's only going to work with primary or secondary alcohols. And you will get inversion of configuration if you have a chirality center present in your final product. So let's take a look at the mechanism. And we'll start with our alcohol. And so the oxygen is going to have to leave somehow. But by itself OH is not the best leaving group. And so we're going to react this alcohol with thionyl chloride to convert it into a better leaving group. And so if we draw the dot structure for thionyl chloride, we would have to sulfur double bonded to an oxygen here. And then the sulfur is also bonded to chlorine. So I'll go ahead and put in those lone pairs of electrons on the chlorines, like this. And if you count up your valence electrons, it turns out you need two more. And those go on the sulfur. It's OK for sulfur to violate the octet rule, since it is in the third period now. So a lone pair of electrons on oxygen is going to form a bond with our sulfur atom, which would therefore kick these electrons in here off onto the top oxygen. So if we go ahead and draw what we get from that first step of our mechanism, now our oxygen is bonded to our sulfur. The oxygen is also a bonded to a hydrogen. One lone pair of electrons formed that new bond, so one pair of electrons is left behind. Which would give this oxygen a plus 1 formal charge. Connected to the sulfur, this top oxygen here had two lone pairs of electrons. Picked up one more lone pair, which gives it a negative 1 formal charge. And this sulfur is still bonded to chlorine. So we can go ahead and draw those in. And we can go ahead and draw that lone pair of electrons on that sulfur like that. And so in the next step of the mechanism, we're going to reform the double bond between oxygen and sulfur. So these electrons are going to kick in here. And these electrons would kick off on to that chlorine. So when we draw the next intermediate here, we would now have our oxygen, still bonded to a hydrogen, still with a plus 1 formal charge like that. And now our sulfur is double bonded to our oxygen again with two lone pairs of electrons on the oxygen. The sulfur is also bonded to one chlorine now, so one of the chlorines left. And we can go ahead and draw in that chlorine. So one of the chlorines left here. It's a negatively charged chloride anion. And then still there's a lone pair of electrons on our sulfur like that. So at this part of the mechanism, the pyridine comes along. So if we go ahead and draw the dot structure for pyridine. It's a base, and so it looks like a benzene ring, except we have a nitrogen here instead. And there'd be a lone pair of electrons on this nitrogen. And so that lone pair of electrons going to function as a Bronsted-Lowry base and take this proton here on the oxygen. And that would kick these electrons back off onto this oxygen. So when we go ahead and draw that-- let's go ahead and get some more room here-- so what would we get? We would now have our carbon bonded to our oxygen. Our oxygen now has two lone pairs of electrons around it. And we have our sulfur, and our chlorine, and our lone pair of electrons on the sulfur. And now we've made a better leaving group. So this is a better leaving group than the OH was in the beginning. And if we think about an SN2 type mechanism now, we know that the bond between carbon and oxygen is polarized, right? Oxygen being more electronegative, it will be partially negative. And this carbon here be partially positive. And so now we can think about our SN2 type mechanism. Our nucleophile will be this chloride anion up here that we formed in the mechanism. So that's going to be the nucleophile, and it's going to attack our partially positive carbon. An SN2 type mechanism. So as the chloride attacks, this stuff on the right is going to leave. So the electrons in magenta are actually going to move in here, and then these electrons are going to kick off onto that chlorine. So when we draw the product, we can go ahead and show the chlorine has now added on to our carbon on the left. And on the right, if you follow the movement of those electrons, they're going to form sulfur dioxides. So SO2. And also the chloride anions, so the Cl minus, like that. And so we've done it. We've substituted our chlorine atom for the OH and formed an alkyl halide. So this is just a better way of forming an alkyl chloride from an alcohol. So if we look at an example, we'll just take something like ethanol here. And if we react ethanol with thionyl chloride, SOCl2, and we had some pyridine as our base. We're going to replace the OH with our chlorine like that. And so once again, if we look at our alcohol, this is a primary alcohol. And so primary alcohols work the best because there's decreased steric hindrance. And we don't have to worry about stereochemistry, since we don't have any chirality centers in our product. Let's look at a way to form alkyl bromide. So we just formed an alkyl chloride. Let's look at the general reaction for forming an alkyl bromide here. So I go ahead and have my alcohol. And I react that with phosphorus tribromide, PBr3. The OH group is going to leave and I'm going to put a bromine in its place. And once again, this mechanism is an SN2 type mechanism. So primary or secondary alcohols only. And possible inversion of configuration for your products, depending on whether chirality centers are present or not. So another SN2 mechanism. And again, we need to use phosphorus tribromide because the OH group is not the best leaving group. When we look at this mechanism here, let's go ahead and show that lone pair of electrons better, like that. We have phosphorus tribromide. So I'm going to go to draw the dot structure. So we would have these bromines here with lone pairs of electrons. And there's three of them. So I'll go ahead and put in those bromines. And we still have two more valence electrons to account for, and those would go on our phosphorus like that. So the first step, it's analogous to our previous mechanism. Lone pair of electrons on oxygen are going to for a bonds with phosphorus. And that would kick these electrons off onto one of the bromines, so I just chose that one. It doesn't matter which one you choose. And so when we show the result of that, we would now have our oxygen bonded to a phosphorus. The oxygen is still bonded to a hydrogen. There's still a lone pair of electrons left behind on this oxygen, which gives this oxygen a plus 1 formal charge. And the phosphorus is now bonded to only two bromines. So we can show the phosphorus bonded to only two bromines here. So I can go and put those in. And I can put in the lone pair of electrons on phosphorus as well. And so we lost one of our bromines, and that formed a bromide anion. So we go ahead and draw in our bromide anion here. And once again, we've made a better leaving group. So all the stuff here on the right is a better leaving group than the OH. So we could think about our SN2 type mechanism, where once again, this carbon right here is going to be electrophilic. So it once it wants negatively charged electrons, which it could get from the bromide anion. So nucleophilic attack, and then that would kick these electrons in here off on to your oxygen. And we can go ahead and show our products. So when we draw the product, all we need to do now is show we have substituted. The bromine is now attached to the carbon like that. And our other product would be a hydrogen attached to an oxygen, and the oxygen would be attached to a phosphorus, and the phosphorus is attached to two bromines like that. So we've formed our alkyl bromide. So if we just show a quick example. Once again, we'll start with ethanol. If we wanted to convert ethanol into ethyl bromide, a all we would have to do is add phosphorus tribromide like that. And of course we're going to replace the OH with a bromine. And so that's one way to prepare alkyl halides from alcohols.