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Voiceover: Here's a general structure for an acyl chloride, also called an acid chloride, and it's a carboxylic acid derivative, so we can form them from carboxylic acid, so if we start with a carboxylic acid and add thionyl chloride, we can form our acyl chlorides, and we would also form a sulfur dioxide and HCl in this reaction. Let's look at the structure of thionyl chloride. Here we have the dot structure right here and we could draw a resonance structure for this, so we could show these electrons in here moving off onto the oxygen, so let's go ahead and draw what we would form from that. Our oxygen would now have three loan pairs of electrons on it, giving it a negative formal charge. Our sulfur would still be barred to these chlorines here. It would solve a loan pair of electrons and it would get a plus one formal charge like that. This is a major contributor to the overall structure. Oxygen is more electronegative than sulfur, and if you think about this pie bond in here, there's ineffective overlap of those p orbitals, and that's because sulfur and oxygen are in different periods on the periodic table, so sulfur is in the third period, so it has a larger p orbital than oxygen. Oxygen's in the second period. You get ineffective overlap of these orbitals here, and so that's another reason why this is going to contribute to the overall structure. Plus, you have these chlorines here, withdrawing some electron density from the sulfur, so chlorine is more electronegative than the sulfur, so that the end result of all this is going to make this sulfur very electrophilic right here, and so therefore, our carboxylic acid is able to act as a nucleophile, and so if these electrons move into here, these electrons can attack our sulfur so the nucleophile attacks our electrophile, and then these electrons kick off onto the oxygen, so let's go ahead and show that. We would have our R group bonded to our carbon, and then that's bonded to an oxygen. The oxygen has two loan pairs of electrons on it. the oxygen formed a bond with the sulfur, and now the sulfur's bonded to this oxygen, which gets a negative one formal charge. This sulfur is bonded to two chlorines, so we draw in our chlorines here with all the loan pairs of electrons, and there's still a loan pair of electrons on our sulfur. We now have a double bond between the carbon and this oxygen, and one loan pair of electrons on this oxygen gives it a plus one formal charge. Following some of our electrons, these electrons in magenta move in here, they'd form our double bond, and then we can think about these electrons in blue, forming the bonds between the oxygen and the sulfur. Finally, we could think about these electrons in here in green moving off onto our oxygen like that. So for the next step, we could think about these electrons moving in here to form our double bond between oxygen and sulfur. That would kick off the chloride anion as a leaving group, and we know the chloride anion is an excellent leaving group because it's stable on it's own. When we draw the results of that, we would have our group. We would have our carbon, we would have that carbon double-bonded to our oxygen here with a loan pair of electrons plus one formal charge, and then we would have our oxygen with two loan pairs of electrons, and our sulfur is now double-bonded to this oxygen. All right, there's still a loan pair of electrons on that sulfur. Only one chlorine bonded to this sulfur now, so we lost the chloride anion. Let's go ahead and show some of those electrons. Let's say these electrons in red here move in to reform the dull bond between oxygen and sulfur, and then we had some electrons kick off onto the chlorine, so we have the chloride anion that forms. Let's go ahead and show that as well. These electrons in here come off onto chlorine, then we have the chloride anions. Let's draw in the chloride anion. Let's get some more space down here. We also have the chloride anion that forms, so we draw that in here like that. A negative, more informal charge. At this stage, we need to consider whether the chloride anion is going to function as a base or a nucleophile, and it could do both, so let's first think about the chloride anion functioning as a base. If it functions as a base, it could take this proton, leaving these electrons behind on the oxygen, so let's go ahead and draw the lewd form. Lewd form are carbonile with two loan pairs of electrons on the oxygen, and then we would have this oxygen here with two loan pairs, and then this sulfur, right when it would still would be double-bonded, one loan pair, and then the chlorine, like that. We're saying that in this acid-based reaction, these electrons in here move back onto here to form the carbonile. You can consider this to be the intermediate for this mechanism. You'll see some versions of this mechanism take this intermediate and continue on to form your product. I'm going to show the chloride anion functioning as a nucleophile over here for this. It's an acid-based reaction, so it's possible to, once again, protonate your carbonile. That's going to activate it. This carbon right here becomes more electrophilic, and that means that the chloride anion can function as a nucleophile and attack our electrophile since the chloride anion isn't a great nucleophile on it's own, so if the chloride anion attacks here, that would push these electrons off onto the oxygen, and we could go ahead and draw what we would form, so we would have our R group, we would have our carbon, we would have this oxygen with two loan pairs of electrons bonded to hydrogen here, so let's show those electrons in blue. If these electrons in blue kick off onto the oxygen, we could say that those are these electrons. Then we could also say that these electrons here in green, on the chloride anion are going to form a bond between the chlorine and the carbon, so we can go ahead and draw in the bond between the chlorine and the carbon like that. We still have our oxygen right here bonded to our sulfur, double-bonded to this oxygen, and then we have our loan pair of electrons here. Then we have our chlorine. What all this does is make a much better leaving group than what we started off with so you could think about this as our leaving group. This is a leaving group, and then the next step of the mechanism, and if we go back up here, that's a much better leaving group than the OH that we started out with. If that's going to be our leaving group, we could show these electrons in here moving into form our double bond which would kick these electrons off onto our oxygen. Let's go ahead and show what we would make from that. We would now form our R group, our carbon would be double-bonded to an oxygen like this, loan pair of electrons plus one formal charge, and then we'd have our chlorine over here with all of its electrons too. Let's go ahead and show where those electrons came from so if these electrons move in here, that would reform our carbonile, so I can go ahead and draw that in. Then we still have this bond in green, here, just to clarify. Then we would have some electrons, and now let's make a magenta, so these electrons in here are going to kick off onto the oxygen, so we could go ahead and draw that. We would have our oxygen. So we'd have three loan pairs of electrons. One of them would be the magenta ones, so we could go ahead and do that. That gives our oxygen a negative one formal charge because it's bonded to the sulfur here, the sulfur's still double-bonded to this oxygen, and once again, still a loan pair of electrons and a chlorine over here like that. The reason why this is a good leaving group is because we can push these electrons into here, push these electrons off onto here, and we can form sulfur dioxide as a gas. If we go ahead and draw the result of that, we would now have this oxygen double-bonded to this sulfur, double-bonded to this oxygen, so draw in loan pairs of electrons here, and we still have a loan pair of electrons on our sulfur. If these electrons in red here move in to form this bond, we've now formed sulfur dioxide, and we also formed a chloride anion, so it once again follows some electrons. If these electrons move off onto chlorine, we have the chloride anion, so now we have our chloride anion which could function as a base and in the last step of our mechanism, deprotonate, so take this proton, leave these electrons behind and finally, we formed our acyl chloride. We have our carbon double-bonded to an oxygen with two loan pairs of electrons, and we have our chlorine over here, like that. If we follow those final electrons, these electrons in blue, move off onto your carbonile. You would also form HCl by this, so you form HCl, which is also a gas, so the formation of these gases, the formation of sulfur dioxide and HCl drives the reaction to completion here. That was a long mechanism. Once again, you'll see some slight variations on this, so if I go back up to here, you could show this step, the nucleophile attacking the electrophile, and directly form here if you wanted to, so as your nucleophile attacks your sulfur, these electrons could kick off onto chlorine to form the chloride anion at this step, and at this step, and then you would get this at this point. Once again, I talked about the possibility of using this as your intermediate. Depending on which textbook you look in, you might see some slight variations on this mechanism. It is a little bit of a long one here. All right, let's look at two other ways to make an acyl chloride, so starting with a carboxylic acid, you could add phosphorus, a pentachloride, or a phosphorus trichloride, and both of those will give you an acyl chloride as well. The mechanism is pretty similar and also we could think about, instead of phosphorus trichloride, we could add a PBr3, so a phosphorus tribromide instead of putting a chlorine here. That would put a bromine, and so we could form an acyl bromide this way, and we'll see a use for this reaction in a later video.