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Current time:0:00Total duration:10:23

Video transcript

here's the general structure for an ACL 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 thymio chloride we can form our acyl chlorides and we would also form a sulfur dioxide and HCl in this reaction so let's look at the structure of thionyl chloride so 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 on to the oxygen so let's go ahead and draw what we would form from that our oxygen would now have three lone pairs of electrons on it giving it a negative 1 formal charge our sulfur would still be bonded to these chlorines here it would solve a lone pair of electrons and it would get a +1 formal charge like that so this is a this is a major contributor to the overall structure oxygen is more electronegative than sulfur and if you think about if you think about this PI bond in here there's ineffective overlap of those P orbitals 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 oxygens in the second period and so 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 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'd have our R group bonded to our carbon right and then that's bonded to an oxygen the option has two lone pairs of electrons on it the oxygen formed a bond with the sulfur and now this sulfur is bonded to this oxygen which gets a negative 1 formal charge this sulfur is bonded to two chlorines alright so we draw in our chlorines here with all the lone pairs of electrons and there's still a lone pair of like on our sulfur we now have a double bond between the carbon and this oxygen and one lone pair of electrons on this oxygen gives it a +1 formal charge so following some of our electrons if these electrons in magenta move in here they form our double bond and then we can think about these electrons in blue right forming the bond between the oxygen and the sulfur and then finally we could think about these electrons in here in green moving off on to our oxygen like that and so for the next step right we could think about these electrons moving in here to form our double bond between oxygen and sulfur and 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 its own so when we draw the result of that we would have our R group we would have our carbon we would have that carbon double bonded to our oxygen here all right with the lone pair of electrons plus one formal charge and then we would have our oxygen alright with two lone pairs of electrons and our sulfur is now double bonded to this oxygen alright there's still a lone pair of electrons on that sulphur only one chlorine bonded to the sulfur now all right so we lost we lost the chloride anion so let's go ahead and and show some of those electrons so let's say let's say these electrons in red here move in to reform the double bond between between oxygen and sulfur and then we have some electrons kick off onto the chlorine so we have the chloride anion that forms let's go ahead and show that as well so if these electrons in here right come off on to chlorine then we have the chloride anion so let's draw in the chloride anion let's get some more space down here alright so we also have the chloride anion that forms so we draw that in here like that so a negative one formal charge so 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 and so 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 what we would form we would form our carbonyl with two lone pairs of electrons on the gin and then we would have this oxygen here with two lone pairs and then this sulfur right would it still happy double bonded one lone pair and then the chlorine like that so we're saying that in this acid-base reaction these electrons in here right move back on here to form the carbonyl and so you can consider this to be the intermediate for this mechanism and you'll see some versions of this mechanism take this intermediate and and continue on to form your product I'm going to show the the chloride anion functioning as a nucleophile over here for for this right so it's an acid-base reaction so it's possible to once again protonate your carbonyl that's going to activate it right so 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 its own so if the chloride anion attacks here right 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 lone pairs of electrons bonded to hydrogen here so let's show those electrons in blue so if these electrons in blue kick off on to the oxygen we could say that those are these electrons alright and then we could we could also say that these electrons here in green right on the chloride anion are going to form a bond right between the chlorine and the carbon so we can go ahead and draw in the bond between the chlorine and the carbon like that and then we still have our oxygen right here right bonded to our sulfur double bonded to this oxygen and then we have our lone pair of electrons here and then we have our chlorine right so what all this does is make a much better leaving group than what we started off with so you could think about you know this as our leaving group right this is a leaving group and the next step of the mechanism and if we go back up here right that's a much better leaving group than the O H that we started out with and so if that's going to be our leaving group right we could show we could show these electrons in here moving in to form our double bond which would kick these electrons off onto our oxygen so let's go ahead and show we would make from that all right we would now form our R group our carbon we double bonded to an oxygen all right like this lone pair of electrons plus one formal charge and then we'd have our chlorine over here with all of its electrons too so let's go ahead and show where those electrons came from so if these electrons move in here all right that would reform our carbonyl so I can go ahead and draw that in and then we still have this bond and green here just to clarify and then we would have some electrons and let's make a magenta so these electrons in here are going to kick off on to the oxygen so we can go ahead and draw that we would have our oxygen alright so we would have three lone pairs of electrons one of them would be the magenta ones so we go ahead and do that that gives our oxygen a negative one formal charge because it's bonded to the sulfur here the sulfur is still double bonded to this oxygen and once again still a lone pair of electrons and a chlorine over here like that the reason why this is a good leaving group is because we can we can push these electrons into here push these electrons off onto here and we can form sulfur dioxide as a gas alright so 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 alright so draw our lone pairs of electrons here and and we still have a lone pair of electrons on our sulfur so if these electrons in red here move in to form this bond right we've now formed sulfur dioxide and we also form the chloride anion alright so once again once we can follow 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 D protonate so take this proton leave these electrons behind and finally we formed our acyl chloride so we have our carbon double bonded to an oxygen with two lone pairs of electrons so we have our chlorine over here like that so if we follow that final those final electrons these electrons in blue move off onto your carbonyl and so you would also form a see L buy this right so you form HCl and which is also a gas so the formation of these gases right the formation of sulfur dioxide and HCL drives the reaction to completion here so that was a that was a long mechanism and once again you'll see some slight variations on this so so if I go back up to here right you could you could show this this step alright the nucleophile attacking the electrophile and directly form here if you wanted to so as your nucleophile attacks yourself or these electrons could kick off onto chlorine to form the chloride anion at this step and then you would get you would get this at this point so and then once again I talked about the possibility of using this as your intermediate so depending on which textbook you look in you might see some slight variations on this mechanism and it is a bit of a long one here alright let's look at two other ways to make an acyl chloride so starting with a carboxylic acid you could add phosphorus pentachloride or phosphorus trichloride and both of those will give you an acyl chloride as well and the mechanism is pretty similar and also we could think about instead of phosphorus trichloride we could ask we could add a pbr3 so phosphorus tribromide and instead of putting a chlorine here right that would put a bromine and so we could form an ACL bromide this way and and we'll see a use for this reaction in a later video