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Acid chloride formation

Acetic acid to acetyl chloride mechanism. Can be generalized to forming any acid halide from a carboxylic acid. Created by Sal Khan.

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  • orange juice squid orange style avatar for user http://facebookid.khanacademy.org/1700211
    At , why does the Chloride ion give an electron to the carbonyl carbon and not to the hydrogen bonded to the positively charged oxygen atom?
    (10 votes)
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    • blobby green style avatar for user rjonesmusicproject
      I think the carbonyl carbon is the most susceptible to nucleophililc attack in this situation, as it is double bonded to one oxygen atom and bonded also to a positively charged oxygen atom, both of which are very electronegative and so will be pulling electrons away from the carbonyl carbon (especially the positively charged oxygen, which will be pulling electrons extra close to try and balance the charge out) giving it a fairly strong partial positive charge, making it an ideal target for a nucleophile such as Cl-.
      On the other hand, the hydrogen is having its lone electron pulled on only by the positively charged oxygen, so the partial charge is perhaps not so great.
      Also perhaps if the chlorine bonded to the hydrogen, there would be only two possible ways to finish the reaction - the Cl could regain its electron, taking us back to the same situation, or HCl could be removed, balancing the charge on the oxygen, and leaving us with a different product molecule entirely.

      I don't know about the finer details, I'm just having an educated guess here, I hope this helped and sorry for rambling!
      (10 votes)
  • blobby green style avatar for user abedini1
    at , when the Chloride attaches to the carbon, why can't the carbon give an electron to the single-bonded oxygen instead of the double bonded oxygen? It seems to me that the step where the top Oxygen breaks the double bond, becomes negative, then reforms the double bond again is pretty pointless. If the other oxygen gets the electron, you can knock out a whole step. Why can't that happen?
    (7 votes)
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    • blobby green style avatar for user WilssonLowe
      (this isn't really an answer, sorry Abedini!)
      I just wanted to say that I completely agree with this and that, as it seems to me anyway, the first step () raises the same issue. What speaks against the assumption that the sulfur just gives its electron to the Cl directly instead of having to go to the trouble of breaking and then reforming the pi bond with the O?

      Maybe it has something to do with the differences between pi bond electrons and sigma bond electrons?

      I'd also like to add the question, if we somehow know that these seemingly pointless steps happen, what method was used to acquire this knowledge?
      (5 votes)
  • leaf green style avatar for user dbzabhiram
    Can anybody help me in giving easy definition of Equilibrium?? i'm bit confused :(
    (4 votes)
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    • leafers ultimate style avatar for user slartibartfast92
      From Wikipedia: Chemical equilibrium is the state in which both reactants and products are present at concentrations which have no further tendency to change with time. Usually, this state results when the forward reaction proceeds at the same rate as the reverse reaction.
      (4 votes)
  • blobby green style avatar for user nj.nicole14
    At , what happened to the lone pair that was on sulfur
    (5 votes)
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  • blobby green style avatar for user pslattery522
    In each Org. Chem. mechanism video:
    Why are these reactions spontaneous?
    How are the orbitals changing with new bonds?
    Why other possible reactions I am contemplating wouldn't react instead of the stated mechanism.
    Where is the intuitive explanations we are so spoiled with?

    Respectfully,
    Patrick
    (6 votes)
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  • leaf green style avatar for user Danny Nguy
    Why is SO2 a better leaving group compared to chlorine? Chlorine seems to be more polarizable and more electronegative than sulfur and oxygen.
    (2 votes)
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    • spunky sam blue style avatar for user Ernest Zinck
      The HOSOCl is a better leaving group because
      a. The O already has a formal positive charge, so the electrons in the C-O bond are relatively close to the O.
      b. The O. S, and Cl are all highly electronegative atoms. This puts even more positive charge on the O atom and weakens the C-O bond even further.
      c. Up to this point, all the reactions are reversible. But the decomposition of the HOSOCl into HCl and SO2 is irreversible, because SO2 is a gas and escapes from the solution. This is what drives the reaction to completion.
      (3 votes)
  • leaf green style avatar for user Anubhav Basu
    is the HCL in the end product ionic or covalent ??
    (2 votes)
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  • blobby green style avatar for user laragahan
    How come the acetic acid donates an electron to the thionyl chloride in the first step? Isn't it really acidic, and therefore it would rather be donating a proton or accepting an electron?
    I just don't understand how this occurs initially. Thank you so much for your help.
    (2 votes)
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    • leaf green style avatar for user Dave M
      They reaction is typically run in pyridine (pyr.) in most textbooks. It is a solvent like benzene except it is a weak base. Therefore, there is no acetic acid....you have ammonium acetate salt floating in pyridine. Ethanoate attacks thionyl chloride, NOT ethanoic acid.
      (2 votes)
  • leaf orange style avatar for user Harshit Jindal
    Is it Acyl Chloride or Acetyl Chloride? I'm so confused :/
    (2 votes)
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  • male robot johnny style avatar for user Emjae Dela Cruz
    what if we use the O in the Carbonyl group to bond with SOCl2 instead?
    (2 votes)
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Video transcript

In this video, we're going to explore how we can start with acetic acid. And acetic acid looks like this. And end up with acetyl chloride that looks like this. And it's going to be done in the presence of thionyl chloride. And thionyl chloride, we haven't seen it before, but it just looks like this. It's a sulfur double bonded to an oxygen. And then, it has single bonds to two separate chlorines. And then, if you want to care, sulphur has six valence electrons. So it has another lone pair right over here. And we'll focus on this. Because this is really the simplest reaction of starting with a carboxylic acid and forming a acyl halide. And it can be generalized very easily. If you turn this methyl group into just a larger chain, and then this will be a larger chain right here. And, instead of having a chlorine here, you could do it with other halides as well. So let's think about how this might occur. So let's have our acetic acid right over here. And then, you have this oxygen bonded to a hydrogen. Now, this guy has got two lone pairs. And then the thionyl chloride-- I'll redraw the thionyl chloride right over here. And I'll set it up in the right position. The thionyl chloride looks like this. And this has a lone pair of electrons right over here. And all of these molecules, chlorine, and oxygen-- or all of these atoms-- chlorine and oxygen-- we can look at a periodic table up here. You see sulphur over here. And then, to the right of sulphur is chlorine. And above sulphur is oxygen. So both chlorine and oxygen are more electronegative than sulfur. So sulphur is going to have a partial positive charge there, even though sulphur is a reasonably electronegative atom in its own right. But these other guys are going to be sucking electronic away from it. So you could imagine this entire molecule could act as a nucleophile, with this oxygen, right over here, giving an electron to the sulphur. Sulphur is an electronegative atom, but it has a partial positive charge in just this thionyl cholride. So it will be attracted there. And then this guy is going to have to give up an electron. And he'll give it back to this oxygen up there. And so that will be-- and this reaction could go in either direction. So, once again, I'll draw it in equilibrium. But right after that happens, we'll have this, I guess we could call it a complex, that will look like this. This oxygen, it had two lone pairs. It had one pair, two pairs. And now, it has a third pair, because it got this electron. It always had that other electron in the covalent bond. And now, it has a negative charge. And now, the sulphur is bonded to those two chlorines. And then, it is also bonded to this oxygen over here. So let me draw it over here. So we have this oxygen, which is bonded to this carbonyl carbon, just like that. And then, it is bonded to a hydrogen right over there. It has one lone pair now. And the other lone pair is, now, turned into a covalent bond with the sulfur. And so this guy gave away an electron. So he now has a positive charge. Now, the next step, you could imagine that this oxygen says, hey, I liked having a double bond with the sulfur. The sulfur is still bonded to a bunch of things that are more electronegative to it. It still has a slightly positive charge there. So you could imagine that this electron gets given back to the sulfur. But then, the sulfur needs to bump an electron. Chlorine is pretty electronegative, so one of these chlorines will be able to take away an electron. So this chlorine could take away an electron. So then, after that happens, our situation-- and once again, this is in equilibrium-- our situation will look like this. Let me draw. So let me draw the original acetic acid. Or the part that was part of the original acetic acid. And so you have this oxygen, right over here, bonded to a hydrogen, bonded to-- let me do it in the same color so we can keep track of things-- bonded to this sulphur right over here, which now has a double bond, which, now again, has a double bond with this oxygen up here. So I'll do it in that same color. It's still bonded to that chlorine up there. But now, this chlorine is left. It has taken that electron with it. So the chlorine had 1, 2, 3, 4, 5, 6, 7 valence electrons. It gained this electron, this orange electron. So now, it has a negative charge. And, by the way, this guy never lost his positive charge. Still has a positive charge like that. Now, the chlorine could act as a nucleophile on the carbonyl carbon. This guy is bonded to two oxygens, more elctronegative, partially positive charge. This guy can give an electron to the carbonyl carbon. And right as that happens, this is a nucleophilc attack, the carbonyl carbon can give up an electron to the carbonyl oxygen. And then, we will be in equilibrium. So make sure you realize we're going to the right then down. Now, we're going to the left with this thing. So this was the carbonyl carbon bonded to this oxygen right now, which now, just took another electron. So now, it has a negative charge. It had 1, 2 lone pairs. And now, it will have one more lone pair because it took that electron. It always had this end of that bond. So now it has both of these electrons over here. It has a negative charge. It gained an electron. And then, it is bonded it to this OH group right over here. And then that is bonded to the sulfur, which is bonded to the chlorine, and now double bonded to that other oxygen. Of course, we have this chlorine over here that did the nucleophilic attack. So you have this chlorine over here. And this nucleophilc attack, it gave an electron to what was this carbonyl carbon. It gave an electron. So it is now neutral. And you can kind of imagine that this negative charge got transferred to this oxygen up here. And this oxygen, right here, still has a positive charge. Don't want to forget that. Now, the next step that could happen-- once again, all these can go in either direction-- is that this guy doesn't like having a negative charge. And this guy is still going to have a partially positive charge, because he's bonded to a bunch of more electronegative atoms than itself. So he wants to reform the double bond. And that kicks off all of this business over here. So then this-- let me do this in a new color. I've already used the magenta. I'll use the pink. This electron, right here, gets taken back by this oxygen that had a positive charge anyway. So it would want to take it back. And then we are left with-- and we're getting pretty close to the punchline. We are, then, in equilibrium. This part over here will then look like this. We now have reformed our double bond. It is only going to be bonded to the chlorine, just like this. And this whole part over here on the right has broken off. So you have this oxygen bonded-- let me do it in the same colors. You have this oxygen bonded to a hydrogen. It now gained an electron. So it is now neutral. And it is bonded to this sulfur. The sulphur is in green. It's bonded to the sulfur, which is bonded to a chlorine. And then, it has a double bond to this oxygen right over here. So we've already formed our acetyl chloride. And we've really finished with the hard part of the reaction-- to see that how you could get an acetyl chloride. And then to kind of complete this reaction-- because if you actually perform this in a beaker, you'll end up with some hydrogen chloride and some sulfur oxide as a by-product. You could imagine that this oxygen, right over here, takes back its electron from this proton, gives it to this sulfur. And then, the sulfur, since it got an electron, will allow the electronegative chlorine to take back its electron. And then, if we just focus on this, the acetyl chloride, at this point, isn't doing anything anymore. This would be in equilibrium with something that looks like this. You now have a hydrogen proton floating around. Just a naked proton, really. There's not even any neutrons there. You have this oxygen, which is now double bonded with this sulfur. So I'm going to try to keep all of the colors the same. This, right here, was yellow. And then, that original bond it had with it was magenta. And then, you have the chlorine nabbed that electron. And it already had seven valence electrons. 1, 2, 3, 4, 5, 6, 7. it now has a negative charge. And then, the final step. So we already have our sulphur dioxide. Now, the final step is just the chlorine giving one of its electrons to the hydrogen to form hydrogen chloride. So then, the final step is just this. So when all is said and done, we end up with some acetyl chloride. We end up with some hydrogen chloride. And we end up with sulphur dioxide. Just like that, and we're done. And, once again, you can generalize this to get it starting with any carboxylic acid and forming the acyl halide version of it, or acyl chloride version, if you want to stick with a chloride right over here. Actually, this, right here, is acetyl chloride. Anyway, hopefully, you found that entertaining.