Sulfonation of benzene. Created by Jay.
Want to join the conversation?
- Why does one molecule of sulfuric acid act as a base and other as an acid even though both are equally reactive?
Can such reactions between inorganic reagents be predicted theoretically?(6 votes)
- This is analogous to the autoionization of water — there is an equilibrium with some molecules acting as a base while others act as an acid.(6 votes)
- Why is sulfur electron deficient in both the resonance structures? I thought sulfur is electrically neutral on the right structure since it has no formal charge.(3 votes)
- I tend to agree with you. But you must remember that the actual structure is a resonance hybrid of the two contributors. In the hybrid, the sulfur atom still has a partial negative charge and will still act as an electrophile. Personally, I would have used the structure on the left as the electrophile, because it shows explicitly that the sulfur atom is electron deficient.(10 votes)
- Couldn't you use the HSO4- as the base that takes the proton off of the benzene?(2 votes)
- That is theoretically possible, but less likely than H₂O acting as the base. This is because H₂O is a much stronger base than HSO₄¯.
To understand this ask yourself which is a stronger acid H₂SO₄ or H₃O⁺ — if you don't know that then you can consult a pKa table such as:
Jay talks about this in this video starting at7:17.(3 votes)
- while adding more amounts of sulfuric acid, the equilibrium shifts towards the products side, is that the only consequence or is there a chance for all the 6 hydrogens to be replaced by SO3H?(2 votes)
- Once the SO3H is added it becomes an Electron Withdrawing Group and a benzene deactivator so it is highly unlikely that you would ever multisubstitute a benzene ring this way with SO3H.(2 votes)
- When you use SO3 and let it react with H2SO4 to form the electrophyle, there is no water leaving group. So when the reaction is proceeding, what takes the hydrogen off the benzene? Does the HSO4- act as the base and take the hydrogen off?(2 votes)
- according to my textbook, even sulfuric acid is basic enough to protonate the cyclohexadienyl cation. Its done in oleum I believe which doesn't have water, thus the sulfuric acid doesn't act like it does in water.(1 vote)
- why does the oxygen BONDS with another hydrogen? its valency is just 2,,,(1 vote)
- if we have C6D6 instead of C6H6, then will the reaction rate change?(1 vote)
- No, because attack of the electrophile is the rate determining step. The H(D) doesn't leave until a later step.(2 votes)
- Why is this reaction a reversible one?(1 vote)
- The activation energies for the intermediate formed at6:00going on to products or returning to reactants are about equal.(2 votes)
- In the beginning you said water is formed however this isn't seen at the end of the reaction. Can you please expantiate on that? Thanks(1 vote)
- He showed water being formed at2:45. In the strong acid medium, it gets converted to H₃O⁺.(2 votes)
- i have question: if i say that the sulfonation reaction under the ionic mechanism is correct?(1 vote)
Here's the reaction for the sulfonation of benzene. So over here we have benzene, and to that we add some sulfuric acid. And so we would form benzene sulfonic acid and also water as a byproduct. Since this reaction is at equilibrium, we can shift the equilibrium by using the different concentrations of sulfuric acid. If we use concentrated sulfuric acid, we would of course shift the equilibrium to the right to make it more benzene sulfonic acid. If we use more dilute sulfuric acid, so we have more water here, that would shift the equilibrium to the left. And so the fact that this is a reversible reaction is sometimes used in synthesis problems. So let's take a look at the mechanism to put an SO3H group onto our benzene ring. And so here I have two molecules of sulfuric acid. And one of these molecules is going to function as an acid and one of these is going to function as a base actually. So I'm going to say the one on the right is going to function as an acid. It's going to donate a proton. And the molecule on the left is going to function as a base and accept a proton. So if this lone pair of electrons accepts a proton, takes that proton, that would leave these electrons behind on to that oxygen. So let's go ahead and show the result of that acid base reaction. So we have sulfur double bonded to an oxygen. We have an OH group over here on the left. And then over here on the right, we've now protonated this oxygen over here on the right. So it actually has a plus 1 formal charge now. So if you follow those electrons, these electrons right in here are going to form this bond and give us a plus 1 formal charge on that oxygen. So for the other sulfuric acid molecule, the one that functioned as an acid here, it's going to lose a proton. So we're going to be left with the conjugate base, which is HSO4 minus-- so over here we would have HSO4 minus the negative 1 charge on our oxygen, our OH over here. And so if we follow those electrons, I'm just going to go ahead and make them red. So these electrons right in here are going to come off onto this oxygen, giving us HSO4 minus. So let's go back to the protonated version of sulfuric acid over here. And we can see that there is a water molecule hiding over here that is a very good leaving group. So this is a similar mechanism in that respect to what we saw in the last video for the mechanism on nitration. And so we have water, which is a good leaving group. And so we can have these electrons move in here, which would kick these electrons off onto the oxygen. And of course, now we have H2O as our leaving group. So over here on the right, water's going to come off like that. And once again, we can follow some electrons. I can say that these electrons in here in blue come off onto the oxygen giving us water. And if we think about what else is formed-- so we still have sulfur double bonded to an oxygen and sulfur double bonded to another oxygen. And now actually, we're going to have sulfur double bonded to the third oxygen here. And this third oxygen still has a hydrogen on it. It still has a lone pair of electrons, and it has a plus 1 formal charge on it now. So let's go ahead and follow those electrons. So I'm going to make these in green here. I'm going to say those electrons in green move in here to form a pi bond like that between the sulfur and the oxygen. And so this is really a protonated version of the sulfur trioxide molecule, or SO3. And the sulfur is actually the electrophilic part of this ion. And it's not immediately obvious looking at how I've drawn it here. But if we think about the oxygen being more electronegative than the sulfur, we could think about a resonant structure for this where the electrons in green come back onto the oxygen. And therefore, that's taking a bond away from that sulfur. And so we could draw it like this, where we have the oxygen now having two lone pairs of electrons. And since I took a bond away from that sulfur, that sulfur now has a plus 1 formal charge on it like that. So this is a little bit more obvious that the sulfur is the electrophilic portion. So this is now our electrophile. The sulfur is the one that is electron deficient. And so you could think about this version on the left, or you could think about this version on the right. It doesn't really matter which one you think about. The sulfur is electron deficient in both of those resonant structures. OK. So I'm actually going to use the version on the right here. So I'm actually going to redraw this protonated version of SO3 for the next part of our mechanism. So that's where the protonated version of SO3 reacts with our benzene ring. And so I'm going to go ahead and draw my benzene ring right here. And I'm just going to redraw that protonated form. So here we have our protonated form of sulfur trioxide. And once again, we know that the sulfur is the electrophilic portion of that ion. So we're going to have the pi electrons function as a nucleophile. And that nucleophile is going to attack our electrophile. So that's the sulfur, which would kick these electrons in here back off onto our oxygen. So if we show the results of that nucleophilic attack-- so we would have these pi electrons in our ring. We would have a hydrogen on our ring. And now we would have a bond to our sulfur. The sulfur is double bonded to some oxygens. And there's also an OH over here on the right like that. So let's once again follow our electrons. These pi electrons in magenta function as a nucleophile and form a bond between that carbon and that sulfur. Taking a bond away from this carbon, of course, means that carbon gets a plus 1 formal charge. And so that is the cation that results. Now, I could draw two more resonance structures for this cation, and we've seen how to do that in some of the earlier videos. And so just to save time, I'm not going to draw those resonant structures. But this, of course, is going to represent our sigma complex. And so the next step in electrophilic aromatic substitution is deprotonation of your sigma complex. And so you could think about this water molecule functioning as a base. So this lone pair of electrons coming along and taking that proton, leaving these electrons behind to move in here to reform your benzene ring. And so let's go ahead and show the result of that. So we would reform our aromatic ring here like that. And now we would, of course, have SO3H as our group coming off of our benzene ring. And so we're done with our reaction. If we show those electrons, I'm going to go ahead and make those blue. So I'm saying these electrons in blue here move in here to reform your benzene ring, to form benzene sulfonic acid as your product. Now, I showed the water molecule functioning as a base, take away that proton. Sometimes you'll see textbooks-- if I go up here-- showing the HSO4 minus anion functioning as a base to take that proton away to regenerate your sulfuric acid catalyst, if you will. So that's just something to think about. Water is probably a little bit better based than HSO4 minus. So that's why I've shown that in this example. But this reaction is very dependent upon the conditions of the reaction. And so let's say you were to do this reaction by adding some more sulfur trioxide in. So instead of the protonated version of sulfur trioxide being generated from the sulfuric acid molecules itself, you could go ahead and start the reaction with some more sulfur trioxide. So let me just go ahead and show the sulfur trioxide molecule. So I'm showing it double bonded to all of these oxygens here. And if you add that to some sulfuric acid, the sulfuric acid would, of course, donate a proton to the sulfur trioxide. So let's go ahead and show that as our first step here. So this lone pair of electrons is going to pick up this proton from sulfuric acid, leaving these electrons behind on that oxygen. And so we can show this acid base reaction where the sulfur trioxide is now protonated. So we now have a protonated version of sulfur trioxide like that. So once again, following our electrons, these electrons right in here-- I'm showing those taking this proton for our protonated version of sulfur trioxide. And that would of course form HSO4 minus as our conjugate base to sulfuric acid. And then you could think about this now reacting in our mechanism exactly how we've shown it up here. So it doesn't really matter how you get your protonated sulfur trioxide. This is just another way to do it. And if you start with some sulfur trioxide already present, you're pretty much going to shift the equilibrium all the way to the right. And so let's go ahead and show that real fast here. So we have-- we're just going to react benzene with sulfuric acid. But this time, we're going to add in some sulfur trioxide as well. So we have SO3 and we have H2SO4 as our catalyst to produce the protonated SO3. And that adds on to your benzene ring to form benzene sulfonic acid as your product. So once again, the addition of the extra SO3 means that you would form a higher yield of your product here. Now, in the mechanism that I've shown you, I've showed you the protonated SO3 functioning as your electrophile. But the sulfonating agent might just be SO3 on its own and not protonated SO3, depending on different reaction conditions. And so you'll see different mechanisms for sulfonations in different textbooks. And so obviously, you should use the one that your professor wants you to use on any exam. And I've just shown you one or two examples in this video.