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Current time:0:00Total duration:10:24

Video transcript

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 sulphonic acid and also water as a by-product since this reactions at equilibrium we can shift the equilibrium by the using different concentrations of sulfuric acid if we use concentrated sulfuric acid we would of course shift the equilibrium to the right to make more benzene at sulphonic acid if we use more dilute sulfuric acid right 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 OHA 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 we follow those electrons right these electrons right in here alright are going to form this bond and give us a plus 1 formal charge on that oxygen so for the 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 one charge on our oxygen ROH 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 HS 4 minus so let's go back to the the protonated version of sulfuric acid right 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 a similar 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 is going to come off like that and once again we can follow some electrons I can say that these electrons in here and blue come off onto the oxygen giving us water and if we think about what else is formed right 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 hydrogen on it it still has a lone pair of electrons and has a +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 let's say those electrons in green move in here to form a PI bond like that between the sulfur and the oxygen 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 looking at how I've drawn it here but if we think about the oxygen being more electronegative than the sulfur right we could think about a we can think about a resonance 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 +1 formal charge on it like that so this is a little bit more obvious that the sulfur is the electrophilic portion right so this is now our electrophile the sulfur is the one that is electron deficient and so you could think of out 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 resonance structures okay so I'm actually going to use the 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 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 right 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 all right so we would have these pi electrons in our ring we would have a hydrogen on our ring and now we would have a a bond to our sulfur the sulfur is double bonded to some oxygens and there's also an OHA 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 +1 formal charge and so that is the cation that results now I could draw a 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 resonance structures but this is course 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 band zyne ring and so let's go ahead and show the result of that so we would we would reform our aromatic ring here so like that and now we would of course have so3h all right 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 right move in here to reform your benzene ring to form a benzene sulphonic acid as your product now I showed the water molecule function is it functioning as a base take away that proton sometimes you'll see textbooks if I go up here showing 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 - so that's why I've shown that in this example but this this reaction is 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 you could go ahead and start the reaction with some more sulfur trioxide so let me just let me just go ahead and show the sulfur trioxide molecule so I'm showing it double bonded to all these oxygens here and if you add that to some sulfuric acid alright 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 all right so we now have a protonated version of sulfur trioxide like that so once again following our electrons these electrons are right in here all right those I'm showing those taking this proton for our protonated version of sulphur trioxide and out of course form hso4 - as our conjugate base - sulfuric acid and then you could think about this now reacting in our mechanism exactly how we've shown it up here alright so it doesn't really matter how you get your protonated sulphur trioxide this is just another way to do it and if you 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 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 h2 so4 as our catalyst alright to produce the protonated so3 and that adds on to your benzene ring to form benzene sulphonic 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 self anything 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 sulphonation 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