Phenol is synthesized from cumene via the scheme shown. Sugges...

Question

Phenol is synthesized from cumene via the scheme shown. Suggest the mechanism for each step of the reaction shown. [The oxygens have been screened in blue and orange to assist you.]

Schematic showing the synthesis of phenol from cumene, highlighting reaction steps and oxygen atoms in blue and orange.

Accepted Answer

Hi, everyone. Welcome back. Here's our next problem. The steps for the synthesis of P cresol from P sing are shown below, propose a mechanism for each step in a series of reactions. The oxygens are color coded for tracking. So we're given a series of four reactions. So this is not the entire mechanism. This is just kind of some of the intermediate steps along the way because this is somewhat of a complicated reaction. Here we start with P syne which is a benzene ring, two substituents in the pair position a methyl group and an isopropyl group step one, it just says it reacts with an initiator. So this would be just some form of molecule with a radical. So we have the addition of a radical and the that produces our original molecule, but now it's lost to hydrogen. That would be implied in that structure, draw that hydrogen on that central carbon and instead has a radical there, this radical reaction step two with diatomic oxygen with 02. And we see and then the product of that step eventually will be that central carbon of the isopropyl group in place of the radical electron it now has ooh added on. So we clearly added the oxygens from our 02 molecule as well as a hydrogen. So clearly, there's some extra steps going on there. Reaction three has that molecule reacting with hydro num ion and we end up with a carbo Cron, but there's clearly been a Carbocaine rearrangement because the connectivity is now different. So before we had a bond from the benzene ring to a carbon of the isopropyl group. Now the benzene ring has a bond to an oxygen, we formed an ether and then that oxygen has a bond to the isopropyl group which now has a positive charge on its central carbon. So we've got a Carbocaine that has rearranged the bonds here. Then in reaction four, that molecule reacts with water. So in the carbo cion, we have a red oxygen, the ether oxygen that came originally from that oxygen molecule, the water is depicted with a blue oxygen. And then we see that our final product is our P cresol. So it's a benzene ring, two substituents in the pas position, the methyl group and now an alcohol group instead of the isopropyl group, where do those three carbons go? The other by-product is acetone. So three carbon molecule with a carbon and the oxygen on the acetone is blue that came from the water, the oxygen in the cresol is red, it came from the molecule, the the reactant molecule. So let's think about what the more detailed mechanism would be. So we'll start with step one and to save space, I'm going to call in my original molecule, the benzene ring with its methyl substituent. I'm just calling R, it's going to be our R group. I'm focusing on the isopropyl group as that's where all the action is happening. So for my mechanism for step one, I'll draw R with that isopropyl group and then I'm specifically indicating the hydrogen on that central carbon since I have to have some action going on with the radical there. So step one is pretty straightforward, my original molecule is reacting with a radical. So I've just drawn it as R prime since I've already used R with a single electron and the hydrogen and one electron from that ch bond are going to come over and meet with the single electron on the radical initiator and form a new bond. So it'll be h bonded to R prime note that we've drawn single headed arrows because we're showing the movement of just one electron, not a pair. And then since our hydrogen is coming with one electron, there'll be one electron left and that will go to the carbon. It's a little hard to draw that accurately. There's not much space. So our carbon will now have a lone electrometer. So our original molecule is now a radical as we expected in step one. So now we move on to step two, we know this is a reaction with oxygen diatomic oxygen. So how do we end up with that? Ooh added on one way to erase the 02 and put it more in a structural form which will be O double bonded to O there. So we've shown it's reacting with 02 in the form of O double bondage to an O. Note that in the previous step, I've shown hr prime coming off as a and then my product of this reaction with the radical and 02, I'm going to have the R group with attached to a carbon with two methyl groups and then O bonded to an O with a single electron. So now we have a radical on our oxygen. So how does this happen? We have our single electron that was on our sine coming and reacting with one of the electrons from the pi bond between the two oxygens forming a new bond between our R molecule and the one of the oxygens which is still joined to the other oxygen by a sigma bond. Then the remaining electron for that pi bond will go on to the oxygen. So we've produced r bonded to carbon with two methyl groups and then a bond, single bond to the oxygen, which is a single bond to the second oxygen. And a it's, that's a free radical with a single electron in the oxygen. So now you can see how it could get proton fairly easily. And in fact, it can react with the original PS Iine to regenerate a PS Iine radical. So as we know, these radical reactions often propagate for many cycles. So we show the single electron from the oxygen reacting with the hydrogen and one of its electrons in that carbon hydrogen bond and then the remaining electron from that bond going on to the central carbon. And we've ended up with a product from step two, which is our, our group to a carbon with two methyl groups. And then ooh scroll down to make a little more room rather than move, making a down arrow as there is in my, what I'm given in the problem just going to start a new line to make it a little clearer. So now I have step oops that should be step three, not step two, we've already done step two. So in step three, let's start with this molecule and we know that it's going to rack with a hydro num ion. And of course, hydro ions being a acid being acidic, very good at proton things. So if we put our lone pair on the second oxygen, the one farther out, we show this lone pair or this is a double headed electron, it's a pair of electrons grabbing a proton from our hydro ion. Of course, those electrons from the ho bond going onto the oxygen to produce water. And then we have this, our group, this carbon with the two methyl groups and now we have O and then o and that oxygen has two hydrogens and a positive charge. It doesn't like that. So now we're going to see this rearrangement to generate a much more stable carbo cat I, so what we will see is actually the electrons from the carbon oxygen bond, they are coming from that carbon. That was an iso purple group or excuse me, I've gotten a little bit lost here, the electrons from the R group carbon bond. So a carbon carbon bond between the R group and the carbon that was an isopropyl group, those electrons are going to come over and form a new bond with the closest oxygen that will eventually be the ether oxygen in the next product as that oxygen gets more electrons, the electrons from the oo bond, we'll go over and on to that positively charged oxygen. It now leaves as a water, it was converted into a good leaving group by being proton. And we now have a new rearrangement of the bonds in our original molecule. So that's going to produce this much more stable carbo cion and we'll draw it out together just so you can see how the new connections are. So I drew the R now instead of having a bond to another carbon, it has a bond directly to the oxygen, but the oxygen still has a bond to the carbon. That was the central carbon of the isopropyl group. So it now has a bond to the carbon which has two methyl groups. So it's as if it has an iso the isopropyl group is now attached to the oxygen. We've now made an ether, our, our group directly attached to the oxygen which is attached to the isopropyl group. And now that carbon that was in the isopropyl group has a positive charge. And that's because we look back at it, it had four bonds. So the R group, the two methyl groups and the oxygen by breaking the bond to the R group, it's left short a set of electrons, it's got a positive charge. So there we have our tertiary carbo cion much more stable than having a positive charge on oxygen. So that's part three of this reaction. So now we'll scroll down, make a little more space and get to part four. Again, I'm not going to directly do an arrow going down just in answer of being a little less confusing. So I'm just going to rewrite this molecule my Carbocaine to start another row for step four. Well, we've got some water floating around in our solution. We know that we're reacting with water in step four. So we definitely can have water attack this positively charged carbon. So the lone pair on the oxygen of our water molecule comes in too, that positively charge Carmen. And now we have a new molecule which is R bonded to an oxygen bonded to a carbon with two methyl groups and an 02 plus group. And now there again, we have water present. So the lone pair from a molecule of water can grab one of those protons, the electrons from the oh bond go on to the positively charged oxygen. And we turn that into an alcohol and we know that that ether oxygen will eventually have to be an alcohol. So we'll go ahead and proton that e oxygen sort of essentially creating a leaving a good leaving group. So one of the lone pairs on the ether oxygen will come over and grab a proton from some more of the regenerated hydro ion leaving the ho electrons to go on the positively charged oxygen. We have water leaving note I drew and hydro ion being produced in the last step. So now we've essentially converted this into a way to break up our ether and generate our pre P cresol. So let's scroll on down to keep going. And since I'm still in step four, I'll use a down arrow this time. Now, we actually don't need another reactant molecule this time, just rearrangement here. So we end up having the loan pair from our oh at the end, the terminal oh goes on to form a double bond with its carbon. And then the electrons from the oc bond from the ether bond on the right side, there then go on to that positively charged oxygen, thereby breaking that oc bond. And there you can see we've generated our two products. We're almost there. Not quite, but almost, we've generated the P cresol because we now have roh we've split the molecule into two. We almost have our acetone. We have carbon, the double bond to an O and then two methyl groups. But that oxygen still has a hydrogen on it from being an alcohol. So the oxygen has a positive charge. So that's basic enough to generate our final product. That way water can de proton that carbon oxygen. So just draw the lone pairs on the oxygen. And those lone pairs will come over and grab that hydrogen causing the electrons from the oh bond to go on to the positively charged oxygen. And we end up with our final product of ROH which is RP cresol. And then we have the acetone. So there's the three carbons, the central carbon and a carbon bond. So there we have it, we've got the detailed mechanism for all four steps. In step one, we had this radical reaction to produce the radical. We needed to keep going. In step two, we reacted with oxygen to add on that ooh group. In step three, we ended up generating our tertiary carbo Cron by a Carbocaine rearrangement. And in step four, our carbo cion, we attached to water rearranged our molecule and ended up with our final two products. So that's our detailed reaction for getting from P thymine to P Cresol. See you in the next video.

Phenol is synthesized from cumene via the scheme shown. Suggest the mechanism for each step of the reaction shown. [The oxygens have been screened in blue and orange to assist you.]

Key Concepts

Electrophilic Aromatic Substitution

Radical Mechanisms

Hydrolysis of Peroxides

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