 Let's look at the Riemer team and reaction in which we start with a phenol and our end product is this phenol with a CHO added here which is called salicyl aldehyde. So let's see what the reagents are and how this reaction takes place. Before I show you the reaction and the steps and the mechanism, let me first give you an overview of the entire process. So we'll start with the preparation. The reagent we're going to use is this X and to prepare this we're going to use chloroform and we're going to react it with any of which. That's one part that is the preparation step and the step one that I've written here is the actual reaction itself where we have the phenol, we are reacting it with NaOH. So basically we're getting this phenol ready for the reaction with this X and then when we add this X in the presence of some base you're going to see the CHO added at the end. So there are two things to know here. One, there is one preparation involved which involves CHCl3 which is chloroform and once we have prepared this reagent X using this process. In our second step this X has some role to play in how this CHO comes about. So let's see how all of this happens. We start our preparation reaction by taking this CHCl3 which is also known as chloroform and we're going to react it with a base. So now what happens is all of these chlorines are pulling away electrons from this carbon and in effect what is happening is this hydrogen can be easily taken off. It's highly acidic. So this base is going to take off this hydrogen and will be left with a negative charge on this carbon. But then we also know something about this chlorine. It is going to leave taking the electrons with it. So what is going to happen is we end up with this CCL2 which is called dichlorocarbin. Now the point to remember here is that this carbine is actually a neutral species. So there is no positive charge on the carbon here and we know that in electrophilic substitution reactions the electrophile is generally diluted by an E+. Although there is no positive charge on this carbon it is still acting as an electrophile. So how is that happening? So the reason is that this carbon is electron deficient. If we look at the Lewis structure of CCL2, carbon has four valence electrons which are these white ones and each of the chlorines have seven valence electrons one of which from both of the chlorines are shared with the carbon to form the bond. So now if we look at this carbon it has these six electrons and to complete its octet it needs two more electrons which is why it is electron deficient and in this case it will act like an electrophile. And of course because it has these two electrons here it can actually also act as a nucleophile if it decides to share these electrons. But in the case of the Riemann reaction as you'll see in the next few steps this CCL2 will act as an electrophile. And also one more thing to remember is that although this is the preparation step, CCL2 cannot actually be isolated. It is a reaction intermediate. So it is only formed when we initiate the first part of our reaction where we take chloroform in the presence of a base. So let's see how the reaction proceeds after that. So as we saw before we start our reaction with NaOH and chloroform which is CHCl3. And what this does is first it takes off this hydrogen. So we have a negative charge that comes here. And the second thing is because of this reaction between NaOH and CHCl3 we have a reaction intermediate which is CCL2 which is formed here. So now in the next step because this negative charge is going to be shared with the ring the pi bond shifts and you have an increased electron density and then this ring goes and attaches itself to the CCL2. And so now we have a double bond O here and the CCL2 added here. It results in this formal negative charge. So to get rid of this the hydrogen shifts to this carbon and this double bond shifts to this position because of which you have two of these chlorines on this carbon. Now after the step what is going to happen is each of these chlorines is going to leave with electrons. So if you think of this chlorine which is leaving with its electrons. So we get a double bond here and a double bond here. Now we have this OH negative which is going to attach itself to this carbon. This double bond will shift here and you get a negative charge on the oxygen which leads to this where we have the negative charge here. The OH is now on this carbon with a chlorine and a hydrogen here. Now just like this chlorine left this chlorine will also leave. So let me just clear this and we'll see how this continues. So we were here at the end of the last step and what's going to happen here is this chlorine is going to leave with its electrons and again after this step there's going to be a shifting of double bonds. So this oxygen has a lone pair that is going to share with this carbon. This pi bond will shift here. This double bond will shift to the oxygen giving it a negative charge. So you can see we are almost at the end of the reaction where now we have this carbon with a double bond O and a hydrogen attached to this carbon. So we have our CHO here. Now in the earlier step when we formed the double bond by sharing this lone pair with the carbon this hydrogen was removed. So all that is left is for that hydrogen to get attached to this oxygen and finally we have that hydrogen getting added here forming our product and there's one more thing to note here. The product that we've written down here is the ortho product where the CHO is getting added to the ortho position and this will be a major product. The reason for that is in one of the earliest steps of the reaction we had CCL2 approaching at the ortho position and what is happening at this stage is that this negative charge along with Na plus from the base is sort of aligning this chlorine in this direction which makes it easier for this carbon to be attached to the ring which is why we get an ortho major product. So now to bring everything together we started the reaction by our preparation step in which we had chloroform reacting with NaOH to give us our reactant which is CCL2 or dichlorocarbin and then in the first step when we reacted the phenol with NaOH we got this phenoxide anion and then when we reacted this with our dichlorocarbin from our earlier step what happened was this sodium helped align this carbon at the ortho position by sort of pulling the negatively charged chlorine and preferably placing it near the ortho position and then we had the chlorine leaving with the electrons a lot of resonance and intermediate steps after which we finally got product of our Riemer team and reaction.