 In this video, we will discuss a few problems based on the chemical reactions of ethers. We already know that ethers are the most unreactive functional groups. In fact, ethers are so unreactive that we usually use them as solvents in many chemical reactions. They are highly inert in nature and do not interfere with the reactive species that are formed during a chemical reaction. And therefore, for these ethers to undergo a chemical reaction, we need to provide extremely drastic conditions, like using very strong acids like HI or HBR in large amounts, and also changing the reaction conditions like providing higher temperature, and so on. And a classic reaction of ethers is the breaking of CO bond and forming a substituted product. So let's now look at a few examples of such reactions. The first question is to identify the products formed when methyl isopropyl ether is treated with concentrated HI. Let's first draw the structure of methyl isopropyl ether. It looks like this. We have a methyl and an isopropyl group and when it is treated with acid, the ether gets protonated. The lone pair of electrons on the oxygen atom abstracts the hydrogen and then here we have the protonated ether. Now in the next step, we have an SN2 attack of the iodide ion which is the nucleophile. Now the question is, we have an I- here. Now which carbon would it attack? Would it attack the methyl group or would it attack the isopropyl group? We already know that SN2 mechanism is highly driven by steric factor. So the I- would attack that part of the ether which is least hindered. So obviously the nucleophile would attack the less hindered methyl group and eliminate this as the alcohol. So our final products would be methyl iodide and isopropanol. Ok, so let's look at the next question. In this question, we need to identify the products that are formed in this particular reaction. So here we have an ether and we are treating it with concentrated HBr. In the previous question, we saw that the ether had a methyl and a secondary group. But here we have a tertiary group and a primary group. Now the first step is the same where the protonation of the ether takes place. The lone pair of electrons on the oxygen atom abstracts the proton and give us the protonated ether. Now the next step is a nucleophilic attack. The bromide ion would attack this ether and give us the substituted product, right? Now in this ether where we have a tertiary alkyl group, the nucleophilic attack proceeds via SN1 mechanism. Because if this bond breaks, the CO bond breaks, it gives us a more stable tertiary carbocation. So the next step is the formation of the carbocation which is a slow step. And in this step, you can see that the alcohol gets eliminated. So here we have ethanol forming. And the last step is the attack of the nucleophil which is obviously a fast step where the Br- attacks the carbocation and results in the formation of a tertiary alkyl bromide. So you can see that the products formed in this reaction are a tertiary alkyl bromide and ethanol. So as you can see from these two questions, when we have mixed ethers with two different alkyl groups and if the alkyl groups are primary or secondary, the reaction proceeds via an SN2 attack. Whereas when one of the alkyl groups is a tertiary group, in that case the second step follows an SN1 mechanism generating a tertiary carbocation. Alright, so let's look at the last question now. So here we have a phenolic ether which is heated in excess HI. So the first step is obviously the protonation of the ether. Second step is the nucleophilic attack where I- would attack and produce an alkyl iodide. Correct? Now the question is would the nucleophile or the iodide ion attack the phenyl group and eliminate this as the alcohol? Or would it attack the alkyl group and eliminate a phenol? Obviously that bond will be broken which is weaker. Now the bond between oxygen and the ethyl group here is weaker than the bond between the O and phenyl group. The carbon of the phenyl group is sp2 hybridized. Because of this, the bond between this carbon and oxygen will be shorter and stronger as compared to the bond between oxygen and the sp3 carbon here. The CO bond also has a partial double bond character because of the delocalization with the benzene ring that makes the bond stronger and more difficult to break. So because of these reasons, the nucleophilic attack would happen at this carbon eliminating phenol. So the final products are phenol and ethyl iodide. So easy to remember that whenever we treat phenolic ethers with excess of hydrohalic acids like HI or HBR the product is always phenol and corresponding alkyl halide.