 In the previous video, we talked about the mechanism involved when an alkyne is treated with the hydrogen halide. But what we really saw was a terminal symmetrical alkyne being treated. Why don't we make it a little interesting and talk about an unsymmetrical alkyne instead? So let's go. This is another terminal alkyne but but but it isn't symmetrical. And we add excess of HCl and we need to figure out the major product. Why don't you try it yourself before we do it together? HCl being polar breaks as H plus and Cl minus ions. The pi electrons would go and attack this H plus ion. What are the two possibilities? Well, this pi bond can break and we can get a plus here and a negative here. Or this pi bond can break and we can get a negative here and a positive here. So there are two possibilities, right? Let's find out what would we get when we try each of them. So if we go by the first case, the carbocation would look something like this. This carbon grabs a proton and we get this carbocation. While in the second case, we get this carbon to grab the proton and this is my carbocation. Which one of the two would be more stable? Because if we find the stable one, we'll figure out where the reaction would be driven more, right? So let's compare these two carbocatons. In each case, it's a vinylic carbocation. There's no possibility of resonance. The positive charge is present on the carbon that is involved in the formation of a double bond. What else? The left one is a one degree vinylic carbocation while the right one is a two degree vinylic carbocation. It has a methyl group attached and this methyl group has an electron donating power. It donates electron density via sigma bonds that is the plus i effect. So the right carbocation is more stable, right? And Cl minus would attack in each case. It would attack this carbocation here and it would attack this carbocation here and attach itself there and we would get the following products where this is the major product. What next? What happens now? Does the reaction stop here? Well, there's excess of HCl. So this major one would drive the reaction in its formation, right? It would form more and more and more and this would be a little less. So let's find out what this major product would form once it reacts further. This is my major product from the previous step. There's again H plus Cl minus ions. The pi electron density would want to go and grab that proton but how will this bond break? Well, there are two ways again. It might break in a way which gets a positive charge here and a negative charge here and this negative goes and grabs the proton or there might be a negative charge here and a positive charge here and this carbon goes and grabs the proton. What would be the product in each case? In the first case, we get this carbocation while in the second case, we get this carbocation. So the one that is most stable would form the major product. Which of the two would be most stable? Why don't we rewrite them and then compare their stabilities? So these are the carbocations and if I try and compare their stability, I see how the left one is a 2 degree carbocation while the right one is a 1 degree carbocation. There is this methyl group attached. Do you remember how NSP3 hybridized carbon attached to NSP2 hybridized carbon is called an alpha carbon and the hydrogens attached to it were called alpha hydrogens and this can stabilize the carbocation via hyperconjugation and also via inductive. Why not via sigma bonds as well? Yes it can. It is an electron donating group. It stabilizes the electron deficient carbocation. But there is no such group directly attached to you. Wait, wait, wait. But there is a chlorine atom which is directly attached to this carbocation in the left case while it is a little far from the carbocation on the right-hand side and we know that chlorine has a minus i effect. It withdraws electron density via sigma bonds. Yes it does but it also has a lone pair of electrons. It can donate this lone pair of electrons via resonance to this carbocation and stabilize it. So the left carbocation is actually being stabilized due to the plus H of CH3 and the plus R of CL which is why it would form the major product. Right? So let's quickly form that. So the product on the left-hand side would be this CL would directly go and attach itself to the core. We get this while on the right-hand side the product would be this. Which one is the major one? This one which is a gem dihalide while the visceral dihalide is the minor product.