 In today's video, we will talk about the addition of hydrogen halides to an alkyne. In the previous videos, we already talked about the reaction between an alkene and the hydrogen halides, the mechanism involved, the intermediates formed, whether rearrangement occurs or not and also we talked about how adding a peroxide to the same reaction mixture can lead to an entirely different reaction mechanism, right? So let's check out what happens when we add the hydrogen halides like HBr, HCl, HF, etc to a given alkyne. What do we know about alkynes? Well, alkynes are electron rich species. There's a pi electron cloud that's there. So what would happen if this alkyne, acetylene or ethane is reacted with HBr? Let's find out. Well, HBr is a polar molecule. It breaks and forms H plus and Br minus ions. The pi electron cloud being electron dense would want to go and grab this proton and that's what it does. This pi bond breaks. Let's say this carbon gets a positive charge, this gets a negative charge. The negatively charged carbon would go and attach itself to this proton here and we get this carbocation. Well, is there a possibility of resonance in this case? We know how there is resonance when this positive charge is present alternately to the pi bond. Why? Because in that case the p orbitals are parallel to each other and perpendicular to the plane of the molecule and well, they can help each other. But here the positive charge is present on the vinylic carbon. It is present on the doubly bonded carbon atom which is sp2 hybridized. Well, an sp2 hybridized carbon would have a trigonal planar geometry. So this carbon here would have one sigma bond with the adjacent carbon, the other sigma bond with it and there would be sp2 hybridized orbital which is not forming any sigma bond yet. Right? And what about the pi bonds? Well, the p orbital on this carbon and the p orbital on this carbon are parallel and they are forming a pi bond. Since this orbital here is not parallel to the adjacent carbon's p orbital, it will overlap and form pi bond. So there is no possibility of resonance here. In fact, this sp2 hybridized carbon having a positive charge is not really a nice thing because we know how as the person page s character increases, the electronegativity increases, right? And electronegative atoms do not really like to keep a positive charge on themselves. It's not really stable. So what we conclude here is that there is no possibility of resonance and it is not so stable as well. So what would happen next? Well, in the next step, Br minus simply attacks this carbocation and we get the product. If we tend to take one equivalent of HBr, we might be able to stop the reaction here. But usually since there is still pi electron density, it can attack another molecule of HBr, right? That is exactly what it does. There's another H plus Br minus the pi electrons would want to go and attack it. What happens next? What is the possibility now? Well, there's a double bond, right? It can break either way. Either this carbon gets a positive, this gets a negative or this carbon gets a negative, this gets a positive, right? So there are two possibilities. Let's draw the result of both the possibilities. So we either get this carbocation or this carbocation. Which one is more stable? Let's find out. So these are the two carbocations that may form. What do I see here? Well, in the left one, there is a one-degree carbocation and here it is again a one-degree carbocation. What about the bromine atom attached? Bromine has a minus i effect, right? Oh, it is closer to this carbocation. Does this mean that this carbocation would be less stable? What we shouldn't really forget is how bromine has a lone pair of electrons also and it is adjacent to a carbocation and therefore it can donate its electron density via resonance. It shows the minus i effect in either case. It has this tendency to withdraw electron density via sigma bonds in either case but in the right carbocation, there is a possibility of resonance. So it stabilizes this carbocation via resonance and therefore when Br minus attacks these carbocations, we get these products. And which one is the major product? This one. This is major. What is it called? Whenever there are two halides on a single carbon, we call it geminal dihalide. So it is a geminal dibromide and whenever the halides are present on adjacent carbons, we call them visceral dihalides. So this one is a visceral dibromide which is the minor product in this case.