 Okay, so let's show this reaction, notice we have a dihalide here, okay, so both of the halogen atoms, both of the aromines are on the same carbon. So notice this is kind of a different dihalide than what you're used to. Here we've got that sodium spectator ion, which is positively charged. If you can't tell that this is a strong base already, you should see that sodium spectator ion and kind of have it clue you in as the counter ion has a full-on negative charge, okay, so that makes it a strong base. And remember what we said, so this too means that there's two of those, okay, that's kind of giving you a clue as to how the reaction proceeds. But remember when we're doing mechanism, we're going to erase all this stuff and only put the Lewis structure of the active ingredients, right? So what's the active portion? It's going to be that NH2, okay, but it's NH2 with two lone pairs and a negative charge. So let's just expand this, so it looks like that. So that nitrogen doesn't like to have a negative charge, it would prefer to become ammonia, which is a small stable molecule. So if we look here, right, so remember when we have alkyl halide, protons that are on the alpha carbon to that alkyl halide are acidic, you guys recall that? So if we look here, we've got two alpha carbons, but only one of them has protons on them, okay? So these protons here, if you want to explicitly draw them, are going to be acidic. So we have a strong base here and an acid here, what do you think is going to happen? Yeah, it's going to deprotonate it, an acid base, right? Everybody's favorite fast reaction. So how do we draw the deprotonation reaction? Of course, first arrow from the base to the acidic proton and you could have picked either one of those. Then these electrons are going to do what? Flow down into there, right? So this is just like an E2 reaction. In fact, it is an E2 reaction, what we're about to do, kick off that bromine. That's an E2 reaction. That equilibrium is going to lie effectively to the right, so I'm just going to put a small word arrow. So now what do we have? Remember there's still an H there, made ammonia. Remember that was the solvent. If you don't remember, go back and look. And what did we say? There was two moles of sodium amide. If you guys recall, we're on alkyne section, right? So what do you think this second mole of sodium amide is going to do? Protonate. So deprotonate, right? Deprotonate, the acidic proton. You might think that this is a strange acidic proton. It is a strange acidic proton because it's on an alkyne. Normally you wouldn't think that those are too acidic, but in this case they are. You got this atom also on the alkyne, so that halogen makes this proton very acidic, kicks those electrons into the double bond, making a triple bond, knocking that bromine out. And in the end, we get a what? Yeah, an alkyne. So hopefully you predicted the final structure from that mechanism would be this. And then, of course, we get another ammonia molecule from that. We also, of course, get our Br minus in this step, another Br minus here. Alkyne is this? Internal. That's an internal out here. Any questions on how to do that one? OK, pretty straightforward mechanism, hopefully, for you guys.