 Okay, this is a birth reduction, of course, and hopefully you guys can see that we're reducing this benzene ring, we're adding two hydrogens to it, okay, in fact the two hydrogens are being added, one's being added right here, the other one's being added right there, okay, so if you'll notice we're adding sodium to this reaction, this reaction is going to be a radical reaction, okay, because sodium metal has that, it's a radical, right, it's got that one electron, okay, so the one thing you want to know is that alkyl groups, well you already know this, are electron donating, okay, so this is something that we talked about yesterday, so these are electron donating, so since there's a lot of electron density on this carbon, the sodium's electron's not going to go there, it's going to go away from it, okay, so in fact that carbon doesn't get reduced, okay, so that's what you want to know, so I'm going to erase this part, okay, so the first thing you want to think of, like I said, sodium, so a lot of times you think of sodium donating this electron to ammonia first, okay, but I don't mind you thinking of it just donating it directly to the benzene ring, okay, because why would it donate it to ammonia first, because ammonia is the solvent, right, there's ammonia in there and it will take that electron, but you guys can just think of it as donating to that benzene ring directly, that's fine with me, okay, so when you put your sodium in there, remember fissure arrows, okay, so you can think of it going somewhere in the middle to make kind of a temporary ionic covalent bond, okay, I'll show you in a second, remember sodium is just a spectator ion, so it's going to allow that whole negative charge to stay on that carbon, so when we do something like this, right, we're going to get something that looks like this, notice this bond here, nothing happened to it, so it stays the same, now this carbon, we could say is maybe attached to that sodium like that, okay, but we know sodium and carbon doesn't make a covalent bond, right, so what we can effectively say is that since the electronegativity of carbon is much greater than that of sodium, we have this ionic bond here, okay, that's bond, of course we see that we formed a double bond here, hopefully, and here we now have an electron, the other thing I'd like you to remember is that there is a hydrogen atom here and a hydrogen atom here, so we've got to, I mean especially for you guys, probably want to stick those hydrogen atoms on, and in fact I'm drawing them with the quote unquote correct geometry, okay, because remember radicals are effectively trigonal planar, okay, and of course this carbon there is Sv3 hybridized, okay, so you know you kind of want to push that hydrogen out to the edge, so now we've got the solvent in there which is in this case methanol, right, so methanol's got an acidic proton especially when you've got a carb anion, where carb anion is concerned with that proton like that, we do that of course we make that carbon very happy, but this carbon still is a radical, okay, so what we'll need is, so what we're going to need is another sodium atom, okay, so we can again think of it doing that kind of covalent ionic thing, or a lot of times you'll see this single electron transfer, SET, so you can just put it like that, or like I said you can make it be like making a covalent slash ionic, whichever one works best for you please, so when that happens of course, I'm not going to draw those hydrogen so now we've got, we've made this from an Sv2 spike center to an Sv3 center with the deprotonation of the solvent, and again it could potentially be deprotonating ammonia, it could be deprotonating this or that, you know, so most likely that methanol and the single electron is most likely transferred to the molecule. Are there any questions on this one? Okay, so make sure that when you're doing an electron donating group, right, you're keeping that bond there, okay, because why, because that electron doesn't want to go there, because there's a lot of negative charge density there, okay, why, because this alkyl group is donating a lot of negative charge, and electrons are negatively charged, are there any questions?