all right so continue to talk about basically again again thinking about these we're really kind of focusing on nitrogen obviously could be other things but we're just kind of honing in on nitrogen because nitrogen is willing to give up its lone pairs act like a base so two different factors here we just did we talked we talked about resonance now we're going to talk about both substitution and hybridization states let's do a hybridization state first this is a classic one you have a nitrogen that's an sp hybridized nitrogen it's sp2 it actually has sp3 which one of these is going to be the best base the most reactive base is based on what hybrid orbital this lone pair is in so in this case if this is sp hybridized this hybrid orbital is in a sp orbital this lone pair is an sp2 hybrid orbital and this lone pair is an sp3 hybrid orbital knowing the shapes of those hybrid orbitals right sp having you know 50 s 50 p character sp2 you know 30 60 and then sp3 let me get a little bit more there you go much more p character 25 p character 25 f 75 p the lone pairs in each of these right which one's going to be more reactive well the one that's farther away from the nucleus most of the time is sp3 the nucleus has positively charged protons the farther away the electrons are from that the more reactive they're going to be so this will be the most reactive base followed by the amine and then lastly the nitrogen which has a lone pair in an sp orbital right because of where it's located the hybrid orbitals so you know hybridization state all right this is sp hybridized that means it has two p's two p orbitals and two sp orbitals sp2 hybridized one p three sp2s sp3 four sp3s and the hybrid orbitals associated with the lone pairs go okay maybe a simplified version but there's something to think about primary secondary tertiary mean which one's going to be the most basic which one will be the most basic so if i did i'll do this a b c so it actually is c c is the most basic now the better question is why why is it the most basic so there's two things competing here sterics and electronics so somebody could say which is not a wrong argument could say wait a second so i'm looking at this lone pair for all of these this one's a primary amine carbon is bigger than hydrogen this is going to be able to get stuff easier because it's less sterically hindered that's a reasonable argument all right that makes sense what's the other type of argument the other type of argument it's an electronics argument where it says well wait a second each one of these groups has sigma bonds and can do hyper conjugation where helps donate electron density towards that nitrogen so the more hyper conjugation interactions sigma bonds two bonds away the more electron rich that amino be perfect so that's the overriding factor in this one case so this has nine one two three four five six seven eight nine sigma bonds two bonds away that can donate electron density so that's gonna make this the most reactive base now be careful though because there's a limit to this notice how these are just methyls i chose those on purpose if all of a sudden i start doing stuff like this now things get a little more complicated right because now i've really changed the sterics around this nitrogen and now this now this doesn't become so clear what's the most important thing sterics or electronics hyperconjugation or sterics so just be careful right in a general sense the more substituted the more reactive because of hyperconjugation but there's limits to that right and we take advantage of those limits actually to do some different things because so just be careful with that so we've done hybridization state we've done substitution now now we gotta really start thinking about these kind of these are called heterocycles things with nitrogens in the rings and also relates back to aromaticity and what you know about aromaticity so this right both of these nitrogens are one hybridization state they are both sp2 hybridized right same hybridization state so that so it's not so simple now the question is what hybrid orbital are those lone pairs so in this case that's an sp2 hybrid orbital how do i know that because if you're sp2 hybridized what do you have for hybrid orbitals you have a p an sp2 an sp2 and then another sp2 p orbitals are used to do what show me your p orbitals they're they're used to make pi bonds you have one pi bond that p orbital must be used for that pi bond it has to be so that's off the table so it has that lone pair has to be an sp2 orbital if it's in an sp2 orbital can't do resonance no because you only do resonance in p orbital systems so it can't do resonance so that's not even a factor either right here this nitrogen is also sp2 hybridized right so it has the same hybrid orbitals a p an sp2 and sp2 and sp2 same hybrid orbitals the question is where is that lone pair at no cells nitrogen does have a pi bond and this low this nitrogen is next to a pi bond so it wants to do resonance right things want to do resonance so you're going to this nitrogen is going to choose to put this lone pair into a p orbital it's going to choose to be sp2 hybridized when it does that it's allowed to do resonance it's also going to be aromatic in this case so the lone pair is in a p orbital and it's doing resonance and it's in an aromatic ring is this one then gonna be that reactive that basic no because it's doing resonance first of all then also it's in an aromatic ring doing resonance right and we don't want to break up aromatic rings those are really stable all right so understanding hybridization state but also then what hybrid orbital these nitrogens or these lone pairs are in is really key as well