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- Solutions 57

There are only four constitutional isomers with molecular formula C_{4}H_{9}NO_{2} that contain a nitro group (–NO_{2}). Three of these isomers have similar pKa values, while the fourth isomer has a much higher pKa value. Draw all four isomers and identify which one has the higher pKa. Explain your choice.

Alright. So, here we have the molecular formula C4H9NO2 and what we need to do is we have four constitutional isomers that have a nitro group, and we know that a nitro groups looks something like this, right? We got our R, R is just any carbon chain or any group, right? And we've got NO2, this can also be represented, right? Like this N, O negative, double bond O, but now guys notice that O negative, the oxygen here, is not the only thing that has a charge because our nitrogen has four bonds, right? So, it's going to have a positive charge. So even that our functional group is neutral over all, we know that there are actually two oposite charges next to each other. So, what can we do to figure out the constitutional isomers? Well, the first step is to figure out our IHD, we need to follow our formula 2N plus 2 minus H, all divided by 2, where N is the number of carbons and H is equal to, if we have H, right? We know that we can have X is equal to 1, oxygen is equal to 0, so we don't really need to worry about it, right? And nitrogen is equal to negative 1. So, we need to go ahead and factor this all into the equation. So, let's go ahead and do that, right? Here we know that we've got 2 times 4 because we have got four carbons in our molecule plus 2 minus the total number, the number of 8, right? So, it's going to be total number of hydrogens that we have, so 9, plus, right? We know that we've got one nitrogen, so it's going to be plus minus 1, let's go ahead and close that in parenthesis, and guys, what does this give us? Let's go ahead and calculate this, 2 times 4 is 8, plus 2, that's 10, minus 9 minus 1, right? So what we get is 8 here, so that's 10 minus 8 is equal to 2 over 2, so what we get is an IHD of 1, which means that we can do a couple of things, right? We've got rings or we got double bonds but not both, so let's go ahead, scroll down so we can get some more room here, or not.

Okay, so here what we can do is what if we just add our four carbon chain and then our nitrogen group at the end, right? Can we do that? Absolutely, and why can ge do that? Well guys, because we've got this double bond here, right? This nitrogen no longer has the hydrogen that it can have and this oxygen no longer has the oxygen that it can have, the hydrogen that it can have, so they're both going to be missing one hydrogen giving us an IHD of 1. So, let's just go ahead and just give our NO2 here, is it something else we can do? Absolutely, right? Just draw, just move out nitro group around, so we can do in NO2 there, is there anything else we can do? Well guys, not there, right? We can go ahead and do a couple of creative things now, we can start moving around our carbons. So, if we need to have 4 carbons we can do something, right? We've seen a terbutal, right? We know what that looks like, we know it looks something like this with our 4 carbons here, 1, 2, 3, 4, and now guys, what if we just go ahead and put or nitro group over here, right? We can actually continue with the structure and just move our nitro group again. So, let's go ahead and do this, right? And now instead of having a nitro group coming off here, what if we just added our NO2 here and if I clean that up what we know is that it's going to look a little bit different, hold on one second, I need to do this, we know that if we go ahead and clean that up it would look something like this, right? So guys, these are the four nitro groups that we can come up with, you can't get any other ones because the nitro group compounds because if we add a ring, guys guess what? We just added an extra IHD so we have an IHD of 2 instead of one. Alright guys, so let me know if you have any questions, let's move forward.