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A compound with molecular formula C_{5}H_{11}N has no π bonds. Every carbon atom is connected to exactly two hydrogen atoms. Determine the structure of the compound.

Hey guys. So, in this problem we need to go ahead and figure out the constitutional isomer for C5H11N, no pi bonds and a half ll of the carbones are connected to just two hydrogens, okay? And you might be thinking maybe a ring, right? So, just before we go and draw structures to say if they are wrong or right what ever it is we can figure this out using IHD, right? So, let's go ahead and do this, we know that our IHD formula is 2N plus 2 minus H all divided by 2, and here we don't just have carbons and hydrogens we have a nitrogen, right? So, we know that nitrogen is equal to -1 if for H's, so we actually have to factor that in, so we have 2 times 5 plus 2 minus 11 plus negative 1, right? So, what's all that become to simplify 2? Well, 2 times 5 plus 2 we know that that is going to be 12, right? And then 11 minus 1 is going to be 10, so 12 minus 10 is going to be 2 divided by 2, we are going to get and IHD of one, okay? So, yes we definitely do need a ring if we are not going to have any pi bonds, so if we got five carbons, right? We can go ahead and figure this out, let's start with the five carbon ring. So, just draw your Pentagon, right? And now, okay, all of our carbons are connected to two hydrogens now, right? But what if we add this nitrogen, well, now all of a sudden, guys, all of these carbons are connected to hydrogens but this one in blue only has one nitrogen right? So, this doesn't work, so now guys, what we can actually do is incorporate our nitrogen into the ring, so if we do something like this, guess what? No it works, right? If we add our hydrogen here, let's go ahead and see, this one has two hydrogens so that needs so is this one, so is this one, so is this one and so is that one, guys, all of the carbons have two hydrogens directly attached because they are all connected to carbons or a carbon and a nitrogen, right? So, guys that's it for this problem, let's move on.