Write structural formulas for all the constitutionally isomeric compounds having the given moleclular formula.

(a) C_{4}H_{10}

So let's go ahead and come up with some structural formulas for C4H10 and I'm going to go ahead and draw them in bond line and then convert them into structural formula, so guys we know that the easiest way to start off is by actually just using our ihd formula and calculating it, okay? So we know that our ihd formula is 2 n plus 2 where n is the number of carbons minus H, and normally we'd have to worry about something like nitrogen a halide or oxygen but he would just draw hydrogen, so we just count our H's as a total number of hydrogens, okay? Okay? And this is all divided by two, so let's go ahead and plug things in, we know that this is 2 times 4 plus 2 minus, we've got 10 hydrogens. So, 10 divided by 2, so that's 2 times 4 is 8 plus 2, 10 minus 10 is equal to 0, so we know that we have no double bonds or rings, okay? So that's good news, we can do something just very simple like this, right? there we go, we've got 1, 2, 3, 4 carbons there and we should have all of our hydrogens, let's go ahead and count to make sure, we know that we've got three here and then we've got two at these positions, right? So we should have 3, 5, 7, 10, perfect, okay? So I'm going to go ahead and erase these and keep it draw explicitly a bond line and then the other thing we can do is split, right? We can say that we've got three carbons here and then we can just go ahead and add one in the middle, right? if we were to go ahead and add over here or over here it would just be this molecule again, right? So let's just give ourselves this, so now this guy over here on the left is going to be CH3, CH2, CH2, CH3 or you can condense it, which is very helpful for longer chains, you can say CH3 and then you can parentheses CH2,2, CH3, let me go ahead and fix that up, because now this CH2,2 tells me that we've got two of these CH2 groups, right? now this one over here, let me go ahead and shift it over, I've got a CH3, a CH, right? And then I've got this CH3,2, right? Because we've got 2CH2 group, CH3 groups coming off of this carbon, right? I'm going to go ahead and color code now. This carbon is there and then we got two of these methyl groups coming off, right? Another one you can do this, right? if you notice this is also a methyl group, so we can just say we've got CH, we can say that we've got CH, CH3,3, okay? Those are 2 different ways to write this up they're both totally fine. Alright guys, so let me know if you have any questions, if not, let's move on.