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Solution: Draw bond-line structures for all constitutional isomers of ...

Question

Draw bond-line structures for all constitutional isomers of C 5H12.

Video Transcript

Alright guys. So, for this question we are looking for constitutional isomers of our molecular formula C5H12, okay? And just remember that constitutional isomers means that we have the same formula but different connectivity, so the easiest way to start up with this is the first one that I can think of is Jus this 5 carbons like this, so we have 1, 2, 3, 4 and 5, okay? And just for clarity let's take a look at all the hydrogens there, we have three right here, 2, 2, 2 and 3, so remember that these are hydrogens, Rigth? So we have 3 here, 2, 2, 2 and 3, so if we add that up we get 6, 8, 10, 12 hydrogens, okay? But how do we connect these five carbons differently? Well, we could draw some kind of branches, okay? So, another one we could do is, for instance, we have these 4 carbons, let's say we took carbon five and we place it somewhere else on our molecule, so this is what it originally was but we are not going to include it there, right? Let's say we include it over here, well, that's the same compound, so the only two places we could actually add it, is right here or right here, and no matter where we add it it still is going to give us this compound right here. So, now let's actually draw this compound in and take a look at the hydrogens. Remember that it's 3 here, okay, there was three here and remember there is one there, 2 and 3. So, let's just 3 over this, remember we have 3, 3, 3, so that's 9, we have 10 and 11 and 12, okay? So, can you think of any other branch that we can draw, right? Now, I encourage you to look at, we have our one our two, look at 2 redraw it here in 3 and think of any other way looking at the same method that we did before, okay? So we have this compound, now what we can do is just take away another carbon, right? So we just have something like this, now this is where that carbon originally was in 2 but we are not going to draw it there, let's actually include it right here, okay? And notice that this is actually another constitutional isomer and the last one for this problem. Now, remember that what we can do for this one is, lets draw it right underneath, if we had this compound let me show you what the hydrogens would be, again there will be 3, 3, so wrap up to 6, 9 and 12, okay? So all these represent hydrogens and as soon the sooner this carbon is bonded 2 four other atoms so it doesn't have any other hydrogens on it, alright guys? So, again we have a 3, 3, 3Ê and 3 but our only constitutional isomers are the 3 that I drew right on here, okay? So hopefully guys that made sense, let's take a look now at your next question.

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