Organic Chemistry / Alkane Combustion
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Solution: Write a balanced chemical equation for the combustion of eac...

Question

Write a balanced chemical equation for the combustion of each of the following compounds:

(d) Cyclopentylcyclopentane 

Video Transcript

Hey everyone in this question we're looking at cyclo penta cyclopentane. Now, to draw that out we know that is going to be a cyclopentane, right? So, five carbon ring and then we're going to have an alkyl group that also cyclopentane group, so what this is going to look like is we're going to have to 5 membered rings right next to each other, okay? And let's redraw this one, so this is going to be our compound and we need to write out their combustion reaction. So, although we drew it out, we need to come up with a molecular formula for it. Now, it's actually a little bit easier than counting up every carbon and hydrogen because we have some different formulas that we can follow, one being CnH2n right now CnH2n is for say an alkene, right? But what we have here we have rings but CnH2n can be for either an alkene or can be for a structure that has one ring, right? We can say this is for one ring but we have two rings in our structure, so we're going to use a formula for an alkyne, CnH2n minus 2 because it's going to reduce the number of hydrogens to make a cyclic molecule. So, here we can say this is for an alkyne but the other ones we can use are one double bond one ring or even two rings, so this is going to apply to two rings. Now, how many total carbons are we going to have? we're going to have 10 because they're going to be 5 in this ring, 5 in this ring, so it's going to be C10 and if we use our formula plugging in N equals to 10 you should get how many hydrogens? Well, is got to be 20 minus 2. So, 18. So, our formula is going to be C10H18, okay? And now we're going to be drawing a combustion reaction for this molecular formula, our hydrocarbon. Now, in combustion where you need to remember that the products are always going to be carbon dioxide and water, okay? For a combustion reaction, so we can say the products are going to be CO2 and water. So, H2O, but we're still missing one molecule here, we need to add oxygen for our combustion, okay? So it's a hydro carbon plus oxygen equals CO2 and water, so now hopefully this looks a little bit familiar, this is probably going to take you back to your chemistry days and remember we need to balance out all of our atoms, so this isn't going to be that hard right? all we need to do is look at our carbons, we have 10, go to the other side write in 10 where we see our carbon, alright?

Now go ahead and try to do the same thing for hydrogen. Notice that there is going to be 18 hydrogens, so we look at the other side of our formula, we see hydrogen located over here, but now do you notice anything different? Well, there's this little number 2 next to it, so whatever you write in there, we can't just write an 18, because there is going to be 18 times 2 and we would get 36, so we're going to write in 9, so that when we multiply it by 2 we get 18, okay? Does that make sense, alright? Now wait a second. Now, we need to balance out our oxygens as well. So, sometimes it takes, it's going to be good to keep track of our atoms, so we have CH and O, so far we have 10, 18 and over here we also have 10 carbons and of course 18 hydrogens but now how many oxygens do we have? Well, this 10 is been applied to this oxygen but we need to multiply that 10 by 2, so we should expect 20 oxygens in that compound and then the other one we have 9, so we're going to say 20 plus 9 that's going to give us 29 total oxygens, so what do we do? do we write in 29? do we divide by 2? what do we do now to fill in this blank? Well, the easy way to do this is to do 29 divided by 2, write that in here and then since we're I get a whole number, we're going to simply multiply this whole formula by 2. Now, what that's going to do is first of all I'll take this 29 over 2, we're going to multiply it by 2 over 1, these will cancel out and we'll be left with a number of 29, so let's do the same thing for the other ones, 10 times 2 is going to equal 20, 9 times 2 is going to equal 18, we can cross out these numbers now and then are we done? ou number is going to be 29, 20, and 18 or are we missing something? Well, we're missing something, we have an imaginary 1 there, right? You can't forget about that because it's going to be one to that and one to that, it's not a zero there, right? So that's two is going to apply to this one, this is going to become then the number two, so as you can get different numbers that correspond to all of our compounds 2, 29 on the left and then 20 and 18 on the right, so now our formula will be balanced for our combustion reaction and our molecule that we started out with was just cyclopentyl cyclopentane, okay? So I hope this makes sense and now the next time you are asked to write a combustion reaction you'll be able to do exactly what we just did following the same steps, okay? So, let me know if you have any questions.