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## Naming alkanes, cycloalkanes, and bicyclic compounds

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# Heats of combustion of alkanes

## Video transcript

- [Voiceover] Alkanes are very unreactive, but they do undergo combustion reactions. So if you take methane
and react it with oxygen, you'll get carbon dioxide
and water as the products. And for all hydrocarbons,
you're gonna get CO2 and H2O as the products of a combustion reaction. Delta H zero for this reaction is negative 890 kiloJoules per one mole of methane which means one mole of methane, if you combust
one mole of methane, you're gonna get 890 kiloJoules of heat. And so this is an exothermic
reaction, heat is given off. So the heat of combustion
is the heat that's released on the complete combustion
of one mole of a substance. In this case we're talking about methane. Let's talk about hexane next. Let me draw out the structure for hexane. So we have six carbons and notice we have two CH3 groups, so here's
a CH3 and here's a CH3, so CH3 and CH3. And then we have four CH2 groups. So here's one, two, three, and four. So that's why we have a four right here. So there are four CH2 groups in hexane, and here we have the heat of combustion. And now we're gonna talk
about the heat of combustion as the negative change in the enthalpy. So negative delta H zero in
terms of kiloJoules per mole. That gives us a positive
value here, so for hexane it's 4,163 kiloJoules for every one mole of hexane that we combust. Notice what happens as we
move on to heptane here. We're increasing in one CH2 group. We're going from four CH2
groups to five CH2 groups. So now we're at five CH2 groups. We get an increase in
the heat of combustion, so more heat is released. And that makes sense, if you
increase the number of carbons, you're gonna increase in
the heat of combustion. So if you increase by one CH2 group, how much do we increase in
terms of the heat of combustion? Well, if you take 4,163 and subtract that from 4,817 that's a difference of 654 kiloJoules. The pattern continues
when we move on to octane. So octane, now we have 6 CH2 groups. So we've added one more. We've increased our heat
of combustion to 5,471 kiloJoules per one mole. And that is also an
increase of 654 kiloJoules. So for nonane, with one more CH2, we could predict, we could estimate the heat of combusion. And we could just add 654 to 5,471. So we can go ahead and
set that up over here. So we have 5,471. We're adding 654 to that. So that would give us a five here, and then seven and five is 12, so we carry the one, and that gives us 11, so we get 6,125. So if I right that in here, 6,125 kiloJoules per mole
is a pretty good estimation for the heat of combustion of nonane. Heats of combustion are really
useful when you're trying to compare the stability of isomers. For example, here we have three isomers: octane, 2-methylheptane,
and 2,2-dimethylhexane. So all these molecules have
the molecular formula C8H18. So if we write out the
combustion reactions, so we have plus O2, we know the products are CO2 and H2O. So we have CO2 plus H2O. Let's go ahead and balance this. We have eight carbons on the left side, so we need an eight
over here on the right, we'll put that in front of the CO2. And now let's look at hydrogens. There's 18 on the left, and
so we need 18 on the right. And we can do that by putting a nine here, cause nine and two give us 18 hydrogens. Now let's see how many
oxygens are on the right side. Well we have eight and 2
give us 16 oxygens here, and then nine gives us nine for the water. So that's a total of 25
oxygens on the right side. So we need 25 oxygens on the left side. Now if we pretend like this
two isn't here for a second, we could write a 25 right here, but there is a two, so
we need to divide by two. And if we're talking
about one mole of C8H18, we're gonna leave this as a fraction, so 25 over two is the amount
of oxygen that's necessary to completely react
with one mole of C8H18. So now we have our balanced equation for our combustion reaction, and if we compare these three isomers, octane, 2-methylheptane,
and 2,2-dimethylhexane in terms of their heats of combustion, we can figure out which one
is the most stable isomer. Here we have an energy
diagram for our three isomers. So on the left here, we'll put
increasing energy this way. All three of our isomers would
produce the same products if you combust one mole of them. Each one would give us
eight CO2 and nine H2O. So there's an energy level
associated with our products. And we'll say this is the energy level associated with our products. It's the same for all
three of our isomers. Next we look up the
heat of combustion data for the isomers, and this is
determined experimentally. For example, octane has
a heat of combustion of about 5,471 kiloJoules per mole, so that's how much heat is released, 5,471 kiloJoules per mole. And if we know the energy
level of the products, and we know how much energy was released, we can find the energy
level of our reactants. So there is the energy level for octane. Next let's think about 2-methylheptane. So again, the energy level
of the products is here, and if we look up the heat of combustion for 2-methylheptane we'll find a value somewhere around 5,466. So that's how much energy is released when you combust one
mole of 2-methylheptane, 5,466 kiloJoules per mole. Since the energy level of
the products is the same as the one before, but we have a smaller heat of combustion this
time, that must mean the energy level for
2-methylheptane is lower than the energy level for octane. And finally, we move on
to 2,2-dimethylexane. The energy level of the
products is the same as before. The heat of combustion is 5,458, so somewhere around
5,458 kiloJoules per mole is released, right? That's how much energy,
or heat, is released when you combust one mole
of 2,2-dimethylhexane. And since we have a smaller value for the heat of combustion, that must mean the energy is lower
for 2,2-dimethylhexane. Now we can finally compare the stabilities of our isomers. The higher the energy, the
less stable the compound. So octane has the highest energy level, has the highest heat of combustion. So this is the least
stable out of these three, this is the least stable. The lower the energy for
our starting compound, the more stable it is. So that must meant that
2,2-dimethylhexane is the most stable isomer out of these three. So now we can think about trends. As we go from octane to 2-methylheptane to 2,2-dimethylhexane we're increasing in branching. So we're increasing in the
amount of branching that we have. What happened to the heats of combustion? Alright, if we define heat of combustion as the negative of the
change in the enthalpy, the heats of combustion decrease. We went from 5,471 to 5,466 to 5,458. So there was a decrease
in the heat of combustion. And what about stability? We got more stable as we branched more. So increased branching in general means increased stability. So it's important to be able to analyze heat of combustion data. Just remember the lower the energy, the more stable the compound. So the compound that had the
highest heat of combustion was the least stable. And the compound with the
lowest heat of combustion was the most stable. So branched alkanes are lower in energy, or more stable, than
straight chain alkanes.