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  • - [Instructor] So I have these two molecules here,

  • propane on the left and acetaldehyde here on the right.

  • And we've already calculated their molar masses for you,

  • and you see that they have very close molar masses.

  • And so based on what you see in front of you,

  • which of these, you think,

  • would have a higher boiling point,

  • a sample of pure propane

  • or a sample of pure acetaldehyde?

  • Pause this video, and think about that.

  • All right, well, in previous videos,

  • when we talked about boiling points

  • and why they might be different,

  • we talked about intermolecular forces.

  • Because you could imagine, if you have a bunch of molecules,

  • let's say, in a liquid state,

  • the boiling point is going to be dependent

  • on how much energy you need to put into the system

  • in order for the intermolecular forces between the molecules

  • to be overcome so that molecules could break free

  • and enter into a gaseous state.

  • And so when we're thinking about

  • which might have a higher boiling point,

  • we really just need to think about

  • which one would have higher intermolecular forces.

  • Now, in a previous video,

  • we talked about London dispersion forces,

  • which you can view as random dipoles forming

  • in one molecule, and then that can induce dipoles

  • in a neighboring molecule.

  • And then the positive end, even temporarily positive end,

  • of one could be attracted to the temporarily

  • negative end of another and vice versa,

  • and that whole phenomenon can domino.

  • And we said that you're going to have more

  • of those London dispersion forces

  • the more polarizable your molecule is,

  • which is related to how large of an electron cloud it has,

  • which is related to its molar mass.

  • And when we look at these two molecules,

  • they have near identical molar masses.

  • So you might expect them

  • to have near identical boiling points,

  • but it turns out that that is not the case.

  • The boiling point of propane

  • is negative 42.1 degrees Celsius,

  • while the boiling point of acetaldehyde

  • is 20.1 degrees Celsius.

  • So what makes the difference?

  • Why does acetaldehyde have such a higher boiling point?

  • Why does it take more energy

  • for the molecules in liquid acetaldehyde

  • to be able to break free of each other

  • to overcome their intermolecular forces?

  • Well, the answer, you might imagine, is other things are

  • at play on top of the London dispersion forces.

  • And what we're going to talk about in this video is

  • dipole-dipole forces.

  • So you might already imagine where this is going.

  • In the video on London dispersion forces,

  • we talked about a temporary dipole

  • inducing a dipole in a neighboring molecule

  • and then them being attracted to each other.

  • Now we're going to talk about permanent dipoles.

  • So when you look at both of these molecules,

  • which one would you think has a stronger permanent dipole?

  • Or another way of thinking about it is

  • which one has a larger dipole moment?

  • Remember, molecular dipole moments are just the vector sum

  • of all of the dipole moments of the individual bonds,

  • and the dipole moments are all proportional

  • to the differences in electronegativity.

  • When we look at propane here on the left,

  • carbon is a little bit more electronegative than hydrogen

  • but not a lot more electronegative.

  • So you will have these dipole moments on each of the bonds

  • that might look something like this.

  • So you would have these things that look like that.

  • If that is looking unfamiliar to you,

  • I encourage you to review the videos on dipole moments.

  • But as you can see, there's a symmetry to propane as well.

  • So if you were to take all of these arrows that I'm drawing,

  • if you were to take all of these arrows

  • that I'm drawing and net them together,

  • you're not going to get much of a molecular dipole moment.

  • You will get a little bit of one,

  • but they, for the most part, cancel out.

  • Now what about acetaldehyde?

  • Well, acetaldehyde, there's a few giveaways here.

  • One is it's an asymmetric molecule.

  • So asymmetric molecules are good suspects

  • for having a higher dipole moment.

  • Another good indicator is you have some character here

  • that's quite electronegative.

  • In this case, oxygen is quite electronegative.

  • And even more important,

  • it's a good bit more electronegative than carbon.

  • So right over here, this carbon-oxygen double bond,

  • you're going to have a pretty significant dipole moment

  • just on this double bond.

  • It might look like that.

  • And all of the other dipole moments

  • for all of the other bonds aren't going

  • to cancel this large one out.

  • In fact, they might add to it a little bit

  • because of the molecule's asymmetry.

  • And so net-net, your whole molecule

  • is going to have a pretty significant dipole moment.

  • It'll look something like this,

  • and I'm just going to approximate it.

  • But we're going to point towards the more negative end,

  • so it might look something like this,

  • pointing towards the more negative end.

  • And I'll put this little cross here

  • at the more positive end.

  • And so you would expect a partial negative charge

  • at that end and a partial positive charge at this end.

  • And so what's going to happen

  • if it's next to another acetaldehyde?

  • Well, the partially negative end of one acetaldehyde

  • is going to be attracted to the partially positive end

  • of another acetaldehyde.

  • And so this is what people are talking about

  • when they say dipole-dipole forces.

  • We are talking about a permanent dipole

  • being attracted to another permanent dipole.

  • And so acetaldehyde is experiencing that

  • on top of the London dispersion forces,

  • which is why it has a higher boiling point.

  • Now some of you might be wondering,

  • hey, can a permanent dipole

  • induce a dipole in a neighboring molecule

  • and then those get attracted to each other?

  • And the simple answer is yes, it makes a lot of sense.

  • You can absolutely have a dipole

  • and then induced

  • dipole interaction.

  • And we might cover that in a few examples in the future,

  • but this can also occur.

  • You can have a temporary dipole

  • inducing a dipole in the neighbor,

  • and then they get attracted to each other.

  • And you could have a bit of a domino effect.

  • You can have a permanent dipole interacting

  • with another permanent dipole.

  • They get attracted to each other.

  • And you could have a permanent dipole inducing a dipole

  • in a neighboring molecule.

- [Instructor] So I have these two molecules here,

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