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This is the answer to chapter three, problem number 35 from the smith organic chemistry textbook.
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And in this problem, we're asked to rank three sets of three molecules each in order of increasing melting point.
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And so remember that a higher melting point is going to be correlated with stronger intermolecular forces.
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And so we just really need to identify the intermolecular forces present, and that's going to help us to rank these.
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And so we're asked to rank them in order of an increasing melting point, so i'm going to do one is going to be lowest mp, and three will be highest mp.
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Okay? okay.
00:59
And so then looking at a, essentially we are just looking at the three types of intermolecular forces that we've talked about in this chapter.
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So for a, the alcohol is going to have vanderwals forces, dipole, dipole interactions, and h bonds.
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The middle molecule here is going to have just vanderwall's forces.
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And the third molecule here, acetone is going to have vanderwals forces as well as dipole -diple interactions.
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And so once we can see all that information, it becomes pretty straightforward to say this middle molecule has just vanderwals forces, so it's going to have the lowest melting point.
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Acetone will be in the middle with vanderwals forces and dipole -dipole interactions.
02:03
And then isopropanol here, the alcohol, is going to have hydrogen bonds as well, and so that's going to have the highest melting point.
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So then looking at b, we need to consider the polarizability of the halogens that are attached to the methyl groups here.
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And so so this is actually going to increase in the order that it's written.
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Iodine is the most polarizable of these halogens, followed by chlorine.
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And then lastly, fluorine...