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Hello everyone.
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Today we're doing chapter 14, problem 16.
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And this problem asks us how many peaks are present in an nmr signal of each of the indicated protons? so the question has some arrows pointed to which protons it's interested in, but i just drew them in blue for this question.
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And the very last one is interested in all protons in the molecule.
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So these are in black.
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So let's start with the most left one, that being a.
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So we're only interested in this blue proton here.
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And if you remember, there's two rules we need to consider.
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So the number one is the splitting pattern for your proton is equal to your number of neighboring protons plus one.
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But now, what happens if you have two neighboring protons that are chemically unique from one another? so if you have one neighboring proton that's different from the other neighboring proton, then you have a kind of modified rule being that your splitting pattern equals the number of neighboring protons from your adjacent carbon a, from adjacent carbon a plus one.
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And that's multiplied by the number of neighboring protons from your adjacent carbon b.
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So your other adjacent carbon plus one.
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So we'll see when this second formal comes in handy, but let's start with a.
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So in a, we're interested only in this blue proton here, and we see that this blue proton is bound into a central item here and it's neighboring several things.
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So it's neighboring this terminal methyl group.
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So this methyl group has three protons and also the central carbon with this blue proton of interest is also neighboring this methyl group that has three protons.
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But then also the central carbon is bounded to this carbonyl carbon, but this carbonal carbon has zero neighboring protons.
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So we have to ask ourselves, are these two neighboring protons identical or different? if they're identical, we use the first equation.
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If they're different, from each other, then we have to use a second equation.
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But these are identical because they're both methyl groups.
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So here we have six neighboring protons plus one.
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So we will have seven peaks for this single one proton here.
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Now let's look at the second problem here.
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So the second problem has identified the unique protons.
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So we have a terminal group here.
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And then we have this ch2 here and another ch2 here.
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So there's no internal plane of symmetry in this molecule.
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And if you see here that this ch2 is bounded to a terminal methyl group and an apropyl group, while this ch2 on the right is bound to an ethyl group and another ethyl group.
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So what you have essentially is two unique, chemically distinct protons here.
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So now let's look at the blue protons.
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So the blue protons, we have this determined the splitting pattern, number of peaks here.
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So the blue protons, we're interested in these three protons.
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How would they look in the chemical endemar spectra? so these three protons are bound to central carbon, which is neighboring this ch2 here.
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The ch2 has, like it mentioned, two protons.
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So you have two neighboring plus one.
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So you have two plus one to give you three peaks for these three protons here.
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And now let's look at these protons here.
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So these protons at the very end, the red protons.
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So these red protons, remember, these red protons are adjacent to the ch2 here and also the ch2 here, but this proton here, if you see we have one, two, three, four, five protons, this is exactly in the middle of this molecule.
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So this ch2 on the right side is identical to the ch2 on the left side.
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So in essence, we have the same adjacent protons on either side of this molecule, so we can still use the first equation here.
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So we have four neighboring protons plus one, which gives us five peaks for these red protons.
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But now here's where the interesting case comes in.
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So i save this for the last one of this green proton here.
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So this green proton is it bound to the central carbon.
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And the central carbon is bound to two adjacent carbons.
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One being this terminal methyl group on the left, one being the ch2 bound on the right.
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So obviously this terminal methyl and the ch2 are different types of protons.
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So this is a terminal methyl three protons.
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This is a ch2, two protons, which is also adjacent to another ch2 -ch3 bond here.
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So these are two unique neighbors...