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Hello everyone.
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Today we're doing chapter 17 problem 48 and this problem asks us to consider both these molecules and ask us to determine which hydrogen atom is most acidic and which hydrogen atom labeled is least acidic so we have two options here.
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We have the hydrogen labeled in black and blue black and blue in both of these molecules so how we would do something like this is what we need to determine is a stability of the conjugate base of these acids.
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So what we know about acids is that they are proton donors.
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So they give up their proton.
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So we have two options.
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For, for example, this molecule, we can give up the black proton, in which case we would get a carbon ion in this carbon location.
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Or we can give up the blue proton, and then we would get a carbon ion in this location.
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So i'll say this is minus.
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The h, the blue hydrogen.
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This is minus the black hydrogen.
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So let's determine the stability of these two compounds.
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Well, we know that an aromatic compound is unusually stable.
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And we know that an anti -aromatic compound following the 4n rule is unusually unstable.
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So if something, whatever conjugate base is more stable, that is going to mean that the acid or the starting material is more acidic.
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So we need to determine the stability and we need to determine the aromaticity of these two possible conjugate bases.
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If they follow the aromatic rule of 4m plus 2 or the anti -eromatic rule of 4n.
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Well, let's determine the number of pi electrons.
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So we have two pi electrons here, two here.
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But then this one, these two lone pairs exist in the purebital that are in the plane of the carbon -carbon bond.
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They don't exist in the purebital that goes up and down.
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So they're actually not able to delocalize around the ring.
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So in essence, we only have 4 pi electrons.
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So this follows the anti -aromatic rule of 4n.
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Now what about this structure here? well, this structure, we have those two electrons in each of those alkyne -level bonds.
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Then we see that all the bonds on this carbon ion are actually in the plane of the carbon -carbon bonds.
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So that means this lone pair electrons must be in the purebidle that is going up and down.
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Therefore, yes, it can delocalize around the ring...