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This is the answer to chapter 20, problem number 22 from the mcmurray organic chemistry textbook.
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This problem is basically the opposite of the previous problem.
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So this time we are given eight iupac names and we are asked to draw the corresponding structures.
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Okay.
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So for a, we have cis 1 -2 cyclohexane di -carboxylic acid.
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Okay, and so cyclohexane di -carboxylic acid is sort of the root here.
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So we know that this is going to be a cyclohexane derivative.
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So that's a good place to start.
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We know that this is going to be a di -carboxylic acid.
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And we're told the position in stereochemistries of those substituents by the cis 1 -2 part.
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And so we can put the carboxylic acids on either the wedge or the dash, but since they are cis, we need to make them both the same.
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So i'm going to make them both wedged, and we're told one, two, so they're going to be on carbons one and two.
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And that is all that there is for a.
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Okay, so for b, we have heptane -dioic acids.
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So seven carbons with a carboxylic acid on each end.
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So one, two, three, four, five, six, seven.
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And then we just need to make each end carbon into a carboxylic acid.
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All right.
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So there is b.
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For c, we have two hexene four ionoic acid.
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Okay, so that we have a double bond and a triple bond and a carboxylic acid here.
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So, and then also six carbons in total.
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So i'm going to just use letter notation for this.
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It's probably the easiest way to do this.
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So we're going to have one methyl group.
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Well, yeah.
02:52
Okay, so we'll have one.
02:58
Okay, yeah.
02:59
So this would be carbon six.
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So then we would have carbon five here with the triple bonds to carbon four, since it's iron oic, four ionoic acid.
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So there is our four -ign.
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So then we have a two -n, so a double bond at carbon two.
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So that was carbon four.
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So here is carbon three with a double bond, as we said.
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Carbon 2, and then carbon 1 is our carboxylic acid.
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Okay, so there we go.
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There's c.
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For d, we have 4 -ethyl -2 -prople octinoic acid.
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Okay, so octanolic acid is the root.
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So 1, 2, 4, 5, 6, 7, 8.
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Put our carboxylic acid here on carbon number one.
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So there is our octanolic acids.
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Now we need an ethyl group at carbon four and a propyl group at carbon two...