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This is the answer to chapter four, problem number 41, from the smith organic chemistry textbook.
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And in this problem, we are given many iupac names and asked to draw the molecule for each of these.
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And so just a heads up, this is going to be a bit of a longer answer than usual.
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There's quite a few names that we're given.
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And so i am just going to start drawing them out.
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And so remember the way to approach this, the end of an iupac name tells us either the longest carbon chain or the biggest ring.
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If there's a ring that's bigger than any of the carbon chain is present.
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And so that's a good place to start.
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Start with that base name.
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Again, that's the end of the name.
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And then use the rest of the name to fill in some information about substituents.
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So for a, we have three ethyl -2 methyl -methal.
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Hexane.
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And so, again, the end of the name tells us this is a hexane derivative.
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So we can go ahead.
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Six.
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There's a hexane.
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And then again, it's three ethyl two methyl.
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So one, two, ethel, two methyl.
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So there we go.
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There's three ethel, two methyl hexane for a.
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And so that's that's the way that i'm going to solve these.
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I'm going to look at the end of each iupac name.
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Use that to determine.
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The base chain or the base ring, and then use the rest of the name to fill in the substituents.
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Okay, so that was a.
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B, we have secbutylopentane.
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So cyclopentane is the base name here.
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So five -membered ring, cyclopentane.
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And then on any of these carbons, we can draw a sec -butal group.
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So i don't like the way that looks.
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So there we go.
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Better secbutyl group.
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Okay.
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And again, it could be any carbon because we're not given a number or anything.
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And really, there's no way to number it because all the carbons are equivalent until we put a substituentine.
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So that's secbutyl cyclopentane.
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So then looking at c, we have four isopropyl two, four, five, trimethal heptane.
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Okay.
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And so again, heptane is the base name here.
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So one, two, three, four, five, six, seven.
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Then let's see, four, isopropyl, two, four, five, trimethyl heptane.
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So i like to go left to right whenever possible.
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So this will be carbon two.
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There's a methyl.
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Carbon four, there's a methyl.
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Carbon five, there's a methyl.
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And then four isopropyl.
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So actually, let me back up a little bit.
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So there's one on carbon five.
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There's one on carbon four.
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And then here is our is our isoproble group also on carbon four.
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Okay, and that's four isoproble 245 trimethal heptane.
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So then d is going to be cyclobutyl cyclo -heptane.
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Okay, so this is a seven -membered ring with a four -membered ring attached to it.
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Okay, seven -membered rings are always a struggle to draw.
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I like to pretend that i'm drawing a six -membered ring.
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And then instead of bringing the last two sides together, as i would for a six -membered ring, i bring each of them down only a little bit and then put a connecting piece between them.
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So that's how i do it.
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That's not a great looking.
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There we go.
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That's a little better.
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So there's our cyclo -heptane.
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And then anywhere we can put a cyclobutane.
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So there we go.
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There's d.
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E is going to be 3 -ethyl -1 -1 -dymethyl cyclohexane.
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So we need to draw a cyclohexane to get started.
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So draw a hexagon.
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And then it's 3 -ethyl -1 -1 -dymethyl.
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So we can pick any of these carbons to call 1.
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So i'll start here.
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So this is carbon 1.
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That makes this carbon 3...