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This is the answer to chapter 20, problem number 34 from the mcmurray organic chemistry textbook.
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This problem is sort of the opposite of the problem before it, number 33.
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In that, in number 33, we were asked to start from butanogac acid and convert that into five different things.
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But here in number 34, we are asked to start from five different starting materials and convert each of them into butenoic acid.
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Okay.
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So looking at a, we have one butanol here, and we want to get to butenoic acid.
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And so we can do this simply by using a strong oxidizing agent.
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So chromium trioxide is a good one.
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And then we would do an acidic work up there.
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And that is going to get us to our butytoxic acid.
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Okay.
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And then for b and c, we're actually going to try to get to one butanol, and then again, use chromium trioxide to oxidize that to butanogac acid.
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Okay.
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So for b, as i said, we need to start by getting from bromobutane, one bromobutane to one butanol.
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And so we can do that just by using sodium hydroxide.
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And so that is going to get us to one butanol, and then again, just like in a, chromium trioxide, and some acid gets us to butanolic acid.
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Okay, and then for c, we will start again by getting to one butanol, and to do that in c, we need to do an anti -marcognacobhydration.
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So we need to add the elements of water in an anti -marcovnikov fashion.
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And so the way to do that is to use a boron reagent.
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So bh3 works.
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And we always write that we include, you know, that we do that in thf.
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And then we need to follow that with hydrogen peroxide and some base.
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So h2o2.
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And then just hydroxide ion is usually how that's written.
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And so that is, again, going to get us to one butanol.
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And then from there, just as in a and b, chromium trioxide, and acid.
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Okay...