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Hello everyone.
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Today we're doing chapter 11 problem 60 and this problem asks us to generate these two products a and b using only this starting material, organic starting material, um in the process.
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So what we have is for you need to start with this one, two, three, oh, h.
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So we need to start with this alcohol starting material and we need to somehow oxidize this to an alkyne with addition of carbon.
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So we can say that in our product, a, we want one, two, three, so one, two, three.
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So one, two, three, four, five, six carbons.
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One, two, three, four, five, six.
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And we need to retro synthesize this from the starting material.
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So let's break it down to how we can go about generating something like this.
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So we know we need to add some carbon.
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So right now we already have three.
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We only have three carbons and we need to add an extra three.
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So we know that if we were to break apart this acetyline or this alkyne, if we were to break this apart, then we could get something that has three carbons and then something that also has three carbons.
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So this has, let me draw something that has three carbons with three carbons.
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So we have this alkyne here.
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With this acidic proton that can be deprotonated to make a carb anion that can act as a nucleophile.
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And this nucleophile can attack electrophilic alkaliid with three carbons and a leaving group.
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For example, we can say bromine.
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So this is probably going to be the last step.
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So from this alkyne, we were going to nucleophilically attack this alkali to kick off our halide being bromine to give our final product.
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So this is going to be the last step when we go into the...
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In the correct direction.
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But now, how can we generate these two components? well, our starting material has three carbons, and each of these components also has three carbons.
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So let's start with the easier one.
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So how do we get at halide, specifically bromine or chlorine, from an alcohol? well, working backwards, we know that we get this three carbons from our starting material by using pvr3.
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And remember, if, for example, instead of bromine, it was chlorine, then i can use sokol.
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So slcl2, or slco3, sorry.
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So now we know how to get this starting material from, or this intermediate from our starting material using kbr3, but how do we get this from this? so let's first oxidize this to an alkyne, so the one level oxidation.
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So we know we can oxidize this by using some sort of strong acid.
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We can just use h3o plus.
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So what we'll do, protonate this to protonated water, if you guys remember.
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And now this wants to leave, but in order to leave, you need to do two things.
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Well, this protonated, this got protonated, this alcohol got protonated to water by picking up a proton from here.
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So now we have h2o, so the conjugate base of our acid catalyst.
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But now we need to reform our acid catalyst.
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So this h2o needs to pick up a proton.
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So we can pick up this proton to reform h3o plus and at the same time form our alken.
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Now from an alkyne we need to go to an alkyne.
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So we've done this before in the previous example.
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So if we add diatomic halide, we know that we will get addition on either side, the halide.
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And then if i add some sort of base, two equivalents of base actually, so 2 in h2 for example, just like in previous example.
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So this is everything we've done already in previous examples.
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This is why i'm moving pretty fast.
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So we make a new bond.
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When you make a bond, you break a bond, make a bond, you break a bond.
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Surrey halite falls out, and you're going to do this two times.
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You'll pick up this proton now, make a bond.
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Okay, off this bromine.
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And then guess what? you made your final intermediate here to get to this step.
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So overall, let's write this nicely.
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So overall, we're going to start with this product.
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We're going to add acidic water to get an alkyne.
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From alkyne, we're going to add diatomic halides.
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We're going to be br2, cl2, anything but diatomic, to get trans addition across.
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And then we're going to add some sort of base.
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So i'll write the actual salt version of it this time.
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And we need two equivalents, so two equivalents, to get now the starting material.
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Now from here, we need to react it with this...