How could you synthesize isopropyl propyl ether, using isoprop...

Question

How could you synthesize isopropyl propyl ether, using isopropyl alcohol as the only carbon-containing reagent?

Accepted Answer

All right. Hello everyone. So for this question, let's go ahead and provide a reaction scheme in which 133 dimeth butane, two oxy, 33 dimethyl butane is synthesized using 33 dimeth butane, 20 as the only carbon source. OK. So the name itself is pretty complex. So let's take some time to analyze what's being given here. The presence of an alcoy substituent indicates that this compound is an ether. So the parent chain is 33 dimethyl butane. So that's a four carbon chain with two metal groups on position three. And the alcoy group from the ether is connecting to carbon one of that parrot chain. So from carbon one, I would draw oxygen and then from oxygen, I would draw the art group of the Al Coxie substituent. So that's the 33 dimeth butane, two ill group. So the substituent here is another molecule of 33 dimeth butane or another substituent that resembles 33 Dimetane. But the connection to the oxygen is on carbon too. So that's four carbon chain connecting to oxygen from carbon two with three, excuse me, two methyl groups on carbon three. So now this is the intended final product. And our starting material is 33 dimethyl butane tool. So we have a four carbon chain with a hydroxy group on carbon too and two methyl groups on carbon three. Now recall here that we can actually go ahead and use the Williamson Ether synthesis to generate this molecule. Recall that the Williamson Ether synthesis is an SN two reaction between an oxide ion and an oy Halide. But Williamson ether synthesis cannot create any ether. Because of that SN two requirement, an ether cannot be produced from a relatively hindered oy Halide, for example, no tertiary o highlight. So in this case, let's analyze the R groups that are on either side of the intended ether product to the left of oxygen is a primary carbon which is unhindered. And on the right side is a secondary carbon which is not quite as hindered as a tertiary carbon, but the primary carbon is definitely more suitable for SN two. So this means that our halogen our leaving group must be on that first carbon. But there is a slight issue because our starting material 33 dimethyl butane 20 has the hydroxy group on carbon two. So there has to be a way to not only convert this alcohol into an O Halide but also to make sure that it's in the right position. So to understand this, let me go ahead and redraw our starting material on the bottom here. Recall that an elimination reaction followed by addition is a very common way of not only changing the functional group but changing the position if necessary. So first, we can start with an elimination reaction that's going to convert this alcohol into an Aine. And we can do this using acid catalyzed dehydration. So that's our alcohol here with H two. So four in the presence of heat, this would create a double bond in between the carbon atom that contains the hydroxy group and any one of the two carbon atoms directly adjacent to it. But in this case, our only possible product is 33 dimeo bute one in because right next to the carbon atom that's next to the hydroxy group is a ordinary carbon that has no proton that can be removed. So now we have at this 0.33 dimethyl butte one in and now we can use an addition reaction to create an ocho Halide in which the halogen is on position one. Now notice how position one is actually the less substituted carbon of the two from this all clean here. This means that the addition reaction necessary must be anti Markov recall that the addition of HBR or HCL for that matter in the presence of a peroxide such as hydrogen peroxide installs bromine onto the less substituted carbon. So that makes the addition of HBR and time are nico. So that becomes one bromo 33 dimethyl butane, which we can then use in our S and two reaction during the Williamson ether synthesis. But in addition to the cul Halide, we do in fact need an oxide ion. So now separate to the formation of this o Halide, we also need an Aloxi ion. Once again, going back to the structure of the intended ether product. Notice how the al Coxie group on the right side of the structure resembles are alcohol starting material. So because of that, the conjugate base of 33 dimethyl butane 20 is going to be the oxide necessary for this SN two process. So scrolling back down here to my synthesis, we can take another molecule of our alcohol, move it off to the side a little bit for visibility and we can go ahead and detonate it using a base such as sodium hydride, hydride, deproteinate oxygen and we get the appropriate oxide ion. So now we have our strong nucleo which combined with a primary cul Halide undergoes SN two and then ultimately yields are intended ether product and there you have it. So here is our completed reaction synthesis. And with that being said, thank you so very much for watching. And I hope you found this helpful.

How could you synthesize isopropyl propyl ether, using isopropyl alcohol as the only carbon-containing reagent?

Key Concepts

Ethers and Their Synthesis

Dehydration Reaction

Catalysis in Organic Reactions

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