Show how you would synthesize the following compound from alky...

Question

Show how you would synthesize the following compound from alkyl halides, vinyl halides, and aryl halides containing no more than six carbon atoms.

(c) trans-oct-3-ene

Accepted Answer

Hey everyone and welcome back in today's video, we are going to solve the following problem, proposing synthesis to create the following substance using alcohol. Hey lights, vinyl Hey lights and Ariel Hey lights that contains no more than six carbon atoms as a starting material. Trans hex two in. According to the problem, we understand that we are synthesizing trans hex to in and we are ready to draw a six member chain. The position of the double bond starts at carbon number two. So he may essentially say that We are placing our double bond at this carbon atom right between Carbons two and 3. So this is our final material. And we want to propose a synthesis for at using either and alcohol highlight vinyl. Hey light or errol Hey light now because we are limited to these three possibilities. One of the possible ways is to use a coupling reaction between let's suppose we start with an alcohol hey libra which would have to react with an organic cooperate region. Basically what we're going to do here, we're going to illustrate the general reaction scheme for such a reaction. So let's suppose that our alcohol, hey light as an alcohol group are bonded to the leaving group for the halogen acts right, this is our alcohol. Hey light. Now the general form of organic separate agendas are two C U. L. I. Or basically to alcohol group responded to copper lithium. So we are going to use a different color for that. I suppose there is our group and there are two of them. So that that would be R T Copper lithium and the products of that reaction are simply Are bonded to are the blue are bonded to the Green one. So that's how we are able to produce our final product. We that's how we can essentially make our chain longer. Right? We are essentially combining two out kill groups together into a longer chain. And now if we think about our starting material because we want to use an L. Kill hey light with no more than six carbon atoms, we may just essentially break it Symmetrically with three carbon atoms on the left and three carbon atoms on the right. Now we will understand that the left one might act as one R. And the right one might act as another R. So let's see how that works for us. Suppose that the alcohol hey light that we're choosing will essentially come from the right part. So if we use the right part we may just redraw it. There are three carbon atoms, right? 1, 2, 3 in the position of the double bond starts over here and we want to make sure that there is a good leaving group bonded to that carbon atom. So we may essentially add it over here. Let's suppose that the leaving group that we choose is iodine, even though we can choose blooming if you want or chlorine because there's still good leaving groups. Right? So this is our alcohol Hey light and what we're adding is the green are bonded to copper lithium. So once again we have three carbon atoms that will be proposed groups. So we can say that it would react with a proposed group. Or basically here let's add parenthesis because there are two proposed groups. So that would be 123. We can add a subscript of two, indicating that there are two groups bonded to copper lithium and this would essentially give our final product the world. Last thing we want to think about is how to form the organic cooperate region itself. And this has done the following way. We have to understand the nature of the alcohol substitution. And this is a proposal group. So we may start with another alcohol hey light. So that'd be one bromo propane. Right? And we want to use a set of regions. There will be two reactions, recall that number one is addition of lithium and number two addition of copper one iodide. This would essentially produce us with the given product Right now. We understand our scheme. We want to start with one bromo propane. We want to treat it with lithium and copper one iodide to produce our organic cooperate region. And then that organic cooperate region would react in a coupling reaction with our other region which is E one Idol probe one in and then we get our desire product which is trans hex two in So we can simply summarize our findings. This is our first starting material, basically the very initial step of our reaction is to use one bromo propane so we may just draw one bromo propane. We understand that in the first stop we're adding lithium. In the second step we are adding copper one iodide to form the organic grape. And when the organic operators form, we're going to add our final region to form our final product. So the reaction scheme is the following one. We may simply label this as our final reaction scheme and we can say that the correct answer choice to this multiple choice question is B However, if you have any questions, don't hesitate to leave your comment down below in the comment section and thank you for watching.

Show how you would synthesize the following compound from alkyl halides, vinyl halides, and aryl halides containing no more than six carbon atoms.
(c) trans-oct-3-ene

Key Concepts

Alkyl Halides

Elimination Reactions

Stereochemistry

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