Show how you would synthesize the following compounds from any...

Question

Show how you would synthesize the following compounds from any starting materials containing no more than six carbon atoms.

(b)

Accepted Answer

All right. Hello everyone. So this question is asking us to provide a synthetic pathway for the below given compound using starting materials that contain seven or fewer carbon atoms. Ignore stereo chemistry. All right. So first and foremost, when examining the structure of the intended product, we can see that it has a total of 14 carbon atoms. One approach that we can take is by combining two molecules that have seven carbons each. And so what that implies is that we have to somehow create a new carbon carbon bond to join together. These hypothetical starting materials recall that organo metallic compounds are very, very useful when it comes to combining our groups and extending carbon chains. One convenient approach that we can take is using a green yard reagent. Now, organo meal compounds including green yard reagents are very useful in that they make carbon nucleic. Now, the nucleic carbon can attack an electro pilic carbon allowing for a new bond between the two atoms. So here a very common reaction used involving a green yard reagent is between a green yard and some kind of carbonel compound. For example, an aldehyde or a ketone. But where do we set up this reaction right? Where is the car bal and where is the Green Yard reaction or Green yard reagent? To understand this a little bit better. We're going to take a retro synthetic approach in which we try to create the structure of the starting materials based on the structure of the product. Now, the APOC side is a good place to start recall that oxides can be created using Aines. So if we want to take a step backwards, and we add a double bond between the two carbons of the oxide, we now know the structure of an All Keen intermediate. So here I am essentially just going to go ahead and remove that oxygen and add a double bond between the two carbons of the oxide ring. This gives us an intermediate that we will come across during the actual synthesis. So now let's pause and go back to our discussion about the green yard reagent and the car bono compound recall also that the combination of a granard and an aldehyde or a ketone is going to produce an alcohol that alcohol can subsequently be eliminated to produce an alchem. So the question is now considering where the double bond in the intermediate is, where can we best place the hydroxy group of this alcohol? Well, recall our seven carbon maximum. And first, let's imagine what potential intermediate can be generated by placing the hydroxy group on either carbon of the Keen intermediate. All right. So here I briefly went ahead and paused the recording to go ahead and draw out both possible alcohols based on the positioning of the all keen, we have two possibilities for the alcohol, either it's a secondary alcohol or a tertiary alcohol for our purposes. The secondary alcohol is better suited because it can be prepared using a seven carbon Green Yard Regent and a seven carbon aldehyde. This means that we can now separate the structure of the secondary alcohol into the structures of the Green yard reagent and the aldehyde. And that separation is going to be based off of where the carbon atom containing the hydroxy group happens to be. And I should actually go ahead and move this downwards just so that we can continue with our retro synthetic approach. All right. So from here, we can separate our 14 carbon intermediate into the 27 carbon fragments that we're going to use to combine. One of the fragments is a six member ring, a cyclohexane ring with one additional carbon, which is where the hydroxy group is because that's where the hydroxy group is. That fragment is going to be our aldehyde. So here is our seven carbon aldehyde, which means that the other r group on the other side of that separation must have been our greener reagent. So that's a benzene ring with one metal substituent and a magnesium bromide group on the opposite side of the methyl group. Now, the green yard reagent can be synthesized from an al bromide. So that's where we're going to start and I should scroll down to make space for this whole process here. All right. So our starting material for the purposes of this synthetic scheme is going to be the Ocu bromide that will produce the necessary green yard reagent. So here are a starting material is just going to have bromine attached to the carbon of the benzene ring instead of the magnesium bromide group. So the starting material is one bromo for methyl benzene. And after reacting this with eom masi in the presence of ether, we get our green yard. We so so far our intermediate is para Talal magnesium bromide. Now we introduce our bring your reaction the alde Heide in question, which will add in the first step of the sequence here or rather this part of the overall sequence is cyclohexane carb Hyde. It is worth mentioning here that during the actual green yard reaction, it has to be divided into two phases. After the green yard reagent attacks the car bono carbon, an oxide ion is produced because the pi electrons of the carbo go towards the Carboni oxygen. Because of this, it's important to add some acid represented here as H 30 positive to proton that oxide ion and produce an alcohol. So from here, we have 14 carbons in total and the secondary alcohol that was discussed previously. So from here, we can eliminate this alcohol, right. We can eliminate this hydroxy group using acid catalyzed dehydration to produce an alky. And so the reagent for that, if you recall is sulfuric acid, that's H two. So four in the presence of heat. So here's our all keen and last but not least is oxidation recall that the reagent for or the most common reagent for the oxidation of an Keen is meta chloro paroxy Benzo acid or MCP B for short. And there you have it. Now, we have the oxide and our synthesis is complete just to go ahead and summarize our steps here. The first step was the preparation of a greeny reagent using an cul bromide and magnesium in the presence of ether. The second step was then the reaction between the green yard reagent and an aldehyde that produces a secondary alcohol after potent nation. From there step three was acid catalyzed dehydration of the secondary alcohol to produce an alkene. And then the fourth step was oxidation of the Keen with MC PB A. So if you watch this video all the way through, thank you so very much for doing so. And I hope you found this helpful.

Show how you would synthesize the following compounds from any starting materials containing no more than six carbon atoms.
(b)

Key Concepts

Functional Groups

Reaction Mechanisms

Retrosynthetic Analysis

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