What other product is formed in this reaction?

Question

Chemical reaction diagram showing nucleophilic aromatic substitution with KOH in water, resulting in an amine and carbonyl compound.

Accepted Answer

All right. Hello everyone. So this question is asking us to provide a synthetic scheme including the necessary catalysts, inorganic reagents and carbon containing compounds with no more than three carbons demonstrating the formation of three ethyl, two methyl hex three in from hex three E. So here we have the line angle formulas of the starting material and the product as well as the condensed formula of hex three E for the sake of showing the synthesis later on. Now, when we go ahead and analyze the starting material and the product, the main feature that we can note here is that the carbon chain or rather the number of carbons present in the product has increased compared to the starting material. So this means that at some point, a new carbon carbon bond must have formed. Recall that organo melic compounds are great ways to add new carbon chains to a starting material. However, not only do we have new carbons or new carbon carbon bonds, we also have a double bond at the end in our products still. So in this case, if we're going to use an organo metallic compound, that is itself nucleic, that means we have to create some electro pilic site in our starting material. Recall that all keens like the one we have in the beginning can be converted into oxides using a perox acid from there. The organo melic compound can attack one of the carbons of the oxide which is going to establish a new carbon carbon bond. So the first step here involving hex three in would use a perox acid such as ethane perox OIC acid to convert the double bond into an oxide ring. Recall that this ends up happening because the bond between the two oxygen atoms in the perox acid is generally considered to be very weak. So as a result, the pi bond between the two carbons breaks and we have one oxygen atom connecting to both carbons from the starting double bond. So now that's 23 di ethyl oxy R. So now that we've established an electro pilic site with our oxide carbons, we can go ahead and introduce a green yard reagent, the organo metallic, the go ahead and attack. Now, the question is what the R group of the organo metallic must look like. Let me actually move this down so much if we consider the structures of the starting material and the product, and we focus on the product. In particular, we can see the addition of what looks to be an isopropyl substituent. So the next part of our synthesis is going to involve a green yard reagent with that same exact isopropyl group to go ahead and open the oxide ring. Now recall that after the green yard reagent attacks the oxide, the ring is going to open and that creates an oxide ion because oxygen would have a negative charge because of this, a second step is necessary to go ahead and proton that oxide ion. So first, we have the addition of isopropyl magnesium bromide, that's her green yard reagent. And this species is the nucleo file that opens the oxide ring and creates an oxide ion afterwards, though the addition of an acid such as HCL would be necessary in this case to make sure that the oxide ion is proton to produce an alcohol. Now, before I go ahead and draw out the final product, I do want to point out here that normally the nuclei in this case, the green yard reagent would attack the less hysterically hindered carbon of the starting oxide. However, this particular oxide is symmetrical. So both carbons are equally substituted, but just keep that in mind for other cases. But ultimately, a study material becomes an extended carbon chain with an isopropyl group. Now attached as well as an ethel group still present and a hydroxy group after potent nation, this creates four ethyl, two methyl hexane 30. Now, at this point, we're almost there, but we do have to somehow eliminate the hydroxy group and produce the all keen that we noticed in our final product. This means that the final step of the synthesis must be an acid catalyzed dehydration of an alcohol. Now recall that acid catalyzed dehydration is in fact elimination. So water is removed across two carbons to create a double bond in between them. Now recall also that this particular elimination reaction dehydration follows Zaitsev's rule which means that the most substituted double bond would be the major or desired product. If the hydroxy group here is on carbon three of the parent chain, then the double bond can be formed either between carbons, two and three or in between carbons three and four. The double bond in between carbons three and four is not only our desired product, but it's also the zits and therefore the major product. So here upon the addition of a strong acid such as sulfuric acid that's H two. So four in the presence of heat, then our intended product is going to be produced and there you have it. Now keep in mind before ending this video that for the purposes of answering this problem, I'm only showing the major or sites of product after the dehydration step. But keep in mind that both products are in fact produced. But with that being said, here is the completed synthesis. So if you watch this video all the way through, thank you so very much. And I hope you found this helpful

What other product is formed in this reaction?

Key Concepts

Reaction Mechanism

Functional Groups

Stoichiometry

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