A, a compound with molecular formula C_6

Question

A, a compound with molecular formula C₆H₁₀, contains three methylene units. A reacts with one equivalent of H₂ over Pd/C to yield B. A reacts with aqueous acid to form a single product, C, and undergoes hydroboration/oxidation to form a pair of enantiomers, D and E. Ozonolysis of A followed by reaction with dimethyl sulfide forms F with molecular formula C₆H₁₀O₂. Provide structures for A–F.

Accepted Answer

Welcome back everyone. Compound A with the molecular formula C seven H 12 has an ethyl group and a cyclic ring. It reacts with one equivalent of hydrogen in the presence of palladium over carbon to produce b. The hydro oxidation reaction of A produces a pair of in animous C and D. After the osis of a, a reaction with dimethyl sulfide produces E with the chemical formula C seven H 12 which contains one aldehyde and one ketone functional group. Draw the structures for A two E. Our first goal in this problem is to calculate the hygiene deficiency index to visualize any sort of uns saturation present in the given structure. According to the formula. We're going to multiply two by the number of carbons. Then we're going to add to subtract the number of hydrogen atoms, add the number of nitrogen atoms and subtract any halogens in the form X then divide everything by two. In this case, we have two, we have seven carbon atoms. So that be two multiplied by seven plus two minus 12 hydrogen atoms. There are no nitrogen atoms or allergens. So there is no necessity to continue our numerator with any zeros, we can simply say that we have 14 plus two, that's 16 minus 12, that gives us four divided by two. That is two. The problem already implies that there is one ring in the structure. So the additional degree of uns saturation or the hydrogen deficiency index corresponds to a pi bond or an additional ring, but it's not likely right. So we already have one ring in the structure. So it must be one pi bond and one ring because each degree of in saturation corresponds to either a pi bond or a ring. Now, we have an ethel group and the molecular formula has seven carbon atoms, an ethyl group has two carbon atoms. So this leaves us with a five member ring. Let's go ahead and draw a five member which has an ey Groupon tip. Now, because there's an additional pi bond, we actually have three possibilities due to the symmetry of the ring. So where can we actually add our pi bonds? Well, we can actually add a pie bond between number one and two or we can shift it clockwise two times and we will end up with three possible structures for a definitely, we don't yet have sufficient information analyzed to state where the double bond is exactly. But first of all, let's just think about the three possibilities. And now we are ready to move on. The most important part is actually the Ozols part here to understand the position of the double bond, we have to recall that if we're forming an ad hide and a key, that means a one sp two hybridized carbon atom will be monos substituted and the other one will be di substituted. So the structure of a is actually the very first alkin that we got notice that for the two other possibilities that we had previously, we have sp two carbons that each have hydrogen atoms. So after the cleavage of the pi bond, we will get two aldehyde instead of an aldehyde and a Keone. But for the structure that we have chosen, there's one sp two carbon that is di substituted and breaking the double bond, we will end up with a Keone well done. So now we know the structure of A and we can just proceed with the other steps. We want to produce product B. So let's redraw structure A and let's think about the reaction that we're performing or actually, we can just continue, it reacts with one equivalent of hydrogen gas in the presence of palladium over carbon. We are dealing with the catalytic hydrogenation. And what we're going to do is just add two hydrogen atoms across our double bond to produce a psycho alkin or ethyl cyclo pense. That'll be our structure B. Then according to the context of the problem, A undergoes hydro ration oxidation to produce a pair of an anti or cnd. So we can draw another arrow from a, let's simply add our reactants needed for hydration. Oxidation. Number one, we want to make sure that we add Bora in the presence of tetra hydro. Number two, the second step of the reaction is the addition of hydrogen peroxide and an aqueous solution of hydroxide. OK. According to this reaction, we have to recall that it follows the anti marconi of re chemistry so that the hydrogen atom is added to the more substituted carbon and the alcohol group is added to the less substituted carbon. And of course, we will have sin addition, meaning both hygiene and the alcohol group will be pointing in the same direction relative to the ring. So let's break our double bond first and then let's consider the required stereo chemistry because we are forming a pair of inan humors first. Let's add a hydrogen and the alcohol group are both pointing down because we have some addition and then both pointing up, right. So those are our two possibilities. Now, we just want to make sure that all of our other bonds are correct. So if the hydrogen atom is pointing down, that means the ethyl group will be pointing up, right. Hydrogen is pointing down the group is pointing up. So we want to adjust our group and we actually wanted to make sure that we draw a wedge. Let's make it slightly neater. So we can understand there's an L group pointing up, we will not show the implicit hydrogen atom. And now for the other structure, if the hydrogen atom is pointing up, then the ethyl group is pointing down well done. So we got the required structures for C and D, let's label them as C and D. And now we want to identify the remaining ones. So what else do we have? We want to find out the file structure. E after the o analysis of A, we are producing a compound which has two oxygen atoms added to it, right? So to form e, we simply have to recall the osis rules. First of all, we are using ozone at negative degrees Celsius. Number two, we're using the methyl sulfide for the reductive World Cup. And what we want you to hear is just break our cc double bond. Let's do that as a separate step. So we can visualize everything very clearly. It should not be difficult, but we want to make sure everything is consistent and we simply want to enumerate our carbon atoms. So let's suppose that we go from the ethyl group as being carbon number one, then we have number two, number three, number 456 and seven according to, to the OS analysis rules. First of all, we're breaking our double bond. So we're opening the ring in this case and we're going to draw our seven number chain. Let's go ahead and do that. So we can start from left to right and we will have 123456 and seven. Let's add our numbers. So that'd be 123456 and seven. What we noticed is that carbon number seven has had a double bond. After we break the double bond and open the ring, we still end up with two double bonds. So we're going to add a double bond to carbon number seven. And now, of course, carbon number seven had an implicit hydrogen. So it is clear to us now that we are indeed forming an aldehyde because during the ossis reaction, we're breaking our double bond, we end up with two separate double bonds. So we also have a double bond on carbon number three. And there will be simple one to add oxygen atoms to each of the two double bonds formed. This is how we get a one aldehyde and one Keone in the ossis product. E so let's label it as product e great job. So we got all of our products. So the starting material A that was our, we determined it from the hydrogen deficiency index as well as the results of the osmosis reaction. Based on the fact that there's an aldehyde and a ketone in the structure. We got b using basic rules of catalytic hydrogenation. CND were obtained using the rules of hydration oxidation. In addition, and the anti marconi of free geochemistry and for the product, E we simply use the ossis reaction breaking the double bond opening the ring and assigning oxygen atoms to each of the newly formed double bonds. Those will be our final answers. And thank you for watching.

Key Concepts

Molecular Formula and Structure

Reactions of Alkenes

Stereochemistry and Enantiomers

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