Draw the products of the following reactions, including their ...

Question

Draw the products of the following reactions, including their configurations:

Chemical reaction diagram illustrating hydrogenation and addition reactions with reactants and products labeled.

Accepted Answer

Hey, everyone and welcome back, determine the major products of the reactions given below and includes the chemistry of applicable. The starting material is an alkene, specifically 33 dimethyl cycle hex one in and we have three reactions. A addition of hygiene and the presence of palladium over carbon B addition of hyen and the presence of sulfuric acid and reaction C addition of hyen bromide. Let's start with reaction. A let's read through the starting material. So we have our six member ring which has a double bond and two metal groups at the adjacent carbon. And we are treating it with hygiene and the presence of podium over carbon. So that corresponds to a hydrogenation reaction, right? Let's remember that this reaction is a hydrogenation because we're using hydrogen gas in the presence of podium over carbon. And this allows us to add two hydrogen atoms to our unsaturated carbon atoms in sin addition, right? We have to keep in mind that this reaction corresponds to sin addition. But because we will not have any chiral centers formed, we can ignore that concept even though we have to remember that the two hydrogen atoms are added from the same phase of the ring. However, this is not relevant because now that we are adding two hydrogen atoms, we get two CS two groups. And that means the carbon atoms are not chiral because they do not have four different substitutions. So that's our product, which is an a cyclo ALCA. Now, for b we're using hyen bromide and let's remember that this corresponds to an electrolytic edition mechanism. So we are going to take our starting material and we have to remember that the very first step is addition of hygiene from the acid. So we are using hyen bromide and now we have two equally substituted carbon atoms. So the more cognitive rule states that hygiene is going to be added to a less substituted carbon. But here because we're interested in the major product, we're going to add it at carbon number one. And soon we will understand why this is the case for. Now, let's just take it as granted and let's see what we get. So for the major product, we get the following carbo intermediate, which has a positive charge at carbon number two within our king, since we have added hydrogen to position one. And the reason why this carbo ion is preferred is because we can simply stabilize it. We have our secondary Carbocaine entering one comma two metal shift. We are going to produce a more stable tertiary carbo ion. So we definitely want to do that. That means we get a more stable tertiary carbo ion. And now that the tertiary carbo formed, we can think about the nucleic attack, stop the bromite and iron previously formed can attack the carbo on from the front side or the back side because we are going to end up with a chiral carbon atom. So you will have two products for this one. The first one will have a six membered ring. Now we have one product in which bromine will be pointing up and the metal group will be pointing down as a result. Now, the other product will essentially be very similar, but we will have our backside attack. And for that purpose, we are going to switch bromine and methyl. Now you may notice that we haven't added our metal group and that's because we have an additional chiral position. And this is really important during the one comma two metal shift because there's a pre bond rotation, we can specifically perform that shift from either side of the ring as well. So one possibility for us is to have that metal group at the adjacent position pointing up. So we are going to add methyl group pointing up in each case. And then the other two products, we'll have that muscle group pointing down. So we will have a total of four products, let's stroll all of them. OK. Right. So once again, this problem has a total of four products. This is really important because specifically, you may notice how one comma two metal shift corresponds to a formation of a new chiral center. And during that shift, that metal group can be added, added from either side of the ring. So that's why we have a total of four products. And finally, we are ready to move on to part C which is quite similar to part B. So the problem begins with a protonation step, we can use the two steps that we have previously indicated. But instead of hyen bromide, we're using sulfuric acid. So we are going to clearly state that. So that'd be H OS 03 H to clearly indicate our leaving group. And now, of course, when we form our carbo ion, we will have a similar scenario, we will have our one comma two metal shift. And then when we form our carbo ion, so let's draw it out a more stable tertiary carbo, we will have our nucleic attack stopped by water, right? Because that's our nuclei in the reaction, we have an acid catalyzed hydration of an king. And once again, it can attack the carbon from the front side or the back side. And similarly to what we saw before the metal group during the shift can be pointing up or down. So once again, we will have a total of four products and actually, we can easily draw them if we just think about the four products, products that we farmed previously simply changing to our alcohol group. So we will have to keep into account the fact that we will also have our deep protonation stuff so that we form an O age group instead of H2O. And now we will have our four products. So now, as we know, we can consider each center separately. So if we start with the first center, the metal group can be pointing up, right. So let's suppose that in each case, the metal group is pointing up. So that means the wage group will be pointing down. So that'll be our backside attack. For the front side attack, we will have our O H pointing up and the muscle group pointing down, keeping our adjacent chiral position fixed with a muscle group pointing up, right. Another possibility is to perform the same steps. If the adjacent position, carbon number two has our methyl group winding down, then we can have O H pointing up and O H winding down, right? Actually, it's opposite based on our assignment in the first structure O H is pointing down in the second structure O H is pointing out. So once again, we have a total of four products. And that's because we have two chiral positions, right? Let's remember that if we have two chiral positions, theoretically, the number of serial isomers would be two squared, which is four. And in practice, this also corresponds to what we have. OK. So this problem was a bit lengthy let's just label our final answers for reaction A we had our hydrogenation product. It did not have any chiral position. So we only had one product for reaction B we ended up with a total of four alkyl bromides, right? And then for reaction C which is very similar to reaction B we got a total of four products. Thank you for watching. And I hope to see you in the next one.

Draw the products of the following reactions, including their configurations:

Key Concepts

Hydrogenation

Electrophilic Addition

Regioselectivity

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