a. How many alkenes could you treat with H₂, Pd/C t...

Question

a. How many alkenes could you treat with H₂, Pd/C to prepare methylcyclopentane?

b. Which of the alkenes is the most stable?

c. Which of the alkenes has the smallest heat of hydrogenation?

Accepted Answer

Alright. Hi, everyone. So for this question, we have three parts to it. Part one is asking how many all kids can react with hydrogen and palladium slashes C catalyst to produce metal cycle vaccine. Part two is asking to identify the most stable, all keen in part one and part three is asking to identify the all keen with the smallest heat of hydrogenation. So let's begin by drawing out our starting sorry, our product, our method cyclo hexane. So we're going to have a six member ring with a method substitution attached. Now recall that in catalytic hydrogenation, right, you are converting an all keen into an all cane. So if we have this kind of generic, all keen on the bottom here. Yeah, with any substitutes really, when you go ahead and you react, you're all keen with elemental hydrogen in the presence of a catalyst. What you're doing is you're replacing that carbon carbon double bond with two carbon hydrogen bonds instead, and your substitutions are going to remain the same because it's important to realize that hydrogenation is never going to change the sigma bond framework of your starting material. It's never going to change your carbon backbone or your carbon skeleton, so to speak. So when we're trying to figure out all keens that can react with hydrogen and a catalyst to produce methyl cyclo hexane. What we're doing here is we're essentially converting all Kane within the carbon backbone of methyl cyclo hexane to a double bond because it's important to realize that in a hydrogenation reaction that would not change. So for part one here, so I'm gonna draw in blue For part one here. What I'm gonna do is I'm gonna redraw our product and I'm just gonna go ahead and replace some of these single bonds with double bonds. So the first one I can do is replace this methyl group with a methylene group, right with a double bond in step. The second possibility is that I can keep my method group, but I can have a double bond on the carbon that has that method group. Mhm. And then another possibility is to go ahead and move my double bond, one carbon over inside the ring. And then my last possibility is to go ahead and move that 11 carbon over so that it's farthest away from my metal group. Mhm. So it's important to realize that at this point, if I go ahead and I draw any more possible, all keane's, I'm just going to be reiterating structures that have already drawn due to the symmetry of this molecule. So I'm going to stop here and this means that we have four potential all keens to work with. Yeah. So now I'm gonna go ahead and scroll down so that we can address parts two and three. Now part two was asking to identify the most stable, all keen in part one and recall that stability for all keen's depends on how substituted it is, right? So If I go ahead and I start my part two in green here, this first all keen that I've drawn, it has just to all kill substitue ints, right? So this is going to be di substituted And I 100% did not spell that rate. My bad guys. There we go. So this is a di substituted. All Keen because we have to all code groups attached to it. Now, the next Tolkien has three alcohol groups associated to it. So this is gonna be tri substituted and now the remaining two Alkins that we have here, they both have to occupy groups associated to them. So they're both di substituted. So now recall that the more substituted you're all keen is the more stable it's going to be right. So what that means is that our tri substituted? All Keen here is going to be the most stable because it is the most substituted. So this is going to be our most stable. All Keen and now we can address the final part of the question, right? And if I scroll up to kind of recap it's asking us to identify the all keen with the smallest heat of hydrogenation. So before we continue, let's recall really quickly, what heat of hydrogenation refers to write your heat of hydrogenation refers to the change in entropy that corresponds to the hydrogenation of an all keen to an all came. So heat of hydrogenation is always going to be negative. So what it's asking here is it's asking us to give us or rather to figure out the smallest heat of hydrogenation. So recall that when we're talking about heat of hydrogenation, the most stable all keen mhm is going to have the smallest heat of hydrogenation. So because we discussed in part two that our tri substituted, all keen is the most stable. It's also going to have the smallest heat of hydrogenation. It's going to have the most negative heat of hydrogenation. So your answer for parts two and three is actually going to be the same molecule. It's going to be that tri substituted. All Keen. So this is going to be it for this question, right? We have four possible all Keen starting materials that can react to form um method, cyclo hexane and our tri substituted all Keen starting material, not only is the most stable, it therefore has the smallest heat of hydrogenation. So your answer here is going to be option B in the multiple choice. So if you stuck around to the end, thank you so much for watching. And I hope you found this helpful.

a. How many alkenes could you treat with H₂, Pd/C to prepare methylcyclopentane?
b. Which of the alkenes is the most stable?
c. Which of the alkenes has the smallest heat of hydrogenation?

Key Concepts

Alkene Hydrogenation

Stability of Alkenes

Heat of Hydrogenation

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