Student Doctor Network Forums > Pre-Medical Forums > MCAT Discussions > MCAT Study Question Q&A > Heat of Hydrogentaion and stability of alkenes PDA View Full Version : Heat of Hydrogentaion and stability of alkenes lilietta200003-23-2008, 06:52 PMSo, the heat of hydrogenation is the amount of heat released when Hs are added to the alkene. The more stable the alkene, the lower the heat of hydrogenation. Right? What I don't understand is this: if it's more stable, then more energy should be required to to break the pi bonds and add the Hs not less...:confused: tncekm03-23-2008, 07:03 PMIf an alkene is stable, its in a "low energy state", so when it becomes more stable it will be in a "lower energy state". So, given that, if something goes from a high energy state to a low energy state it will have a higher heat of hydrogenation because its energy released was large as it went from high to low. And, if something was in a low energy state and goes into a lower energy state, the heat of hydrogenation will be smaller because it went from low to lower. I believe heat of hydrogenation is an enthalpy value, so think of it in terms of enthalpy and I think it will make more sense. ΔH = ΔE + PΔV - if we assume that there is no change in volume. ΔH = ΔE = E_final - E_initial So, when something is unstable it has a "high internal energy" and when it is stable it has a "low energy". So, the former case I presented in the first paragraph is indicative of a large, negative ΔE (Efinal << Einitial), and the latter case is indicative of a small, negative ΔE (Efinal < Einitial, but not "much"), so their heats of hydrogenation are large, and small, respectively. Hope that helps. lilietta200003-23-2008, 08:17 PMIf an alkene is stable, its in a "low energy state", so when it becomes more stable it will be in a "lower energy state". So, given that, if something goes from a high energy state to a low energy state it will have a higher heat of hydrogenation because its energy released was large as it went from high to low. And, if something was in a low energy state and goes into a lower energy state, the heat of hydrogenation will be smaller because it went from low to lower. I believe heat of hydrogenation is an enthalpy value, so think of it in terms of enthalpy and I think it will make more sense. ΔH = ΔE + PΔV - if we assume that there is no change in volume. ΔH = ΔE = E_final - E_initial So, when something is unstable it has a "high internal energy" and when it is stable it has a "low energy". So, the former case I presented in the first paragraph is indicative of a large, negative ΔE (Efinal << Einitial), and the latter case is indicative of a small, negative ΔE (Efinal < Einitial, but not "much"), so their heats of hydrogenation are large, and small, respectively. Hope that helps. Thanks.I think I understand now. It's basically a matter of comparison...it's still a little fuzzy though...if a trisubstitited alkene is lower E than a mononsubstituted, meaning the trisub. is more stable than the monosubs., why does it take less E to break the bond and add Hs as a more stable compound would have more resistance in giving up its stable state than a less ?..I guess I'm being stupid right now:):o unsung03-28-2008, 12:06 AMSo, the heat of hydrogenation is the amount of heat released when Hs are added to the alkene. The more stable the alkene, the lower the heat of hydrogenation. Right? What I don't understand is this: if it's more stable, then more energy should be required to to break the pi bonds and add the Hs not less...:confused: Remember there's 2 parts of a reaction diagram. There's the "going uphill" part of going from a reactant to the transition state (highest energy pt of the process). For an exothermic reaction, the energy difference between the reactant and the transition state is the transition energy. Then there's the "sliding downhill" stage of forming the product moving forward from the transition state. How easy it is to reach the transition state (ie break the bonds, etc.) depends on the activation energy and is distinct from how likely the product is to form (i.e. how stable the end product is relative to the starting reactant- the energy difference between the two is the enthalpy). So when they're talking about how stable the product is, that's a thermodynamic concept, which has to do with the enthalpy. When you talk about how difficult it is to get the thing to react in the first place, that deals with kinetics, which has to do with the transition state. That's how I understand it, at least. Sometimes a reaction can be thermodynamically unfavorable, but kinetically favorable so it'll happen anyway. That's why the less substituted alkene (Hoffman product, I think? Instead of Zaitsev product) can be favored sometimes. Jorje28603-30-2008, 04:47 PMDo we have alkenes on the test? I thought we don't. The AAMC topics webpage explicitely mentions that alkenes are not on the test... tncekm03-30-2008, 06:19 PMYeah, there are no alkenes. But, a little background knowledge doesn't hurt ;) BerkReviewTeach03-31-2008, 09:35 AMDo we have alkenes on the test? I thought we don't. The AAMC topics webpage explicitely mentions that alkenes are not on the test... Be careful on stuff like this. While alkene reactions are not on the exam, it doesn't mean they can't use an alkene as a molecule for a question that focuses on a different topic. Heats of hydrogenation questions can be asked about any compound, so if an alkene is one of the molecules, there won't be any questions on its reactivity. But conjugation and energetics are still fair game. And hydrogenation is a biochemistry topic too, so it can show up.