Suzy Tran Posted November 6, 2015 Share Posted November 6, 2015 Hi guys, I need help with this E1 reaction's mechanism. I just need someone to explain what do I have to move around after the leaving group leaves. Thanks! PS. You can use this site to draw it and take a screenshot if you can: https://epoch.uky.edu/ace/public/mechmarvin.jsp Link to comment Share on other sites More sharing options...
Fuzzwood Posted November 7, 2015 Share Posted November 7, 2015 Are you familiar with the terms kinetic and thermodynamic product? If so, what is more stable: a terminal alkene or an internal alkene? Link to comment Share on other sites More sharing options...
Suzy Tran Posted November 7, 2015 Author Share Posted November 7, 2015 No I am not familiar, I didn't learn that yet. Link to comment Share on other sites More sharing options...
BabcockHall Posted November 7, 2015 Share Posted November 7, 2015 Have you learned anything about how carbocations can rearrange? Link to comment Share on other sites More sharing options...
Fuzzwood Posted November 7, 2015 Share Posted November 7, 2015 No I am not familiar, I didn't learn that yet. And.......... what exactly is preventing you from looking it up? Link to comment Share on other sites More sharing options...
Suzy Tran Posted November 7, 2015 Author Share Posted November 7, 2015 (edited) Have you learned anything about how carbocations can rearrange? Yes, I've learned it and I actually tried to do a rearrangement as well but was unsure which bond i should move. And.......... what exactly is preventing you from looking it up? I just looked it up, but I am still confused about it how this can be applied to the mechanism part. I think the product from this reaction is has an internal alkene since the C=C pi bond is not at the end of the carbon chain. Also the internal alkene is more stable than the terminal alkene due to the greater number of carbons connected to the double bond. Edited November 7, 2015 by Suzy Tran Link to comment Share on other sites More sharing options...
BabcockHall Posted November 7, 2015 Share Posted November 7, 2015 I suggest drawing the first carbocation in the reaction and asking what kind of carbocation it is (primary, secondary, or tertiary). Carbocation rearrangements occur when an electron pair from a sigma bond moves to form a different bond. They are often seen when they lead to a carbocation of greater stability in the product than in the reactant. Link to comment Share on other sites More sharing options...
Suzy Tran Posted November 7, 2015 Author Share Posted November 7, 2015 I suggest drawing the first carbocation in the reaction and asking what kind of carbocation it is (primary, secondary, or tertiary). Carbocation rearrangements occur when an electron pair from a sigma bond moves to form a different bond. They are often seen when they lead to a carbocation of greater stability in the product than in the reactant. Well I tried drawing the mechanism and this is what I did, I think my main problem is that I can't visualize the structure of the cyclopentane turning into cyclohexane properly. Link to comment Share on other sites More sharing options...
BabcockHall Posted November 7, 2015 Share Posted November 7, 2015 When the electrons of the sigma bond leave the carbon atoms, that carbon atom should have a formal positive charge, which I do not see in your drawing. 1 Link to comment Share on other sites More sharing options...
Suzy Tran Posted November 7, 2015 Author Share Posted November 7, 2015 When the electrons of the sigma bond leave the carbon atoms, that carbon atom should have a formal positive charge, which I do not see in your drawing. Like this? Link to comment Share on other sites More sharing options...
BabcockHall Posted November 7, 2015 Share Posted November 7, 2015 That's better. The third from the last structure is a six-membered ring with a positive charge, and there is no reason to draw it looking so distorted. I don't understand what is happening between the third-from-last and the second to last structure. Finally, a base (which could be water) is needed to accept a proton to form the alkene product. The water molecule you drew is doing something that I do not understand. 1 Link to comment Share on other sites More sharing options...
Suzy Tran Posted November 7, 2015 Author Share Posted November 7, 2015 That's better. The third from the last structure is a six-membered ring with a positive charge, and there is no reason to draw it looking so distorted. I don't understand what is happening between the third-from-last and the second to last structure. Finally, a base (which could be water) is needed to accept a proton to form the alkene product. The water molecule you drew is doing something that I do not understand. Ah thank you, I was trying to draw the distorted figure so I forgot about the steps after that. But here's what I did now. Link to comment Share on other sites More sharing options...
BabcockHall Posted November 7, 2015 Share Posted November 7, 2015 That looks fine to me. 1 Link to comment Share on other sites More sharing options...
Fuzzwood Posted November 8, 2015 Share Posted November 8, 2015 That was exactly what I was getting at. That rearrangement costs energy, although it is the most stable product. At low temperatures, water leaves the molecule, giving you 1-methyl-1-vinylcyclopentane. At higher temperatures, the ring will rearrange to give the more stable 1,2-dimethylcyclohexene. Link to comment Share on other sites More sharing options...
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