Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: Milan1899 on January 20, 2025, 10:07:01 AM
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Hi, I don't speak english very well so please be patient :-[.
The reaction is the addition of an alcohol to an alkene: 1-methylcyclopentene. The alcohol is methanol. I obtain two intermediate carbocations: one tertiary and one secondary. However, the secondary one undergoes a 1,2 hydride shift and becomes identical to the tertiary one. I will therefore obtain two products: the main one is 1-methyl-1-methoxycyclopentane and the secondary one is 2-methyl-1-methoxycyclopentane. Is the reasoning correct? Thank you.
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If you only get one carbocation, why do you get 2 different products? Can you draw the structures?
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I think there will be only 1 product. 1-methyl-1-methoxycyclopentane.
The Markovnikov product, because the tertairy carbocation is more stable.
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Thanks to both of you for your answer. I thought that a very small amount of the secondary carbocation remains in the solution long enough to react with the nucleophile. So that it will form a secondary product. Because the reaction follows the Markovnikov's rule, and in agreement with this rule the main product is formed, but I must also consider the addition of the electrophile to the less hydrogenated carbon, which leads to the formation of a secondary carbocation (I wouldn’t have considered this if it had formed a primary carbocation because the latter is too unstable).
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The question then is if this is practically useful? If you did the reaction, would you be able to isolate any of the product from the secondary carbocation? Theoretical, there probably will be some. With a strong acid and a cyclic alkene, there probably be a few molecules of other exotic by-products as well
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The question then is if this is practically useful? If you did the reaction, would you be able to isolate any of the product from the secondary carbocation? Theoretical, there probably will be some. With a strong acid and a cyclic alkene, there probably be a few molecules of other exotic by-products as well
Thank you very much rolnor
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If you only get one carbocation, why do you get 2 different products? Can you draw the structures?
I have drawn the structures and the reaction mechanism, and I have left out the deprotonation of the protonated ether for simplicity.
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I gave you a "Mole Snack"