Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: carbocation on September 21, 2016, 10:08:17 PM
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How do I know where the proton will attach itself to? It could be either of the circled carbons according to Markovnikov's rule. Am I missing something? Is this question wrong? I've looked everywhere and I haven't seen any with an asymmetrical alkene like this :(
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Try this:
http://lmgtfy.com/?q=markovnikov%27s+rule
That will tell you which carbon the proton will add to. Hint, something else could also happen.
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Markovnikov doesn't help because both carbons in the double bond have the same number of hydrogens, and both are secondary.
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You are correct, and the sarcastic linking to "let me google that for you" really wasn't appropriate in this thread IMO. You'd already pointed out why Markovnikov's rule doesn't apply.
I haven't seen one with an asymmetrical alkane like this either, normally they ensure that one of the carbon atoms does have more hydrogens. You don't give the full question, I assume this IS a question where the alkene is reacting with an acid or water, because if it isn't then Markovnikov's rule doesn't apply anyway.
Assuming that you are reacting that alkene with HX, I would assume you'd get an equal mixture as I can see no reason in the mechanism why either intermediate would be favoured.
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Continue drawing the mechanism to the next intermediate with the proton going to either carbon (draw both). Which do you think is more likely/stable? Why?
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You are correct, and the sarcastic linking to "let me google that for you" really wasn't appropriate in this thread IMO. You'd already pointed out why Markovnikov's rule doesn't apply.
I haven't seen one with an asymmetrical alkane like this either, normally they ensure that one of the carbon atoms does have more hydrogens. You don't give the full question, I assume this IS a question where the alkene is reacting with an acid or water, because if it isn't then Markovnikov's rule doesn't apply anyway.
Assuming that you are reacting that alkene with HX, I would assume you'd get an equal mixture as I can see no reason in the mechanism why either intermediate would be favoured.
yes, it's reacting with acid in water
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Continue drawing the mechanism to the next intermediate with the proton going to either carbon (draw both). Which do you think is more likely/stable? Why?
I don't know, both seem the same to me. Unless the carbocation switches to to the quaternary carbon.
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I don't know, both seem the same to me.
If you are only thinking about primary vs secondary vs tertiary, then yes they are the same. However, there is something called inductive effect (https://en.wikipedia.org/wiki/Inductive_effect) which will come into play here.
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I don't know, both seem the same to me.
If you are only thinking about primary vs secondary vs tertiary, then yes they are the same. However, there is something called inductive effect (https://en.wikipedia.org/wiki/Inductive_effect) which will come into play here.
I feel like you're trying to tell me that the right carbon would be more likely to get the hydrogen since the left one has that very stable quaternary carbon next to it?
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I feel like you're trying to tell me that the right carbon would be more likely to get the hydrogen since the left one has that very stable quaternary carbon next to it?
Perhaps, but if I am, its not because a t-butyl group is "stable". t-Butyl groups are many things, but stable (almost always in acid) is not one of them... typically on part because of this reason.
Also should ask for future steps in the plausible mechanism: have you covered hydride/methyl shifts in your lectures yet?
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Also should ask for future steps in the plausible mechanism: have you covered hydride/methyl shifts in your lectures yet?
Yeah, I know a bit about those.
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check for carbocation stability
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I think orgopete is on to something. Something else could happen.
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You are correct, and the sarcastic linking to "let me google that for you" really wasn't appropriate in this thread IMO. You'd already pointed out why Markovnikov's rule doesn't apply.
I haven't seen one with an asymmetrical alkane like this either, normally they ensure that one of the carbon atoms does have more hydrogens. You don't give the full question, I assume this IS a question where the alkene is reacting with an acid or water, because if it isn't then Markovnikov's rule doesn't apply anyway.
Assuming that you are reacting that alkene with HX, I would assume you'd get an equal mixture as I can see no reason in the mechanism why either intermediate would be favoured.
Let me apologize for that. I simply misread it. I thought the reaction showed 2,2-dimethyl-1-butene, hence my sarcasm. I had simply looked at the example and anticipated the next reaction that was to occur and did not examine the post properly. Again, my error and my apology.
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@OP,
Can one of the carbocations rearrange?
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You are correct, and the sarcastic linking to "let me google that for you" really wasn't appropriate in this thread IMO. You'd already pointed out why Markovnikov's rule doesn't apply.
I haven't seen one with an asymmetrical alkane like this either, normally they ensure that one of the carbon atoms does have more hydrogens. You don't give the full question, I assume this IS a question where the alkene is reacting with an acid or water, because if it isn't then Markovnikov's rule doesn't apply anyway.
Assuming that you are reacting that alkene with HX, I would assume you'd get an equal mixture as I can see no reason in the mechanism why either intermediate would be favoured.
Let me apologize for that. I simply misread it. I thought the reaction showed 2,2-dimethyl-1-butene, hence my sarcasm. I had simply looked at the example and anticipated the next reaction that was to occur and did not examine the post properly. Again, my error and my apology.
Fair enough. :)