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Topic: Nucleophilic Substitution Mechanisms  (Read 3620 times)

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Offline waycoo

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Nucleophilic Substitution Mechanisms
« on: October 19, 2009, 01:43:34 AM »
Hello, I am practicing with the following set of problems
http://www.cem.msu.edu/~reusch/VirtualText/Questions/MechPrb/mechprb.htm

For number 3, I don't understand why the mechanism is E2 rather than E1 given that Methanol is protic and I was taught that this usually results in an E1 reaction

Also for 5 and 6, I understand that the nucleophile is weak compared to the leaving group but I don't understand why 5 works and 6 doesn't.

Thanks.

Offline renge ishyo

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Re: Nucleophilic Substitution Mechanisms
« Reply #1 on: October 19, 2009, 02:14:06 AM »
For #3 the methanol is the solvent, not the nucleophile. The nucleophile is the methoxy ion (-OCH3)which is very strong. If the methoxy ion reacts with the solvent it simply generates another methoxy ion, so the only way for the reaction to go forward is for the methoxy ion to react with the ringed compound. Since the halogen is attached to a secondary carbon and not a tertiary carbon (tertiary is what E1's favor), and the nucleophile is strong, going with an E2 is the best guess.

Offline jinclean

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Re: Nucleophilic Substitution Mechanisms
« Reply #2 on: October 19, 2009, 07:39:45 AM »
Hello!
For the 3 question,I think when E1 occurs in this compound the onion of the  cyclohexane isn't a very stable ion compared with the t-butyl ,sfuthermore the CH3O- is a very stong base ,the conditions is more favorable to the E2 conditions.
The sixth question you can find in my figure ,the steric hindrance will prevent the cyanide ion to close to the center carbon .

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