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Topic: Elimination reactions  (Read 7790 times)

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Offline JohnTan

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Elimination reactions
« on: August 09, 2010, 10:08:04 PM »
Hi everybody,
I'm just wondering something about elimination reactions...So I understand that Sn1 reactions are favoured by tertiary alkylhalides and Sn2 reactions are favoured by methyl or primary alkylhalides, but I am wondering if something similar can be said for E1 and E2. What does the nature of the substrate tell us about the reaction that will take place in elimination reactions?
Thank you

Offline Jorriss

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Re: Elimination reactions
« Reply #1 on: August 09, 2010, 10:34:46 PM »
Elimination reactions in general are favored at high temperature, can you answer why? That's important.

In general though, a hindered substrate and a strong base/nucleophile leads to E2 - especially at high temperature.

A moderate to weak base or nucleophile and a substrate that can support a relatively stable carbocation (tertiary, benzyllic, allylic, etc, etc) with a good LG will undergo Sn1 or E1. There's generally more E1 at high temperature.

Tert-butyl bromide reacts with hydroxide. What can this react by?  Highlight your thought process.

Offline JohnTan

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Re: Elimination reactions
« Reply #2 on: August 10, 2010, 01:02:41 AM »
I would think it would proceed by E2 based on your description (strerically crowded substrate and strong base), however, part of me wants to say E1, as the carbocation would be stable and it would be difficult for hydroxide to bond with a hydrogen from the molecule when the reaction is concerted. Is this only really relevant for nucleophilic substitution reactions?

Offline Jorriss

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Re: Elimination reactions
« Reply #3 on: August 10, 2010, 02:46:22 AM »
You're right, it undergoes E2. There won't be any E1 or Sn1 (even though E1 gives the same product, none will be by that path). There are two things you need to remember, a tertiary carbocation is stable by carbocation standards but few carbocations are legitimately stable (ie you can isolate them). A stable carbocation won't outweigh the importance of having a strong base.

I think of it this way - a strong base likes to get things done, hard and fast. A carbocation takes it time forming, just kickin' back waiting for the leaving group to, well, leave. A strong base won't wait around for that carbocation.



But all of that is unimportant under the next principle - hydroxide is a base! Under basic conditions we won't have positive charge (other than counter ions)! That means no carbocations in basic conditions.


If you suspect E1 or Sn1 look for acidic conditions, or near neutral.


If you want more practice in this, I highly recommend renting or buying Grossmans book, 'The Art of Writing Reasonable Organic Reaction Mechanisms.'


Lastly, these rules 'mostly' apply for nucleophilic substitution and elimination but if you really begin to see the connections, their implications are far more reaching. For example, why does high temperature favor elimination? Because elimination produces more products than there are reactants. Why does that matter? Because more products increases entropy, and according to Gibbs free energy, G=H-TS (I omitted the deltas, idk how to insert that character), an increase in temperature increases spontaneity alongside entropy!

Then there is the one we just said, that can extend to any reaction really - in basic conditions, don't think positive charge and in acidic conditions don't have negative charge (other than counter ions and conjugate acid/bases).

Just think globally - analyze every aspect of the problem - solvent, conditions, hows my substrate? hows my nucleophile? what IS my nucleophile and electrophile? and so on and so fourth.

These will take you far in organic.



Offline JohnTan

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Re: Elimination reactions
« Reply #4 on: August 10, 2010, 07:56:13 AM »
Thanks for all the info. Much appreciated!

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