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Topic: Elimination reactions  (Read 8828 times)

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Offline deutdeut

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Elimination reactions
« on: February 09, 2007, 10:10:48 PM »
In a primary, secondary or tertiary halide react with weak bases, elimination reactions would also occur, but to a less extent than using a strong base, is it right?

Offline enahs

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Re: Elimination reactions
« Reply #1 on: February 09, 2007, 10:21:59 PM »
I might be a little confused on the wording.

Strong bases tend to favor elimination reactions. Except with a primary, just a strong base typically does not favor elimination, it requires a hindered strong base (large, bulky, sterically hindered base)

Offline deutdeut

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Re: Elimination reactions
« Reply #2 on: February 09, 2007, 10:31:44 PM »
If say , a primary haloalkane reacts with aqueous KOH, then SN2 is in dominance over elimination, right?

Offline enahs

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Re: Elimination reactions
« Reply #3 on: February 09, 2007, 10:38:34 PM »
If say , a primary haloalkane reacts with aqueous KOH, then SN2 is in dominance over elimination, right?

No. In order for a primary substrate to react by a SN2 mechanism, the base will have to be a large, bulky, sterically hindered base.

If the base is not a large bulky base, it can easily approach the carbon bearing the leaving group.

Offline english

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Re: Elimination reactions
« Reply #4 on: February 09, 2007, 10:39:27 PM »
If say , a primary haloalkane reacts with aqueous KOH, then SN2 is in dominance over elimination, right?

Right.  Your alkene formed is not stable enough to yield a high percent of product...compared to forming the alcohol.

If say , a primary haloalkane reacts with aqueous KOH, then SN2 is in dominance over elimination, right?

No. In order for a primary substrate to react by a SN2 mechanism, the base will have to be a large, bulky, sterically hindered base.

If the base is not a large bulky base, it can easily approach the carbon bearing the leaving group.

enahs you have it completely backwards.  Bulky bases favor elimination.  The bulkier the base...the more E2 is favored over SN2.  The reactant's bulkiness has an effect as well.  Increasing the bulkiness of the halide will increase the number of groups of the alkene formed by elimination...so the E2 will win the competition for a sterically hindered primary alkyl halide.

Offline deutdeut

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Re: Elimination reactions
« Reply #5 on: February 09, 2007, 10:42:49 PM »
But in case if I use a less polar solvent, like ethanol, then the yeild of elimination products would be higher than using water as in the above example?

Offline enahs

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Re: Elimination reactions
« Reply #6 on: February 09, 2007, 10:48:15 PM »
If say , a primary haloalkane reacts with aqueous KOH, then SN2 is in dominance over elimination, right?
enahs you have it completely backwards.  Bulky bases favor elimination.  The bulkier the base...the more E2 is favored over SN2.  The reactant's bulkiness has an effect as well.  Increasing the bulkiness of the halide will increase the number of groups of the alkene formed by elimination...so the E2 will win the competition for a sterically hindered primary alkyl halide.

You are right, in this case.
I got confused onto where/what I was posting, talking about the same thing in two different threads is confusing.

Offline english

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Re: Elimination reactions
« Reply #7 on: February 09, 2007, 10:54:17 PM »
deutduet, you must separate your conditions.

First, define it as a substitution or elimination reaction.

Second, define your nucleophile.  Good (strong)?  Poor?

Third, define the solvent.  Polar solvents will slow down SN2 and E2 reactions.  Aprotic polar solvents will speed them up.  Nonpolar solvents have no effect on the rate.

Polar solvents speed up SN1 and E1 reactions. 

Offline deutdeut

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Re: Elimination reactions
« Reply #8 on: February 09, 2007, 11:00:14 PM »
Say, this time I use tertiary halide to react with aqueous KOH, this is a strong base, hence at high temperarature, the products would be almost 100% from elimination. But if I operate it under a relatively lower temperature, more SN1 products are produced than before. Is it right?

Offline english

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Re: Elimination reactions
« Reply #9 on: February 09, 2007, 11:05:22 PM »
Say, this time I use tertiary halide to react with aqueous KOH, this is a strong base, hence at high temperarature, the products would be almost 100% from elimination. But if I operate it under a relatively lower temperature, more SN1 products are produced than before. Is it right?

A strong base does not produce an SN1 mechanism.

Offline deutdeut

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Re: Elimination reactions
« Reply #10 on: February 09, 2007, 11:27:11 PM »
Since when the halides react with bases, both substitution and elimination takes place, when writing equations, do we need to show them all, or just show the reaction that's in dominace?

Offline enahs

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Re: Elimination reactions
« Reply #11 on: February 09, 2007, 11:30:42 PM »
Since when the halides react with bases, both substitution and elimination takes place, when writing equations, do we need to show them all, or just show the reaction that's in dominace?

That depends on your teacher and the question. If they ask which one is the predominate, just it. If however the test says the most likely "compound(s)" you must include both the predominate substitution and predominate elimination; and label which one overall is the most predominate..

Offline deutdeut

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Re: Elimination reactions
« Reply #12 on: February 09, 2007, 11:35:06 PM »
But the questions in my book just ask me to complete the equation, then I should just write down the dominant reaction?

Offline enahs

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Re: Elimination reactions
« Reply #13 on: February 09, 2007, 11:38:05 PM »
But the questions in my book just ask me to complete the equation, then I should just write down the dominant reaction?

This again depends on how picky your teacher will grade your homework. The real answer is to include both (or however many there are); however in many cases you can just include the main product and then say "others via E1 mechanism" or whatever it happens to be.
It all depends on the teacher.

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