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Offline Fischer

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Mechanism Question
« on: November 02, 2013, 12:09:06 PM »
Hi there everyone. Been lurking the forums for a while, decided to make an account.

Anyways, I have a question regarding the mechanism for the reaction

4-chloro but-1-ene + NaI   :rarrow: 4-Iodo but-2-ene + NaCl
                                     Acetone

Now I understand the mechanism, but why can't this reaction go backwards? Why can't the Cl- substitute the Iodine and reform 4-chloro but-1-ene?

Offline spirochete

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Re: Mechanism Question
« Reply #1 on: November 02, 2013, 12:19:19 PM »
Good question. The carbon Iodine bond is weaker than the carbon chlorine bond so at first we might expect the equilibrium to favor the starting material. In this case it does not, however, defying our expectations.

Google Finklestein reaction. Basically we can exploit the differential solubility of the side products to pull the reaction forward by Le Chatlier's principle.

Offline Fischer

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Re: Mechanism Question
« Reply #2 on: November 02, 2013, 12:23:30 PM »
Thanks for your reply.

So what happens is that since NaCl is not soluble in acetone, it precipitates out and thus the reaction is favored for products instead of the reverse reaction. But why is Cl- or Br- not soluble in acetone whereas I- is?

Offline spirochete

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Re: Mechanism Question
« Reply #3 on: November 02, 2013, 12:41:35 PM »
Thanks for your reply.

So what happens is that since NaCl is not soluble in acetone, it precipitates out and thus the reaction is favored for products instead of the reverse reaction. But why is Cl- or Br- not soluble in acetone whereas I- is?

I can't give you a definite answer but I'd guess it's because the chloride anion is smaller than the iodide anion, which allows Na+ to form a stronger ionic bond to it. The Born Haber cycle, I'm pretty sure, gives a detailed quantitative treatment of solubility issues. Organic chemists don't usually worry too much about why certain salts are soluble in different solvents, they just exploit the differences for synthetic gain.

Offline Fischer

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Re: Mechanism Question
« Reply #4 on: November 02, 2013, 12:43:07 PM »
So the main reason why this reaction won't go in the reverse is because the NaCl precipitates out due to solubility differences. Would I be given full marks for this answer?

Offline spirochete

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Re: Mechanism Question
« Reply #5 on: November 02, 2013, 12:44:04 PM »
Probably yes. Mention Le Chatlier's principle.

Offline Fischer

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Re: Mechanism Question
« Reply #6 on: November 02, 2013, 12:51:26 PM »
The reaction would be favored in the forward direct (favoring the products) since NaCl is insoluble in acetone. Therefore NaCl would precipitate out and according to Le Chatelier's principle, the decrease in concentration on the products side would push the reaction to the right (ie, the products) and more NaCl would thus be formed.

Thanks a lot spirochete, really appreciate it buddy!

Offline zsinger

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Re: Mechanism Question
« Reply #7 on: November 02, 2013, 01:07:28 PM »
Spirochete-
I do understand, and agree with your postulate.  However I did want to know how you can explain the reaction if you put it under thermo-conditions (such as creating an enolate ion when there is more than one possibility).  Perhaps heating this rxn is the driving force, instead of connectivity of the bonds in solution breaking and re-forming (After all, were not talking Wittig :)).
"The answer is of zero significance if one cannot distinctly arrive at said place with an explanation"

Offline Fischer

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Re: Mechanism Question
« Reply #8 on: November 02, 2013, 01:09:00 PM »
So heating this reaction would allow for the reverse reaction to occur?

Offline zsinger

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Re: Mechanism Question
« Reply #9 on: November 02, 2013, 01:46:29 PM »
Spirochete beat me to it.  Whoever that man/woman is knows ALOT ALOT ALOT!!!!  Thanks for the free knowledge spiro!
"The answer is of zero significance if one cannot distinctly arrive at said place with an explanation"

Offline zsinger

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Re: Mechanism Question
« Reply #10 on: November 02, 2013, 01:47:33 PM »
Fischer-
Check the DeltaH of Rxn :).
"The answer is of zero significance if one cannot distinctly arrive at said place with an explanation"

Offline spirochete

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Re: Mechanism Question
« Reply #11 on: November 02, 2013, 04:57:31 PM »
So heating this reaction would allow for the reverse reaction to occur?

That would depend on whether heating the reaction caused all of the salt byproducts to go into solution. Certainly using a different solvent where the products/byproducts/starting materials were soluble would change the equilibrium.

Offline spirochete

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Re: Mechanism Question
« Reply #12 on: November 02, 2013, 05:00:37 PM »
Spirochete-
I do understand, and agree with your postulate.  However I did want to know how you can explain the reaction if you put it under thermo-conditions (such as creating an enolate ion when there is more than one possibility).  Perhaps heating this rxn is the driving force, instead of connectivity of the bonds in solution breaking and re-forming (After all, were not talking Wittig :)).

I don't think heating is the driving force per se, because when people say driving force they are usually talking about a thermodynamic driving force. Heating to some degree I assume would be required to reach the transition state needed to make the reaction happen, but I don't know the specific conditions for a Finklestein.

But here it is really the thermodynamics I was refering to when I was talking about differential solubility and le chatlier's principle.

Sorry for the double post I forgot to combine both quotes into one post!

Offline zsinger

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Re: Mechanism Question
« Reply #13 on: November 02, 2013, 08:46:12 PM »
Gotcha….that Finklestein Rxn. is extremely interesting.
"The answer is of zero significance if one cannot distinctly arrive at said place with an explanation"

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