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Topic: Conversion of alcohols to haloalkanes  (Read 6257 times)

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Offline dddap

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Conversion of alcohols to haloalkanes
« on: November 26, 2010, 03:11:51 AM »
I was taught to consider acidity/basicity of systems when dealing with reaction mechanisms. I'm having trouble applying that concept for the attached reaction (sorry for a crude MS Paint job).

pKa of HX would be around -7, and that of hydronium -2. That would make X- a better leaving group than water. However, in step 2, X- (pKa -7) is attacking the protonated alcohol to expel a worse leaving group, water (the conjugate base of hydronium -- pKa -2). I want to know why the substitution of a protonated alcohol with a halide ion is a thermodynamically favored process.

Offline mschelthoff

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Re: Conversion of alcohols to haloalkanes
« Reply #1 on: November 28, 2010, 12:21:52 AM »
Keep in mind that you're expelling a neutral molecule; water.  Now ask yourself whether a protonated alcohol is more or less stable than a neutral water molecule.

Also, it isn't always a thermodynamically favored process; taking the example you provided, would a primary alcohol yield an alkyl halide without any external energy? Things to consider.

cupid.callin

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Re: Conversion of alcohols to haloalkanes
« Reply #2 on: December 06, 2010, 03:09:38 AM »
you also use some other agents like anhy. ZnCl2 so that the water formed doesn't react back

Offline orgopete

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Re: Conversion of alcohols to haloalkanes
« Reply #3 on: December 06, 2010, 01:52:56 PM »
This is one of those really good questions that we can sometimes forget to ask ourselves. Let's start with something that is similar to what the poster was asking. In the equilibrium of HBr + H2O, the equilibrium of the reaction shifts to the right. HBr is a stronger acid than hydronium so bromide is the weakest base. That drives the equilibrium.
       

So, in the reaction of butanol with HBr, we could envision the following equilibria.
       

We could write the following equilibrium expression without knowing the equilibrium constant. However, if the reaction were entirely equilibrium controlled (it isn't), then we could ask, "What effect will increasing the amount of acid have on the equilibrium?
       

It can have two effects, one is to drive the bromide concentration lower, but because HBr would be the strongest acid, this will have limited effectiveness. The other compensation is to drive the butanol concentration lower. That can only happen by increasing the butyl bromide concentration. So, adding acid will drive an equilibrium to the right or toward butyl bromide as the product.

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