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  •  » Why does butanol/HBr equilibrium favor butyl bromide product?

#1 2010-12-06 13:52:05


Why does butanol/HBr equilibrium favor butyl bromide product?


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.

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.

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. Why doesn't it shift to the left similar to the ionization of HBr?

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.

However, we need to look at how the reaction is performed. Is this an equilibrium reaction? Typically, the reaction mixture is heated. This will increase the reaction rate. As butyl bromide is formed, it forms a separate layer, essentially removing it from equilibrium. The back reaction of butyl bromide with water will be further decreased because water is in a separate layer. The acid present also ties up the electrons of water making it a poorer nucleophile. This will further slow the back reaction. Lastly, the rate of the back reaction with water is slow. Even though a mixture of water and butyl bromide can produce butanol + HBr, the activation energy for the reaction is probably reasonably high slowing the kinetic rate of the reaction.

I expect this kinetic vs thermodynamic effect is operational in substitution reactions of iodide. Iodide is both a good leaving group, because it is the weakest base and a good nucleophile. Because HI is such a strong acid, meaning its equilibrium greatly favors ionization, then why shouldn't any reaction with iodide reverse? The difference is the kinetics of the reaction. In the ionization of HI, this is entirely thermodynamic. There is virtually no barrier to the ionization process. Reaction of an alkyl iodide may be thermodynamically favored, but kinetically controlled.


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