Why is t-butyl chloride favored over t-butanol at 0°C? [closed]








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To make t-butyl chloride from t-butanol, you need to bubble HCl gas into a 0°C ether solution of t-butanol.


To make t-butanol from t-butyl chloride, you can dissolve t-butyl chloride into water (not sure the temperature).


Why are the products favored in these two conditions? What is the significance of using a 0°C solution in the first case?




This is basically an esterification equilibrium:


$\ce{t-Butanol + HCl = t-butyl chloride + H2O}$


The term esterification is not limited to esters of carboxylic acids.
You are shifting the chemical equilibrium according to [Le Chatelier's principle](https://en.wikipedia.org/wiki/Le_Chatelier%27s_principle).


When talking about chemical thermodynamics and chemical equilibrium, the following equation should come to your mind:


$\Delta G = \Delta H - T\Delta S$


This is often referred to as the **Gibbs-Helmholtz-equation**.


Lower temperatures will cause the -TΔS term and favor the formation of t-butyl chloride + H2O over the formation of t-butanol + HCl. This is because I suggest that the entropy is greater for the t-butanol + HCl side. By using an excess of HCl gas, you further shift the equilibrium.


In the second case, you hydrolyse t-butyl chloride (the ester) by applying an excess of H2O. If the entropy is greater for the t-butanol + HCl side as I suggest, an increased temperature will favor the hydrolysis.


I think for the E1 of tert-butanol, harsher conditions like conc. H2SO4 would be necessary.


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