Preparation of 2-chloro-2-methylpropane

In this experiment we produce a halogenoalkane from an alcohol. The concentrated hydrochloric acid used contains both the hydrogen ion to convert the OH group into the better leaving group (water) and the chloride ion for substitution:

(CH₃)₃COH + HCl ⇒ (CH₃)₃CCl + H₂O

In this case we see many of the techniques used to purify an organic liquid. We first separate the organic and aqueous layers using a separating funnel. The halogenoalkane product is largely in the organic layer, but will be contaminated with hydrochloric acid, water and unreacted alcohol. Sodium hydrogencarbonate reacts with any acid, and the salt formed will be removed in the aqueous layer. Carbon dioxide is another product of this reaction, so care needs to be taken as the pressure in the separating funnel builds up rapidly. This pressure must be released on a regular basis. Water must now be removed from the product. The presence of small droplets of water in the oily product forms an emulsion that makes it look milky. We add the drying agent, anhydrous sodium sulfate. This removes the water from the product so the milkiness disappears and a clear liquid results. The solid sodium sulfate is filtered off and the resulting liquid is distilled. Distillation is the final stage in purifying an organic liquid. This should separate our product from any unreacted alcohol.

Video - The preparation of 2-chloro-2-methylpropane

You will have heard in the video that 12.32 g of product was obtained in this experiment. Use this information to calculate the percentage yield obtained in this case. You can check your answer by clicking here.