Question

The Williamson ether synthesis involves treatment of a haloalkane with a metal alkoxide. Following are two reactions intended to give benzyl tert-butyl ether. One reaction gives the ether in good yield, and the other reaction does not. Which reaction gives the ether? What is the product of the other reaction, and how do you account for its formation?

Short answer

Answer: Reaction A successfully provides the desired product, benzyl tert-butyl ether. The product of Reaction B is styrene and tert-butyl chloride, formed through an E2 elimination mechanism.

Step-by-step solution

Understand Williamson ether synthesis

Williamson ether synthesis is a reaction that forms an ether by treating a haloalkane with a metal alkoxide. The reaction occurs via an SN2 mechanism, where the alkoxide nucleophile attacks the electrophilic carbon of the haloalkane, displacing the halide.

Examine the given reactions

We are given two reactions intended to give benzyl tert-butyl ether: Reaction A: Benzyl chloride + potassium tert-butoxide Reaction B: tert-Butyl chloride + potassium benzyloxide

Analyze the reactions

In Reaction A, benzyl chloride is the haloalkane, and potassium tert-butoxide is the metal alkoxide. Since benzyl chloride has a good leaving group (chloride) and the benzyl carbon is electrophilic, it can undergo an SN2 reaction with the tert-butoxide ion, which acts as a nucleophile. This reaction should give benzyl tert-butyl ether in good yield. In Reaction B, tert-butyl chloride is the haloalkane, and potassium benzyloxide is the metal alkoxide. Although tert-butyl chloride has a good leaving group (chloride), the carbon it is bonded to is a tertiary carbon, making it sterically hindered. This greatly reduces the rate of an SN2 reaction, as it is difficult for the nucleophile to approach the electrophilic carbon. As a result, this reaction does not yield significant amounts of benzyl tert-butyl ether.

Identify the products of Reaction B and explain their formation

Due to the difficulty of an SN2 reaction occurring in Reaction B, an alternative reaction pathway becomes more favorable; the nucleophile (benzyloxide ion) can undergo an E2 (elimination) reaction with the sterically hindered haloalkane, resulting in the formation of a double bond and the release of the halide ion. The product, therefore, would be styrene (C6H5-CH=CH2) and tert-butyl chloride. In conclusion, Reaction A gives the desired product, benzyl tert-butyl ether, in good yield. Reaction B does not give the desired ether, but instead forms styrene and tert-butyl chloride through an E2 elimination mechanism.