Answer: C) tert-Butyl phenyl ether
Step-by-step explanation:
n Williamson ether synthesis, an alcohol is treated with an alkyl halide in presence of a strong base such as sodium hydride (NaH), to form an ether. Initially, an alkoxide ion is formed by the reaction of the alcohol with sodium hydride (NaH). The alkoxide ion then acts as a nucleophile and attacks alkyl halide to form an ether. The reaction is represented as follows:
\text{ROH}\xrightarrow{\text{NaX}}\text{R}{{\text{O}}^{-}}\text{ N}{{\text{a}}^{+}}\xrightarrow{\text{R }\!\!'\!\!\text{ }-\text{X}}\text{R}-\text{O}-\text{R }\!\!'\!\!\text{ }
The R’-X used in the given reaction is generally a primary alkyl halide, which suggests that one of the hydrocarbon chains in the product ether will be a primary alkyl group. In the given ether, the two alkyl groups are the isopropyl group and the methyl group. The methyl group is a primary alkyl group, which suggests that the given ether can be prepared by Williamson ether synthesis. As a result, option A is incorrect.
In the given ether, the two alkyl groups are the tert-Butyl group and the methyl group. The methyl group is a primary alkyl group, which suggests that the given ether can be prepared by Williamson ether synthesis. As a result, option B is incorrect.
In the given ether, one is tert-Butyl group and the other is phenyl group. None of the two hydrocarbon groups is a primary alkyl group, which suggests that the given ether can not be prepared by Williamson ether synthesis. As a result, option C is the correct choice.
In the given ether, one is methyl group and the other is phenyl group. The methyl group is a primary alkyl group, which suggests that the given ether can be prepared by Williamson ether synthesis. As a result, option D is incorrect.