Question

Which of the following is the best method to prepare phenyl-t-butyl ether?
(A) $ {{(C{{H}_{3}})}_{3}}C-{{O}^{-}}N{{a}^{+}}+{{C}_{6}}{{H}_{5}}Br $
(B) $ {{C}_{6}}{{H}_{5}}ONa+{{(C{{H}_{3}})}_{3}}CCl $
(C) $ {{(C{{H}_{3}})}_{2}}C=C{{H}_{2}}\xrightarrow[{{C}_{6}}{{H}_{5}}OH]{Hg{{(OCOC{{H}_{3}})}_{2}}}\xrightarrow[{}]{NaB{{H}_{4}}} $
(D) None of these.

Answer 1

Hint :To answer this we must know the use of Williamson’s synthesis. Williamson’s synthesis is used for the synthesis of symmetrical and unsymmetrical ethers. The synthesis of ethers is done using alkyl halides which react with sodium alkoxide.

Complete Step By Step Answer:
We know that Williamson’s synthesis is used for the synthesis of symmetrical and unsymmetrical ethers. In Williamson’s synthesis, alkyl halide $ (R'-X) $ reacts with sodium alkoxide $ (R{{O}^{-}}N{{a}^{+}}) $ to produce ethers $ (R-O-R') $ . Sodium halide $ (N{{a}^{+}}{{X}^{-}}) $ is the by-product of the reaction.
Williamson’s synthesis is a nucleophilic substitution reaction. In alkyl halides, a positive charge develops on the carbon atom attached to the halogen atom. This is because the carbon halogen bond is polar in nature due to the high electronegativity of the halogen atom. As a result the carbon atom becomes electrophilic in nature. The alkoxide ion acts as a nucleophile. The alkoxide ion substitutes the halide ion from the alkyl halide. The reaction is a bimolecular nucleophilic substitution $ {{S}_{{{N}^{2}}}} $ reaction i.e. reaction.
Thus here $ {{(C{{H}_{3}})}_{3}}C-{{O}^{-}}N{{a}^{+}}+{{C}_{6}}{{H}_{5}}Br $ will give by product as phenyl-t-butyl ether and the following chemical reaction is given by;
 $ {{(C{{H}_{3}})}_{3}}C-{{O}^{-}}N{{a}^{+}}+{{C}_{6}}{{H}_{5}}Br\to {{C}_{12}}{{H}_{18}}O $

Note :
Williamson's synthesis follows $ {{S}_{{{N}^{2}}}} $ mechanism i.e. bimolecular nucleophilic substitution mechanism. It is a one-step reaction. In the reaction, the bond breaking and bond making occurs simultaneously. No intermediate is formed in the reaction. The nucleophile attacks the alkyl halide from the back side and thus, the product formed has a reversed configuration.