Question

Which reaction conditions would be the best for the synthesis of isobutyl sec-butyl ether ${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CH}}{\left( {{\text{CH}}} \right)_{\text{3}}}{\text{ - O - C}}{{\text{H}}_{\text{2}}}{\text{CH(C}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{2}}}$?
A.${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{2}}}{\text{CHC}}{{\text{H}}_{\text{2}}}{\text{OH + }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + heat}}$
B.${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CH}}\left( {{\text{C}}{{\text{H}}_3}} \right){\text{OH + }}{{\text{H}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}{\text{ + heat}}$
C.${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CH}}\left( {{\text{C}}{{\text{H}}_3}} \right){\text{ONa + }}{\left( {{\text{C}}{{\text{H}}_3}} \right)_2}{\text{CHC}}{{\text{H}}_{\text{2}}}{\text{Br}}$
D.${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{3}}}{\text{CHC}}{{\text{H}}_{\text{2}}}{\text{ONa + C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CH}}\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right){\text{Br}}$

Answer 1

There are different methods for the preparation of the ethers. Among them, it depends upon the product that is required, that which method will be followed.

Complete step by step solution:
In the isobutyl sec-butyl ether, the first part of the ether has the isobutyl group while the second part has the sec-butyl group. So the reactants must have these functional groups.
Among the reactants, the first one is the 1 –methyl butanol which on being heated in presence of concentrated sulphuric acid would give rise to alkene while the dilute acid would give the ether that is not desired.
Coming to the second compound, Butan-2-ol, this product when heated in presence of dil. Sulphuric acid also does not give the required product.
The reactants in the third option will react by Williamson-ether synthesis to form the desired product. This is because the starting material is sodium iso-butoxide which reacts with sec-butyl bromide to form the desired product.
The reactants in the fourth option would not form the same, the functional group sec-butyl is not present.
The reaction will be as:
${\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right)_{\text{3}}}{\text{CHC}}{{\text{H}}_{\text{2}}}{\text{ONa + C}}{{\text{H}}_3}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CH}}\left( {{\text{C}}{{\text{H}}_{\text{3}}}} \right){\text{Br}} \to {\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{CH}}{\left( {{\text{CH}}} \right)_{\text{3}}} - {\text{O}} - {\text{C}}{{\text{H}}_{\text{2}}}{\text{CH(C}}{{\text{H}}_{\text{3}}}{{\text{)}}_{\text{2}}}$

Hence, the correct answer is option C.

Note:
Williamson –Ether Synthesis: The formation of ether from sodium salt of alcohols and organic halides is called the Williamson –Ether Synthesis. It was developed by Alexander Williamson in 1850. This involves the reaction of an alkoxide ion with primary alkyl halide via an ${{\text{S}}_{\text{N}}}{\text{2}}$ reaction.