A) \[{{\left( C{{H}_{3}} \right)}_{3}}C-Br+NaOC{{H}_{3}}\]
B) \[C{{H}_{3}}Br+NaOC{{\left( C{{H}_{3}} \right)}_{3}}\]
C) \[C{{H}_{3}}C{{H}_{2}}Br+NaOC{{\left( C{{H}_{3}} \right)}_{2}}\]
D) \[{{\left( C{{H}_{3}} \right)}_{2}}C-Br+NaOC{{H}_{2}}C{{H}_{3}}\]
Correct Answer: B
Solution :
Tertiary halide can involve elimination of HX to give alkene in presence of a base. Hence, a tertiary alkoxide and primary alkyl halide is taken for the reaction. \[C{{H}_{3}}Br+\underset{sod.\,t-butoxide}{\mathop{NaOC{{\left( C{{H}_{3}} \right)}_{3}}}}\,\to \underset{methyl\,t-butyl\,ether}{\mathop{C{{H}_{3}}O-C{{\left( C{{H}_{3}} \right)}_{3}}+NaBr}}\,\]
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