A) dimethyl ether
B) di-tert-butyl ether
C) tert-butyl methyl ether
D) isobutylene
Correct Answer: D
Solution :
Tertiary halides on treatment with base, such as sodium meth oxide, readily undergo elimination resulting in the formation of alkenes. (Williamson?s Synthesis)\[C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|}{\mathop{\overset{C{{H}_{3}}}{\mathop{\overset{|}{\mathop{C}}\,}}\,}}\,}}\,-Cl+N\overset{+}{\mathop{a}}\,\overset{-}{\mathop{O}}\,C{{H}_{2}}C{{H}_{3}}\xrightarrow{{{F}_{2}}}C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,=C{{H}_{2}}\]
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