A) Isobutane
B) Isobutylene
C) Sodium t-butoxide
D) t-butyl methyl ether
Correct Answer: B
Solution :
\[C{{H}_{3}}-\underset{C{{H}_{3}}}{\overset{C{{H}_{3}}}{\mathop{\underset{|\,\,\,\,\,\,}{\overset{|\,\,\,\,\,\,\,}{\mathop{C\,-\,}}}\,}}}\,Br+C{{H}_{3}}ONa\xrightarrow{\text{Elimination}}\]\[\underset{\text{Isobutylene}}{\mathop{C{{H}_{3}}-\overset{C{{H}_{3}}}{\mathop{\overset{|\,\,\,\,\,\,}{\mathop{C\,\,=\,}}\,}}\,C{{H}_{2}}}}\,+C{{H}_{3}}OH+NaBr\] \[C{{H}_{3}}ONa\to C{{H}_{3}}{{O}^{-}}+N{{a}^{+}}\] methoxide ion \[(C{{H}_{3}}{{O}^{-}})\] is a strong base, therefore it abstract proton from \[{{3}^{o}}\]alkyl halide and favours elimination reaction.
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