A) \[LiAl{{H}_{4}}\]
B) \[AlC{{l}_{3}}\]
C) \[NaB{{H}_{4}}\]
D) \[Zn/HCl\]
Correct Answer: B
Solution :
\[\underset{n-\text{butane}}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}}}\,\underset{\Delta }{\mathop{\xrightarrow{\text{Anhyd}\text{.}\,AlC{{l}_{3}}}}}\,\underset{\text{iso}\,\text{butane}}{\mathop{C{{H}_{3}}-\underset{\underset{\,\,C{{H}_{3}}}{\mathop{|\,\,\,\,\,}}\,}{\mathop{CH}}\,-C{{H}_{3}}}}\,\]
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