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