A) \[Zn,\text{ }HCl\]
B) \[LIAl{{H}_{4}}\]
C) \[Mg,\]Anhydrous ether,\[{{H}_{2}}O\]
D) \[{{B}_{2}}{{H}_{6}}\]in THF
Correct Answer: D
Solution :
\[Zn+HCl,LiAl{{H}_{4}}\]and\[Mg,\]anhydrous ether, \[HgO\]etc convert n-butyl chloride into\[n-\] butane but\[{{B}_{2}}{{H}_{6}}\]in THF does not convert\[n-\] butyl chloride into n-butane. \[{{C}_{4}}{{H}_{4}}Cl\xrightarrow[{}]{Zn+HCl}\underset{n-bu\tan e}{\mathop{{{C}_{4}}{{H}_{10}}}}\,\] \[{{C}_{4}}{{H}_{9}}Cl+2H\xrightarrow[-2HCl]{LiAl{{H}_{4}}}\underset{n-bu\tan e}{\mathop{{{C}_{4}}{{H}_{10}}}}\,\] \[{{C}_{4}}{{H}_{9}}Cl+Mg\xrightarrow[ether]{Anhydrous}{{C}_{4}}{{H}_{9}}MgCl\xrightarrow[{}]{{{H}_{2}}O}\]
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