A) \[Zn-HCl\]
B) \[Sn-HCl\]
C) \[Zn-Hg\]
D) \[Pd/{{H}_{2}}\]
Correct Answer: D
Solution :
\[\underset{butene-1}{\mathop{C{{H}_{2}}=CH}}\,-C{{H}_{2}}C{{H}_{3}}\xrightarrow[\Delta ,\,\,pressure]{Pd/{{H}_{2}}}\] \[\underset{\text{butane}}{\mathop{C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{3}}}}\,\] other reagents are successful with polar double bonds.
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