A) \[C{{H}_{3}}CH=C{{H}_{2}}\xrightarrow[2.\,\,AgN{{O}_{3}}/NaOH]{1.\,\,{{B}_{2}}{{H}_{6}}}\]
B) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Cl\xrightarrow[2.\,{{H}_{2}}{{O}_{2}}]{1.\,Mg/ether}\]
C) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}I\xrightarrow[{}]{HI/\Delta 150{}^\circ C}\]
D) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}COONa\xrightarrow[\Delta ]{NaOH(CaO)}\]
Correct Answer: C
Solution :
Alkyl halides undergo reduction with red phosphorus and hydrogen iodide and result in the formation of alkane. \[\underset{propyl\,\,iodide}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}I}}\,\xrightarrow[150{}^\circ C]{\operatorname{Re}d\,P/HI}\underset{propane}{\mathop{C{{H}_{3}}C{{H}_{2}}C{{H}_{3}}}}\,\]
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