A) Propane
B) Propanoic acid
C) Propene
D) Propyne
Correct Answer: C
Solution :
\[C{{H}_{3}}OH\underset{{{I}_{2}}}{\mathop{\xrightarrow{NaOH}}}\,\]\[\underset{C{{H}_{2}}-I}{\overset{C{{H}_{2}}\ \ \,}{\mathop{\underset{|\,}{\overset{||}{\mathop{C}}}\,H\ \ \ }}}\,\xrightarrow{HI}\underset{C{{H}_{2}}-I}{\overset{C{{H}_{3}}\ \ \,}{\mathop{\underset{|\,}{\overset{||}{\mathop{C}}}\,H-I}}}\,\xrightarrow[-{{I}_{2}}]{}\underset{C{{H}_{2}}\ \ }{\overset{C{{H}_{3}}\ \ \,}{\mathop{\underset{||}{\overset{|\,}{\mathop{C}}}\,H\ \ \ }}}\,\,\]
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