A) \[C{{H}_{3}}CH=C{{H}_{2}}\]
B) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{3}}\]
C) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{3}}\]
D) \[C{{H}_{3}}-C{{H}_{3}}\]
Correct Answer: B
Solution :
Key Idea: It is Kolbes electrolysis \[\underset{\text{sodium}\,\text{salt}\,\text{of}\,\text{acid}}{\mathop{2RCOONa+{{H}_{2}}O}}\,\xrightarrow{\text{Electricity}}\underset{\text{alkane}}{\mathop{R-R}}\,+2C{{O}_{2}}\] \[+\,2\,NaOH+{{H}_{2}}\] \[\underset{\text{sodium}\,\text{propionate}}{\mathop{2C{{H}_{3}}C{{H}_{2}}COONa}}\,\,+{{H}_{2}}O\xrightarrow{\text{Electricity}}\] \[\underset{\text{butane}}{\mathop{C{{H}_{3}}C{{H}_{2}}-C{{H}_{2}}-C{{H}_{3}}}}\,+2C{{O}_{2}}+2NaOH+{{H}_{2}}\]\[\therefore \]butane is formed by electrolysis of sodium propionate
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