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 Kolbe's electrolysis \[\underset{sodium\,\,salt\,\,of\,\,acid}{\mathop{2RCOONa+{{H}_{2}}O}}\,\xrightarrow{Electricity}\underset{alkane}{\mathop{R-R}}\,+2C{{O}_{2}}\] \[+2NaOH+{{H}_{2}}\] \[\underset{sodium\,\,propionate}{\mathop{2C{{H}_{3}}C{{H}_{2}}COONa+{{H}_{2}}O}}\,\xrightarrow{Electricity}\] \[\underset{butane}{\mathop{C{{H}_{3}}C{{H}_{2}}-C{{H}_{2}}}}\,-C{{H}_{3}}+2C{{O}_{2}}+2NaOH+{{H}_{2}}\] \[\therefore \]butane is formed-'by electrolysis of sodium propionate
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