question_answer

Which of the following yield both alkahe and alkene?

A) Kolbe's reaction

B) Williamson's synthesis

C) Wurtz reaction

D) Sandmeyer reaction

Correct Answer: A

Solution :

Kolbe's method of electrolysis of organic acid can be used to prepare alkane (except methane) alkene and alkyne. This reaction follows free radical mechanism. \[\text{2C}{{\text{H}}_{3}}COOK\xrightarrow[+2{{H}_{2}}O]{electricity}\underset{\underbrace{Ethane}_{at\,anode}}{\mathop{C{{H}_{3}}-C{{H}_{3}}}}\,+2C{{O}_{2}}\uparrow \] \[\underbrace{2KOH+{{H}_{2}}}_{at\,cathode}\uparrow \]                 For ethyne \[\underset{\underset{Pot.furnerate}{\mathop{CH\,COOK}}\,}{\overset{CH\,COOK}{\mathop{||}}}\,+2{{H}_{2}}O\xrightarrow{electrolysis}\underset{Ethyne}{\mathop{\underset{CH}{\overset{CH}{\mathop{|||}}}\,}}\,\] \[+2C{{O}_{2}}+{{H}_{2}}+2KOH\]                 For ethane \[\underset{Pot.\,succinate}{\mathop{\underset{C{{H}_{2}}COOK}{\overset{C{{H}_{2}}COOK}{\mathop{|}}}\,}}\,\xrightarrow[electrolysis]{HOH}\underset{Ethere}{\mathop{\underset{C{{H}_{2}}}{\overset{C{{H}_{2}}}{\mathop{||}}}\,}}\,+2C{{O}_{2}}+{{H}_{2}}+2KOH\]                 Williamson synthesis \[\underset{\begin{smallmatrix}  Sodium \\  alkoxide \end{smallmatrix}}{\mathop{R-\overset{-}{\mathop{O}}\,}}\,-\overset{+}{\mathop{N}}\,a+\underset{\begin{smallmatrix}  Alkyl \\  halide \end{smallmatrix}}{\mathop{X-R'}}\,\xrightarrow{{}}\underset{Ether}{\mathop{ROR'}}\,+NaX\] Wurtz reaction is used to prepare higher  alkanes \[RX+Na+RX\xrightarrow{dry/ether}R-R+NaX\] \[RX+Na+R'X\xrightarrow{{}}\] \[\underbrace{R-R+R-R'+R'}_{mixture\,of\,alkanes}-R'+NaX\] Sandmeyer reaction is used for preparation of chloroarenes and bromo arenes