A) Williamsons synthesis
B) Clemmensens reduction
C) Fittig reaction
D) Wurtz reaction
Correct Answer: A
Solution :
Williamsons synthesis This reaction is used for the preparation of ether. In this method an alkyl halide is treated with sodium alkoxide. The reaction involves nucleophilic substitution of the halogen atom by the alkoxide group. \[R-{{O}^{-}}N{{a}^{+}}+X-R\xrightarrow{{}}\underset{ether}{\mathop{R-O-R}}\,\,\,+NaX\] This method can be used to prepare symmetrical and unsymmetrical ethers. Note- For preparing unsymmetrical ethers, the halide used should preferably the primary because secondary and tertiary alkyl halides may form alkene as major product due to elimination process, eg, \[\underset{sod.\text{ }te/f-butoxide}{\mathop{C{{H}_{3}}-\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,}}\,-O-Na+}}\,\underset{primary}{\mathop{C{{H}_{3}}C{{H}_{2}}Br}}\,\] \[\underset{ethyl\text{ }tert\text{ }-butyl\text{ }ether}{\mathop{\xrightarrow{Heat}C{{H}_{3}}-\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,}}\,-O-{{C}_{2}}{{H}_{5}}+NaBr}}\,\] \[\underset{tertiary\text{ }halide}{\mathop{C{{H}_{3}}-\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{C}}\,}}\,-Br+}}\,NaO{{C}_{2}}{{H}_{5}}\] \[\underset{2-methyl\text{ }propene}{\mathop{\xrightarrow{{}}C{{H}_{3}}-C=C{{H}_{2}}}}\,+\underset{ethanol}{\mathop{{{C}_{2}}{{H}_{5}}OH}}\,+NaBr\]
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