A) conc. \[{{H}_{2}}S{{O}_{4}}\]
B) Sodium alkoxide
C) Dry silver oxide
D) Grignard reagent
Correct Answer: D
Solution :
\[\underset{\begin{smallmatrix} \text{Methoxymethane} \\ \,\,\,\text{(Lower}\,\text{ether)} \end{smallmatrix}}{\mathop{C{{H}_{3}}OC{{H}_{3}}}}\,\xrightarrow{C{{l}_{2}}/hv}\underset{\alpha -\text{Chlorodimethyl}\,\text{ether}}{\mathop{C{{H}_{3}}OC{{H}_{2}}Cl}}\,\]\[\underset{-MgBr(Cl)}{\mathop{\xrightarrow{C{{H}_{3}}MgBr}}}\,\underset{\begin{smallmatrix} \text{Methoxyethane} \\ \,\,\text{(Higher}\,\text{ether)} \end{smallmatrix}}{\mathop{C{{H}_{3}}OC{{H}_{2}}C{{H}_{3}}}}\,\]
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