Question

The reaction, $ ROH+{{H}_{2}}C{{N}_{2}} $ in the presence of $ HB{{F}_{4}}, $ gives the following product

• A. $ ROC{{H}_{3}} $
• B. $ RC{{H}_{2}}OH $
• C. $ ROHC{{N}_{2}}{{N}_{2}} $
• D. $ RC{{H}_{2}}C{{H}_{3}} $

Answer 1

Answer: (A)

In the presence of tetrafluoroboric acid, alcohols react with diazomethane $ (C{{H}_{2}}{{N}_{2}}) $ to form methyl ethers in excellent yield. $ \underset{alcohaol}{\mathop{ROH}}\,+\underset{\begin{smallmatrix} diazo \\ methane \end{smallmatrix}}{\mathop{C{{H}_{2}}{{N}_{2}}}}\,\xrightarrow{HB{{F}_{4}}}\underset{\begin{smallmatrix} methoxy \\ ethane \\ (ethar) \end{smallmatrix}}{\mathop{ROC{{H}_{3}}}}\,+{{N}_{2}} $