Question

The reaction, \(ROH + H_2CN_2\) in the presence of \(HBF_4\), gives the following product

• A. \(ROCH_3\)
• B. \(RCH_2OH\)
• C. \(ROHCN_2N_2\)
• D. \(RCH_2CH_3\)

Hint:

Diazomethane methylates alcohols and acids. In acidic medium, it gives methyl ether: \(ROH \rightarrow ROCH_3\).

Answer 1

The Correct Option is A

Step 1: Identify the reagent.
\(H_2CN_2\) represents diazomethane (\(CH_2N_2\)).
Diazomethane is a methylating agent.
Step 2: Role of \(HBF_4\).
Strong acid protonates diazomethane to generate \(CH_3^+\)-like species.
This methyl group then reacts with alcohol oxygen.
Step 3: Product formed.
Alcohol is converted into methyl ether:
\[ ROH \rightarrow ROCH_3 \] Final Answer: \[ \boxed{ROCH_3} \]