Question

Methyl alcohol on oxidation with acidified \( \text{K}_2\text{Cr}_2\text{O}_7 \) gives:

• A. \( \text{CH}_3\text{COCH}_3 \)
• B. \( \text{CH}_3\text{CHO} \)
• C. HCOOH
• D. \( \text{CH}_3\text{COOH} \)

Hint:

Methyl alcohol (methanol) is oxidized to formic acid (HCOOH) when reacted with acidified potassium dichromate.

Answer 1

The Correct Option is C

Step 1: Understand the oxidation reaction.
Methyl alcohol (methanol, \( \text{CH}_3\text{OH} \)) undergoes oxidation with acidified potassium dichromate (\( \text{K}_2\text{Cr}_2\text{O}_7 \)), where it is oxidized to formaldehyde (HCHO) and further oxidized to formic acid (HCOOH).
Step 2: Identify the correct product.
When methyl alcohol undergoes oxidation, it is first converted to formaldehyde (HCHO), which is further oxidized to formic acid (HCOOH).
Step 3: Eliminate the incorrect options.
- Option (A) \( \text{CH}_3\text{COCH}_3 \): Incorrect, this is acetone, not the product of methanol oxidation. - Option (B) \( \text{CH}_3\text{CHO} \): Incorrect, this is formaldehyde, an intermediate in the oxidation process. - Option (C) HCOOH: Correct, formic acid is the final product of methanol oxidation. - Option (D) \( \text{CH}_3\text{COOH} \): Incorrect, this is acetic acid, which is produced from ethanol, not methanol.
Final Answer: \[ \boxed{\text{HCOOH}} \]