Q77 / 240 · KCET 2026 · Chemistry
$R-\text{CH}_2\text{OH}$ is converted into $R-\text{CHO}$ by reacting with ________.
- AAlkaline $\text{KMnO}_4$
- B$\text{LiAlH}_4$
- C$\text{Na}/\text{C}_2\text{H}_5\text{OH}$
- Correct answerDPCC (Pyridinium Chlorochromate)
Solution
The conversion of a primary alcohol ($R-\text{CH}_2\text{OH}$) to an aldehyde ($R-\text{CHO}$) requires a mild oxidizing agent.
Alkaline $\text{KMnO}_4$ is a strong oxidizing agent and oxidizes primary alcohols directly to carboxylic acids ($R-\text{COOH}$).
$\text{LiAlH}_4$ and $\text{Na}/\text{C}_2\text{H}_5\text{OH}$ are reducing agents and do not oxidize alcohols.
PCC (Pyridinium Chlorochromate) is a mild oxidizing agent that selectively oxidizes primary alcohols to aldehydes, stopping the reaction at the aldehyde stage.
Answer: PCC (Pyridinium Chlorochromate)
Alkaline $\text{KMnO}_4$ is a strong oxidizing agent and oxidizes primary alcohols directly to carboxylic acids ($R-\text{COOH}$).
$\text{LiAlH}_4$ and $\text{Na}/\text{C}_2\text{H}_5\text{OH}$ are reducing agents and do not oxidize alcohols.
PCC (Pyridinium Chlorochromate) is a mild oxidizing agent that selectively oxidizes primary alcohols to aldehydes, stopping the reaction at the aldehyde stage.
Answer: PCC (Pyridinium Chlorochromate)
Hence, the correct answer is option: 'D'.