Question:
Identify products A and B. $CH_3 - CH(CH_3) - CH_3 \xrightarrow{\text{dil. } KMnO_4, 273 K} A \xrightarrow{CrO_3} B$
Identify products A and B. $CH_3 - CH(CH_3) - CH_3 \xrightarrow{\text{dil. } KMnO_4, 273 K} A \xrightarrow{CrO_3} B$
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Tertiary carbons are the only alkane carbons easily oxidized by $KMnO_4$ to form alcohols.
Updated On: Jan 9, 2026
- A
- B
- C
- D
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The Correct Option is A
Solution and Explanation
Step 1: Alkanes with a tertiary H-atom are oxidized to tertiary alcohols by $KMnO_4$. Thus, isobutane $\to$ tert-butanol (A).
Step 2: $CrO_3$ is an oxidizing agent. However, tertiary alcohols are generally resistant to oxidation under these conditions. *Note: In many exam variants, if the starting material is an alkene, A would be a diol and B a ketone. For isobutane, A is tert-butanol.*
Step 2: $CrO_3$ is an oxidizing agent. However, tertiary alcohols are generally resistant to oxidation under these conditions. *Note: In many exam variants, if the starting material is an alkene, A would be a diol and B a ketone. For isobutane, A is tert-butanol.*
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