JEE Main PYQ | 1-Methylcyclopentene Reaction Sequence | Reaction of Alkene with Dil. KMnO4 & Cr2​O3

Описание: In this video, I provide a detailed step-by-step solution for Question 45 from the JEE Main 2021 exam (Paper I, Chemistry), held on 24th February 2021 (Shift 1).

🧪 The Question: Identify products A and B in the following reaction sequence:

Reactant: 1-Methylcyclopentene

Reagent 1: dil. KMnO4, 273 K → A

Reagent 2: Cr2O3 → B

📝 Solution Breakdown:

Step 1: Formation of Product A (Baeyer's Test)

The reagent Dilute KMnO4 at 273 K (Cold) is known as Baeyer’s Reagent.

Function: It performs Syn-Hydroxylation (addition of two -OH groups) across the double bond. It does not break the ring (unlike hot acidic KMnO4).

Result (A): The double bond breaks to form a vicinal diol: 1-methylcyclopentane-1,2-diol. (Contains one 3∘ alcohol and one 2∘ alcohol).

Step 2: Formation of Product B (Oxidation)

The reagent Cr2O3 (Chromium Oxide) acts as an oxidizing agent.

Selectivity:

Secondary (2∘) Alcohol: Oxidizes to a Ketone (C=O).
Tertiary (3∘) Alcohol: Resists mild oxidation (does not react easily without dehydration).

Result (B): The secondary -OH group converts to a ketone, while the tertiary -OH group remains intact.

Product: 2-hydroxy-2-methylcyclopentanone.

✅ Correct Answer: Option (2).

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