JEE Main 2021 · 24 Feb · Shift 1 · Chemistry Q45

← JEE Main 2021 · 24 Feb · Shift 1

ChemistrySingle correct+4 / -1Question 45 of 90

Identify products A and B.

1-Methylcyclopentene is treated with dil. KMnO $_4$ at 273 K to give A, which with CrO $_3$ gives B.

Scheme: 1-methylcyclopentene → (dil. KMnO4, 273 K) A → (CrO3) B.

A: 1-methylcyclopentane-1,2-diol; B: 2-methylcyclopent-2-en-1-one

A: 1-methylcyclopentane-1,2-diol; B: 2-hydroxy-2-methylcyclopentan-1-one

A: 2-methylcyclopentan-1-ol; B: 2-methylcyclopentan-1-one

A: 5-oxohexanal; B: 5-oxohexanoic acid

Option D: A = OHC-CH2CH2CH2-CO-CH3, B = HOOC-CH2CH2CH2-CO-CH3.

Answer and worked solution

Official answer: B

Steps. Cold dilute KMnO₄ gives the cis-diol, 1-methylcyclopentane-1,2-diol. CrO₃ oxidises the secondary OH to a ketone; the tertiary OH cannot be oxidised: 2-hydroxy-2-methylcyclopentanone.

(B)

Topics:Alkenes Alcohols

← Previous question Next question →

JEE Main 2021 · 24 Feb · Shift 1, Chemistry question 45