The reactions which produce alcohol as the product area : A. \(\mathrm{CH}_{4}+\mathrm{O}_{2} \xrightarrow[\Delta]{\mathrm{Mo}_{2} \mathrm{O}_{3}}\) B. \(2 \mathrm{CH}_{3} \mathrm{CH}_{3}+3 \mathrm{O}_{2} \xrightarrow[\Delta]{\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2} \mathrm{Mn}}\) C. \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{CH} \xrightarrow{\mathrm{KMnO}_{4}}\) D. \(2 \mathrm{CH}_{4}+\mathrm{O}_{2} \xrightarrow{\mathrm{Cu} / 523 \mathrm{~K} / 100 \mathrm{~atm} .}\) E. \(\mathrm{CH}_{3}-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_{3} \xrightarrow{\mathrm{KMnO}_{4} / \mathrm{H}^{+}}\) Choose the correct answer from the options given below :

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Hydrocarbons

The reactions which produce alcohol as the product area :
A.

CH_{4} + O_{2} Mo_{2} O_{3} Δ

2 CH_{3} CH_{3} + 3 O_{2} (CH_{3} COO)_{2} Mn Δ

(CH_{3})_{3} CH KMnO_{4}

2 CH_{4} + O_{2} Cu /523 K /100 atm .

CH_{3} - CH = CH - CH_{3} KMnO_{4} / H^{+}

Choose the correct answer from the options given below :

(A)A and D Only
(B)A, C and E Only
(C)C and D Only✓ Correct
(D)B, D and E Only

Answer: (C)

Solution

Each conversion is judged by the functional group formed. Catalytic air oxidation of methane over

M o_{2} O_{3}

gives methanal, not an alcohol. Manganese(II) acetate catalysed oxidation of ethane gives acetic acid. Oxidation of isobutane,

(C H_{3})_{3} CH

, with

KMn O_{4}

attacks the lone tertiary C-H to give tert-butyl alcohol. Oxidation of methane over copper at 523 K and 100 atm gives methanol. Acidic

KMn O_{4}

cleaves but-2-ene at the double bond to give acetic acid, not an alcohol. Only the isobutane and the copper-catalysed methane reactions yield alcohols.

JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Single correct option

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JEE Main · 2026 · 28 Jan 2026 · Shift 2 · Hydrocarbons

Solution

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