CH3Br→LiAlH4A;CH3CH2Br+Mg→dryetherX→CH3OHB.A and B are respectively

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${CH}_{3} Br \overset{{LiAlH}_{4}}{\to} A$ ; ${CH}_{3} {CH}_{2} Br + Mg \overset{dryether}{\to} X \overset{{CH}_{3} OH}{\to} B$ .A and B are respectively

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C₂H₆ and C₂H₆

CH₄ and C₂H₆

C₂H₆ and CH₄

CH₄ and CH₄

answer is B.

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${CH}_{3} Br \overset{{LiAlH}_{4}}{\to} {CH}_{4}$

${CH}_{3} {CH}_{2} Br + Mg \overset{dryether}{\to} {CH}_{3} {CH}_{2} MgBr (X) \overset{{CH}_{3} OH}{\to} C_{2} H_{6} (B) + {CH}_{3} OMgBr$

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