Convert ethyl bromide into ethane.

Ethyl bromide in the presence of an aqueous solution of KOH will give ethanol as a product and KBr as a side product. Now, reduction of a hydroxyl group(-OH) using lithium aluminum hydride will produce ethane C2H6.

CH₃-CH₂-Br + aq. KOH—->CH₃-CH₂-OH + KBr

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