Methyl iodide & Zn/Cu couple in alcohol produces :

ethane

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methane

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Both A and B

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none of the above

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Solution

The correct option is B methane
$C H_{3} I + 2 [H] Z n / C u c o u p l e - ------- \to a l c o h o l C H_{4} + H I$
In this reaction methyl iodide is reduced by using Zn/Cu couple and alcohol to form methane as a product.

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