Preparation of ethene from Iodoethane is represented as:

C H_{3} C H_{2} I + K O H \to C_{2} H_{2} + K I + H_{2} O

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C H_{3} C H_{2} I + 3 K O H \to C_{2} H_{4} + 3 K I + H_{2} O

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C H_{3} C H_{2} I + K O H \to C_{2} H_{4} + K I + H_{2} O

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C H_{3} C H_{2} I + K O H \to C_{2} H_{6} + K I + H_{2} O

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Solution

The correct option is C $C H_{3} C H_{2} I + K O H \to C_{2} H_{4} + K I + H_{2} O$
Reaction:
$C H_{3} C H_{2} I + K O H \to C_{2} H_{4} + K I + H_{2} O$
The given reaction is $β$ elimination reaction. In this reaction alc. KOH reamoves halide groups from $α$ carbon and removes hydrogen form $β$ carbon.

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C H_{3} C H_{2} I + K O H (a q) \to C H_{3} C H_{2} O H + K I

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