Balance equation for the reaction of conc. sulphuric acid with an alcohol :

C_{2} H_{5} O H + H_{2} S O_{4} (c o n c .) \to C H_{3} - C H_{3} + H_{2} O

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C_{2} H_{5} O H + H_{2} S O_{4} (c o n c .) \to C H_{2} = C H_{2} + H_{2} O

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C_{2} H_{4} O H + H_{2} S O_{4} (c o n c .) \to C H_{2} = C H_{2} + H_{2} O

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C_{2} H_{5} O H + H_{2} S O_{4} (c o n c .) \to C H_{2} = C H_{2} + H_{2} O_{2}

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Solution

The correct option is B $C_{2} H_{5} O H + H_{2} S O_{4} (c o n c .) \to C H_{2} = C H_{2} + H_{2} O$
${C_{2} H}_{5} O H + H_{2} {S O}_{4} (c o n c) \to {C H}_{2} = {C H}_{2} + H_{2} O$
Conc. $H_{2} S O_{4}$ will act as a dehydrating agent.

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H O C l (C l_{2} + H_{2} O)

C H_{3} - C H_{2} - X + K O H_{(} a l c) \to C H_{2} = C H_{2} + K X + H_{2} O

2 C H_{3} - C H_{2} - O H H^{+} - -- \to 413 K C H_{3} - C H_{2} - O - C H_{2} - C H_{3} + H_{2} O

A, B

C

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