The action of bleaching powder on ethyl alcohol gives :

C {C l}_{4}

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C H {C l}_{3}

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C O C l

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Ether

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Solution

The correct option is B $C H {C l}_{3}$
Chloroform $C H C l_{3}$ and calcium formate $C a (C O O H)_{2}$ are formed when bleaching powder reacts with ethyl alcohol.
$C a (O C l) C l + H_{2} O \to C l_{2} + C a (O H)_{2}$
$C H_{3} - C H_{2} - O H + C l_{2} \to C H_{3} - C H O + 2 H C l$
$C H_{3} - C H O + 3 C l_{2} \to C C l_{3} - C H O + 3 H C l$
$2 C C l_{3} - C H O + C a (O H)_{2} \to 2 C H C l_{3} + C a (C O O H)_{2}$

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