Ethylene on addition of hypochlorous acid forms:

H_{2} C = C H_{2}

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H O - C H_{2} - C H_{2} - C l

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H_{2} C = C H - C l

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C l - C H = C H - O H

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Solution

The correct option is C $H O - C H_{2} - C H_{2} - C l$
Ethylene on addition hypochlorous acid forms a product where double bond breaks and electrophile and nucleophile addition takes place.
Hence the product is
$C H_{2} = C H_{2} H O C l - -- \to H O - C H_{2} - C H_{2} - C l$

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Similar questions

C H \equiv C H

C H_{2} = C H_{2}

H_{2} C = C H - C l

H C | C l = C H | C l

C H_{3} - C l | C H | O H - C H_{2} - C H_{2} - C H_{2} - C l

H_{2} C = C H - C H | O H - {C H}_{2} - {C H}_{2} - {C H}_{3}

C H_{3} - C | C l H - C H_{2} - C H_{2} - C | O H H - C H_{3}

A . C H_{3} - C H = C H_{2} + H C l ⟶ C H_{3} - C | C l H - C H_{3}

B . C H_{3} - C H = C H_{2} + H B r ⟶ C H_{3} - C H_{2} - C H_{2} B r

C . C H_{3} - C H = C H_{2} + H B r p e r o x i d e - ---- \to C H_{3} - C H_{2} - C H_{2} - B r

D . C H_{3} - C H = C H_{2} + H I p e r o x i d e - ---- \to C H_{3} - C | I H - C H_{3}

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