Butene when treated with chlorine at about 500 $^{o}$ C forms:

MeCH

_{2}

CHCl - CH

_{2}

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MeCH(Cl)-CH=CH

_{2}

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ClCH

_{2}

_{2}

CH = CH

_{2}

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MeC(Cl

_{2}

)CH=CH

_{2}

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Solution

The correct option is C MeCH(Cl)-CH=CH $_{2}$
Butene when treated with chlorine at about 500 $^{0} C$ forms 3-chloro-1-butene. in this reaction, the allylic H is replaced with Cl.
$C H_{3} - C H_{2} C H = C H_{2} C l - -- \to 500^{o} C C H_{3} - C H C l - C H = C H_{2}$

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