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  4. Propene reacts with Cl 2 at 400-600° C to give

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Q. Propene reacts with $Cl _{2}$ at $400-600^{\circ} C$ to give

Hydrocarbons

Solution:

$CH _{3} \underset{\text{Propene}}{- CH} = CH _{2}+ Cl _{2} \stackrel{400-600^{\circ} C }{\rightarrow} Cl - \underset{\text{Allyl chloride}}{CH _{2}- CH} = CH _{2}$

At high temperature, halogenation takes place at allylic position.