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H2overset(14)(C) = CH - CH3 underset("or...

`H_2overset(14)(C) = CH - CH_3 underset("or highi temp.")overset("low conc. of" Br_2)to (?)`
Product of the above reaction is :

A

`H_2 overset(14)C= CH -CH_2 -Br`

B

`H_2 C =C H - overset(14)(CH_2)-Br`

C

` underset(Br)underset(|)overset(14)(CH_2) - underset(Br) underset(|) (CH)-CH_3`

D

Both (a) and (b)

Text Solution

Verified by Experts

The correct Answer is:
B
low conc. of `Br_2` and high temperature favour substitution reaction, proceed through free radical.
`therefore ` Substitution will be major product.
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MS CHOUHAN-HYDROCARBONS (ALKENES)-LEVEL-2
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