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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_2overset(14)C = CH - CH_2 - Br`

B

`H_2C = CH - 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:
D
Both
`H_2 overset(14)C=CH-CH_2 - Br and H_2C = CH - overset(14)CH_2 - Br`
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