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

H_2overset(14)(C) = CH - CH_3 underset("or highi temp.")overset("low c

1.	H_2overset(14)(C) = CH - CH_3 underset("or highi temp.")overset("low conc. of" Br_2)to (?) Product of the above reaction is :
Answer» `H_2overset(14)C =CH - CH_2 - BR` `H_2C = CH - OVERSET(14)CH_2 - Br ` `underset(Br)underset(\|)overset(14)CH_2 - underset(Br)underset(\|)CH-CH_3` both (a) and (b) SOLUTION :Both `H_2 overset(14)C=CH-CH_2 - Br and H_2C = CH - overset(14)CH_2 - Br`