Question

2. In free radical bromination reaction, 3° alkyl bromide
form as Major Product in ?-
CH,
CH,
(3) CH-CH-CH-CH, (4) All of these​

Answer 1

In free radical bromination reaction, 3° alkyl bromide

In free radical bromination reaction, 3° alkyl bromideform as Major Product in (4) All of the above.

1. In the first option , the compound is methylcyclohexane. Here free radical formation will take place and in order to stabilize the compound it will take place on the 3° carbon ( which is the carbon in ring attached to methyl group). Hence, the bromine free radical will attack and form 3° alkyl bromide.
2. In the second case, the compound is 2- methylcyxlohexene. Here, same steps will be followed as mentioned for option 1.
3. (Note: Here double bond will not be broken as this is free radical substitution reaction and addition doesn't take place)
4. In option 3 the compound is 2-methylbutane. In this scenario, the stable radical will be formed at second carbon position which is a tertiary carbon and 3° alkyl bromide.