Reaction of HBr with propene in the presence of peroxide gives :-

Options

A. isopropyl bomide
B. 3 - bromo propane
C. allyl bomide
D. n - propyl bromide

Answer

D

Explanation

Reaction of `HBr` with propene in the presence of peroxide gives n - propyl bromide. This addition reaction is an example of anti - Markowinkoff addition reaction.
(ie, it is complete in form of free radical addition)
`CH_(3)-CH=CH_(2)+HBr overset("Peroxide")rarr`
`underset("n - propyl bromide")(CH_(3)-CH_(2)-CH_(2)Br)`
Mechanism of this reaction is represented as follows :
Step 1 : Formation of free radical of peroxide by means of decomposition.
`underset("Benzoyl Peroxide")(C_(6)H_(5)-underset(O)underset(||)C-O-O-underset(O)underset(||)C-C_(6)H_(5))overset(Delta)rarr`
`underset("benzoate free radical")(2C_(6)H_(5)-COO)`
Step 2 : Benzoate free radical forms bromine free radical with HBr.
`C_(6)H_(5)COoverset(.)O+HBr rarr C_(6)H_(5)COOH+overset(.)Br`
Step 3 : Bromine free radical attacks on `C=C` of propene to form intermediate free radical.

Hence, `CH_(3)-overset(.)CH-CH_(2)Br` is the major product of this step.
Step 4 : More stable free radical accept hydrogen free radical from benzoic acid and give final product of reaction.
`CH_(3)-overset(.)CH-CH_(2)Br+C_(6)H_(5)COOHrarr`
`underset("n - propyl bromide")(CH_(3)-CH_(2)-CH_(2))Br+C_(6)H_(5)COoverset(.)O`
Step 5 : Benzoate free radical are changed into benzoyl peroxide for the termination of free radical chain.
`C_(6)H_(5)COoverset(.)O+C_(6)H_(5)COoverset(.)Orarr(C_(6)H_(5)CO)_(2)O_(2)`