Question

Question 13.16 Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1-bromopropane. Explain and give mechanism.

Class XI Hydrocarbons Page 397

Answer 1

concept :- in the presence of peroxide , it follows the free radical Mechanism. while in the absence of peroxide , the reaction occurs by cationic machenism .

addition of hydrogen bromide ( HBr) to propene occurs through 'Markownikoff's rule.
HBr provide an electrophilic ( H+) , which attacks double bond to form carbocation.

(i) H ----Br ----> H+ + Br-
(ii) CH3 - CH = CH - CH3 + H+ ---> CH3 - CH2 - CH2+ ( 1° carbocation )
1° carbocation is less stable so, it's arranged by shifting of 1,2-hydride and form 2° carbocation (CH3 - CH⁺ - CH3) which is more stable .

The 2° carbocation is attacked by Br- ion to form the product as follows :
Br⁻ + CH3 - CH⁺ - CH3 ----> CH3 - CH(Br) - CH3 ( 2- bromopropane ) major product.

in the presence of benzoyl peroxide : addition of HBr to propane ( unsymmetrical Alkene) takes place against Markownikoff's rule .

this happens only with HBr but not for HCl or HI. this reaction is known as peroxide or kharash effect or addition reaction anti to Markownikoff's rule.

machenism are :
step 1 :- C6H5-(C=O)-O-O-(C=O)-C6H5 ---(homolysis)-->2C6H5-(C=O)-O::• ---->2C6H5• + 2CO2

step2 :- C6H5• + H-Br ---> C6H6 + Br•

step3 :- CH3 - CH = CH2 + Br• ----> CH3 - CH(Br) - CH2• ( 1° free radical ) less stable
CH3- CH• -CH2 - Br (2° free radical ,more stable)

step 4 :- CH3 - CH• - CH2-Br + H - Br ----(homolysis)--> CH3 - CH2 - CH2-Br + Br•