Question

(a) Addition of HBr to but-1-ene in the presence of benzoyl peroxide yields 1-bromo-butane the major product. Write mechanism also.
(b) How will you out the following conversions:
(i) Propyne to propanone
(ii) Benzene to m-nitrochloro benzene​

Answer 1

Answer:

Explanation:

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•