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 mechanisms.


Addition of HBr to propene is an example of an electrophilic substitution reaction.
Hydrogen bromide provides an electrophile, H+. This electrophile attacks the double bond to form 1° and 2° carbocations as shown:
                H3C3-C2H=C1H2+H-Br                          H+
                           
 H3C-CH2-CH2+Br¯+      Less stableprimarycarvbocation       H3C-CH-CH3++Br¯         Move stable   secondarycarbocatuion
Secondary carbocations are more stable than primary carbocations. Hence, the former predominates since it will format a faster rate. Thus, in the next step, Br– attacks the carbocation to form 2 – bromopropane as the major product
H3C-CH+-CH3+BrCH3-CHBr-CH3
                                                   2-Bromopropane
This reaction follows Markovnikov’s rule where the negative part of the addendum is attached to the carbon atom having a lesser number of hydrogen atoms. In the presence of benzoyl peroxide, an addition reaction takes place anti to Markovnikov’s rule. The reaction follows a free radical chain mechanism as:
C6H5-CO-O-O-CO                Homklysis2C6HS-CO-2C*6H5+2CO2
C6H5+H-BrHomolysisC6H5+BrCH3-CH=CH2+Br                  
                     
CH3-CH|-CH2            Br     Lessstableprimaryfreeradical         CH3-CH-CH2-Br         Morestablesecondaryfreeradical
Secondary free radicals are more stable than primary radicals. Hence, the former predominates since it forms at a faster rate. Thus, 1 – bromopropane is obtained as the major product.
CH3-CH*-CH2Br+H-Br                  
                         CH3-CH2-CH2Br+Br·                   1-Bromoproane                       Majorproduct
In the presence of peroxide, Br free radical acts as an electrophile. Hence, two different products are obtained on addition of HBr to propene in the absence and presence of peroxide