Addition of HBr to propene yields 2-bromopropane, while in the presence of benzoyl peroxide, the same reaction yields 1bromopropane. Explain and give mechanism.
Answers
The example for electrophilic substitution reaction is addition to Hydrogen Bromide. The electrophile H+ was provided by Hydrogen Bromide. 1° and 2° carbocations are formed when the electrophile violates the double bond. The formation of carbocations is shown in first diagram.
Primary carbocations are less stable than the secondary carbocations. It will form in a very fast manner so that the former holds sway. Thus, in the next step the major product 2-bromopropane was formed when Br- violates the carbocation as shown in second diagram.
Carbon atom which has the lesser number of hydrogen atoms is connected with the adverse part of the addendum is formed. This reaction was followed by the rules 'Markovnikov's'.
With the presence of benzyl peroxide the reaction is added to convert anti to rule of Markovnikov's. The reaction is followed by the mechanism known as free radical chainasshown in third and fourth diagram.
Primary radicals are less stable than secondary free radicals. Hence it will form in a very faster manner so that the former holds sway. The major product is obtained as the result known as 1-bromopropane as shown in fifth diagram.
Br free radical acts as an electrophile because of the presence of peroxide. As a result, two different products are obtained on addition to Hydrogen Bromide to propene in the absence and presence of peroxide.
Answer:
Addition of HBr to propene is an ionic electrophillic addition reaction in which the electrophile i.e., H+ first add to give a more stable 2° carbocation. In the 2nd, the carbocation rapidly attached by the nucleophile Br-ion to give 2-Bromopropane.
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In the first step Br• radical add to propene in such a way so as to generate a more stable 2°free radical. In secon step, the free radical thus obtained rapidly abstracts a hydrogen atom from HBr to give 1Bromopentane
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