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.

Asked by Pragya Singh | 1 year ago |  57

Solution :-

Addition of HBr to propene is an example of an electrophilic addition reaction. Hydrogen bromide provides an electrophile, $$H^{+}$$ . This electrophile attacks the double bond to form primary and secondary carbocations as shown:



Secondary carbocations are comparatively more stable than primary carbocations. Thus, the former predominates since it will form at a faster rate. Thus, now $$Br^{-}$$ attacks the carbocation to form 2 – bromopropane as the major product.

This reaction follows Markovnikov’s rule. Now, In the presence of benzoyl peroxide, an additional reaction takes place by anti-Markovnikov’s rule. The reaction follows a free radical chain mechanism as;

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In the presence of peroxide, Br free radical acts as an electrophile. Therefore, two different products are obtained in addition of HBr to propene according to the absence and presence of peroxide.

Answered by Abhisek | 1 year ago

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