Question

When Propene is treated with $\mathrm{HBr}$ in the presence of peroxides, the major product formed is:}

• A. 2-Bromopropane
• B. 1-Bromopropane
• C. 1,2-Dibromopropane
• D. Peroxypropane

Hint:

Peroxide effect is observed only with HBr. \[ \text{Alkene} + \mathrm{HBr} \xrightarrow{\text{Peroxide}} \text{Anti-Markovnikov Product} \] HCl and HI do not show peroxide effect.

Answer 1

The Correct Option is B

Concept: Addition of HBr to an unsymmetrical alkene normally follows Markovnikov's rule. However, in the presence of organic peroxides, the reaction proceeds through a free radical mechanism known as the Kharasch effect or Peroxide effect. In this case, HBr adds according to anti-Markovnikov orientation.

Step 1: Write the structure of propene. \[ \mathrm{CH_3-CH=CH_2} \] Propene is an unsymmetrical alkene.

Step 2: Understand the role of peroxide. Peroxides generate free radicals. \[ ROOR \rightarrow 2RO^\bullet \] These radicals initiate a chain reaction involving HBr.

Step 3: Apply anti-Markovnikov addition. In radical addition, bromine radical adds first to the double bond in such a way that the more stable carbon radical is formed. Hence bromine attaches to the terminal carbon atom. \[ \mathrm{CH_3CH=CH_2} \xrightarrow[\text{Peroxide}]{\mathrm{HBr}} \mathrm{CH_3CH_2CH_2Br} \]

Step 4: Identify the product. The product formed is \[ \mathrm{CH_3CH_2CH_2Br} \] which is 1-Bromopropane. Therefore, \[ \boxed{\text{Option (2)}} \]