Question

Propylene on treatment with HBr/H$_2$O$_2$ provides

• A. 1-bromopropane
• B. 2-bromopropane
• C. 1,2-dibromopropane
• D. 1-bromopropan-2-ol
• E. 2-bromopropan-1-ol

Hint:

Peroxide effects only apply to the addition of HBr, not HCl or HI. Without peroxide, the major product would be 2-bromopropane (Markovnikov's rule).

Answer 1

The Correct Option is A

Concept: The addition of HBr to an unsymmetrical alkene in the presence of peroxides follows the Anti-Markovnikov rule due to a free radical mechanism, also known as the Kharasch effect.

Step 1: {Identify the structure of propylene.} Propylene (prop-1-ene) has the chemical formula $\text{CH}_3\text{CH}=\text{CH}_2$. It is an unsymmetrical alkene with a terminal double bond.

Step 2: {Apply the effect of hydrogen peroxide (H$_2$O$_2$).} In the presence of peroxides, the addition of HBr proceeds via a radical mechanism. The bromine radical attacks the less substituted carbon atom to form a more stable secondary radical.

Step 3: {Determine the final product.} The Bromine atom attaches to the terminal carbon ($\text{C}_1$), and the Hydrogen atom attaches to the middle carbon ($\text{C}_2$). $$\text{CH}_3\text{CH}=\text{CH}_2 + \text{HBr} \xrightarrow{\text{H}_2\text{O}_2} \text{CH}_3\text{CH}_2\text{CH}_2\text{Br}$$ The resulting product is 1-bromopropane.