Question

Identify the major product of following reaction.
\(\text{Propene}\rightarrow _{\text{Peroxide}\,}^{\text{HBr}\,}\,?\)

• A. 1 - Bromopropane
• B. 2 - Bromopropane
• C. 3 - Bromopropene
• D. 2 - Bromopropan-1-ol

Hint:

With peroxide, HBr adds to propene by a free radical path against Markovnikov, giving the 1-bromo product.

Answer 1

The Correct Option is A

Step 1: Understand the reaction
Propene, \(\text{CH}_3\text{CH=CH}_2\), reacts with HBr in the presence of a peroxide. This is the peroxide (Kharasch) effect, which changes the usual orientation of addition.

Step 2: Mechanism
The peroxide produces a bromine radical \(\text{Br}\cdot\) from HBr. The radical adds to the terminal carbon \(\text{CH}_2\) of the double bond so that the more stable secondary radical \(\text{CH}_3\dot{\text{C}}\text{H}-\text{CH}_2\text{Br}\) forms.

Step 3: Complete the step
The secondary radical removes H from another HBr molecule, giving \(\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}\) and a new \(\text{Br}\cdot\), so the chain continues.

Step 4: Check the options
The product is 1-bromopropane (A). 2-Bromopropane (B) is the Markovnikov product without peroxide. 3-Bromopropene (C) needs substitution at the allylic position, and (D) needs an OH group, which is not available in this reaction.

Final Answer:
Anti-Markovnikov addition gives 1-bromopropane. This is option (A). \[ \boxed{\text{(A) }\text{1-Bromopropane}} \]