JEE Mains · Chemistry · STD 11 - 9. Hydricarbon
\(Ph - CH = CH _2 \xrightarrow[ HBr ]{( PhCOO )_2}\) Product
Consider the above reaction
A. The reaction proceeds through a more stable radical intermediate.
B. The role of peroxide is to generate \(\dot{ H }\) (Hydrogen radical).
C. During this reaction, benzene is formed as a biproduct.
D. 1-Bromo-2-phenylethane is fanned as the minor product.
E. The same reaction in absence of peroxide proceeds via carbocation intermediate.
Identify the correct statements. Choose the correct answer from the options given below:
- A A & E Only
- B A, B & D Only
- C C, D & E Only
- D A, C & E Only
Answer & Solution
Correct Answer
(D) A, C & E Only
Step-by-step Solution
Detailed explanation
\(Ph - CH = CH _2 \xrightarrow[ HBr ]{( PhCOO )_2} Ph - CH _2- CH _2- Br\) Anti Markovnikov addition • Reaction follow radical addition in presence of peroxide • In absence of peroxide follow carbocation mechanism • Benzene also formed
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