question_answer

Reaction of HBr with propene in the presence of peroxide gives:                                                     

A) isopropyl bromide

B) 3-bromo propane

C) allyl bromide             

D) n-propyl bromide

Correct Answer: D

Solution :

Reaction of HBr with propene in the presence of peroxide gives n-propyl bromide. This addition reaction is an example of Anti-Markownikoff addition reaction. (i.e., it is completed in form of free radical addition) \[C{{H}_{3}}-CH=C{{H}_{2}}+HBr\xrightarrow{Peroxide}\] \[C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}Br\] \[n\]-propyl bromide Mechanism of this reaction is represented as follows: Step 1. Formation of free radical of peroxide by means of decomposition. \[\underset{\text{Benzoyl}\,\,\text{peroxide}}{\mathop{{{C}_{6}}{{H}_{5}}-\underset{\begin{smallmatrix} || \\ O \end{smallmatrix}}{\mathop{C}}\,-O-O-\underset{\begin{smallmatrix} || \\ O \end{smallmatrix}}{\mathop{C}}\,-{{C}_{6}}{{H}_{5}}\xrightarrow{\Delta }}}\,\] Step 2. Benzoate free radical forms bromine free radical with HBr. \[{{C}_{6}}{{H}_{5}}COO+HBr\xrightarrow{{}}{{C}_{6}}{{H}_{5}}COOH+Br\] Step 3. Bromine free radical attacks on\[C=C\] of propene to form intermediate free radical. Hence, \[\text{C}{{\text{H}}_{\text{3}}}\text{CH}\text{C}{{\text{H}}_{\text{2}}}\text{Br}\]is the major product of this step. Step 4. More stable free radical accept hydrogen free radical from benzoic acid and give final product of reaction. \[C{{H}_{3}}-\overset{\bullet }{\mathop{C}}\,H-C{{H}_{2}}Br+{{C}_{6}}{{H}_{5}}COOH\xrightarrow[{}]{{}}\] \[\underset{n-\text{propyl bromide}}{\mathop{C{{H}_{3}}C{{H}_{6}}C{{H}_{2}}Br+{{C}_{6}}{{H}_{5}}CO\overset{\bullet }{\mathop{O}}\,}}\,\] Step 5. Benzoate free radicals are changed into benzoyl peroxide for the termination of free radical chain. \[{{C}_{6}}{{H}_{5}}CO\overset{\bullet }{\mathop{O}}\,+{{C}_{6}}{{H}_{5}}CO\overset{\bullet }{\mathop{O}}\,\xrightarrow{{}}{{({{C}_{6}}{{H}_{5}}CO)}_{2}}{{O}_{2}}\]