Question

Which of the following compound is produced when \[{\rm{C}}{{\rm{H}}_{\rm{2}}} = {\rm{CH}} - {{\rm{(C}}{{\rm{H}}_{\rm{2}}}{\rm{)}}_5}{\rm{COOH}}\]reacts with HBr in presence of peroxides
A) \[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CH(C}}{{\rm{H}}_{\rm{2}}}{{\rm{)}}_{\rm{5}}}{\rm{COOH}}\]
B) \[{\rm{BrC}}{{\rm{H}}_2}{\rm{C}}{{\rm{H}}_2}{{\rm{(C}}{{\rm{H}}_{\rm{2}}}{\rm{)}}_{\rm{5}}}{\rm{COOH}}\]
C) \[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{{\rm{(C}}{{\rm{H}}_{\rm{2}}}{\rm{)}}_{\rm{5}}}{\rm{COOH}}\]
D) \[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{BrC}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{COOH}}\]

Answer 1

Hint: The peroxide effect says that the reaction of an alkene with HBr in the medium of peroxide gives anti-Markovnikov's product. Here, we have to find the anti-Markovnikov product of the reaction.

Complete Step by Step Answer:
Let's understand Markovnikov's rule first. The rule says that, when an alkene reacts with hydrogen halide (HCl, HBr, HI, etc.), the hydrogen is bonded to that carbon atom of the double bond to which most numbers of atoms of hydrogen are bonded and the halogen atom is bonded to another carbon atom. For example, propene undergoes a reaction with HCl as follows:
\[{{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} + {\rm{HCl}} \to {{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH(Cl)}} - {\rm{C}}{{\rm{H}}_3}\]

From the above reaction, it is clear that the Cl atom is bonded to that carbon atom of the double bond to which fewer numbers of atoms of hydrogen are attached. Here, the question is about the reaction of an alkene with HBr in presence of peroxide. So, the product obtained is the anti-Markovnikov product. So, the bromine atom is bonded to that carbon atom of the double bond to which more numbers of hydrogen atoms are bonded.
\[{\rm{C}}{{\rm{H}}_{\rm{2}}} = {\rm{CH}} - {{\rm{(C}}{{\rm{H}}_{\rm{2}}}{\rm{)}}_5}{\rm{COOH}} + {\rm{HBr}} \overset{Peroxide}{\rightarrow}{\rm{C}}{{\rm{H}}_{\rm{2}}}({\rm{Br}}) - {\rm{C}}{{\rm{H}}_2} - {{\rm{(C}}{{\rm{H}}_{\rm{2}}}{\rm{)}}_5}{\rm{COOH}}\]
Hence, the answer is option B.

Note: It is to be noted the peroxide effect is not applicable in the other hydrogen halides such as HI, HCl, and HF. In these halogen compounds, the product is formed by the Markovnikov addition. So, the peroxide effect is only in the case of the reaction of an alkene with HBr.