
Addition of HCl to propene in presence of peroxide gives
A) 1-Chloropropane
B) 2-Chloropropane
C) 3-Chloropropane
4) Propene dichloride
Answer
164.1k+ views
Hint: The peroxide effect says that the reaction of an alkene with HBr in the medium of peroxide gives anti-Markovnikov's product. But, the peroxide effect is not applicable to other hydrogen halides such as HI, HCl etc.
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 hydrogen atoms 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 Cl atom is bonded to that carbon atom of the double bond to which fewer numbers of atoms of hydrogen are attached.
Here, we have to find out the product obtained due to the reaction of propene with HCl in presence of peroxide. We know the peroxide effect gives anti-Markovnikov products but only in the case of HBr. So, when propene reacts with HCl in the presence of peroxide, Markovnikov product is obtained. So,
\[{{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} + {\rm{HCl}}\overset{Peroxide}{\rightarrow}\mathop {{{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH(Cl)}} - {\rm{C}}{{\rm{H}}_3}}\limits_{{\rm{2 - chloropropene}}} \]
Therefore, answer is option B.
Note: Always remember that the rule of anti-Markovnikov product is only formed when the alkene undergoes a reaction with HBr in the medium of peroxide. In other cases like HI, HCl, HF when undergoing a reaction with alkene in peroxide medium gives Markovnikov product.
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 hydrogen atoms 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 Cl atom is bonded to that carbon atom of the double bond to which fewer numbers of atoms of hydrogen are attached.
Here, we have to find out the product obtained due to the reaction of propene with HCl in presence of peroxide. We know the peroxide effect gives anti-Markovnikov products but only in the case of HBr. So, when propene reacts with HCl in the presence of peroxide, Markovnikov product is obtained. So,
\[{{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} + {\rm{HCl}}\overset{Peroxide}{\rightarrow}\mathop {{{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH(Cl)}} - {\rm{C}}{{\rm{H}}_3}}\limits_{{\rm{2 - chloropropene}}} \]
Therefore, answer is option B.
Note: Always remember that the rule of anti-Markovnikov product is only formed when the alkene undergoes a reaction with HBr in the medium of peroxide. In other cases like HI, HCl, HF when undergoing a reaction with alkene in peroxide medium gives Markovnikov product.
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