
The name of the product obtained by the addition of HI to propene in presence of peroxide catalyst is
A) Isopropyl iodide
B) 2-Iodopropene
C) 2-Iodopropane
D)1-Iodopropane
Answer
161.7k+ views
Hint: Peroxide is a catalyst used in the conversion of alkene to haloalkane. The reaction of an alkene with HBr in the presence of catalyst peroxide gives a haloalkane formed by following the anti-Markovnikov rule.
Complete Step by Step Solution:
Let's first discuss Markovnikov's rule. The rule says, when an alkene undergoes a reaction with Hydrogen halides namely HCl, HI, HBr, etc. The hydrogen atom is added to that carbon atom of the double bond to which most atoms of hydrogen are bonded and the halogen atom is added to another carbon atom. For example,
\[{{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} + {\rm{HCl}} \to {{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH(Cl)}} - {\rm{C}}{{\rm{H}}_3}\]
Let's discuss the anti-Markovnikov addition. This rule is valid for only the reaction of an alkene with HBr in presence of the peroxide catalyst. According to this rule, the halogen atom is added to the carbon atom to which most numbers of hydrogen atoms are bonded and the hydrogen atom is added to the other carbon atom. For example,
\[{{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} + {\rm{HBr}} \overset{Peroxide}{\rightarrow} {{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{C}}{{\rm{H}}_2} - {\rm{C}}{{\rm{H}}_2}{\rm{(Br)}}\]
Here, we have to find out the product of the reaction of HI and propene in the presence of catalyst peroxide. The reaction follows Markovnikov's addition because the peroxide effect is applicable for only HBr. So, the reaction is,
\[{{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} + {\rm{HI}} \overset{Peroxide}{\rightarrow} \mathop {{{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH(I)}} - {\rm{C}}{{\rm{H}}_3}}\limits_{{\rm{2 - Iodopropane}}\,\,{\rm{or}}\,\,{\rm{isopropyl}}\,{\rm{chloride}}} \]
So, 2-iodopropane or isopropyl chloride is obtained.
Hence, options A and C are correct.
Note: Students might get confused with the peroxide catalyst. It is a point of mistake that students might think the product obtained in the reaction of propene with HI in the presence of catalyst peroxide gives an anti-Markovnikov product.
Complete Step by Step Solution:
Let's first discuss Markovnikov's rule. The rule says, when an alkene undergoes a reaction with Hydrogen halides namely HCl, HI, HBr, etc. The hydrogen atom is added to that carbon atom of the double bond to which most atoms of hydrogen are bonded and the halogen atom is added to another carbon atom. For example,
\[{{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} + {\rm{HCl}} \to {{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH(Cl)}} - {\rm{C}}{{\rm{H}}_3}\]
Let's discuss the anti-Markovnikov addition. This rule is valid for only the reaction of an alkene with HBr in presence of the peroxide catalyst. According to this rule, the halogen atom is added to the carbon atom to which most numbers of hydrogen atoms are bonded and the hydrogen atom is added to the other carbon atom. For example,
\[{{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} + {\rm{HBr}} \overset{Peroxide}{\rightarrow} {{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{C}}{{\rm{H}}_2} - {\rm{C}}{{\rm{H}}_2}{\rm{(Br)}}\]
Here, we have to find out the product of the reaction of HI and propene in the presence of catalyst peroxide. The reaction follows Markovnikov's addition because the peroxide effect is applicable for only HBr. So, the reaction is,
\[{{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} + {\rm{HI}} \overset{Peroxide}{\rightarrow} \mathop {{{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH(I)}} - {\rm{C}}{{\rm{H}}_3}}\limits_{{\rm{2 - Iodopropane}}\,\,{\rm{or}}\,\,{\rm{isopropyl}}\,{\rm{chloride}}} \]
So, 2-iodopropane or isopropyl chloride is obtained.
Hence, options A and C are correct.
Note: Students might get confused with the peroxide catalyst. It is a point of mistake that students might think the product obtained in the reaction of propene with HI in the presence of catalyst peroxide gives an anti-Markovnikov product.
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