
In the reaction, \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} + {\rm{H}} - {\rm{X}} \to {\rm{Product}}\] . What is the product?
A) \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}} - {\rm{C}}{{\rm{H}}_{\rm{3}}}\]
B) \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{C}}{{\rm{H}}_2}{\rm{X}}\]
C) \[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}} - {\rm{CHX}} - {\rm{C}}{{\rm{H}}_{\rm{3}}}\]
D) \[{\rm{C}}{{\rm{H}}_3} - {\rm{C}}{{\rm{H}}_2}{\rm{X}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_2}\]
Answer
164.4k+ views
Hint: Here, the reaction of an alkene with a hydrogen halide (HI, HCl, HBr) is given. So, the Markovnikov rule is to be followed while answering this question.
Complete Step by Step Solution:
Let's first understand what the Markovnikov rule states. According to this rule, when an alkene undergoes a reaction with hydrogen halide, the double bond of the alkene breaks and the hydrogen atom forms a bond with that carbon atom of the double bind to which most numbers of atoms of hydrogen are bonded. And the halogen atoms get attached to another carbon atom of the double bond. For example, the reaction of propene and hydrogen chloride gives 2-iodo-propane.
\[{{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} + {\rm{HCl}} \to {{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH(Cl)}} - {\rm{C}}{{\rm{H}}_3}\]
In the same way, the given alkene undergoes a reaction with HX. Markovnikov's rule is to be followed here also.
\[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} + {\rm{H}} - {\rm{X}} \to {{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{CH(X)}} - {\rm{C}}{{\rm{H}}_3}\]
Therefore, option C is right.
Additional Information: Let's discuss what Markovnikov's rule is. This rule is the opposite of Markovnikv's rule. This rule applies to the reaction of an alkene with HBr in the medium of peroxide. According to this rule, the halogen atom forms a bond with that carbon atom of the double bond to which most numbers of hydrogen atoms are bonded, and the hydrogen atom forms a bond with another carbon atom of the double bond.
\[{{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} + {\rm{HBr}} \overset{Peroxide}{\rightarrow} {{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{C}}{{\rm{H}}_2} - {\rm{C}}{{\rm{H}}_2}{\rm{(Br)}}\]
Note: It is not to confuse the reaction of other hydrogen halides except HBr with alkene in presence of catalyst peroxide. Only, with HBr, the anti-Markovnikov product is obtained. With other hydrogen halides, Markovnikov product is obtained with peroxide.
Complete Step by Step Solution:
Let's first understand what the Markovnikov rule states. According to this rule, when an alkene undergoes a reaction with hydrogen halide, the double bond of the alkene breaks and the hydrogen atom forms a bond with that carbon atom of the double bind to which most numbers of atoms of hydrogen are bonded. And the halogen atoms get attached to another carbon atom of the double bond. For example, the reaction of propene and hydrogen chloride gives 2-iodo-propane.
\[{{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} + {\rm{HCl}} \to {{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH(Cl)}} - {\rm{C}}{{\rm{H}}_3}\]
In the same way, the given alkene undergoes a reaction with HX. Markovnikov's rule is to be followed here also.
\[{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} + {\rm{H}} - {\rm{X}} \to {{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{5}}}{\rm{CH(X)}} - {\rm{C}}{{\rm{H}}_3}\]
Therefore, option C is right.
Additional Information: Let's discuss what Markovnikov's rule is. This rule is the opposite of Markovnikv's rule. This rule applies to the reaction of an alkene with HBr in the medium of peroxide. According to this rule, the halogen atom forms a bond with that carbon atom of the double bond to which most numbers of hydrogen atoms are bonded, and the hydrogen atom forms a bond with another carbon atom of the double bond.
\[{{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{CH}} = {\rm{C}}{{\rm{H}}_{\rm{2}}} + {\rm{HBr}} \overset{Peroxide}{\rightarrow} {{\rm{H}}_{\rm{3}}}{\rm{C}} - {\rm{C}}{{\rm{H}}_2} - {\rm{C}}{{\rm{H}}_2}{\rm{(Br)}}\]
Note: It is not to confuse the reaction of other hydrogen halides except HBr with alkene in presence of catalyst peroxide. Only, with HBr, the anti-Markovnikov product is obtained. With other hydrogen halides, Markovnikov product is obtained with peroxide.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Main 2025 Session 2: Exam Date, Admit Card, Syllabus, & More

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
Types of Solutions

Degree of Dissociation and Its Formula With Solved Example for JEE

Classification of Drugs

Instantaneous Velocity - Formula based Examples for JEE

JEE Main Chemistry Question Paper with Answer Keys and Solutions

JEE Main Chemistry Online Mock Test for Class 12

Other Pages
Haloalkanes and Haloarenes

Alcohol Phenol and Ether Class 12 Notes: CBSE Chemistry Chapter 7

Coordination Compounds Class 12 Notes: CBSE Chemistry Chapter 5

NCERT Solutions for Class 12 Chemistry In Hindi Chapter 10 Haloalkanes and Haloarenes In Hindi Mediem

Total MBBS Seats in India 2025: Government College Seat Matrix

NEET Total Marks 2025: Important Information and Key Updates
