
In which of the following reactions, Markownikoff's rule is violated?
A) $C{{H}_{3}}-O-CH=C{{H}_{2}}\xrightarrow[CC{{l}_{4}}]{HBr}$
B) $C{{H}_{3}}-NH-CH=C{{H}_{2}}\xrightarrow[CC{{l}_{4}}]{HBr}$
C) $C{{H}_{3}}-S-CH=C{{H}_{2}}\xrightarrow[CC{{l}_{4}}]{HBr}$
D) ${{O}_{2}}N-CH=C{{H}_{2}}\xrightarrow[CC{{l}_{4}}]{HBr}$
Answer
510.6k+ views
Hint: We know that we should know that Markovnikov’s rule is an empirical rule which is used to predict regioselectivity of electrophilic addition reactions of alkenes and alkynes. Keeping the purpose in mind, we need to describe the mechanism of Markovnikov's rule.
Complete answer:
The Markovnikov’s rule states that with the addition of a protic acid HX to an asymmetric alkene, the acid hydrogen \[\left( H \right)\] gets attached to the carbon with more hydrogen substituents, and the halide \[\left( X \right)\] group gets attached to the carbon with more alkyl substituents. Let us explain Markovnikov's rule with the help of a simple example.
When a protic acid \[HC\text{ }\left( X=Cl,\text{ }Br,\text{ }I \right)\]is added to an asymmetrically substituted alkene, addition of acidic hydrogen takes place at the less substituted carbon atom of the double bond, while \[X\]is added to the more alkyl substituted carbon atom. In other words, hydrogen is added to the carbon atom with more number of hydrogen atoms attached to it and halide is added to the carbon with the least number of hydrogen atoms.
We should know that the driving force behind the reaction is the formation of carbocation in addition of H to the alkene, in the first step of the reaction. The most stable carbocation is formed when H is added to carbon having more number of hydrogen atoms already attached, due to factors like induction and hyperconjugation, which gives the major product with Br added to less hydrogen rich carbon.
Since an electron withdrawing group is present here, it doesn't follow Markovnikov's Rule.
${{O}_{2}}N-CH=C{{H}_{2}}\xrightarrow[CC{{l}_{4}}]{HBr}{{O}_{2}}N-C{{H}_{2}}-C{{H}_{2}}-Br$ , Since, \[-N{{O}_{2}}~\] group destabilizes the carbocation.
Therefore, the correct answer is option D.
Note: Remember that there are various practical applications of Markovnikov’s rule. Some of them are Halohydrin formation (in alcohol and water), Oxymercuration and Demercuration and Acid catalyzed hydration. These reactions are the laboratory usages of Markovnikov’s rule and are used in a variety of chemical processes.
Complete answer:
The Markovnikov’s rule states that with the addition of a protic acid HX to an asymmetric alkene, the acid hydrogen \[\left( H \right)\] gets attached to the carbon with more hydrogen substituents, and the halide \[\left( X \right)\] group gets attached to the carbon with more alkyl substituents. Let us explain Markovnikov's rule with the help of a simple example.
When a protic acid \[HC\text{ }\left( X=Cl,\text{ }Br,\text{ }I \right)\]is added to an asymmetrically substituted alkene, addition of acidic hydrogen takes place at the less substituted carbon atom of the double bond, while \[X\]is added to the more alkyl substituted carbon atom. In other words, hydrogen is added to the carbon atom with more number of hydrogen atoms attached to it and halide is added to the carbon with the least number of hydrogen atoms.
We should know that the driving force behind the reaction is the formation of carbocation in addition of H to the alkene, in the first step of the reaction. The most stable carbocation is formed when H is added to carbon having more number of hydrogen atoms already attached, due to factors like induction and hyperconjugation, which gives the major product with Br added to less hydrogen rich carbon.
Since an electron withdrawing group is present here, it doesn't follow Markovnikov's Rule.
${{O}_{2}}N-CH=C{{H}_{2}}\xrightarrow[CC{{l}_{4}}]{HBr}{{O}_{2}}N-C{{H}_{2}}-C{{H}_{2}}-Br$ , Since, \[-N{{O}_{2}}~\] group destabilizes the carbocation.
Therefore, the correct answer is option D.
Note: Remember that there are various practical applications of Markovnikov’s rule. Some of them are Halohydrin formation (in alcohol and water), Oxymercuration and Demercuration and Acid catalyzed hydration. These reactions are the laboratory usages of Markovnikov’s rule and are used in a variety of chemical processes.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

How much time does it take to bleed after eating p class 12 biology CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

