
Markovnikov's rule provides guidance of addition of \[HBr\] on
(a) \[C{H_2} = C{H_2}\]
(b) \[C{H_3} - C{H_2} - C{H_3}\]
(c) \[C{H_3} - CH = CH - C{H_3}\]
(d) \[C{H_2} = CHBr\]
Answer
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Hint: Markovnikov's was used to define the addition of hydrogen halide (such as\[HBr\]) molecules over the unsymmetrical carbon-carbon double bond. This rule was formulated by Russian chemist Vladimir Markovnikov in the year of 1870.
Complete Step by Step Answer:
Markovnikov’s rule states that, when a hydrogen halide (\[HX\]) reacts with the carbon-carbon double bond of an unsymmetrical alkene. The hydrogen atom of hydrogen halide (\[HX\]) attacks the end of the carbon-carbon double bond that has the highest number of the hydrogen atom. Whereas the halide part prefers to bind the end of the carbon-carbon double bond which has the least number of hydrogen atoms.
Markovnikov’s rule can be best understood by taking the example of the addition of \[HBr\]on a propene molecule (an unsymmetrical alkene) as a result formation of 2-bromopropane (major product) and 1-bromopropane occurs (minor product).
\[C{H_3} - CH = C{H_2} + HBr \to C{H_3} - CHBr - C{H_3}\] (2-bromopropane)
In the above reaction the addition of \[HBr\] over propene takes place via the addition of hydrogen to the end of the carbon-carbon double bond that possesses a maximum number of the hydrogen atoms. Whereas the bromine atom goes to that end of the carbon-carbon double bond which contains a smaller number of the hydrogen atom.
Similarly, the given question can also be described. As we know, Markovnikov’s rule is applicable only to the addition of \[HBr\]on an unsymmetrical alkene. Therefore, we can easily predict that options (a), (b) and (c) are the incorrect answers. Because these options contain symmetrical alkene and alkane molecules.
Therefore from the above explanation we can say that option (d) will be the correct option because it follows all the conditions which are required for Markovnikov’s rule:
Note: Anti-Markovnikov’s rule: When the addition of \[HBr\]to the unsymmetrical alkene takes place in the presence of peroxide then this reaction is called as anti-Markovnikov rule or peroxide effect or sometimes it is also known as the Kharasch effect. In such cases, the major product the derived from less stable carbocation.
Complete Step by Step Answer:
Markovnikov’s rule states that, when a hydrogen halide (\[HX\]) reacts with the carbon-carbon double bond of an unsymmetrical alkene. The hydrogen atom of hydrogen halide (\[HX\]) attacks the end of the carbon-carbon double bond that has the highest number of the hydrogen atom. Whereas the halide part prefers to bind the end of the carbon-carbon double bond which has the least number of hydrogen atoms.
Markovnikov’s rule can be best understood by taking the example of the addition of \[HBr\]on a propene molecule (an unsymmetrical alkene) as a result formation of 2-bromopropane (major product) and 1-bromopropane occurs (minor product).
\[C{H_3} - CH = C{H_2} + HBr \to C{H_3} - CHBr - C{H_3}\] (2-bromopropane)
In the above reaction the addition of \[HBr\] over propene takes place via the addition of hydrogen to the end of the carbon-carbon double bond that possesses a maximum number of the hydrogen atoms. Whereas the bromine atom goes to that end of the carbon-carbon double bond which contains a smaller number of the hydrogen atom.
Similarly, the given question can also be described. As we know, Markovnikov’s rule is applicable only to the addition of \[HBr\]on an unsymmetrical alkene. Therefore, we can easily predict that options (a), (b) and (c) are the incorrect answers. Because these options contain symmetrical alkene and alkane molecules.
Therefore from the above explanation we can say that option (d) will be the correct option because it follows all the conditions which are required for Markovnikov’s rule:
Note: Anti-Markovnikov’s rule: When the addition of \[HBr\]to the unsymmetrical alkene takes place in the presence of peroxide then this reaction is called as anti-Markovnikov rule or peroxide effect or sometimes it is also known as the Kharasch effect. In such cases, the major product the derived from less stable carbocation.
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