
Which of these does not follow anti-markovnikov's rule:
A. 2-butene
B. 1-butene
C. 2-pentene
D. 2-hexene
Answer
163.2k+ views
Hint: There are two ways by which addition of halogen acid takes place:- Markovnikov rule and anti-markovnikov's rule. According to the Markovnikov rule, the negative part of the halogen is added to the double bonded carbon atom having least number of hydrogen atoms whereas anti-markovnikov rule states that the negative part of the halogen acid is added to the double bonded carbon atom having more number of hydrogen atoms.
Complete Step by Step Answer:
Both markovnikov rule and anti-markovnikov are valid only on unsymmetrical alkenes. The example of markovnikov rule is as follows:-
$CH_{3}-CH=CH_{2}{+}HBr\rightarrow CH_{3}-CHBr-CH_{3}$
In the above reaction, the halogen acid is HBr ($H^{+}$ and $Br^{-}$) the negative part $Br^{-}$ is added to the carbon having the least amount of hydrogen. In the above example, propene was used which is an unsymmetrical alkene.
Now an example of addition of HBr to an unsymmetrical alkene using anti-markovnikov is as follows:-
$CH_{3}-CH=CH_{2}{+}HBr\rightarrow CH_{3}-CH_{2}-CH_{2}Br$
In the above reaction propene, an unsymmetrical alkene is used in which the $Br^{-}$ get attached with the carbon atom having more number of hydrogen atoms. While applying the above rules, we have to make sure that the negative part is added to that carbon which is double bonded as well as having the least and more number of hydrogen atoms respectively. Now in the above question 2- butene is a symmetrical alkene that is it has equal number of carbon atoms on the both the sides of the double bond due which neither Markovnikov nor anti-markovnikov rule can be applied.
Thus, the correct answer is option A:- 2-butene.
Note: In contrast to the Markovnikov addition, the Anti-Markovnikov addition proceeds via a free radical, giving rise to different products. It's vital to remember that the Kharasch effect only appears in combination with HBr.Because it will start the reaction, peroxide is a crucial ingredient in this reaction. Because it is an unstable chemical, peroxide readily forms radicals.
Complete Step by Step Answer:
Both markovnikov rule and anti-markovnikov are valid only on unsymmetrical alkenes. The example of markovnikov rule is as follows:-
$CH_{3}-CH=CH_{2}{+}HBr\rightarrow CH_{3}-CHBr-CH_{3}$
In the above reaction, the halogen acid is HBr ($H^{+}$ and $Br^{-}$) the negative part $Br^{-}$ is added to the carbon having the least amount of hydrogen. In the above example, propene was used which is an unsymmetrical alkene.
Now an example of addition of HBr to an unsymmetrical alkene using anti-markovnikov is as follows:-
$CH_{3}-CH=CH_{2}{+}HBr\rightarrow CH_{3}-CH_{2}-CH_{2}Br$
In the above reaction propene, an unsymmetrical alkene is used in which the $Br^{-}$ get attached with the carbon atom having more number of hydrogen atoms. While applying the above rules, we have to make sure that the negative part is added to that carbon which is double bonded as well as having the least and more number of hydrogen atoms respectively. Now in the above question 2- butene is a symmetrical alkene that is it has equal number of carbon atoms on the both the sides of the double bond due which neither Markovnikov nor anti-markovnikov rule can be applied.
Thus, the correct answer is option A:- 2-butene.
Note: In contrast to the Markovnikov addition, the Anti-Markovnikov addition proceeds via a free radical, giving rise to different products. It's vital to remember that the Kharasch effect only appears in combination with HBr.Because it will start the reaction, peroxide is a crucial ingredient in this reaction. Because it is an unstable chemical, peroxide readily forms radicals.
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