Answer
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Hint: The Anti-Markovnikov’s addition of $HBr$ is only observed in unsymmetrical alkenes because the unsymmetrical alkenes have a different number of substituents around the double bond.
Complete step by step solution:
The Anti-Markovnikov’s addition is also called the peroxide effect or the Kharasch effect. The Anti-Markovnikov’s rule states that, in the presence of peroxide like benzoyl peroxide (${{C}_{6}}{{H}_{5}}CO-O-O-CO{{C}_{6}}{{H}_{5}}$), the addition of $HBr$ to unsymmetrical alkenes takes place contrary to the Markovnikov’s rule.
In Markovnikov’s rule, the negative part of the addendum gets attached to the carbon atom in the double bond which has a lesser number of a hydrogen atom or more substituted carbon atom. So, in Anti- Markovnikov’s rule, the negative part of the addendum gets attached to the carbon atom in the double bond which has more number of hydrogen atoms.
(A)- Propene
The formula of Propene is $C{{H}_{2}}=CH-C{{H}_{3}}$. This is an unsymmetrical alkene. So according to the Anti- Markovnikov’s rule, the negative part will attach the carbon atom with 2 hydrogen atoms.
(B)- But-1-ene
The formula of But-1-ene is $C{{H}_{2}}=CH-C{{H}_{2}}-C{{H}_{3}}$. This is an unsymmetrical alkene. So according to the Anti- Markovnikov’s rule, the negative part will attach the carbon atom with 2 hydrogen atoms.
(C)- 2-Butene
The structure of 2-Butene is $C{{H}_{3}}-CH=CH-C{{H}_{3}}$. This is a symmetrical alkene and the Anti- Markovnikov’s rule will not be observed in 2-Butene.
(D)- 2-Pentene
The structure of 2-Pentene is $C{{H}_{3}}-CH=CH-C{{H}_{2}}-C{{H}_{3}}$. This is an unsymmetrical alkene. So according to the Anti- Markovnikov’s rule, the negative part will attach the carbon atom with the methyl group because it is less substituted.
Hence, the correct answer is an option (c)- 2-Butene.
Note: If a reaction is given and we can differentiate whether the reaction will follow Markovnikov’s rule or Anti- Markovnikov’s rule by the presence of peroxide in the catalyst. If peroxide is present in the catalyst then the reaction will follow Anti- Markovnikov’s rule.
Complete step by step solution:
The Anti-Markovnikov’s addition is also called the peroxide effect or the Kharasch effect. The Anti-Markovnikov’s rule states that, in the presence of peroxide like benzoyl peroxide (${{C}_{6}}{{H}_{5}}CO-O-O-CO{{C}_{6}}{{H}_{5}}$), the addition of $HBr$ to unsymmetrical alkenes takes place contrary to the Markovnikov’s rule.
In Markovnikov’s rule, the negative part of the addendum gets attached to the carbon atom in the double bond which has a lesser number of a hydrogen atom or more substituted carbon atom. So, in Anti- Markovnikov’s rule, the negative part of the addendum gets attached to the carbon atom in the double bond which has more number of hydrogen atoms.
(A)- Propene
The formula of Propene is $C{{H}_{2}}=CH-C{{H}_{3}}$. This is an unsymmetrical alkene. So according to the Anti- Markovnikov’s rule, the negative part will attach the carbon atom with 2 hydrogen atoms.
(B)- But-1-ene
The formula of But-1-ene is $C{{H}_{2}}=CH-C{{H}_{2}}-C{{H}_{3}}$. This is an unsymmetrical alkene. So according to the Anti- Markovnikov’s rule, the negative part will attach the carbon atom with 2 hydrogen atoms.
(C)- 2-Butene
The structure of 2-Butene is $C{{H}_{3}}-CH=CH-C{{H}_{3}}$. This is a symmetrical alkene and the Anti- Markovnikov’s rule will not be observed in 2-Butene.
(D)- 2-Pentene
The structure of 2-Pentene is $C{{H}_{3}}-CH=CH-C{{H}_{2}}-C{{H}_{3}}$. This is an unsymmetrical alkene. So according to the Anti- Markovnikov’s rule, the negative part will attach the carbon atom with the methyl group because it is less substituted.
Hence, the correct answer is an option (c)- 2-Butene.
Note: If a reaction is given and we can differentiate whether the reaction will follow Markovnikov’s rule or Anti- Markovnikov’s rule by the presence of peroxide in the catalyst. If peroxide is present in the catalyst then the reaction will follow Anti- Markovnikov’s rule.
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