Question

Which one of the following statements about the Markonikov rule is not true?
A. In Markovnikov additions, the alkene ${\text{C}}$-atom with the most ${\text{H}}$ atoms will be protonated.
B. When ${\text{HX}}$ adds to an unsymmetrical alkene, the halogen attaches to the least substituted alkene ${\text{C}}$-atom.
C. When ${\text{HX}}$ adds to an unsymmetrical alkene, the halogen attaches to the most substituted alkene ${\text{C}}$-atom.
D. When ${\text{HX}}$ adds to an unsymmetrical alkene, the protonation leads to the most stable carbocation intermediate.

Answer 1

Hint: The Markovnikov rule states that when an asymmetrical alkene reacts with an asymmetrical reagent, the negative part of the reagent gets attached to that carbon atom which has less number of hydrogen atoms.

Complete solution:
We know that the Markovnikov rule states that when an asymmetrical alkene reacts with an asymmetrical reagent, the negative part of the reagent gets attached to that carbon atom which has less number of hydrogen atoms.
Example of Markovnikov addition is as follows:
The reaction of but-1-ene with hydrogen bromide will give two products. The product obtained by Markonikov rule will be the major product while the other is the minor product.
Thus, the reaction of but-1-ene with hydrogen bromide is as follows:
${\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {\text{CH}} = {\text{C}}{{\text{H}}_2} + {\text{HBr}}\xrightarrow{{{\text{ROOR (peroxide)}}}}{{\text{C}}_2}{{\text{H}}_5} - {\text{CH(Br)}} - {\text{C}}{{\text{H}}_3}{\text{ (Major)}} + {{\text{C}}_2}{{\text{H}}_5} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2}{\text{Br (Minor)}}$
The major product obtained by the Markovnikov addition will be ${{\text{C}}_2}{{\text{H}}_5} - {\text{CH(Br)}} - {\text{C}}{{\text{H}}_3}$ and the minor product obtained will be ${{\text{C}}_2}{{\text{H}}_5} - {\text{C}}{{\text{H}}_2} - {\text{C}}{{\text{H}}_2}{\text{Br}}$.
Consider the protonation of but-1-ene:
\[{\text{C}}{{\text{H}}_3} - {\text{C}}{{\text{H}}_2} - {\text{CH}} = {\text{C}}{{\text{H}}_2} + {{\text{H}}^ + } \to {{\text{C}}_2}{{\text{H}}_5} - {}^ + {\text{CH}} - {\text{C}}{{\text{H}}_3}{\text{ (Major)}} + {{\text{C}}_2}{{\text{H}}_5} - {\text{C}}{{\text{H}}_2} - {}^ + {\text{C}}{{\text{H}}_2}{\text{ (Minor)}}\]
The major product obtained by the protonation will be \[{{\text{C}}_2}{{\text{H}}_5} - {}^ + {\text{CH}} - {\text{C}}{{\text{H}}_3}\] (secondary carbocation) and the minor product obtained will be \[{{\text{C}}_2}{{\text{H}}_5} - {\text{C}}{{\text{H}}_2} - {}^ + {\text{C}}{{\text{H}}_2}\] (primary carbocation). The secondary carbocation is more stable.

> From the reaction, we can conclude that when ${\text{HX}}$ adds to an unsymmetrical alkene, the alkene ${\text{C}}$-atom with the most ${\text{H}}$ atoms will be protonated, the halogen attaches to the least substituted alkene ${\text{C}}$-atom and the protonation leads to the most stable carbocation intermediate.
Thus, the statement ‘when ${\text{HX}}$ adds to an unsymmetrical alkene, the halogen attaches to the most substituted alkene ${\text{C}}$-atom’ is not true.

Thus, the correct option is, (C) when ${\text{HX}}$ adds to an unsymmetrical alkene, the halogen attaches to the most substituted alkene ${\text{C}}$-atom.

Note: When alkenes react with hydrogen bromide in the presence of peroxide occurs contrary to Markonikov rule. This is known as peroxide effect. The products obtained by anti-Markovnikov addition are reverse of these.