Question: How can we apply Markovnikov’s rule in \[C{{H}_{3}}C{{H}_{2}}CH=C{{H}_{2}}+HBr\to ~\] to determine w...

Question

How can we apply Markovnikov’s rule in $C{{H}_{3}}C{{H}_{2}}CH=C{{H}_{2}}+HBr\to ~$ to determine which carbon the hydrogen bonds with and which carbon the bromine atom bonds with?

Answer 1

Solution

Markovnikov’s rule is used to predict the selectivity of electrophilic addition reactions of alkenes and alkynes. Hydrogen halides include compounds like etc.

Complete step by step solution:
Electrophilic addition reactions include the addition of an electrophile or a positive charged species to an organic compound like alkanes or alkenes. The Markovnikov’s rule can be stated as:
When an unsymmetrical alkene reacts with a hydrogen halide to give an alkyl halide, the hydrogen adds to the carbon of the alkene that has the greater number of hydrogen atoms. The halogen part of the hydrogen halide goes to the carbon atom having fewer number of hydrogen atoms in the alkene. This rule can be depicted with a simple example as follows;
$C{{H}_{3}}C{{H}_{2}}CH=C{{H}_{2}}+HBr\to ~C{{H}_{3}}CH(Br)C{{H}_{3}}$
We know that in the above alkene, the carbon atom on the right side of the double bond has a greater number of hydrogen atoms than the other. So, by applying the Markovnikov’s rule we can see that the hydrogen atom in the Hydrogen Bromide goes to the carbon atom on the right side of the double bond and the bromide ion to the carbon atom on the left side of the double bond.
This will be the major product of the reaction and the minor product will be that when the hydrogen atom goes to the carbon atom with a smaller number of hydrogen atoms and the product will be; the reason is that the ${{H}^{+}}$ from the $HBr$ adds to the carbon that will give the more stable carbocation.
If the ${{H}^{+}}$ adds to $C-1$ , the intermediate is the more stable $2{}^\circ$ carbocation.
If the ${{H}^{+}}$ adds to $C-2$ , the intermediate is the less stable $1{}^\circ$ carbocation

Note: Hydrogen halides undergo heterolytic cleavage before attacking the compound. It becomes ${{H}^{+}}$ and ${{X}^{-}}$ then attacks the compound in the reaction mixture. The more positive part of the reagent adds to the less substituted carbon atom of the alkene.