Explain the reaction of propene with $HBr$.

(N/A) The reaction of propene $(CH_3CH=CH_2)$ with $HBr$ is an electrophilic addition reaction that follows Markovnikov's rule.
$1$. Markovnikov's rule states that when an unsymmetrical reagent adds to an unsymmetrical alkene,the negative part of the addendum attaches to the carbon atom of the double bond that has the lesser number of hydrogen atoms.
$2$. In the reaction of propene with $HBr$,the $Br^-$ ion (negative part) attaches to the central carbon atom (which has one $H$ atom),while the $H^+$ ion attaches to the terminal carbon atom (which has two $H$ atoms).
$3$. This leads to the formation of $2$-bromopropane as the major product $(90\%)$ and $1$-bromopropane as the minor product $(10\%)$.
The reaction is: $CH_3CH=CH_2 + HBr \rightarrow CH_3CH(Br)CH_3$ ($2$-bromopropane).