## Organic Chemistry (8th Edition)

The reaction follows the following steps: 1) Protonation of the pi bond forms a carbocation at the most substituted carbon. So the arrow originating from the double bond points towards H. Hydrogen attaches to the least substituted carbon. Because of this attack, Bromine leaves with the electrons involved in the bonding. (In this molecule both the carbons involved in double bond are equally substituted so there is no preference). 2) Then the halide attacks on the carbocation. So here $Br^-$ attacks on the carbocation to give the Markovnikov products, 1-bromo-3-methylcyclohexane and 1-bromo-4-methylcyclohexane.