In the given reaction, the alkene reacts with excess HBr, leading to an electrophilic addition.
The double bond in \( \text{CH}_2 = \text{CH} - \text{CH}_2 - \text{OH} \) breaks and the hydrogen from HBr adds to the most substituted carbon (according to Markovnikov's rule).
The bromine atom (Br⁻) adds to the other carbon.
Since there is excess HBr, both carbons of the double bond will be attached to bromine atoms, leading to the major product being a dibromide.