Mechanism: Electrons from the C1-C2 π bond attack the hydrogen of HBr, expelling the bromide anion and leading to the formation of a carbocation (Step 1, arrows A and B). Note here that the carbocation preferentially forms on C2 (secondary) and not C1 (primary) since secondary carbocations are more stable.

First, if an excess of HI (or HBr) is used in the cleavage reaction, the alcohol formed is converted by a nucleophilic substitution reaction to the appropriate alkyl halide: \[\ce{ROH + HI → RI + H2O} \nonumber\] In view of this substitution, some textbooks simplify the overall cleavage process as:

Addition due to excess HBr present. Most Hydrogen halide reactions with alkynes occur in a Markovnikov-manner in which the halide attaches to the most substituted carbon since it is the most positively polarized. A more substituted carbon has more bonds attached to 1) carbons or 2) electron-donating groups such as Fluorine and other halides.

First, if an excess of HI (or HBr) is used in the cleavage reaction, the alcohol formed is converted by a nucleophilic substitution reaction to the appropriate alkyl halide: \[\ce{ROH + HI → RI + H2O} \nonumber\]

Addition of HBr to alkenes in the presence of ROOR (peroxides) and light / heat (of course) results in "anti-Markovnikov" addition of HBr. But why? Read on!