91Ó°ÊÓ

We first look at the reaction of 1-butene with \( \mathrm{HBr} \). The alkene has a double bond between the first two carbons. According to the Markovnikov rule, the hydrogen atom (H) of \( \mathrm{HBr} \) will add to the carbon in the double bond of 1-butene, which has more hydrogen atoms already attached to it. Hence, Hydrogen attaches at the terminal carbon and Bromine attaches to the second carbon. Therefore, the resultant product, in this case, will be 2-bromobutane.

Now we move to the reaction of 2-butene with \( \mathrm{HBr} \). Here, the double bond is in between the second and third carbons of the butene chain. In spite of the location of the double bond, the Markovnikov Rule still applies. The Hydrogen atom (H) will add to the carbon that has more hydrogen atoms attached, and the Bromine (Br) will add to the other carbon. This results in the formation of 2-bromobutane again, same as the product of step 1. It's vital to note that in a case when a symmetric alkene like 2-butene reacts with \( \mathrm{HBr} \), it doesn't matter where the hydrogen ends up, because either carbon in the double bond has the same number of hydrogens to begin with, hence there is no 'rich-gets-richer' point of differentiation.