C)H2C = CH- CH3 + HBr -----> CH3 - CH - CH3 | Br

The reaction you've provided is an example of an electrophilic addition reaction, which is typical for alkenes. C)H2C = CH- CH3 (propene) reacts with HBr (hydrogen bromide) to form CH3-CH-CH3 | Br (2-bromopropane). The reaction goes through the following steps:

1. The pi bond (double bond) electrons of the alkene (C)H2C=CH-CH3) attack the hydrogen atom of HBr. This is because the double bond has a high electron density and acts as a nucleophile (electron pair donor), while the hydrogen in HBr is partially positive and acts as an electrophile (electron pair acceptor).

2. The bond between hydrogen and bromine in HBr breaks, resulting in the formation of a carbocation intermediate (positively charged carbon) and a bromide anion (Br-). The bromine atom leaves with an electron pair, becoming negatively charged.

3. The positively charged carbocation intermediate is stabilized through the inductive effect, where the methyl group (CH3-) can donate electron density toward the positive charge. This intermediate is preferably formed at the more substituted carbon atom (secondary carbocation), following Markovnikov's rule.