Markovnikov's rule states that *in the addition of a protic acid HX or other polar reagent to an asymmetric alkene, the acid hydrogen (H) or electropositive part gets attached to the carbon with more hydrogen substituents, and the halide (X) group or electronegative part gets attached to the carbon with more alkyl substituents* (wikipedia).
This rule implies that the hydrogen of HBr will be attached to C-1 and the carbocation will be on C-2. Remember that the order of stability of carbocations is tertiary > secondary > primary > methyl. A hydride shift can yield a tertiary carbocation.
C-3 is a tertiary carbon atom. If the hydride on carbon 3 shifts to carbon 2, a tertiary and more stable carbocation is formed. This accounts for the major product in the reaction.
When acids are mixed with water the following reaction takes place:
H^+(aq) + H2O(l) ------> H3O^+(aq)
Hence when acids are added to water, acids donate a proton to water to form the oxonium ion H3O^+ by coordinate covalent bonding. Note that acids contain the hydrogen ion H^+
Electrons are found in shells or orbitals that surround the nucleus of an atom. Protons and neutrons are found in the nucleus. They group together in the center of the atom.
