Final answer:
The molecule CH₃CHBrCH(OH)CH₃ has 4 possible stereoisomers as it contains two chiral centers and does not form meso compounds.
Step-by-step explanation:
The molecule CH₃CHBrCH(OH)CH₃ contains two chiral centers, which are the carbon atoms bonded to the Br and the OH groups, respectively. The maximum number of stereoisomers possible for a compound with n chiral centers can be calculated using the formula 2¹⁵, where n is the number of chiral centers. In this case, since there are 2 chiral centers, the maximum number of stereoisomers is 2² which equals 4. However, the actual number may be less due to potential meso compounds, which are achiral, but this specific molecule does not have any plane of symmetry and thus, does not form a meso compound. Therefore, the molecule CH₃CHBrCH(OH)CH₃ has 4 possible stereoisomers, including 2 pairs of enantiomers.