**Total number of stereoisomers of compound CH3-CH(OH)-CH(Br)-CH3**

To determine the total number of stereoisomers of the compound CH3-CH(OH)-CH(Br)-CH3, we need to consider the presence of chiral centers and the possibility of different stereoisomers.

1. Identify the chiral centers: Chiral centers are carbon atoms that are bonded to four different groups. In this compound, the carbon atom bonded to the hydroxyl group (OH) and the carbon atom bonded to the bromine atom (Br) are the chiral centers.

2. Determine the number of stereoisomers for each chiral center: Each chiral center can have two possible stereoisomers, known as enantiomers. Enantiomers are mirror images of each other and cannot be superimposed.

3. Calculate the total number of stereoisomers: To calculate the total number of stereoisomers, we multiply the number of stereoisomers for each chiral center. In this case, there are two chiral centers, so we multiply the number of stereoisomers for each chiral center (2) by itself.

Total number of stereoisomers = 2 * 2 = 4

Therefore, the compound CH3-CH(OH)-CH(Br)-CH3 has a total of 4 stereoisomers.

It is important to note that the compound does not have any meso compounds. Meso compounds are stereoisomers that have chiral centers but are superimposable on their mirror images, thus not exhibiting optical activity.

In conclusion, the compound CH3-CH(OH)-CH(Br)-CH3 has a total of 4 stereoisomers.

There will be total of 4 stereoisomers possible for this compound as it has 2 stereocentre(here chiral centre) and according to the formula , no. of stereoisomers for asymmetrical molecule =2^n ( where , n = no. of stereocentre)