1. Write the reaction

Trans-but-2-ene is: $\text{CH}_3-CH=CH-CH_3$ with the two $\text{CH}_3$ groups on opposite sides of the double bond.

On addition of bromine, electrophilic addition occurs across the double bond: $\text{CH}_3-CH=CH-CH_3 \xrightarrow{Br_2} \text{CH}_3-CHBr-CHBr-CH_3$ So the product is 2,3-dibromobutane.

2. Stereochemical nature of bromine addition

Addition of $Br_2$ to an alkene proceeds through a bromonium ion intermediate, followed by anti addition.

Thus, the two bromine atoms add on opposite faces of the double bond.

3. Apply this to trans-but-2-ene

For trans-but-2-ene, anti addition gives:

• one structure with configuration $(2R,3S)$
• and its mirror representation $(2S,3R)$

But these are actually the same meso compound, because 2,3-dibromobutane with opposite configurations at the two centers has an internal plane of symmetry.

So the product formed is meso-2,3-dibromobutane.

4. How many stereoisomers are formed?

Although attack can occur from either face of the alkene, both lead to the same meso product. Hence the two products are identical mesomers (older terminology: meso forms).

Therefore, the stereoisomers formed are: $\boxed{\text{2 identical mesomers}}$

5. Check options

• A: 2 enantiomers and 2 mesomers — incorrect
• B: 2 identical mesomers — correct
• C: 2 enantiomers — incorrect
• D: 1 racemic and 2 enantiomers — incorrect

Hence, the correct option is: $\boxed{B}$