The stereoisomers that are formed by electrophilic addition of bromine to trans-but-2-ene is/are:

We have trans-but-2-ene, that is $$\mathrm{CH_3\!-\!CH=CH\!-\!CH_3}$$ with the two $$\mathrm{CH_3}$$ groups on opposite sides of the double bond. During the electrophilic addition of bromine, the reaction always proceeds through a cyclic bromonium ion. The first step is

$$$\mathrm{CH_3\!-\!CH=CH\!-\!CH_3 + Br_2 \;\longrightarrow\;} \mathrm{CH_3\!-\!CH\!\!\overset{+}{\!\!\!\!Br}\!-\!CH\!-\!CH_3} \;+\;Br^-$$$

Now, bromide ion $$Br^-$$ opens the three-membered ring from the side opposite to the $$\overset{+}{Br}$$ (anti-addition). So the two bromine atoms always enter from opposite faces of the original double bond plane.

To see the stereochemical result clearly we assign the substituents around each sp3 carbon after addition. Let us choose the configuration in which on carbon-2 ($$C_2$$) the incoming nucleophilic $$Br^-$$ attacks from the front while on carbon-3 ($$C_3$$) it attacks from the rear. We can write the skeletal formula obtained:

$$$\mathrm{CH_3\,(front\;Br)\,CH\!-\!CH\,(back\;Br)\,CH_3}$$$

The molecule so obtained possesses two chiral centres, $$C_2$$ and $$C_3$$. We assign priorities by the C.I.P. rules (stated briefly: atom with higher atomic number gets higher priority). On each centre

Priority order: $$$Br(1) > CH_2CH_3(2) > CH_3(3) > H(4).$$$

Because the bromines are on opposite faces, one centre becomes $$R$$ while the other automatically becomes $$S$$. Explicitly, if $$C_2$$ is $$R$$, $$C_3$$ is $$S$$; the mirror image would have $$C_2$$ as $$S$$ and $$C_3$$ as $$R$$. However these two descriptions represent the same molecule: it has an internal mirror plane passing through the middle of the C-C bond and bisecting the molecule. Hence the compound is superimposable on its mirror image and is optically inactive. Such a stereoisomer is called a meso form.

If, in the bromonium opening step, we imagine the attack sequence reversed—i.e. bromide coming from the rear on $$C_2$$ and from the front on $$C_3$$—we draw

$$$\mathrm{CH_3\,(back\;Br)\,CH\!-\!CH\,(front\;Br)\,CH_3}$$$

Again, priority assignment shows $$C_2$$ to be $$S$$ and $$C_3$$ to be $$R$$, producing exactly the same $$RS$$ (meso) configuration as above. Thus the two routes give two structures that are identical, not different stereoisomers. No $$RR$$ or $$SS$$ combination can arise because anti-addition on a trans alkene never places the two bromines on the same face, a prerequisite for those diastereomeric pairs.

Therefore, electrophilic anti-addition of $$Br_2$$ to trans-but-2-ene furnishes only one unique stereoisomer, viz. the meso-2,3-dibromobutane. The phrase “2 identical mesomers” in the option list simply recognises that the reaction pathway can be visualised in two ways but both lead to the same meso product.

Hence, the correct answer is Option A.