ISSUE:

Hello mates.
Carrying out some test-type questions, the following question QIR appears to me: What product/s are obtained by adding Br2/CCl4 to (E)-2-butene?: 1. 2,3-dibromobutane (racemic mixture).

2. Meso-2,3-dibromobutane.

3. (E)-1-bromo-2-butene.

4. Mixtures of 2,3-dibromubutanes (meso and pair of enantiomers).

I don't understand why the correct answer is 2, if when dealing with an E alkene the alkyl groups in position 2 and 3 would be oriented in the opposite way and when removing the configuration of the chiral carbons it would give 2S 3S.
Isn't the meso form supposed to be 2R3S? In the attached pdf I show how I have taken the configurations in case there are any errors... Thank you. All the best.

Hello, indeed it gives the meso form.
As you well say, the methyls are oriented to opposite sides, but the bromines also enter anti. If you turn that molecule that has everything on opposite sides, you will see that all the groups end up on the same side and that is why it is a meso form. All the best!!