Hello,

42 can be tricky.

For this question, the challenge is that the molecule is flipped over in the answer in addition to the displacement taking place. Keep in mind that an Sn2 is an inversion of the stereocenter, and when you flip over a molecule, you change hash bonds to wedge bonds and vice versa.

Although D looks like there has a been retention of steroechemistry overall, if you flip the molecule over to match the question stem, that hash bond will turn to a wedge (because if you flip something over that has parts pointing away from you, they will come towards you again)

For 44, you have to rotate the ends of the Fischer projection. For the rotation, if I take the top stereocentre, I can rotate the Br to where the CH3 is, the CH3 to the H, and the H to the Br.

This results in the rotated B as

____H
____|
Br--C--CH3
____|
Br--C--CH3
____|
____H

If you look at this molecule, it has a mirror plane through it as follows:


____H
____|
Br--C--CH3
____|
xxxxxxxxxxx
____|
Br--C--CH3
____|
H

Making B meso.

Keep in mind that you can only rotate the ends of a Fischer projection, not the middle bits. However, here there is no middle bit so no worries here.