I don't understand why on the problem a. B and C are enantiomers, while on the problem b. B is identical to the last molecule.
$\begingroup$ In the 1,3-disubstituted cyclohexanes, A can jump over the mirror and superimpose on its mirror image. It has a plane of symmetry. B and C cannot superimpose on one another. No symmetry plane in B or C. $\endgroup$– user55119May 12, 2021 at 23:17
Ask yourself this: in problem $a$, are B and C superimposable on each other? You'll find that that is not the case. So those two are enantiomers. In problem $b$, B and its mirror image are perfectly superimposable on each other, hence identical. This also shows that in problem $a$, B is am optically active molecule, while in problem $b$ it is not.
$\begingroup$ How I understand that B and C (in problem a) are not superimposable? $\endgroup$– Silvia95May 12, 2021 at 14:30
1$\begingroup$ If you try to do it, you would see that turning the plane of the mirror image(C) would cause the wedges and dashes to reverse, hence its impossible for it to superimpose with B. $\endgroup$ May 12, 2021 at 14:38