Question

How many stereoisomers of 2-chloro-3-methylbutane exist?

Answer 1

Four.

Simply put, there are two configurations per stereocenter on a compound with all `sp^3` carbons; R or S. So, we rack up stereoisomers according to `2*2*2*[...]*2 = 2^n` where `n` is the number of stereocenters.

Also, although meso isomers could reduce the number of stereoisomers, there are none here because clearly, `"Cl" ne "CH"_3`, i.e. there is no plane of symmetry. Therefore, we can ignore meso isomers.

As a result, there are `2^2 = 4` stereoisomers.

And if you worked it out, you can see that you only have permutations of (R, R), (R, S), (S, R), and (S, S).