Question

How to draw all stereoisomers of 2-bromo-4-methylpentane?

Answer 1

Well, you got the one chiral centre...

Explanation:

For `H_3C-stackrel("*")CHBrCH_2CH(CH_3)_2` the starred carbon is the chiral centre...if you draw this out in the familiar wedge, dash fashion, the interchange of either 2 substituents about the chiral centre gives the enantiomer. Interchange again, and you get the enantiomer of an enantiomer, i.e. the original compound.

See here for background. And please remember that to demonstrate chirality properly you really should have a model in front of you, one that you can play with and consider...