How to draw 4 isomers of c4h9br

Butanediol isomers?

CH₂OH-CHOH-CH₂-CH₃ buta-1,2-diol
CH₂OH-CH₂-CHOH-CH₃ buta-1,3-diol
CH₂OH-CH₂-CH₂-CH₂OH buta-1,4-diol
CH₃-CHOH-CHOH-CH₃ buta-2,3-diol

These are the four positionally isomeric butadiols. There is nothing more - if you tried a 2.4 position, for example, then 1.3 would come out again.

But what you can do is frameworkisomerism: You can record 2-methylpropa-1,2-diol and 2-methylpropa-1,3-diol, these are really new molecules and they are isomeric to the butadiols, but they are not butadiols themselves (because they are different be called). I suppose you meant these two even if you did penta instead of propa wrote.

On the other hand, the 2-methylpropa-1,1-diol that you also proposed is not a stable molecule because it immediately splits off water and converts into 2-methylpropanal:

CH₃-CH (CH₃) -CH (OH) ₂ ⟶ CH₃-CH (CH₃) -CO-H + H₂O

This is what all alcohols with two OH groups on the same carbon atom do, except when there is a special situation (CCl₃-CH (OH) ₂, for example, is stable).

There are also others Functionalisomers of the formula C₄H₁₀O₂, namely those in which one or two O atoms do not occur in OH groups but in ether bridges. In this case, two O atoms are allowed to hang on the same C.