# How are there only 8 structural isomers of C5H11Br?

Everywhere I see mentioned, that there are only 8 structural isomers of $\ce{C5H11Br}$. That is, the simple $\ce{C5}$ straight chain, $\ce{C4}$ with substituted methyl and $\ce{C3}$ with two substituted methyl groups, giving us a total of 8 isomers. But what about substituted ethyl groups? Or substituted bromo ethyl group like $\ce{C2H4Br}$? Am I missing something? Also is there a general formula to know the number of such structural isomers of haloalkanes?