[psychoNOT]

Working on this problem.. it says to write as many isomers of bromoalkanes with formula C4H9Br as possible.

I have

H₃C-CH₂-CH₂-CH₂-Br

H₃C-CH₂-CH₂(Br)*-CH₃ *the Br is attached to the 2nd to last C, not the last

H₃C-CH(CH₃)*-CH₃-Br *same idea

H₃C-CH(CH₂-Br)-CH₃

H₃C-C(CH₃)(Br)-CH₃

A little hard to follow but I can't really draw it on here..
The book answer has the first 3 and the 5th but not the 4th.  Why is the 4th wrong?  From my reading the only thing I know is that haloalkanes just need to be a halogen attached to a carbon which has 3 other R-groups... and there are no double bonds anywhere so they're all alkanes and they all have the right formula (I think.. checked like 3 times).  What am I doing wrong?

[Oxygen]

Working on this problem.. it says to write as many isomers of bromoalkanes with formula C4H9Br as possible.

I have

H₃C-CH₂-CH₂-CH₂-Br

H₃C-CH₂-CH₂(Br)*-CH₃ *the Br is attached to the 2nd to last C, not the last

H₃C-CH(CH₃)*-CH₃-Br *same idea

H₃C-CH(CH₂-Br)-CH₃

H₃C-C(CH₃)(Br)-CH₃

A little hard to follow but I can't really draw it on here..
The book answer has the first 3 and the 5th but not the 4th.  Why is the 4th wrong?  From my reading the only thing I know is that haloalkanes just need to be a halogen attached to a carbon which has 3 other R-groups... and there are no double bonds anywhere so they're all alkanes and they all have the right formula (I think.. checked like 3 times).  What am I doing wrong?

When I drew out the structures you had posted it's clear that two of your structures are the same.  The second and the fourth are the same structure although you have them draw oppositely from each other, that doesn't constitute a different molecule. 

[psychoNOT]

Ah.  When I drew them out not in condensed structure it was easier to see... but I believe you mean that the third and 4th structures are the same... just hard to understand exactly what I meant the way I wrote them.  I'll edit the post to make it easier to understand.

Here's what was wrong:

H3C-CH(CH3)*-CH2-Br *same idea: this molecule has a methyl on the second carbon.  I wrote the formula wrong in the first post... it should be CH2, not CH3 as that would give 10 H.

H3C-CH(CH2-Br)-CH3: this one meant on the second carbon from the left was attached CH2-Br as one R-group and a methyl group as the other... if this molecule is straightened out it is evident that it is the same as the previous one.

Also.. why can I edit this post but not the first one?

[Oxygen]

lol yes, that's what I meant....I do ochem not math!