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?
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You'll probably need to draw all the possible isomers to see it, but with only 5 carbons to work with the ethyl isomers are redundant with the methyl isomers: i.e. 1-bromo-2-ethyl-propane is the same thing as 1-bromo-2-methyl-butane.
Remember we are only talking about structural isomers, and ignoring stereoisomers.
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$\begingroup$ Okay, I see now. I repeated the same structure in different ways. Is there any short way to determine the number of non-redundant structures? $\endgroup$ – Arka Jul 18 '18 at 7:06
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$\begingroup$ There's probably some complicated group theory that can, but in general, figuring out the number is too open ended. $\endgroup$ – Zhe Jul 18 '18 at 15:16
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$\begingroup$ Draw n-pentane. 3 places to put Br. Take n-pentane. Clip of a methyl and add it to the remaining n-butane. Only one answer: 2-methyl butane. Add Br to it. 4 monobromo cmpds. Take n-pentane. Clip off 2 methyls and add them to the remaining propane. Only one way, 2,2-dimethylpropane and only one mono bromo cmpd. Does it come out to 8 structural (constitutional isomers? I'm doing this in my head. $\endgroup$ – user55119 Jul 18 '18 at 19:22
