I need to figure out the chain and positional isomers of C4H7Cl. I am not looking for someone to directly give me the answer, I just want to know what steps are needed to be able to figure them out for a given compound.

This is different to my prior question. Here I am asking for the technique, not the direct answer. Thanks.

  • 2
    $\begingroup$ Possible duplicate of What are some examples of chain and positional isomers of C4H7Cl? $\endgroup$
    – Mithoron
    Jan 29, 2016 at 22:13
  • $\begingroup$ No. Please re-read. $\endgroup$
    – Tom
    Jan 29, 2016 at 22:15
  • $\begingroup$ You should have edited your old question, don't ask second time about the same topic. $\endgroup$
    – Mithoron
    Jan 29, 2016 at 22:18
  • $\begingroup$ Would it have reappeared on the main question page if I edited it? $\endgroup$
    – Tom
    Jan 29, 2016 at 22:19
  • $\begingroup$ It surely would. $\endgroup$ Jan 29, 2016 at 22:20

1 Answer 1

  1. Is the parent hydrocarbon unbranched? That would be n-butane, $\ce{C4H10}$. Replacing one hydrogen atom by chlorine would lead to $\ce{C4H9Cl}$. Apparently, you're not having that!

  2. Move a methyl group in n-butane, yielding 2-methylpropane, $\ce{C4H9}$. Here, replacing one hydrogen atom by chlorine gives $\ce{C4H8Cl}$. You're not having that either!

  3. Does your target molecule have $\ce{C=C} $double bonds? Calculate the number of double bond equivalents in $\ce{C4H7Cl}$! I count one.

Work your way down from here.


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