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For the above molecule, I know that the longest carbon chain has 7 carbons, so the backbone is heptane.

The two substituents are at the 4 position on the heptane.

One of the substituents is a methyl group. That much is obvious.

The other appears to be a complex substituent and is a 1-chloroethyl group.

Thus, would the name be:


I think the issue is what happens when we have different substituents with the same loctant. I've looked in one book and at all its practice problems, and this scenario never occurs.

I know that 4-methylheptane is a real compound and that its name is correct. So I am leaning toward 4,4-(1-chloroethyl)methylheptane as being a correct name; the complex substituent is correctly named; both locatants are present; everything is alphabetized; my only issue is whether I am missing some convention (e.g. should there be a dash somewhere, a comma somewhere, or what?)


1 Answer 1


Close. But each (different type of) group added to the heptane backbone is numbered separately: 4-(1-chloroethyl)-4-methylheptane. Also, you have an asymmetric carbon, so the chirality needs to be specified (I will leave that to you).

  • $\begingroup$ Actually we are both wrong. The longest, most substituted chain in this compound is the 6-carbon chain (hexane). $\endgroup$
    – Dissenter
    Jun 4, 2014 at 19:34
  • $\begingroup$ This 6 carbon chain as a 2-chloro substituent, a 3 methyl substituent, and a 3 propyl substituent; therefore the name is (2S)-chloro-3-methyl-3-propylhexane. $\endgroup$
    – Dissenter
    Jun 4, 2014 at 19:35
  • $\begingroup$ Actually I'm not sure anymore about the above name (the one my professor gave). I just read in Wade that we use the longest chain; if there are two chains of equal length, use the one that is more substituted. The longest chain is unequivocally the 7-carbon chain. $\endgroup$
    – Dissenter
    Jun 4, 2014 at 22:48
  • 1
    $\begingroup$ @Dissenter I believe this answer is as correct as it gets. And your last comment also substantiates this claim. Maybe it is time to accept this answer ;D (And you could clear some of your comments) $\endgroup$ Aug 17, 2014 at 7:29

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