In the following ASCI structure (single bond under "$\ce{CH}$" containing $\ce{CH2,CH2, CH3}$)

$$ \begin{align*} \ce{CH3-CH2-CH2-&CH-CH3}\\ &|\\ \ce{&CH2-CH2-CH3} \end{align*} $$

The correct way to name it is "4-methlyheptane".

But when I do it the way I have been taught I get the following "4-propylpentane". Here are my systematic steps:

Firstly, I count the longest carbon chain. In this case its 5 carbons together so suffix -pentane. Then I look to see if there are any branched chains and number it from left to right. In this case the branched chain is on number 4. The branched chain contains $\ce{CH2-CH2-CH3}$, resp. $\ce{C3H7}$, aka propyl.

Ergo, 4-propylpentane, however, the correct answer as mentioned above is 4-methylheptane.

How is it 'heptane' ? The longest chain has FIVE carbons. I got the '4-' part correct, but how is $\ce{CH2-CH2-CH3}$ Methyl? Is this a mistake in my book?


2 Answers 2


You need to find the longest continuous carbon chain of:

$$ \begin{align*} \ce{CH3-CH2-CH2-&CH-CH3}\\ &|\\ \ce{&CH2-CH2-CH3} \end{align*} $$

Stretch that out to $\ce{CH3CH2CH2CH(CH3)CH2CH2CH3}$ (with the $\ce{-CH3}$ coming off of the $\ce{CH}$) and you'll see that the book is correct. The "side chain" of $\ce{-CH2CH2CH3}$ is actually part of the main carbon chain. As far as the numbering, counting from either side give the 4- position due to the symmetry of the molecule.


  • $\begingroup$ If anybody knows how to better jury-rig the diagrams, I'm open to an edit. $\endgroup$
    – jonsca
    Oct 11, 2014 at 11:21
  • $\begingroup$ Just draw it on paper and take a picture. The built in package is too limited to draw a compound of even this low level of complexity. $\endgroup$
    – jerepierre
    Oct 11, 2014 at 17:20
  • 3
    $\begingroup$ @jonsca I found this web.chemdoodle.com/demos/sketcher to be of use when no local chemeditor is available. $\endgroup$
    – permeakra
    Oct 11, 2014 at 21:27

The structure is misleadingly drawn. The longest carbon chain is not this:

$$ \begin{align*} \color{red}{\ce{CH3-CH2-CH2 -}}&\color{red}{\ce{CH-CH3}}\\ &|\\ \ce{&CH2-CH2-CH3} \end{align*} $$

but this:

$$ \begin{align*} \color{green}{\ce{CH3-CH2-CH2 -}}&\color{green}{\ce{CH}}\ce{-CH3}\\ &\color{green}{|}\\ &\color{green}{\ce{CH2-CH2-CH3}} \end{align*} $$

That's 1, 2, 3, 4, 5, 6... 7 carbons. Thus, this is a substituted heptane. The single remaining carbon forms a methyl group, which is attached to the fourth carbon (from either end) in the heptane chain. Thus, 4-methylheptane.


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