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Currently I am having an issue due to intersection between substituent species off of a branched alkane that I want to make in a 'complete structure' form. I figured increasing the vertical bond length of the substituent bottom carbon by adding an extra dash between the central carbon (cbmethyl) and [6] in the following code would work:

\documentclass[12pt, reqno]{amsart}
\usepackage{chemfig}
\definesubmol{\lmethyl}{C(-[2]H)(-[4]H)(-[6]H)}
\definesubmol{\rmethyl}{C(-[2]H)(-[6]H)-H}
\definesubmol{\tmethyl}{C(-[2]H)(-[4]H)-H}
\definesubmol{\bmethyl}{C(-[4]H)(-[6]H)-H}
\definesubmol{\cmethyl}{C(-[2]H)(-[6]H)}
\definesubmol{\cbmethyl}{C(-[2]H)}
\begin{document}
\chemfig{!\lmethyl-!\cbmethyl(-[6]!\bmethyl)(-!\rmethyl)}
\end{document}

enter image description hereenter image description here

Instead however the code moves the substituent carbon along horizontally. The only other solution is to increase both horizontal bond lengths of the other carbons, but that makes the compound look a bit bloated in my opinion. Any help would be much appreciated.

2 Answers 2

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If you want to go with Couper/Butlerov style and not to use a modern representation, I see three possibilities:

Increased bond length

Note that there is no need to create a phantom atom since you can pass a numerical value for the bond length as a second argument: [,<bond length>,,,]. Personally, I don't like this method as this makes one particular C-C bond to stick out for no good reason as all of them should have pretty much the same length.

enter image description here

\documentclass[12pt, reqno]{amsart}
\usepackage{chemfig}
\definesubmol{\lmethyl}{C(-[2]H)(-[4]H)(-[6]H)}
\definesubmol{\rmethyl}{C(-[2]H)(-[6]H)-H}
\definesubmol{\tmethyl}{C(-[2]H)(-[4]H)-H}
\definesubmol{\bmethyl}{C(-[4]H)(-[6]H)-H}
\definesubmol{\cmethyl}{C(-[2]H)(-[6]H)}
\definesubmol{\cbmethyl}{C(-[2]H)}
\begin{document}
\chemfig{!\lmethyl-!\cbmethyl(-[6,2]!\bmethyl)(-!\rmethyl)}
\end{document}

Altered geometry

In order to keep C-C bond length consistent, one can tweak the geometry of methyl groups by using ∠H–C–H = 60°. Also, there is no need to define left/right/bottom methyl groups — one is enough if you define it using relative angular parameters: [::<angle>,,,,].

enter image description here

\documentclass[12pt, reqno]{amsart}
\usepackage{chemfig}
\definesubmol{\chfmethyl}{C(-[::0]H)(-[::60]H)(-[::-60]H)}
\begin{document}
\chemfig{C(-[:90]H)(-!\chfmethyl)(-[:180]!\chfmethyl)(-[:-90]!\chfmethyl)}
\end{document}

Shorter C−H bonds

To preserve the orthogonal arrangement, one can also decrease all C−H bond lengths. Considering that the average C(sp³)−C(sp³) length for is 1.54 Å, and C(sp³)−H length is 1.10 Å (data from Wikipedia), and, fixing chemfig's C−C bond at 1, the chemfig's C−H bond length should be reduced to 1.10 Å/1.54 Å ≈ 0.7. This should give enough space around each methyl group so they don't overlap:

enter image description here

\documentclass[12pt, reqno]{amsart}
\usepackage{chemfig}
\definesubmol{\chfmethyl}{C(-[::0,0.7]H)(-[::90,0.7]H)(-[::-90,0.7]H)}
\begin{document}
\chemfig{C(-[:90,0.7]H)(-!\chfmethyl)(-[:180]!\chfmethyl)(-[:-90]!\chfmethyl)}
\end{document}
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Okay yeah wasn't too difficult at all, and really added a lot of clarity to the situation. Just add a phantom point with a -[6] next to where the substituent carbon will be.

\documentclass[12pt, reqno]{amsart}
\usepackage{chemfig}
\definesubmol{\lmethyl}{C(-[2]H)(-[4]H)(-[6]H)}
\definesubmol{\rmethyl}{C(-[2]H)(-[6]H)-H}
\definesubmol{\tmethyl}{C(-[2]H)(-[4]H)-H}
\definesubmol{\bmethyl}{C(-[4]H)(-[6]H)-H}
\definesubmol{\cmethyl}{C(-[2]H)(-[6]H)}
\definesubmol{\cbmethyl}{C(-[2]H)}
\begin{document}
\chemfig{!\lmethyl-!\cbmethyl(-[6]-[6]!\bmethyl)(-!\rmethyl)}
\end{document}

enter image description here

If anyone has some neater solutions I'd still like to have a look, the spacing is still not the greatest.

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