6

I am trying to draw the following compound in tex using chemfig but I am only getting part of it and do not know how to finish the compound.

Expected figure

This is what I got using

\documentclass{article}
\usepackage{chemfig}

\begin{document}
    \chemfig{*5((-NH)-S-Cu(*5(-N=(-CH_3)-(-CH_3)=N)) -N-N=)}
\end{document}

Cu-ATSM]![using chemfig

Any help would be appreciated.

  • 1
    Welcome to TeX SE! Please post the part you've got rather than expecting others to begin by redoing that work before they can even start helping you. You don't even say which part you need help with! – cfr Mar 7 '15 at 4:23
2

Another idea using no rings but chemfig's ?-syntax (and a very little bit of maths). The angles are not the same as the ones in your picture. I only used 30° steps for the outer ring so I could use 45° and a bondlength of \sqrt{2} to reach copper. The interesting bit is

(-[:-45,1.414,,2]^{64}|Cu?[a]?[b])

You'll notice 1.414 (second parameter of the bond) which is \approx\sqrt{2} for the bondlength. The atom is ^{64}|Cu where the | divides it into two parts, the first being ^{64} and Cu being the second. The fourth parameter of the bond is a 2 which means »connect to the second part of the following atom«. The part ends with the two anchors ?[a]?[b] which connect to the corresponding anchors elsewhere in the formula (the nitrogen and sulfure atoms).

\documentclass{scrbook}

\usepackage{chemfig}

\begin{document}

\chemfig{
  -[:-30]\chembelow{N}{H}-[:30](-[:-30]S?[a])
  =_[:90]N-[:30]N(-[:-45,1.414,,2]^{64}|Cu?[a]?[b])
  =_[:60](-[:120])-(-[:60])
  =_[:-60]N?[b]-[:-30]N
  =_[:-90](-[:-150]S?[b])
  -[:-30]\chembelow{N}{H}-[:30]
}

\end{document}

enter image description here

4

I think I found a solution for your problem. It is quite a hack, but it seems to work:

\documentclass{article}
\usepackage{chemfig}

\begin{document}    
\chemfig{
    [:-18]*5(-\chembelow{N}{H}-
    *5(-S-{^{64}Cu}
        *5([::-108]-S-(*5(-\chembelow{N}{H}-))=N-N-)
        *5(-\phantom{N}=(-)-(-)=\phantom{N}-)
    -N-N=))
}
\end{document}

result

I tried to solve everything with rings, so we don't need to specify all angles. Further, I tried to group the code nicely, so it's easier to understand. For the upper ring, I used phantom boxes (\phantom{N}) as the connections and Ns are already drawn by the lower two rings.

I hope this helps. If anything is unclear, try understanding the code by compiling only parts and see what has which effect. Sorry, it is quite difficult to describe/document this kind of command.

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