1

I'm trying to draw Fehling's reagent with chemfig. Here what I come up with

\documentclass{standalone}
\usepackage{chemfig}

\begin{document}
\chemfig[atom sep=20pt]{[:-30]Cu*7(-[:30]O-(=[:-70]O)-[:-30](-OH)-[:-90](-OH)-[:150](=[:70]O)-[:180]O-)}
\end{document}

enter image description here

How do I draw the second cycle (symetric from the first one on cupper atom) ?

2 Answers 2

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You can use two identical rings in sequence, just change the angle so that the second is opposite the first

\documentclass{article}
\usepackage{chemfig}

\begin{document}
    \chemfig{Cu*7([:-25.7]-O-(=O)-(-OH)-(-OH)-(=O)-O-)*7([:180]-O-(=O)-(-OH)-(-OH)-(=O)-O-)}
\end{document}

enter image description here

2
  • -25.7143 degrees to turn it horizontal. Jul 19, 2022 at 7:50
  • 1
    I had just corrected the angle when I saw your message. I didn't remember how to calculate the sum of the interior angles of a regular heptagon, I had to look it up. Thanks.
    – CrocoDuck
    Jul 19, 2022 at 7:53
0

I have a very nasty solution:

\documentclass{article}
\usepackage{chemfig}

\begin{document}
\chemfig[atom sep=20pt]{[:-30]Cu*7(-[:30]O-(=[:-70]O)-[:-30](-OH)-[:-90](-OH)-[:150](=[:70]O)-[:180]O-)}
\hskip-5.35cm\hflipnext%
\chemfig[atom sep=20pt]{[:-30]Cu*7(-[:30]O-(=[:-70]O)-[:-30](-OH)-[:-90](-OH)-[:150](=[:70]O)-[:180]O-)}
\end{document}

If someone has a better solution, I am interested in

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