2

I want to draw the following molecule (see figure below), but I don't know how to rotate the above pentagon:

Enter image description here

Here's my attempt to do draw it:

\documentclass[12pt, a4paper]{article}
\usepackage{chemfig}
\usepackage{mhchem}
\begin{document}

\chemfig{
[:10]*6((-HO)-(-\ce{R3})=(-OH)-(-*5(O-([:-30]-OH)-(*6(-=-=-=))--(=O)-))=-(-\ce{R2})=)
}

\end{document}

And here's its result:

Enter image description here

3 Answers 3

5

Would this be the answer your after?

\documentclass[12pt, a4paper]{article}

\usepackage{chemfig}
\usepackage{mhchem}

\begin{document}
    \chemfig{
        [:10]*6((-HO)-(-\ce{R3})=(-OH)-([1]-[::10]*5(([:-45]-OH)-(*6(-=-=-=))--(=O)-O-))=-(-\ce{R2})=)
        \chemmove{
        \node(end) [at=(cyclecenter3),shift=(30:1.75cm)] {\printatom{\ce{R1}}};
        \draw[-, shorten <= 0.5cm, dashed] (cyclecenter3) -- (end);
        }
    }
\end{document}

You can add an angle before the pentagon, but it needs to be formatted as [::angle]. I adjusted the start of the pentagon too and added the R1 in the last benzene ring.

First attempt

Edit Redid the first attempt to get the structure as depicted in the provided figure. This code will provide that:

\documentclass[12pt, a4paper]{article}

\usepackage{chemfig}
\usepackage{mhchem}

\begin{document}
\chemfig{
    [:54]*5(([:-36]-OH)(-[::-196]*6(=-(-\ce{R2})=(-HO)-(-\ce{R3})=(-OH)-))-(*6(-=-=-=))--(=O)-O-)
    \chemmove{
        \node(end) [at=(cyclecenter3),shift=(18:1.75cm)] {\printatom{\ce{R1}}};
        \draw[-, shorten <= 0.5cm, dashed] (cyclecenter3) -- (end);
        }
    }
\end{document}

The resulting structure is: Second attempt

4
  • That's not the desired molecule, but thanks ! Commented Feb 4, 2023 at 17:31
  • 1
    I give it another shot later tonight. Guess you have to start with the pentagon part first and add the left benzene structure like you did with the right one. Doing that you can alter the starting angle of the pentagon to what you want and the rest of the structure will follow that orientation.
    – alchemist
    Commented Feb 4, 2023 at 17:52
  • Okay I'm gonna try it.. and yeah don't hesitate to make an attempt later Commented Feb 4, 2023 at 18:28
  • 1
    Done @Court_Compilation You may want to adjust the angles around the quarternairy carbon atom a further. They are now multiples of 18 (or 9) degrees as a pentagon had inner angles of 72 degrees.
    – alchemist
    Commented Feb 4, 2023 at 19:07
2

I started the molecule at the central quaternary carbon and adjusted the parts of the molecule around it.

\documentclass[border=5mm]{standalone}
\usepackage{chemfig}
\usepackage[version=4]{mhchem}

\begin{document}
    
\chemfig{([:54]*5(-(*6(-=-=-=))--(=O)-O-))(-[:330]OH)(-[:210]*6(-=(-\ce{R2})-(-HO)=(-\charge{330=${}_3$}{R})-(-OH)=))}
\chemmove{
    \node [at=(cyclecenter2),shift=(20:1.5cm)](end){\printatom{R_1}};
    \draw[-,dashed](cyclecenter2)--(end);
}   
\end{document}

enter image description here

0

The original structure has its atoms printed in sans serif and slightly thicker bonds. On a personal note I would like my substituents a bit closer to the rings. So this would result in this code:

\documentclass[12pt, a4paper]{article}

\usepackage{chemfig}
\renewcommand*\printatom[1]{\ensuremath{\mathsf{#1}}}
\setchemfig{bond style={line width=1pt},atom sep=2.1em}

\usepackage{chemmacros}
\chemsetup{formula=chemformula}
\chemsetup[chemformula]{format=\sffamily}


\begin{document}
\chemfig{
    [:54]*5(([:-27,0.8]-OH)(-[::-205,0.8]*6(=-(-[,0.9]\ch{R2})=(-[,0.8]HO)-(-[,0.8,,,xshift=2mm]\ch{R3})=(-[,0.8]OH)-))-(*6(-=-=-=))--([,0.8]=O)-O-)
    \chemmove{
        \node(end) [at=(cyclecenter3),shift=(18:1.50cm)] {\printatom{\ch{R1}}};
        \draw[-, shorten <= 0.3cm, dashed] (cyclecenter3) -- (end);
        }
    }
\end{document}

resulting in: Personal file

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