Salicylic acid with chemfig

Question

For my students I need to draw the structural formula of salicylic acid showing the carbon and hydrogen atoms but without the hydrogen-carbon bond.

For this I use the chemfig package and I already have this formula

\chemfig{*6(-=-(-O-[::-60](-[::-60])=[::+60]O)=(-(=[::+60]O)-[::-60]OH)-=)}

But when I put in the hydrogen and carbon

\chemfig{*6(CH-CH=CH-C(-O-[::-60]C(-[::-60]CH_3)=[::+60]O)=CH(-C(=[::+60]O)-[::-60]OH)-HC=)}

I obtain some asymmetry in the cycle.

Does anybody know how to remove this asymmetry?

Answer 1

A ring should be attached to a molecule, i.e. the first CH should be outside the ring definition: CH*6(-...=). To obtain a regular shape, as shown in Section 10.5 of the Chemfig manual, the departure and arrival atoms of the last bond need to be manually set: =[,,2,2]:

\chemfig{CH*6(-CH=CH-C(-O-[::-60]C(-[::-60]CH_3)=[::+60]O)=CH(-C(=[::+60]O)-[::-60]OH)-HC=[,,2,2])}

\documentclass{standalone}
\usepackage{chemfig}

\begin{document}

    \schemestart
    \chemfig{CH*6(-CH=CH-C(-O-[::-60]C(-[::-60]CH_3)=[::+60]O)=CH(-C(=[::+60]O)-[::-60]OH)-HC=[,,2,2])}
    \schemestop

\end{document}

Answer 2

Another way to achieve the wanted result is using -H bond as a branch.

\documentclass{article}
\usepackage{chemfig}
\begin{document}
    \chemfig{C*6((-[::-150]H)-C(-[::-60]H)=C(-[::-60]H)-C(-O-[::-60]C(-[::-60]CH_3)=[::+60]O)=C(-C(=[::+60]O)-[::-60]OH)-C(-[::-60]H)=)}
\end{document}

The output:

Answer 3

\documentclass{article}
\usepackage{chemfig,xcolor}
\begin{document}
\chemfig{HC*6(-\chembelow{C}{H}=CH-C(-O-[:-30]C(-[:-90]CH_3)=O)=C(-C(=[:150]O)-[:30]OH)-HC=[,,2]-)}
\bigskip\par
\chemfig{HC**6(-\chembelow{C}{H}=CH-C(-O-[:-30]C(-[:-90]CH_3)=O)=C(-C(=[:150]O)-[:30]OH)-HC=[,,2]-)}\par
\color{red}\vspace{-6.5em}\hspace{1.1em}\chemfig{**6(------)}
\end{document}