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Having successfully received some help for building up a reaction scheme, another problem occurred:

The Arrow from one Hydrogen atom is not pointing directly to the other Hydrogen atom.

I also want it to be fixed:

\documentclass{article}

\usepackage{chemfig}
\usepackage{amssymb}


\begin{document}


\schemestart
\chemfig[atom sep=32pt]{C([2]-H)([4]-H)(-\charge{90=\|,-90=\|}{O}-@{h1}H)([6,2]-C([4,1]-H)([0,1]-\charge{90=\|,-90=\|}{O}-@{h2}H)([6,2]-C([6,1]-@{c3}H)([4,1]-H)([0,1]-\charge{90=\|,-90=\|}{O}-[0,1]@{h3}H)))}
\arrow(@{c3}--)[-90,1]
\chemfig[atom sep=32pt]{C([2]-@{hh1}H)([4]-H)(-\charge{90=\|,-90=\|}{O}-H)([6,2]-C([4,1]-H)([0,1]-\charge{90=\|,-90=\|}{O}-H)([6,2]-C([6,1]-@{c3}H)([4,1]-H)([0,1]-\charge{90=\|,-90=\|}{O}-[0,1]H)))}
\arrow(@{h1}--){0}[0,0.063]\+ 
\chemfig[atom sep=24pt]{C([3,1]=O)([5,1]-H)([1,0.5]--[-1,0.5]-CH_3)} 
\arrow(@{h3}--){0}[0,0.063]\+
\chemfig[atom sep=24pt]{C([3,1]=O)([5,1]-H)([1,0.5]--[-1,0.5]--[-1,0.5]--[-1,0.5]--[-1,0.5]--[-1,0.5]--[-1,0.5]--[-1,0.5]-CH_3)} 
\arrow(@{h2}--){0}[0,0.063]\+
\chemfig[atom sep=24pt]{C([3,1]=O)([5,1]-H)([1,0.5]--[-1,0.5]--[-1,0.5]--[-1,0.5]--[-1,0.5]=-[-1,0.5]--[-1,0.5]--[-1,0.5]--[-1,0.5]-CH_3)} 
\schemestop

\end{document}

Before it was fixed:

enter image description here

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  • 1
    Can you please include a screenshot generated from your posted code?
    – MS-SPO
    Jun 12, 2022 at 12:38

1 Answer 1

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Easy. Add an "invisible" bond (-[4,1.1,,,draw=none]) to the hydrogen on the left in the bottom molecule. This will equal the horizontal size of this molecule, and the hydrogen on top will be in the center of the molecule.

\documentclass{article}

\usepackage{chemfig}
\usepackage{amssymb}


\begin{document}
    
    
    \schemestart
    \chemfig[atom sep=32pt]{C([2]-H)([4]-H)(-\charge{90=\|,-90=\|}{O}-@{h1}H)([6,2]-C([4,1]-H)([0,1]-\charge{90=\|,-90=\|}{O}-@{h2}H)([6,2]-C([6,1]-@{c3}H)([4,1]-H)([0,1]-\charge{90=\|,-90=\|}{O}-[0,1]@{h3}H)))}
    \arrow(@{c3}--)[-90,1]
    \chemfig[atom sep=32pt]{C([2]-@{hh1}H)([4]-H-[4,1.1,,,draw=none])(-\charge{90=\|,-90=\|}{O}-H)([6,2]-C([4,1]-H)([0,1]-\charge{90=\|,-90=\|}{O}-H)([6,2]-C([6,1]-@{c3}H)([4,1]-H)([0,1]-\charge{90=\|,-90=\|}{O}-[0,1]H)))}
    \arrow(@{h1}--){0}[0,0.063]\+ 
    \chemfig[atom sep=24pt]{C([3,1]=O)([5,1]-H)([1,0.5]--[-1,0.5]-CH_3)} 
    \arrow(@{h3}--){0}[0,0.063]\+
    \chemfig[atom sep=24pt]{C([3,1]=O)([5,1]-H)([1,0.5]--[-1,0.5]--[-1,0.5]--[-1,0.5]--[-1,0.5]--[-1,0.5]--[-1,0.5]--[-1,0.5]-CH_3)} 
    \arrow(@{h2}--){0}[0,0.063]\+
    \chemfig[atom sep=24pt]{C([3,1]=O)([5,1]-H)([1,0.5]--[-1,0.5]--[-1,0.5]--[-1,0.5]--[-1,0.5]=-[-1,0.5]--[-1,0.5]--[-1,0.5]--[-1,0.5]-CH_3)} 
    \schemestop
    
\end{document}
    

enter image description here

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  • Thanks again. You have a lot of expertise with this package.
    – Leon
    Jun 12, 2022 at 15:50

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