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I try to show thermal decomposition of Benzoyl peroxide (BPO) with chemfig. But I don't get the coefficients („2“) vertical aligned properly.

\documentclass{standalone}
\usepackage{chemfig}

\begin{document}
\schemestart
  \chemfig{*6(-=-([:30]-(=[2]\lewis{13,O})(-[:-30]\lewis{57,O}-\lewis{13,O}(-[:-30](=[6]\lewis{57,O})(-[:30]*6([:30]=-=-=-)))))=-=)} \arrow{->} 2 \chemfig{*6(-=-([:30]-(=[2]\lewis{13,O})(-[:-30]\Lewis{7.15,O}))=-=)} \arrow{->} 2 \chemfig{*6(-=-(-[,.1,,,draw=none]\Lewis{0.,})=-=)}  \arrow{0}[,0]\+ 2 \chemfig{\lewis{35,O}=C=\lewis{17,O}} 
\schemestop
\end{document}

enter image description here

2
  • Does enclosing 2 in \chemfig{2} work? Not at pc so can't tell ! It should place the first atom/item encountered in \chemfig{} at the baseline...
    – Leeser
    Jan 10, 2017 at 18:58
  • Unfortunately this doesn't work. Jan 10, 2017 at 19:15

1 Answer 1

2

The solution is essentially the same as you did with the \+ component: With the \arrow{0}[,0] arrow, you separate the compound into two separate ones: the coefficient and the formula, which then are aligned correctly.

\documentclass{standalone}
\usepackage{chemfig}

\begin{document}
\schemedebug{true} % remove to disable debug mode
\schemestart
  \chemfig{*6(-=-([:30]-(=[2]\lewis{13,O})(-[:-30]\lewis{57,O}-\lewis{13,O}(-[:-30](=[6]\lewis{57,O})(-[:30]*6([:30]=-=-=-)))))=-=)}
  \arrow{->}
  {2\ } \arrow{0}[,0] % separate coefficient into own compound, and add some horizontal space
  \chemfig{*6(-=-([:30]-(=[2]\lewis{13,O})(-[:-30]\Lewis{7.15,O}))=-=)}
  \arrow{->}
  {2\ } \arrow{0}[,0]
  \chemfig{*6(-=-(-[,.1,,,draw=none]\Lewis{0.,})=-=)}
  \arrow{0}[,0]\+ 2 \chemfig{\lewis{35,O}=C=\lewis{17,O}}
\schemestop
\end{document}

Note how the coefficient is separate from the formula (c2 and c3), and thus aligned correctly.

scheme

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