2

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

  • 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 '17 at 18:58
  • Unfortunately this doesn't work. – Solarflares Jan 10 '17 at 19:15
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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.