2

I'm writing a reaction of hydrolysis, but the glucose isn't aligned with the sucrose nor the fructose. Help.

\documentclass[border=10pt]{standalone}
\usepackage{chemfig}
\begin{document}
\schemestart
\chemfig[cram width=2pt]{HO-[2,.5,2]?<[7,.7](-[2,.5]OH)-[,,,,line width=2.4pt](-[6,.5]OH)>[1,.7](-[:-65,.7]O-[:65,.7]?[b](-[2,.7]CH_2OH)<[:-60,.707](-[6,.5]OH)-[,,,,line width=2.4pt](-[2,.5,,2]HO)>[:60,.707](-[6,.5]CH_2OH)-[:162,.9]O?[b])-[3,.7]O-[4]?(-[2,.3]-[3,.5]HO)}
\arrow{<=>}
\chemfig[cram width=2pt]{HO-[2,0.5,2]?<[7,0.7](-[2,0.5]OH)-[,,,,
line width=2.4pt](-[6,0.5]OH)>[1,0.7](-[6,0.5]OH)-[3,0.7]
O-[4]?(-[2,0.3]-[3,0.5]OH)}
\+
\chemfig[cram width=2pt]{(?[b](-[2,.7]CH_2OH)<[:-60,.707](-[6,.5]OH)-[,,,,line width=2.4pt](-[2,.5,,2]HO)>[:60,.707](-[6,.5]CH_2OH)-[:162,.9]O?[b])}
\schemestop
\end{document}

Current Output

1 Answer 1

5

According to the manual ver 1.41 in section 12, The + sign, it states that you can vertically align the + if you add \arrow{0}[,0] which draws an invisible arrow of zero-length. In your case, you need to put that hack right after the + sign as well, i.e., \arrow{0}[,0]\+\arrow{0}[,0].

Output

\documentclass[border=10pt]{standalone}
\usepackage{chemfig}
\begin{document}
\schemestart
\chemfig[cram width=2pt]{HO-[2,.5,2]?<[7,.7](-[2,.5]OH)-[,,,,line width=2.4pt](-[6,.5]OH)>[1,.7](-[:-65,.7]O-[:65,.7]?[b](-[2,.7]CH_2OH)<[:-60,.707](-[6,.5]OH)-[,,,,line width=2.4pt](-[2,.5,,2]HO)>[:60,.707](-[6,.5]CH_2OH)-[:162,.9]O?[b])-[3,.7]O-[4]?(-[2,.3]-[3,.5]HO)}
\arrow{<=>}
\chemfig[cram width=2pt]{HO-[2,0.5,2]?<[7,0.7](-[2,0.5]OH)-[,,,,line width=2.4pt](-[6,0.5]OH)>[1,0.7](-[6,0.5]OH)-[3,0.7]O-[4]?(-[2,0.3]-[3,0.5]OH)}
\arrow{0}[,0]\+\arrow{0}[,0]
\chemfig[cram width=2pt]{(?[b](-[2,.7]CH_2OH)<[:-60,.707](-[6,.5]OH)-[,,,,line width=2.4pt](-[2,.5,,2]HO)>[:60,.707](-[6,.5]CH_2OH)-[:162,.9]O?[b])}
\schemestop
\end{document}

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