答案1
“原子对原子”解决方案
\documentclass{article}
\usepackage{chemfig}
\begin{document}
\chemfig{
HC?[cc](-[:-120,1,2]O?[oa])-[:-30,1]O-[:-30,1]CH(-[:-120,1,1]O?[oa])
-[:30,2,1]CH(-[,1]O?[ob])-[2,1]O-[2,1]CH(-[,1]O?[ob])
-[:150,2,1,1]CH(-[:120,1,1]O?[oc])-[:210,1]O-[:210,1,1,2]HC?[cc](-[:120,1,2]O?[oc])
}
\end{document}
答案2
这是一个想法。
\documentclass{article}
\usepackage{chemfig}
\begin{document}
\def\0{\phantom{C}}
\definesubmol\x1{-[::-30,,#1]O-[::90,2]O-[::90]\0}
\chemfig[bond offset=4pt]{\0-[::-90,2]CH(!\x1)-[::60,,1]O-[::0]CH-[::60,2,1]CH(!\x2)-[::60,,1]O-[::0]CH-[::60,2,,1]CH(!\x1)-[::60,,1]O-[::0,,,1]CH}
\end{document}