目前它看起来像这样:
这是最少的代码:
\documentclass[a4paper, 12pt]{scrartcl}
\usepackage[ngerman]{babel}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{chemfig}
\begin{document}
\begin{align*}
\schemestart
\chemname{
\chemfig{
\ldots
-[0]C
( -[2]H)
( -[-2]H)
% First C-atom with COOH
-C
( -[2]{CH3})
(-[6]C
( =[7]\charge{360=\|, 270=\|}{O})
( -[5]\charge{315=\|, 135=\|}{O}
( -[6]H)
)
)
-[0]C
( -[2]H)
( -[-2]H)
% Second C-atom with COOH
-C
( -[2]{CH3})
(-[6]C
( =[7]\charge{360=\|, 270=\|}{O})
( -[5]\charge{315=\|, 135=\|}{O}
( -[6]H)
)
)
-[0]C
( -[2]H)
( -[-2]H)
% Third C-atom with COOH
-C
( -[2]{CH3})
(-[6]C
( =[7]\charge{360=\|, 270=\|}{O})
( -[5]\charge{315=\|, 135=\|}{O}
( -[6]H)
)
)
-[0]\ldots
}
}
{Polymer}
\schemestop
\end{align*}
\end{document}
如您所见,右侧的点比左侧旁边的电子连接略低。您知道如何解决这个问题吗?
答案1
一个有技巧的解决方案
\chemfig{
\ldots
-[0]C
( -[2]H)
( -[-2]H)
% First C-atom with COOH
-C
( -[2]{CH3})
(-[6]C
( =[7]\charge{360=\|, 270=\|}{O})
( -[5]\charge{315=\|, 135=\|}{O}
( -[6]H)
)
)
-[0]C
( -[2]H)
( -[-2]H)
% Second C-atom with COOH
-C
( -[2]{CH3})
(-[6]C
( =[7]\charge{360=\|, 270=\|}{O})
( -[5]\charge{315=\|, 135=\|}{O}
( -[6]H)
)
)
-[0]C
( -[2]H)
( -[-2]H)
% Third C-atom with COOH
-C
( -[2]{CH3})
(-[6]C
( =[7]\charge{360=\|, 270=\|}{O})
( -[5]\charge{315=\|, 135=\|}{O}
( -[6]H)
)
)
-\Charge{180=$\,\ldots$}{\vphantom{C}}
}