我想用 chemfig 建立一个方程式,但是我对箭头和酶的名称有问题。
我想要这样的反应:
我是乳胶代码的新手。目前我设法编写了以下代码:
\schemestart
\chemname{\chemfig{H_2N-[,0.5]C(-[2,0.5]H)(-[6,0.5]R)-[,0.5]COOH}}{Acide aminé}
\+
\chemname{\chemfig{H_2O}}{Eau}
\arrow{-U>[$NAD^{+}$][$NADH,H^{+}$][8pt][0.5][90]}
\chemname{\chemfig{R-[,0.5]C^\alpha(-[,0.5]COOH)=[2,0.5]O}}{Acide $\alpha$-cétonique}
\
\
\
\
\
\+
\
\
\
\
\
\chemname{\chemfig{NH_3}}{Ammoniaque}
\schemestop
但是使用 U 箭头我的能力有限。我无法成功,而且我也不知道如何得出这个等式。
对不起,我的英语不好 :'(
答案1
您可以基于已定义的箭头定义自己的箭头。例如:
\documentclass[border=1mm,tikz]{standalone}
\usepackage{chemfig}
\usepackage[version=4,arrows=pgf-filled]{mhchem}
\makeatletter
\definearrow7{-uU>}{%
\CF@arrow@shift@nodes{#5}%
\expandafter\draw\expandafter[\CF@arrow@current@style](\CF@arrow@start@node)--(\CF@arrow@end@node)node[midway](Uarrow@arctangent){};%
\CF@ifempty{#6}
{\def\CF@Uarrow@radius{0.333}}
{\def\CF@Uarrow@radius{#6}}%
\CF@ifempty{#7}%
{\def\CF@Uarrow@absangle{60}}
{\pgfmathsetmacro\CF@Uarrow@absangle{abs(#7)}}% ne prendre en compte que la valeur absolue de l'angle
\edef\CF@tmp@str{[\CF@ifempty{#1}{draw=none}{\unexpanded\expandafter{\CF@arrow@current@style}},-]}%
\expandafter\draw\CF@tmp@str (Uarrow@arctangent)%
arc[radius=\CF@compound@sep*\CF@current@arrow@length*\CF@Uarrow@radius,start angle=\CF@arrow@current@angle-90,delta angle=-\CF@Uarrow@absangle]node(Uarrow@start){};
\edef\CF@tmp@str{[\CF@ifempty{#2}{draw=none}{\unexpanded\expandafter{\CF@arrow@current@style}}]}%
\expandafter\draw\CF@tmp@str (Uarrow@arctangent)%
arc[radius=\CF@compound@sep*\CF@current@arrow@length*\CF@Uarrow@radius,start angle=\CF@arrow@current@angle-90,delta angle=\CF@Uarrow@absangle]node(Uarrow@end){};
\pgfmathsetmacro\CF@tmp@str{\CF@Uarrow@radius*cos(\CF@arrow@current@angle)<0?"-":"+"}%
\ifdim\CF@Uarrow@radius pt>\z@
\CF@arrow@display@label{#1}{0}\CF@tmp@str{Uarrow@start}{#2}{1}\CF@tmp@str{Uarrow@end}%
\else
\CF@arrow@display@label{#2}{0}\CF@tmp@str{Uarrow@start}{#1}{1}\CF@tmp@str{Uarrow@end}%
\fi
\edef\CF@tmp@str{[\CF@ifempty{#3}{draw=none}{\unexpanded\expandafter{\CF@arrow@current@style}},-]}%
\expandafter\draw\CF@tmp@str (Uarrow@arctangent)%
arc[radius=\CF@compound@sep*\CF@current@arrow@length*\CF@Uarrow@radius,start angle=\CF@arrow@current@angle+90,delta angle=\CF@Uarrow@absangle]node(Uarrow@start){};
\edef\CF@tmp@str{[\CF@ifempty{#4}{draw=none}{\unexpanded\expandafter{\CF@arrow@current@style}}]}%
\expandafter\draw\CF@tmp@str (Uarrow@arctangent)%
arc[radius=\CF@compound@sep*\CF@current@arrow@length*\CF@Uarrow@radius,start angle=\CF@arrow@current@angle+90,delta angle=-\CF@Uarrow@absangle]node(Uarrow@end){};
\pgfmathsetmacro\CF@tmp@str{\CF@Uarrow@radius*cos(\CF@arrow@current@angle)<0?"+":"-"}%
\ifdim\CF@Uarrow@radius pt>\z@
\CF@arrow@display@label{#3}{0}\CF@tmp@str{Uarrow@start}{#4}{1}\CF@tmp@str{Uarrow@end}%
\else
\CF@arrow@display@label{#4}{0}\CF@tmp@str{Uarrow@start}{#3}{1}\CF@tmp@str{Uarrow@end}%
\fi
}
\makeatother
\begin{document}
%\schemedebug{true}
\schemestart
\chemname[20pt]{\chemfig{H_2N-[,0.5]C(-[2,0.5]H)(-[6,0.5]R)-[,0.5]COOH}}{Acide aminé}
\arrow{-uU>[\ce{NAD^{+}}][\ce{NADH,H^{+}}][\ce{H_2O}][\ce{NH_3}][15pt][.5][80]}
\chemname[20pt]{\chemfig{R-[,0.5]C^\alpha(-[,0.5]COOH)=[2,0.5]O}}{Acide $\alpha$-cétonique}
\schemestop
\end{document}
得出的结果是(但可能需要一些微调):