Creating ethane using \chemfig

Question

I was tinkering with chemfig package. I created the macros

\documentclass{article}
\newcommand{\lmethyl}{C(-[2]H)(-[4]H)(-[6]H)}
\newcommand{\rmethyl}{C(-[2]H)(-[6]H)-H}
\newcommand{\tmethyl}{C(-[2]H)(-[4]H)-H}
\newcommand{\bmethyl}{C(-[4]H)(-[6]H)-H}
\usepackage{chemfig}
\begin{document}
\chemfig{\lmethyl-\rmethyl}
\end{document}

I thought I would be able to create the code of methane using this. I know the macros are all run at the same time and hence creates the following.

Answer 1

What you're looking for is the \definesubmol{name}{code} command. You can insert a sub-molecule inside a chemfig command with !{name}.

\definesubmol{rmethyl}{C(-[0]H)(-[2]H)(-[6]H)}
\chemfig{R-!{rmethyl}}

There are some other things you can do with that command, such as defining different code depending on what direction the bond comes from (so H_3C and CH_3 can be handled by one sumbol).

For more details, see section 4.7 of the chemfig manual.

Answer 2

You should use chemfig's \definesubmol instead of \newcommand:

\documentclass{article}

\usepackage{chemfig}
\definesubmol{\lmethyl}{C(-[2]H)(-[4]H)(-[6]H)}
\definesubmol{\rmethyl}{C(-[2]H)(-[6]H)-H}
\definesubmol{\tmethyl}{C(-[2]H)(-[4]H)-H}
\definesubmol{\bmethyl}{C(-[4]H)(-[6]H)-H}

\begin{document}
\chemfig{!\lmethyl-!\rmethyl}
\end{document}