8

I'm trying to make a phenyl radical using chemfig, so far I've used this code:

\chemfig{*6(-\Lewis{0.,}=-=-=)}

What I obtain however is this, with a "broken" angle in the phenyl:

wrong

How to obtain a representation like this one?

correct

1 Answer 1

14

I would propose to insert a small invisible bond:

\documentclass{article}
\usepackage{chemfig}
\begin{document}   
\chemfig{*6(-(-[,.1,,,draw=none\Lewis{0.,})=-=-=)}
\end{document}

If you need it more often it is probably a good thing to define a suitable submol:

\documentclass{article}
\usepackage{chemfig}
\definesubmol{e}{-[,.1,,,draw=none]}
\begin{document}
\chemfig{*6(-(!e\Lewis{0.,})=-=-=)}
\end{document}

As for the scheme you can use chemfig's scheming commands (see “part V Reaction Schemes“ of the documentation).

Here is an example:

\documentclass{article}
\usepackage{chemfig}
\definesubmol{e}{-[,.1,,,draw=none]}
\begin{document}

\schemestart
 \chemleft[
  \chemfig{*6(=-=-(-Br)=-)}
 \chemright]
 % \arrow(from--to){->[<above>][<below>]}[<angle>,<length factor>]
 \arrow(a--){->[$k_{f,1}$]}
 % the \arrow is rather complex. all arguments are optional.
 % In this case we name the preceding compound `a' to be able to refer to it
 % later.
 \chemfig{*6(=-=-(!e\Lewis{0.,})=-)}
 % the arrow type `0' is invisible but centers the `phenyl' and the `+':
 \arrow{0}[,0]\+ Br$^-$
 % now we insert the charge and the radical to the first compound, again using
 % an invisible arrow and TikZ' anchors:
 \arrow(@a.north east--){0}[,0] $-$\Lewis{0.,}
\schemestop

\end{document}

enter image description here

1
  • @pygabriel you're welcome :)
    – cgnieder
    Jun 14, 2012 at 9:29

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