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

13

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

  • @pygabriel you're welcome :) – clemens Jun 14 '12 at 9:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.