I'd like to draw a picture like the following with chemfig electron shift visualized by arcs

How can I draw the red arcs (e.g. in H-O) and the red line (e.g. in C=C ).

I startet already:


  \setbondstyle{line width=1pt}
  \setlewis{}{}{line width=1pt}


 \chemmove{\draw[thick,green,-] ([shift=(65:15pt)]a1) arc (65:115:17pt);}
 \chemmove{\draw[thick,green,-] ([shift=(155:15pt)]a1) arc (155:205:17pt);}
 \chemmove{\draw[thick,green,-] ([shift=(245:15pt)]a1) arc (245:295:17pt);}
 \chemmove{\draw[thick,green,-] ([shift=(-25:15pt)]a2) arc (-25:25:17pt);}
 \chemmove{\draw[thick,green,-] ([shift=(65:15pt)]a2) arc (65:115:17pt);}
 \chemmove{\draw[thick,green,-] ([shift=(245:15pt)]a2) arc (245:295:17pt);}

enter image description here

How can I split the C-C bond homgenous, which means to draw a line perpendicular to the bond, like this C-|-C

Besides: I not confident with my solution: If I change the „setatomsep“ I have to adjust all values in the chemmove-commands. Is there a more elegant solution?

enter image description here


Have you thought about using a couple of simple styles and arrow tips?

For example,

  atom sep=2em,
  bond style={line width=1pt},
  lewis style={line width=1pt},
  my arc/.tip={Arc Barb[green,arc=60,length=1.5em]},
  my split/.style={postaction=decorate,decoration={markings, mark=at position .5 with {\arrow[red]{Bar[width=1em]}}}}
  \chemfig{[,,,,-my arc]
    @{a1}C(-[2]H)(-[4]H)(-[6]H)-[,,,,my split]@{a2}C(-[0]H)(-[2]H)(-[6]H)

arrow tips

Note that I updated your obsolete syntax which generates warnings on compilation and will eventually generate errors, according to the manual.

| improve this answer | |
  • 1
    You make my work much simpler. Works perfect for single bonds. But with double bonds I got two arcs (and only one is needed). For example with \chemfig{ C(-[2,,,,-my arc]H)(-[4,,,,-my arc]H)(-[6,,,,-my arc]H)-[,,,,my split]C(=[1,,,,my arc-]O)(-[7,,,,my arc-]O(-[0,,,,-my arc]H))} – Solarflares Jul 30 '18 at 16:06
  • @Solarflares I suggest asking a follow-up and linking this question. I wouldn't expect that, but I maybe the double bonds are drawn as two paths rather than one double. (I only know the chemfig I learnt to answer this question.) If you ask a follow-up explaining that problem, you might get a better solution, especially if you link here and include an MWE. – cfr Jul 30 '18 at 23:25

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