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I'm writing the Eriochrome black T structure

\documentclass[]{article}
\usepackage{chemfig}

\begin{document}
    \chemfig{*6(=-*6(-(-N(=[:-90]N (-[:-90]*6(-=-=-=)) ))=(-OH)-=(-SO^{-}_{3})-=)--=(-HO)-)}
\end{document}

enter image description here

How can I rotate by 30° the aromatic ring at very bottom? The one bound to the last nitrogen atom from top to bottom

  • Can you clarify what you mean by 'rotating by 30 degrees'? I don't suppose you mean to change the angle of the N-N-C bond? – Troy Oct 15 '17 at 8:10
  • Solved: \chemfig{*6(=-*6(-(-N(=[:-90]N (-[:-90]([::-30]*6(-=-(*6(-=-=--))=-(-OH)=))) ))=(-OH)-=(-SO^{-}_{3})-=)--=(-O_{2}N)-)} – user3204810 Oct 15 '17 at 8:44
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Solved:

\chemfig{*6(=-*6(-(-N(=[:-90]N (-[:-90]([::-30]*6(-=-(*6(-=-=--))=-(-OH)=))) ))=(-OH)-=(-SO^{-}_{3})-=)--=(-O_{2}N)-)}

enter image description here

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