Como faço uma caixa colorida em volta do chemfig

Question

Como faço para expandir a caixa sobre ambas as moléculas (espaço centralizado e central ajustável), mas as bordas dentro do enumerateambiente? A imagem é o que eu quero fazer. Tentei fazer o seguinte, mas claramente não é grande o suficiente.

\documentclass{article}
\usepackage{chemfig,chemmacros}
\chemsetup{modules=all}
\usepackage[version=4,arrows=pgf]{mhchem}
\usepackage{amsmath}
\usepackage{tabu}
\usepackage{framed}
\definecolor{SolutionColor}{gray}{0.85}
\usepackage{linegoal}
\newcommand{\mybox}[1]{\colorbox{SolutionColor}{\parbox[t]{\linegoal}{#1}}}
\begin{document}

\begin{enumerate}
\item Are the compounds shown below isomers?
    \begin{center}
    \mybox{
    {\tabulinesep=1.6mm
    \begin{tabu}{cc}
    \chemfig[double bond sep=0.3em,atom sep=2.5em]{H-C(-[2]H)(-[6]H)-C(-[2]H)(-[6]H)-C(-[2]OH)(-[6]H)-C(-[2]H)(-[6]H)-H} & \chemfig[double bond sep=0.3em,atom sep=2.5em]{H-C(-[2]H)(-[6]H)-C(-[2]H)(-[6]H)-C(-[2]H)(-[6]H)-C(-[2]OH)(-[6]H)-H}
    \end{tabu}}
    }
    \end{center}

    Yes or No
    
    Why?

\item Are the following geometric isomers?
      \begin{center}
      \chemfig[double bond sep=0.3em,atom sep=2.5em]{H_3C-[:-60,,2]C(-[:-120]H)=C(-[:-60]CH_3)-[:60]H}
      \hspace{1em}
      \chemfig{@{H}}
      \hspace{1em}
      \chemfig[double bond sep=0.3em,atom sep=2.5em]{H_3C-[:-60,,2]C(-[:-120]H_3C)=C(-[:-60]H)-[:60]H}
      \end{center}
      \begin{tikzpicture}[remember picture, overlay]
      \draw[fill=gray,opacity=0.4]($(H.north east)+(6cm,1cm)$)rectangle($(H.south west)+(-5cm,-1cm)$); 
      \end{tikzpicture}
\end{enumerate}
\end{document}

Answer 1

Não é exatamente uma solução, mas outra abordagem. Com tcolorbox você pode expandir a caixa como desejar.

\documentclass{article}
\usepackage{chemfig}
\usepackage{tcolorbox}

\definecolor{SolutionColor}{gray}{0.85}

\begin{document}
\begin{enumerate}
\item Are the compounds shown below isomers?

\begin{tcolorbox}[colback=SolutionColor, sharp corners, boxrule=0pt, left=2mm, right=2mm, top=4mm, bottom=1cm]
\centering
\setchemfig{double bond sep=0.3em,atom sep=2.5em}
\chemfig{H-C(-[2]H)(-[6]H)-C(-[2]H)(-[6]H)-C(-[2]OH)(-[6]H)-C(-[2]H)(-[6]H)-H}
\hspace{5mm}
\chemfig{H-C(-[2]H)(-[6]H)-C(-[2]H)(-[6]H)-C(-[2]H)(-[6]H)-C(-[2]OH)(-[6]H)-H}
\end{tcolorbox}

Yes or No

Why?
\end{enumerate}
\end{document}

Answer 2

\documentclass{article}
\usepackage{chemfig,chemmacros}
\chemsetup{modules=all}
\usepackage[version=4,arrows=pgf]{mhchem}
\usepackage{amsmath}
\usepackage{tabu}
\usepackage{framed}
\definecolor{SolutionColor}{gray}{0.85}
\usepackage{linegoal}
\newcommand{\mybox}[1]{\colorbox{SolutionColor}{\parbox[t]{\linegoal}{#1}}}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
    
    \begin{enumerate}
        \item Are the compounds shown below isomers?\\
        \begin{center}
        \chemfig[
           double bond sep=0.3em,
           atom sep=2.5em
                ]
           {H-C(-[2]H)(-[6]H)-C(-[2]H)(-[6]H)-C(-[2]OH)(-[6]H)-C(-[2]H)(-[6]H)-@{H}H} 
           \qquad 
         \chemfig[
              double bond sep=0.3em,
              atom sep=2.5em
               ]
            {H-C(-[2]H)(-[6]H)-C(-[2]H)(-[6]H)-C(-[2]H)(-[6]H)-C(-[2]OH)(-[6]H)-H}
        \end{center}
        
        Yes or No
        \\
        Why?
    \end{enumerate}

\begin{tikzpicture}[remember picture, overlay]
    \draw[red,fill=gray,opacity=0.5]($(H.north east)+(6cm,1cm)$)rectangle($(H.south 
    west)+(-5cm,-1cm)$); 
\end{tikzpicture}
    

\end{document}

Edite para abordar a consulta do OP na seção de comentários acima

@jsbibra Desculpe, como posso modificar a âncora tikz @{H} se a molécula não é linear? Anexei-o ao MWE acima. -Dave2627

Escolher @H novamente como uma tag pode confundir o tikz, já que foi usado no primeiro exemplo, basta usar outro alfabeto como @XYZ como uma tag e seguir em frente

    \documentclass{article}
\usepackage{chemfig,chemmacros}
\chemsetup{modules=all}
\usepackage[version=4,arrows=pgf]{mhchem}
\usepackage{amsmath}
\usepackage{tabu}
\usepackage{framed}
\definecolor{SolutionColor}{gray}{0.85}
\usepackage{linegoal}
\newcommand{\mybox}[1]{\colorbox{SolutionColor}{\parbox[t]{\linegoal}{#1}}}
\usepackage{tikz}
\usetikzlibrary{calc}
\begin{document}
    
    \begin{enumerate}
        \item Are the compounds shown below isomers?\\
        \begin{center}
        \chemfig[
           double bond sep=0.3em,
           atom sep=2.5em
                ]
           {H-C(-[2]H)(-[6]H)-C(-[2]H)(-[6]H)-C(-[2]OH)(-[6]H)-C(-[2]H)(-[6]H)-@{H}H} 
           \qquad 
         \chemfig[
              double bond sep=0.3em,
              atom sep=2.5em
               ]
            {H-C(-[2]H)(-[6]H)-C(-[2]H)(-[6]H)-C(-[2]H)(-[6]H)-C(-[2]OH)(-[6]H)-H}
        \end{center}
        
        Yes or No
        \\
        Why?
        \\
        
        \item Are the following geometric isomers?
              \begin{center}
            \chemfig[double bond sep=0.3em,atom sep=2.5em]{H_3C-[:-60,,2]C(-[:-120]H)=C(-[:-60]CH_3)-[:60]@{X}H}
            \hspace{1em}
%           \chemfig{@{H}}
            \hspace{1em}
            \chemfig[double bond sep=0.3em,atom sep=2.5em]{H_3C-[:-60,,2]C(-[:-120]H_3C)=C(-[:-60]H)-[:60]H}
        \end{center}
    \end{enumerate}

\begin{tikzpicture}[remember picture, overlay]
    \draw[red,fill=gray,opacity=0.5]($(H.north east)+(6cm,1cm)$)rectangle($(H.south 
    west)+(-5cm,-1cm)$); 
        \draw[blue,line width=2pt,fill=teal,opacity=0.3]($(X.north east)+(5cm,4pt)$)rectangle($(X.south 
    west)+(-3cm,-5em)$); 
\end{tikzpicture}
    

\end{document}