Nachfolgend noch eine weitere angepasste Lösung.
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%
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\chemfig{H_3C-C(-[:90]CH_3)(-[:-90]CH_3)-@{1}\charge{90:1pt=\:,-90:1pt=\:}{O}-[:-40]H}
\arrow{0}[,0.1] + \arrow{0}[,0.25]
\chemfig{@{4}\charge{0:2pt=$\scriptstyle{\oplus}$}{O}(-[:70,.9,,,line width=3pt,dash pattern=on 0.8pt off 2pt]H)(<[@{3}:110]@{2}H)-[:-90]H}
\arrow{->[][--$\,\mathsf{H_2O}$][1pt][]}[,1.2,-Latex]
\chemfig{H_3C-C(-[:90]CH_3)(-[:-90]CH_3)-\charge{-90:1pt=\:,90:3pt=$\scriptstyle{\oplus}$}{O}(-[:40]H)-[:-40]H}
\schemestop
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