The Acyloin condensation

Introduction: The Acyloin Condensation

Today's post is about the Acyloin condensation.In the Acyloin condensation, two esters are reduced with sodium forming an; α-hydroxy ketone, which is also called an acyloin.

The Mechanism

The Source Code

 Here is the TikZ Source Code of the mechanism:

\documentclass[12pt,landscape]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}

\usepackage{tikz}
\usetikzlibrary{calc}
\usepackage{chemfig}


\begin{document}
%%
%%
\pagestyle{empty}
\begin{tikzpicture}
%% Title
\node at(0.7,1.3){\bf The Acyloin Condensation\rm};

%% Some Styling
%% For reaction arrows:
\tikzstyle{eqArrowSt}=[ thick,->, >={stealth} ]
\tikzstyle{markupStyle}=[ red,thick,dashed,->,>={stealth} ]
\tikzstyle{elMovement}=[ blue,opacity=0.7,->,>={stealth} ]

%% For radicals:
\tikzstyle{radStyle}=[ red ]

%% The reactants:
\node at (0,0) {\chemfig{R-[::30]@{R1E2}(=[::60]O)-[::-60]O|Me}};
\node at (-1,0){2};
\node at (1.5,0) {+};
\node at (3,0) {4~\chemfig{Na^0}};

%% The first reaction step:
\draw[ eqArrowSt ](3.8,0)--(6.2,0) node[ above, midway ]%%
 {-4[\chemfig{Na^\oplus}\chemfig{ ^\ominus OCH_3}]};
 
%% The intermediats:
\node at (9,0) {\chemfig{R-[::45](-[::90]Na^\oplus|O^\ominus)=[::-45](-[::45]O^\ominus|Na^\oplus)-[::-45]R}};

%% The second reaction step:
\draw[ eqArrowSt ](11,0)--(12.5,0) node[ above, midway ]%%
 {\chemfig{H_2O}};
 
%% The final product:
\node at (14.5,0) {\chemfig{%%
R-[::45](=[::90]O)-[::-45](-[::45]O|H)-[::-45]R}}; 

%% Show more detailed steps
%% Two radicals are build:
\draw[ eqArrowSt ](0.0,-1)--(0.0,-2.5) node[ right, midway]%%
{2~\chemfig{Na^0@{R1E1}}};%%

%% First electron movement:
\coordinate (R1r1) at ($(R1E2) + (0.05,-0.1)$);
\chemmove{\draw[ elMovement,rounded corners ](R1E1)--++(0.2,0.2)--(R1r1);}

%% Second electron movement:
\coordinate (R1r2) at ($(R1E2) + (0,0.25)$);
\chemmove{
\draw[ elMovement,rounded corners ]%%
 (R1r2)--++(0.3,0.2)--($(R1r2)+(0.2,1)$);}

%% Draw the first radical
\node at (0,-4.0) {\chemfig{R-[::45](-[::45]O|^\ominus)-[::-90]O|Me}};
\node[ radStyle ] at (-0.1,-4.1) {\Large $\cdot$\normalsize};

%% Draw the second radical with an xshift of 3cm
\begin{scope}[xshift=3cm]
\node at (0,-4.0) {\chemfig{R-[::45](-[::45]O|^\ominus)-[::-90]O|Me}};
\node[ radStyle ] at (-0.1,-4.1) {\Large $\cdot$\normalsize};
\end{scope}

%% Draw the electron movement between the two radical electrons:
\draw[ elMovement,rounded corners=10mm ] ($(-0.1,-4.1)+(0.1,0.1)$) --(1.5,-3.5)--%%
 ($(2.9,-4.1)+(-0.1,0.1)$);

%% Draw the next step of the reaction:
\draw[ eqArrowSt ](0,-6)--(0,-7.5);%%
\node at (0,-8) {\chemfig{R-[::45](-[::90]Na^\oplus|O^\ominus)(-[::135]MeO)-[::-45](-[::45]O^\ominus|Na^\oplus)(-[::0]OMe)-[::-45]R}};

%% Next Step under elimination of OMe:
\draw[ eqArrowSt ](3,-8)--(4.5,-8)node[above,midway]%%
{-[\chemfig{Na^\oplus}\chemfig{^\ominus OCH_3}]};%%
\node at (7.5,-8) {\chemfig{R-[::45](-[::90]Na^\oplus|O^\ominus)(-[::135]MeO)-[::-45](=[::45]O)-[::-45]R}};

%% Electron Movement:
\draw[ elMovement, rounded corners=5mm ] (1.2,-7) --(0.7,-6.8)-- (0.6,-8);

%% Another elimination of OMe:
\draw[ eqArrowSt ](9.5,-8)--(11,-8)node[above,midway]%%
{-[\chemfig{Na^\oplus}\chemfig{^\ominus OCH_3}]};%%
\node at (12.5,-8) {\chemfig{R-[::45](=[::90]O)-[::-45](=[::45]O)-[::-45]R}};

%% Another radical intermediate with two sodium atoms:
%% Next Step under elimination of OMe:
\draw[ eqArrowSt ](12.5,-7.6)--(12.5,-6.5);%%
\node at (12.5,-5.5) {\chemfig{R-[::45](-[::90]Na^\oplus O^\ominus)-[::-45](-[::45]O|^\ominus|Na^\oplus)-[::-45]R}};

%% Draw radical electrons:
\node[ radStyle ] at (13.2,-5.55) {\Large $\cdot$\normalsize};
\node[ radStyle ] at (11.7,-5.55) {\Large $\cdot$\normalsize};

%% Draw the electron movement:
\draw[ elMovement,rounded corners=5mm ] %%
(13.2,-5.55)--(13.0,-5.15)--(12.45,-5.55);
\draw[elMovement,rounded corners=5mm] %%
(11.7,-5.55) --(11.9,-5.15)-- (12.45,-5.55);

%% Reconnecting to the old (overall) path
\draw[ markupStyle ] (10.5,-5.5) -- (7,-5.5)--(7,-1.7);

%% Showing the keto-enol tautomerism
\draw[ thick,dashed,draw=black,>=stealth,-> ] (9,-1.4) -- (9,-3.2) -- (14,-3.2) --(14,-1.4);
\node at ( 11.5,-3.5 ) {keto form is preferred};
\end{tikzpicture} 

\end{document}

Series: Enol/Enolate chemistry

Part II is about the α-substitution of enols/enolates.

Series: Enol/Enolate chemistry

A little series of enol/enolate chemistry will now follow, with the aim to give a short overview of different chemical reactions involving enols/enolates.

Part I is about the keto-enol-tautomerism. 

An NMR task:

The First NMR:

The Second NMR:

The Third NMR:

And the TikZ Source Code:

\documentclass[a4paper,landscape]{article}

\usepackage[english]{babel}
\usepackage[utf8x]{inputenc}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage[colorinlistoftodos]{todonotes}
\usepackage{tikz}
\usetikzlibrary{shadows}
\usepackage{chemfig}

\begin{document}
\pagestyle{empty}
%% A Little NMR Quiz


%% Some Macro definitions
\def\peaka{10}
\def\peakb{7.88}
\def\inta{5.31}
\def\peakc{7.876}
\def\intc{6.11}
\def\peakd{7.873}
\def\intd{2.92}
\def\peake{7.86}
\def\inte{2.4}
\def\peakf{7.858}
\def\intf{7.0}
\def\peakg{7.855}
\def\intg{6.75}

\begin{tikzpicture}
\node[ draw=blue,rounded corners,fill=blue!50!green,drop shadow ] (0,0) {
\begin{minipage}{.7\textwidth}
Find the right NMR spectra for the following three compounds:\\
~\\
\chemfig{[:30]*6(-=-=-(-[::-60](-[::-45]H)=[::45]O)=)}\\
~\\
\chemfig{-[::30]-[::-60](=[::-60]O)-[::60]O-[::-60]-[::60]}
\end{minipage}};\\
~\\
~\\
~\\
\chemfig{-[::45](=[::45]O)-[::-90]}\\
\end{tikzpicture}

\begin{tikzpicture}[xscale=-1]

%% benzaldehyde

%% Background stuff
\draw[fill=green!50!blue!50!white,opacity=0.7,rounded corners,drop shadow]%%
    (-2,-2)rectangle(13,12);

%% Some Styling
\tikzstyle{axesStyle}=[very thick,->,>={stealth}]
\tikzstyle{peakStyle}=[thick]

%% Coordinate Stuff
\coordinate (P0) at (0,0);
\coordinate (Px) at (12,0);
\coordinate (Py) at (0,11);

%% Draw the axes
\draw[axesStyle](P0) -- (Py);
\draw[axesStyle](P0) -- (Px);

%% Labeling of the axes
\node[rotate=-90] at (-0.3,5.5) {Intensity};
\node at (6,-0.7) {Shift[ppm]};

%% Draw the Peaks
\draw[peakStyle](\peaka,10)--(\peaka,0);
\draw[peakStyle](\peakb,\inta)--(\peakb,0);
\draw[peakStyle](\peakc,\intc)--(\peakc,0);
\draw[peakStyle](\peakd,\intd)--(\peakd,0);
\draw[peakStyle](\peake,\inte)--(\peake,0);
\draw[peakStyle](\peakf,\intf)--(\peakf,0);
\draw[peakStyle](\peakg,\intg)--(\peakg,0);
\draw[peakStyle](7.51,6.54)--(7.51,0);
\draw[peakStyle](7.53,4.9)--(7.53,0);

%% 
\foreach \x in {0,1,2,...,11}
{
    \node at (\x,-0.3) {\x};
    \draw[very thick](\x,0.2)--(\x,0);
    \draw(\x+0.5,0.15)--(\x+0.5,0);
}
\end{tikzpicture}

\def\peakone{4.132}
\def\intone{5}
\def\peaktwo{2.32}
\def\inttwo{5}
\def\peakthree{1.26}
\def\intthree{9}
\def\peakfour{1.14}
\def\intfour{9}

\begin{tikzpicture}[xscale=-1]
%% Ethyl propionate

%% Background stuff
\draw[fill=green!50!blue!50!white,opacity=0.7,rounded corners,drop shadow]%%
    (-2,-2)rectangle(13,12);

%% Some Styling
\tikzstyle{axesStyle}=[very thick,->,>={stealth}]
\tikzstyle{peakStyle}=[thick]

%% Coordinate Stuff
\coordinate (P0) at (0,0);
\coordinate (Px) at (12,0);
\coordinate (Py) at (0,11);

%% Draw the axes
\draw[axesStyle](P0) -- (Py);
\draw[axesStyle](P0) -- (Px);

%% Labeling of the axes
\node[rotate=-90] at (-0.3,5.5) {Intensity};
\node at (6,-0.7) {Shift[ppm]};

\draw[peakStyle] (\peakone,\intone) -- (\peakone,0);
\draw[peakStyle] (\peaktwo,\inttwo) -- (\peaktwo,0);
\draw[peakStyle] (\peakthree,\intthree) -- (\peakthree,0);
\draw[peakStyle] (\peakfour,\intfour) -- (\peakfour,0);
%% 
\foreach \x in {0,1,2,...,11}
{
    \node at (\x,-0.3) {\x};
    \draw[very thick](\x,0.2)--(\x,0);
    \draw(\x+0.5,0.15)--(\x+0.5,0);
}


\end{tikzpicture}


\begin{tikzpicture}[xscale=-1]
%% acetone

%% Background stuff
\draw[fill=green!50!blue!50!white,opacity=0.7,rounded corners,drop shadow]%%
    (-2,-2)rectangle(13,12);

%% Some Styling
\tikzstyle{axesStyle}=[very thick,->,>={stealth}]
\tikzstyle{peakStyle}=[thick]

%% Coordinate Stuff
\coordinate (P0) at (0,0);
\coordinate (Px) at (12,0);
\coordinate (Py) at (0,11);

%% Draw the axes
\draw[axesStyle](P0) -- (Py);
\draw[axesStyle](P0) -- (Px);

%% Labeling of the axes
\node[rotate=-90] at (-0.3,5.5) {Intensity};
\node at (6,-0.7) {Shift[ppm]};

\draw[peakStyle] (2.2,7) -- (2.2,0);

%% 
\foreach \x in {0,1,2,...,11}
{
    \node at (\x,-0.3) {\x};
    \draw[very thick](\x,0.2)--(\x,0);
    \draw(\x+0.5,0.15)--(\x+0.5,0);
}


\end{tikzpicture}



\end{document}

A quiz-like exercise to aromatic compounds (Hückel's rule, planarity, conjugation):

Here's the corresponding TikZ code (using chemfig):

\documentclass[12pt,a4paper,landscape]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{tikz}
\usetikzlibrary{shadows}

\usepackage{chemfig}


\begin{document}
\pagestyle{empty}
\begin{tikzpicture}
%% ----------------- A little Quiz ----------------------------
%% Which of the following Compounds are aromatic?
%% 
\node at (3,-1) {\bf Which of the following compounds are aromatic? \rm};

%% Cyclopenta = cyclopentadienylanion!
\coordinate (Cyclopenta) at (0,-3);
\node at (Cyclopenta) {\chemfig{*5(@{C1}-=-=-)}};
\chemmove{
\node[anchor = north] at (C1) {\chemfig{\ominus}};
}

%% Naphtalene
\coordinate (Naphta) at (4,-3);
\node at (Naphta) {\chemfig{[:-60]*6(=-=-=-)-[::-150]=^[::60]-[::60]=^[::60]-[::60]}};

%% Antracene
\coordinate (Antra) at (10,-3);
\node at (Antra) {\chemfig{[:-60]((*6(=-=-(-[::-60]=_[::-60]-[::-60]=_[::-60]-[::-60])%%
=-))-[::-150]=^[::60]-[::60]=^[::60]-[::60])}};

%% Next Line
%% More Molecules

%% Cyclopropanylanion and Cyclopropanylcathion
\coordinate (Propanion) at (0,-6);
\node at (Propanion) {\chemfig{-[::60]@{C2}-[::-120]=[::-120]}};
\chemmove{
\node[ anchor = south ] at (C2) {$\ominus$};
} 

\coordinate (Propcathion) at (2,-6);
\node at (Propcathion) {\chemfig{-[::60]@{C3}-[::-120]=[::-120]}};
\chemmove{
\node[ anchor = south ] at (C3) {$\oplus$};
} 

%% Cyclobutane
\coordinate (Cyclobut) at (4,-6);
\node at (Cyclobut) {\chemfig{*4(-=-=-)}};

%% Cyclobutane-cathion
\coordinate (Cyclobutcathion) at (6,-6);
\node at (Cyclobutcathion) {\chemfig{*4(-=-=)}};
\coordinate (Q) at (5.5,-6.5);
\node[anchor = north east] at (Q) {$\oplus$};

%% Cyclobutane-anion
\coordinate (Cyclobutanion) at (8,-6);
\node at (Cyclobutanion) {\chemfig{*4(-=-=)}};
\coordinate (Q') at (7.5,-6.5);
\node[anchor = north east] at (Q') {$\ominus$};

%% Some more fancy molecules
\coordinate (TwoProp) at (10,-6);
\node at (TwoProp) {\chemfig{@{N}-[::60]=[::-120](-[::120]@{N'}-[::+120])-[::-120]}};
\chemmove{
\node[ anchor=north ] at (N) {$\oplus$};\node[anchor=south] at (N') {$\oplus$};
}

%% Solution:
\node at (-0.6,-8) {\bf Solution:\rm};
\begin{scope}[]
\node[ white,rotate = 180,draw=blue,fill = blue!50!white!50!gray,rounded corners ] at( 3, -11.5 ) {%%
\begin{minipage}{.75\textwidth}
{
\scriptsize
\begin{itemize}
\item 6-$\pi-\mathrm{e^-}$, planar, conjugated$=>$aromatic
\item 10-$\pi-\mathrm{e^-}$, planar, conjugated$=>$aromatic
\item 14-$\pi-\mathrm{e^-}$, planar, conjugated$=>$aromatic
\item 4-$\pi-\mathrm{e^-}$, planar, conjugated$=>$anti-aromatic
\item 2-$\pi-\mathrm{e^-}$, planar, conjugated$=>$aromatic
\item 4-$\pi-\mathrm{e^-}$, planar, conjugated$=>$anti-aromatic
\item 4-$\pi-\mathrm{e^-}$, planar, conjugated$=>$anti-aromatic
\item 6-$\pi-\mathrm{e^-}$, planar, not conjugated$=>$anti-aromatic
\item 2-$\pi-\mathrm{e^-}$, planar, conjugated$=>$aromatic,but not stable!
\end{itemize}
}
\end{minipage}
}
;
\end{scope}
\end{tikzpicture}
\end{document}

Today's Post is about aromatic compounds and benzene:

 Here is the corresponding TikZ source code (using chemfig):

\documentclass[12pt,a4paper,landscape]{article}
\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{tikz}
\usetikzlibrary{shadows}

\usepackage{chemfig}


\begin{document}
\pagestyle{empty}
\begin{tikzpicture}
%% Make a Background:
\draw[ fill = blue!50!white!50!gray, rounded corners, drop shadow ]%%
 ( -1, 1 ) rectangle ( 14, -14 );

%% Some Styling:
\tikzstyle{titleStyle}=[yellow!50!brown]

%% Kekule Structures:
\node[ titleStyle ] at ( 0, 0 ) {Kekule};
\node at ( 2, -2 ){\chemfig{*6(=-=-=-)}};
\node at ( 6, -2 ){\chemfig{*6(-=-=-=)}};
\draw[<->,thick] ( 3.5, -2 ) -- ( 4.5, -2 );

%% So the Kekule Structures give the following net state:
\node at ( 8.5, -2 ) {=};
\node at ( 10.8, -2 ) {\chemfig{*6(-------)}};
\draw[ dashed ] ( 10.8, -2 ) circle ( 1cm );

%% Draw the Dewar-Benzene also:
\node[ titleStyle ] at ( .7, -3.5 ){Dewar-Benzene};
\node at ( 2, -5 ){\chemfig{*4(-?[a]--=)--=^[::90]-[::90]}};


%% Ladenburg-Benzene:
\node[ titleStyle ] at ( 6.5, -4 ){Ladenburg-Benzene}; 
\node[very thick] at( 6, -5 ){\chemfig{%%
    >[::60,,,,line width=.01mm,line join=round]?[a]<[:-60]?[b]-[:180,,,,line width=2.5mm,line join=round]?[c]-%%(-?[a])%%-(-?[b])-(-[?[c]])%%
}};
\begin{scope}[yshift=-0.6cm]
\node[very thick] at( 6, -5 ){\chemfig{%%
    -[::60]?[a](-[::30])-[:-60]?[b](-[::150,,,,,line width=2mm])-[:180,,,,,line width =2mm]?[c](-[::-90,,,,,line width=2mm])-%%(-?[a])%%-(-?[b])-(-[?[c]])%%
}};
\end{scope}

%% Take care of the bad joints in the Ladenburg-Benzene:
%% Define Coordinate Points to colour black:
\coordinate (A) at (5.38,-5.4);
\coordinate (B) at (6.62,-5.4);
\coordinate (C) at (5.38,-6.7);
\coordinate (D) at (6.62,-6.7);

%% Fill The Coordinates With black Circles:
\draw[fill=black] (A) circle (1.2mm);
\draw[fill=black] (B) circle (1.2mm);
\draw[fill = black] (C) circle (1.2mm);
\draw[fill = black] (D) circle (1.2mm);

%% Typical Reactions of Aromatic Compounds
%% 
%% The Friedel-Crafts-Acylation
\node[ titleStyle ] at (1.5,-7.5) {Friedel-Crafts alkylation};

%% Reactants of the Friedel-Crafts-Alkylation
\node at (1,-9) {\chemfig{*6(=-=-=-)}};
\draw[ >=stealth, very thick, -> ] (3,-9)--(5,-9);%%
\node at ( 4, -8.5 ) {\chemfig{[:112.5]R-[::45]-[::45]X}};
\node at (4,-9.5) {\chemfig{AlCl_3}(kat.)};

%% Products of the Friedel-Crafts-Alkylation
\node at (8,-9) {\chemfig{[:180]*6(=-=-=-)-[::210]-[::-45]R }};
\node at (9.2,-9) {+};
\node at (10.7,-9){\chemfig{H-Cl}};

%% Hückel's rule:
\node[fill = blue!50!gray,rounded corners,drop shadow,general shadow={blue!50!black,opacity=0.9}] at ( 10, -6 ){$[\pi - e^\ominus]=4n+2$};

\end{tikzpicture}
\end{document}

Today's post is about the formation of isophorone out of acetone.

And here's the corresponding TikZ source code:
\documentclass[a4paper]{article}

\usepackage[english]{babel}
\usepackage[utf8x]{inputenc}
\usepackage{amsmath}
\usepackage{graphicx}
\usepackage[colorinlistoftodos]{todonotes}
%% Using Tikz:
\usepackage{tikz}
%% ------- Tikz packages ------
%% Having better arrows in chem. equations:
\usetikzlibrary{arrows}
\usetikzlibrary{shadows}

%% -------- End of Tikz packages -----
%% Use chemfig package for chemical stuff:
\usepackage{chemfig}



\begin{document}


%% First Blog Entry
\begin{tikzpicture}


%% TikZ Styles:
\tikzstyle{namingColor} = [white]
\tikzstyle{titleColor} = [namingColor]


%% The Background:
\draw[ draw=blue,fill=blue!50!white!50!gray,drop shadow,rounded corners ]%%
    (-1.4,3.2) rectangle (12,-10);

\node[ titleColor ] at (4.5,2.4) {Intra-molecular \it Micheal\rm-reaction:~The formation of Isophorone};

%% Aldol addition between one acetone molecule and another:
\node at ( 0, 0 ) {%%
    \chemfig{-[::45](=[::45]O)-[::-90]-[::-45]H} };
\draw[ ->,>={stealth},thick ] ( 2, 0 ) -- ( 4, 0 )%%
node [ above, midway ] { + \chemfig{B^\ominus}}%%
node [ below, midway ] { - HB };

%% Electron Arrows:
\coordinate (PBaseStart) at (3,0.5);%% Start of Electron Flow
\coordinate (PBaseEnd) at (0.8,-1.2);%% Control Point for the curve
\coordinate (PBaseControl) at (1.4,1.9);%% End of Electron Flow
\draw[ ->, rounded corners=20mm ] (PBaseStart) -- (PBaseControl) -- (PBaseEnd);

%% Enolate:
\node at (5.5,0.5){%%
    \chemfig {-[::45](-[::45]\lewis{2:,O}|^\ominus)=[::-90]}};
  
%% The Enolate attacks another molecule acetone to form the aldol
%% addition Product:
%% Reaction Arrow:
\draw[ ->,>={stealth},thick ] ( 7, 0 ) -- ( 9, 0 );%%
\draw ( 8, 1.0 ) node {%% Another molecule Acetone
    \chemfig{-[::45](=[::45]O)-[::-90]} } ;

%% The 1. electron flow of the aldol addition reaction
\coordinate (P2BaseStart) at (5.3,1.5);%% Start of Electron Flow
\coordinate (P2BaseEnd) at (5.3,0.7);%% Control Point for the curve
\coordinate (P2BaseControl) at (3.4,1.2);%% End of Electron Flow
\draw[ -<, rounded corners=20mm ] (P2BaseStart) --%%
(P2BaseControl) -- (P2BaseEnd);

%% The 2. electron flow of the aldol addition reaction
\coordinate (P2BaseStart) at (5.8,-0);%% Start of Electron Flow
\coordinate (P2BaseEnd) at (7.8,0.7);%% Control Point for the curve
\coordinate (P2BaseControl) at (7,1.5);%% End of Electron Flow
\draw[ ->, rounded corners=10mm ] (P2BaseStart) --%%
(P2BaseControl) -- (P2BaseEnd);
´

%% The 3. electron flow of the aldol addition reaction
\coordinate (P2BaseStart) at (8.1,1.2);%% Start of Electron Flow
\coordinate (P2BaseEnd) at (8.2,1.8);%% Control Point for the curve
\coordinate (P2BaseControl) at (8.4,1.45);%% End of Electron Flow
\draw[ ->, rounded corners=2mm ] (P2BaseStart) --%%
(P2BaseControl) -- (P2BaseEnd);

%% The Product of the aldol addition
\node at (10.4,0) {%%
\chemfig{-[::45](-[::45]O|^\ominus)(-[::-45])-[::-90]-(=[::90]O)-[::-45]}
};
\node[ namingColor ] at (10,-1) {$\beta$-hydroxy carbonyl};
;

%% The Aldol Condensation Reaction:
\draw[ ->,>={stealth},thick ] ( 9.5, -1.5 ) -- ( 9.5, -3 )%%
node[ left, midway ] {+\chemfig{H^\oplus}}
node[ right, midway] {-\chemfig{H_2O}};

%% The Product of the Aldol Condensation Reaction:
\node at (10,-5.5) {%%
\chemfig{-[::45](-[::-90])=[::45]-[::45](=[::-45]O)-[::90]}};

%% Name the Product:
\node[ namingColor ] at (10,-7.8) {mesityl oxide};

%% The Michael Addition of yet another molecule acetone:
\draw[ ->,>={stealth},thick ] ( 9.0, -5 ) -- ( 7.0, -5 );%%
\node at ( 8, -4.0 ) {\chemfig {-[::45](-[::45]\lewis{2:,O}|^\ominus)=[::-90]}};

%% The 1. electron flow of the Michael Reaction:
\coordinate (P2BaseStart) at (8.2,-4.4);%% Start of Electron Flow
\coordinate (P2BaseEnd) at (10.3,-6.8);%% Control Point for the curve
\coordinate (P2BaseControl) at (8.5,-6.5);%% End of Electron Flow
\draw[ ->, rounded corners=15mm ] (P2BaseStart) --%%
(P2BaseControl) -- (P2BaseEnd);

%% The 2. electron flow of the Michael Reaction:
\coordinate (P2BaseStart) at (10.3,-6.4);%% Start of Electron Flow
\coordinate (P2BaseEnd) at (10.,-5.3);%% Control Point for the curve
\coordinate (P2BaseControl) at (8.5,-6.5);%% End of Electron Flow
\draw[ ->, rounded corners=15mm ] (P2BaseStart) --%%
(P2BaseControl) -- (P2BaseEnd);

%% The Product of the Michael Addition:
\node at (5.5,-4.5) {
\chemfig{-[::45](-[::45]O|^\ominus)=[::-90]-[::-45]([::45]-)([::80]-)-[::-45]-[::-90](=[::45]O)-[::-90]}};

%% Keto-Enol-Tautomerism:
\draw[ ->,>={stealth},thick ] ( 4.7, -6 ) -- ( 4.2, -6.5 )%%
node[ above, midway,rotate=45 ] {+\chemfig{H^\oplus}};

%% The Product of the Michael Addition after the Keto-Enole-Tautomerism
\node at (3.2,-6.8) {
\chemfig{-[::45](=[::45]O)-[::-90]-[::-45]([::45]-)([::80]-)-[::-45]-[::-90](=[::45]O)-[::-90]}};

%% Reaction To The Final Product:
\draw[ ->,>={stealth},thick ] ( 2.6, -6.4 ) -- ( 1.2, -5 )%%
node[ above, midway,rotate=-45 ] {\chemfig{H^\oplus}; -\chemfig{H_2O}};

%% The Final Product:
\node at (0.5,-3){
\chemfig{-[::45](=[::45]O)-[::-90]-[::-45]([::45]-)([::80]-)-[::-45]-[::-90](=[::-45])-[::+90]}};

%% Naming of the Final Product:
\node[ namingColor ] at (0.48, -5.5) {isophorone};

\end{tikzpicture}


\end{document}