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}
