% pdflatex thca.tex
% /home/jmc31337/Downloads/ChemLabLatex/bin/mol2chemfig -w -i direct ''
\documentclass{article}
\usepackage[portrait,margin=1cm]{geometry}
\usepackage{chemfig}
\usepackage{tikz}
\usepackage{graphicx}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{mathtools}
\usepackage{pdfpages}
\usepackage{url}
\usepackage{xcolor}
\usetikzlibrary{calc}
\usetikzlibrary{decorations.markings}
\usetikzlibrary{arrows.meta}
\usetikzlibrary{positioning}
\usetikzlibrary{decorations.pathmorphing}
\usetikzlibrary{arrows.meta, calc}
\usetikzlibrary{matrix}
\newcommand{\mcfcringle}[1]{%
\tikz[baseline=-0.5ex]{
\draw (0,0) circle (#1em);
}%
}
\setchemfig{
charge style={draw=none},
atom sep=1.5em
}
\newcommand{\lp}{\rotatebox{90}{$:$}}
\newcommand{\pka}[1]{\scriptsize p$K_a$=#1}
\newcommand{\pkb}[1]{\scriptsize p$K_b$=#1}
\title{Birch Reduction of THCA and Subsequent Protonation/Aromatization Rearrangement under Anhydrous Ammonia Conditions}
\author{jmc}
\date{August 14, 2026}
\begin{document}
\maketitle
\textit{NOTE: Any differences in bond angles, orientations, or wedge/hash placement between depictions represent alternative 2D projections or conformational views of the same molecular connectivity and stereochemical configuration, and do not necessarily indicate different molecular structures. All SMILES codes were from IBM RXN and converted with mol2chemfig for pdflatex. All blue arrow mechanisms represent 1e- and all red arrows represent 2e-}
% ============================================================
% REACTION SCHEME
% ============================================================
\begin{center}
\begin{tikzpicture}[x=1cm,y=1cm]
\node (THCA_label) at (0,2) {
\textbf{THCA}
};
\node (THCA) at (0,-.5) {
\chemfig{
% 1
-[:330]% 2
-[:30]% 3
-[:330]% 4
-[:30]% 5
-[:330]% 6
-[:270]% 7
(
-[:30,,,,draw=none]\mcfcringle{1.3}% (o)
)
-[:330]% 8
-[:30]% 9
(
-[:90]% 10
(
-[:150]% 12
(
-[:90]% 13
(
-[:150,,,2]HO% 15
)
=[:30]O% 14
)
-[:210]% -> 6
)
-[:30,,,1]OH% 11
)
>:[:330]% 16
-[:30]% 17
=_[:330]% 18
(
-[:30]% 19
)
-[:270]% 20
-[:210]% 21
>[:150]% 22
(
-[:90]% -> 16
)
-[:210]% 23
(
-[:310]% 24
)
(
-[:230]% 25
)
-[:150]O% 26
(
-[:90]% -> 8
)
}
};
% Reaction arrow
\node (arrow) at (3.3,0) {$\longrightarrow$};
% THCA-ANALOG
\node (THCA_Analog_label) at (6.5,2) {
\textbf{THCA-Analog}
};
\node (THCA_Analog) at (6.5,0) {
\rotatebox{-30}{%
\chemfig
{
% 1
-% 2
-[:60]% 3
-% 4
-[:60]% 5
-% 6
=_[:60]% 7
(
-[:120]% 8
(
-[:180,,,2]HO% 10
)
=[:60]O% 9
)
-% 11
(
-[:60,,,1]OH% 12
)
-[:300]% 13
=^[:240]% 14
(
-[:180]% 15
-[:120]% -> 6
)
-[:300]O% 16
-% 17
(
-[:260]% 18
)
(
-[:340]% 19
)
>[:60]% 20
-% 21
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)
=^[:180]% 25
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-[:180]% -> 13
)
(
-[:300]% -> 20
)
}}
};
\end{tikzpicture}
\end{center}
%========================================================================================
%========================================================================================
% ============================================================
% EXPLANATORY TEXT
% ============================================================
\medskip
\noindent
\textbf{Ammonia gas below its boiling temperature of $-33.4\,^\circ\mathrm{C}$ creates a liquified medium where an alkali metal such as Li can release its single outer-valence electron ($e^-$)}
\medskip
\noindent
\textbf{Liquid ammonia is a polar protic solvent. It can act as a base by
accepting protons ($\mathrm{H^+}$), and the N--H bonds make it protic under
the right conditions, allowing it to give up a proton}
\medskip
\noindent
\textbf{The fact that ammonia is basic and capable of accepting
$\mathrm{H^+}$ will be used later. At this point, the ammonia medium
solvates the $\mathrm{Li^+}$ ions and the electrons ($e^-$) indicated by the barbed half-hook blue arrow}
\medskip
\[
\mathrm{Li(s) + NH_3(l) \longrightarrow Li^+ + e^-}
\]
\vspace{-40pt}
\begin{center}
\begin{tikzpicture}
% Reactants
\node (Li) at (0,0) {\charge{0:=$\bullet$}{Li}};
\node (plus) at (0.8,0) {$+$};
\node (NH3) at (2.2,0) {
\chemfig{\charge{180:=$:$}{N}(-[:90]H)(-[:270]H)-H}
};
% Reaction arrow
\node (arrow) at (3.3,0) {$\longrightarrow$};
% Products
\node (Liion) at (4.4,0) {$\mathrm{Li^+}$};
\node (plus2) at (4.9,0) {$+$};
\node (esolv) at (6.3,0) {$e^-_{\mathrm{solv}}$};
% Fishhook
\draw[
line width=0.5pt,
-{Straight Barb[length=5pt,width=1pt]}
]
(Li.east) .. controls +(90:60pt) and +(180:40pt) .. (esolv);
\draw[
blue,
line width=1pt,
-{Straight Barb[length=5pt,width=1pt]}
]
(Li.east)
.. controls +(90:60pt) and +(180:40pt) ..
(esolv);
\end{tikzpicture}
\end{center}
% -\charge{90:=$\bullet$}{C}
\noindent
\textbf{After solvating, we can add in the THCA (Note: we must have pure as possible free THCA oil. Whatever maceration and extraction method chosen must provide a pure THCA oil, without chlorophyll or any added impurities. (Ie., if using butane hash oil, the butane must be fully evaoprated and stripped from the final THCA oil)}
\medskip
\noindent
\textbf{Once THCA oils are added to the ammonia solvated e- medium, ammonia's lone pair could potentially serve as the base and proton acceptor in this medium. The curved arrow notations below show; the e- attacking the benzene ring to possibly form a rearrangment, and, ammonia's lone pair attacking the ($\mathrm{H^+}$) protons of THCA's two sites (carboxylic acid -OH and phenol -OH)}
%=================================================================================================
%=================================================================================================
\medskip
\begin{center}
\begin{tikzpicture}[x=1cm,y=1cm]
\node (THCA_label) at (0,2) {
\textbf{THCA}
};
\node (THCA) at (0,-.5) {
\chemfig{
% 1
-[:330]% 2
-[:30]% 3
-[:330]% 4
-[:30]% 5
-[:330]% 6
-[:270]% 7
(
-[:30,,,,draw=none]\mcfcringle{1.3}% (o)
)
-[:330]% 8
-[:30]% 9
(
-[:90]% 10
(
-[:150]% 12
(
-[:90]% 13
(
-[:150,,,2]HO% 15
)
=[:30]O% 14
)
-[:210]% -> 6
)
-[:30,,,1]OH% 11
)
>:[:330]% 16
-[:30]% 17
=_[:330]% 18
(
-[:30]% 19
)
-[:270]% 20
-[:210]% 21
>[:150]% 22
(
-[:90]% -> 16
)
-[:210]% 23
(
-[:310]% 24
)
(
-[:230]% 25
)
-[:150]O% 26
(
-[:90]% -> 8
)
}
};
\node (plus) at (3.5,0) {\Large $+$};
\begin{scope}[shift={(6,.5)}]
\node (NH3_label) at (0,1.5) {
\textbf{Anhyd. Ammonia Medium}
};
\node (NH3_1) at (0,-.3) {
\chemfig{
\charge{180:=$:$}{N}
(-[:90]H)
(-[:270]H)
-H
}
};
\node (NH3_2) at (0,-2) {
\chemfig{
\charge{180:=$:$}{N}
(-[:90]H)
(-[:270]H)
-H
}
};
\end{scope}
\node (electron_1) at (4.5,1.3) {
$\mathrm{e^-_{(solv)}}$
};
\node (electron_2) at (4.7,-2.5) {
$\mathrm{e^-_{(solv)}}$
};
\node (li_1) at (7.5,1.2) {
\chemfig{\mathrm{Li}^{+}}
};
\node (li_2) at (7.5,-2.5) {
\chemfig{\mathrm{Li}^{+}}
};
% Arrow 1
\coordinate (tail) at (5.5,0.3);
\coordinate (head) at (-.6,1.1);
\coordinate (controlA) at (3.0,2);
\coordinate (controlB) at (1.0,2);
\draw[->, red, thick, >=stealth]
(tail)
.. controls (controlA) and (controlB) ..
(head);
%Arrow 2
\coordinate (tail) at (5.5,-1.4);
\coordinate (head) at (1.4,.3);
\coordinate (controlA) at (3.0,2);
\coordinate (controlB) at (1.5,1);
\draw[->, red, thick, >=stealth]
(tail)
.. controls (controlA) and (controlB) ..
(head);
% e- Arrow 1
\coordinate (head) at (0,0);
\coordinate (tail) at (4.5,-2.2);
\coordinate (controlA) at (1,-2.5);
\coordinate (controlB) at (0,0);
\draw[
blue,
line width=1pt,
-{Straight Barb[length=5pt,width=1pt]}
]
(tail)
.. controls (controlA) and (controlB) ..
(head);
% e- Arrow 2
\coordinate (head) at (.3,-.3);
\coordinate (tail) at (4,1.3);
\coordinate (controlA) at (.5,.5);
\coordinate (controlB) at (.3,.3);
\draw[
blue,
line width=1pt,
-{Straight Barb[length=5pt,width=1pt]}
]
(tail)
.. controls (controlA) and (controlB) ..
(head);
\end{tikzpicture}
\end{center}
%=========================================================================================
%=========================================================================================
\noindent
\textbf{To determine whether or not the Carboxylic and/or Phenolic -OH groups will become deprotonated we must compare the $\mathrm{p}K_a$ of the conjugates (in ammonia solutions not water)}
\noindent
\medskip
\textbf{The fundamental principle is that an acid will donate its proton to a base if the acid is stronger than the conjugate acid of the base. Conversely, a base will accept a proton from an acid if the base is stronger than the conjugate base of the acid. In terms of p$K_a$, the Phenolic acid Ar-OH will deprotonate in the presence of Ammonia if:}
\[
pK_a(\mathrm{Phenol\ Ar-OH}) < pK_a(\mathrm{NH_4^+})
\]
\medskip
\noindent
\textbf{For THCA, there are two acidic hydroxyl groups that must be considered: the phenolic Ar--OH and the carboxylic COOH. Their product favoring proton-transfer equilibria with liquid ammonia are:}
\begin{equation*}
\mathrm{THCA{-}ArOH + NH_3}
\quad
\tikz[baseline=-0.5ex]{
\draw[->] (0,0.08) -- (1.8,0.08);
\draw[<-] (.20,-0.08) -- (1.3,-0.08);
}
\quad
\mathrm{THCA{-}ArO^- + NH_4^+}
\end{equation*}
\begin{equation*}
\mathrm{THCA{-}COOH + NH_3}
\quad
\tikz[baseline=-0.5ex]{
\draw[->] (0,0.08) -- (1.8,0.08);
\draw[<-] (.20,-0.08) -- (1.3,-0.08);
}
\quad
\mathrm{THCA{-}COO^- + NH_4^+}
\end{equation*}
\medskip
\noindent
\textbf{Experimental studies of phenols in liquid ammonia give $pK_a$ values of approximately $3{-}6.2$, with phenol itself reported at $pK_a\approx6.02$. Thus, the THCA phenolic Ar--OH can be reasonably estimated at approximately $pK_a\approx5$ in liquid ammonia. Carboxylic acids are substantially stronger acids in liquid ammonia; using acetic acid as a reference gives an estimated $pK_a$ near $-4.5$. Accordingly, a reasonable working estimate for the THCA carboxylic COOH is $pK_a\approx-4$ to $-5$.}
\medskip
\noindent
\textbf{Therefore, the acidity (lower $pK_a$, higher acidity) of the two THCA hydroxyl groups in liquid ammonia shows:}
\begin{equation*}
\boxed{
\mathrm{THCA{-}COOH}\;(pK_a\approx-4.5)
>
\mathrm{THCA{-}ArOH}\;(pK_a\approx5)
>
\mathrm{NH_4^+}\;(pK_a\approx25)
}
\end{equation*}
\medskip
\noindent
\textbf{Below (as an aside) we see the standardized aqueous water-based relative acidity according to $pK_a$:}
\[
\mathrm{H_3O^+}\;(pK_a\approx-1.74)
>
\mathrm{THCA{-}COOH}\;(pK_a\approx4.5)
>
\mathrm{THCA{-}ArOH}\;(pK_a\approx10)
\]
\medskip
\noindent
\textbf{The acidity does not favor this reaction mechanism when WATER is used as the aqueous medium. However, it does favor deprotonation in liquid ammonia, according to laboratory literature values.}
\medskip
\noindent
\textbf{Birch reduction of aromatic rings also occurs within the medium. See the following scientific literature regarding this mechanism:}
\begin{itemize}
\item \url{https://www.masterorganicchemistry.com/2019/10/17/birch-reduction/}
\item \url{https://www.ch.ic.ac.uk/rzepa/blog/?p=8570}
\end{itemize}
\medskip
\noindent
\textbf{For the Birch intermediate, $NH_3$ functions primarily as the solvent and base. The liquid-ammonia medium is favorable for aromatic-ring Birch reduction. Protonation of the reduced aromatic intermediates will require a separate proton-donating source. In other words, liquid ammonia can deprotonate the two OH substituents, but it will not also provide those protons for protonating the groups back. In the literature example above, EtOH served as the proton-donor quench source (protonation will cbe discussed later)}
\medskip
\noindent
\textbf{Focusing on the reduction of the aromatic ring (assuming the acidic substituent protons have already been deprotonated by ammonia), the mechanism looks like:}
%=================================================================================================
%=================================================================================================
\medskip
\begin{center}
\begin{tikzpicture}[x=1cm,y=1cm]
\node (THCA) at (0,-.5) {
\chemfig{
% 1
-[:330]% 2
-[:30]% 3
-[:330]% 4
-[:30]% 5
-[:330]% 6
-[:270]% 7
(
-[:30,,,,draw=none]\mcfcringle{1.3}% (o)
)
-[:330]% 8
-[:30]% 9
(
-[:90]% 10
(
-[:150]% 12
(
-[:90]% 13
(
-[:150,,,1]
\charge{
175:=$-$,
90:=\rotatebox{90}{$:$},
0:=$:$,
270:=\rotatebox{90}{$:$}
}{O}% 15
)
=[:30]O% 14
)
-[:210]% -> 6
)
-[:30,,,1]\charge{
15:=$-$,
90:=\rotatebox{90}{$:$},
180:=$:$,
270:=\rotatebox{90}{$:$}
}{O}
)
>:[:330]% 16
-[:30]% 17
=_[:330]% 18
(
-[:30]% 19
)
-[:270]% 20
-[:210]% 21
>[:150]% 22
(
-[:90]% -> 16
)
-[:210]% 23
(
-[:310]% 24
)
(
-[:230]% 25
)
-[:150]O% 26
(
-[:90]% -> 8
)
}
};
\node (plus) at (3.5,0) {\Large $+$};
\begin{scope}[shift={(6,.5)}]
\node (NH4_1) at (0,-.3) {
\chemfig{
\charge{45:=$+$}{N}
(-[:90]H)
(-[:0]H)
(-[:270]H)
-[:180]H
}
};
\node (NH4_2) at (0,-2) {
\chemfig{
\charge{45:=$+$}{N}
(-[:90]H)
(-[:0]H)
(-[:270]H)
-[:180]H
}
};
\end{scope}
\node (electron_1) at (4.5,1.3) {
$\mathrm{e^-_{(solv)}}$
};
\node (electron_2) at (4.7,-2.5) {
$\mathrm{e^-_{(solv)}}$
};
\node (li_1) at (7.5,1.2) {
\chemfig{\mathrm{Li}^{+}}
};
\node (li_2) at (7.5,-2.5) {
\chemfig{\mathrm{Li}^{+}}
};
% e- Arrow 1
\coordinate (head) at (0,0);
\coordinate (tail) at (4.5,-2.2);
\coordinate (controlA) at (1,-2.5);
\coordinate (controlB) at (0,0);
\draw[
blue,
line width=1pt,
-{Straight Barb[length=5pt,width=1pt]}
]
(tail)
.. controls (controlA) and (controlB) ..
(head);
% e- Arrow 2
\coordinate (head) at (.3,-.3);
\coordinate (tail) at (4,1.3);
\coordinate (controlA) at (.5,.5);
\coordinate (controlB) at (.3,.3);
\draw[
blue,
line width=1pt,
-{Straight Barb[length=5pt,width=1pt]}
]
(tail)
.. controls (controlA) and (controlB) ..
(head);
\end{tikzpicture}
\end{center}
%=================================================================================================
%=================================================================================================
\medskip
\noindent
\textbf{These nucleophilic e- attack the phenyl/resorcinol-derived aromatic ring. Below, the drawing shows the pre-intermediates' before the e- attacks the ring per literature):}
\medskip
\noindent
\begin{center}
\begin{tikzpicture}
\matrix[column sep=1.5cm] {
\node (THCA_preInt1) {
\rotatebox{-30}{%
\chemfig
{
% 1
-% 2
-[:60]% 3
-% 4
-[:60]% 5
-% 6
=_[:60]% 7
(
-[:120]% 8
(
-[:180,,,1]\charge{90=\:,180=\:,270=\:}{O^-}%10
)
=[:60]O% 9
)
-\charge{270={\raisebox{1pt}{\scalebox{1.5}{$\cdot$}}}}
{\rotatebox{30}{\ensuremath{\mathrm C}}}% 11
(
-[:60,1.3]\charge{90=\:,180=\:,270=\:}{O^-}% 12
)
-[:300]% 13
=^[:240]% 14
(
-[:180]\charge{200=\:,70={\hspace{5pt}\rotatebox{35}{\ensuremath{-}}}}
{\rotatebox{35}{\ensuremath{\mathrm C}}}%15
-[:120]% -> 6
)
-[:300]O% 16
-% 17
(
-[:260]% 18
)
(
-[:340]% 19
)
>[:60]% 20
-% 21
-[:60]% 22
-[:120]% 23
(
-[:60]% 24
)
=^[:180]% 25
>:[:240]% 26
(
-[:180]% -> 13
)
(
-[:300]% -> 20
)
}
}
};
&
\node {\text{\bfseries or}};
&
\node (THCA_preInt2) {
\rotatebox{-30}{%
\chemfig
{
% 1
-% 2
-[:60]% 3
-% 4
-[:60]% 5
-% 6
=_[:60]% 7
(
-[:120]% 8
(
-[:180,,,1]\charge{90=\:,180=\:,270=\:}{O^-}%10
)
=[:60]O% 9
)
-% 11
(
=[:60,1.3]\charge{90=\:,180=\:}{O}% 12
)
-[:300]% 13
=^[:240]% 14
(
-[:180]\charge{200=\:,70={\hspace{5pt}\rotatebox{35}{\ensuremath{-}}}}
{\rotatebox{35}{\ensuremath{\mathrm C}}}%15
-[:120]% -> 6
)
-[:300]O% 16
-% 17
(
-[:260]% 18
)
(
-[:340]% 19
)
>[:60]% 20
-% 21
-[:60]% 22
-[:120]% 23
(
-[:60]% 24
)
=^[:180]% 25
>:[:240]% 26
(
-[:180]% -> 13
)
(
-[:300]% -> 20
)
}
}
};
\\
};
\end{tikzpicture}
\end{center}
%=================================================================================================
%=================================================================================================
\medskip
\noindent
\textbf{Note: The image above initially led me to believe that a lone pair on the carboxylate oxygen would form a $\mathrm{C=O}$ $\pi$ bond (see above right image) and interfere with reduction of the ring when e- attack it. However, based on the Birch-reduction mechanism, the electron distribution is better represented as a carbanion ($\mathrm{C^-}$) at one end of the ring, with a two-electron lone pair, while the electron involved in the initial one-electron attack gives rise to a carbon radical ($\mathrm{C^{\bullet}}$) at the carbonyl site. With this radical intermediate present, the lone pair on $\mathrm{O^-}$ cannot form a $\mathrm{C=O}$ $\pi$ bond without simultaneously accounting for the unpaired radical carbon. The would be C=O bond gets delocalized back to an O with 3 lone pairs}
\textbf{Either of the above two pre-intermediates, will form the true-intermediate resonances below:}
\begin{tikzpicture}[x=1cm,y=1cm]
\node (THCA) at (0,-.5) {
\chemfig{
% 1
-[:330]% 2
-[:30]% 3
-[:330]% 4
-[:30]% 5
-[:330]% 6
-[:270]% 7
(
-[:30,,,,draw=none]\mcfcringle{1.3}% (o)
)
-[:330]% 8
-[:30]% 9
(
-[:90]% 10
(
-[:150]% 12
(
-[:90]% 13
(
-[:150,,,1]
\charge{
175:=$-$,
90:=\rotatebox{90}{$:$},
0:=$:$,
270:=\rotatebox{90}{$:$}
}{O}% 15
)
=[:30]O% 14
)
-[:210]% -> 6
)
-[:30,,,1]\charge{
15:=$-$,
90:=\rotatebox{90}{$:$},
180:=$:$,
270:=\rotatebox{90}{$:$}
}{O}
)
>:[:330]% 16
-[:30]% 17
=_[:330]% 18
(
-[:30]% 19
)
-[:270]% 20
-[:210]% 21
>[:150]% 22
(
-[:90]% -> 16
)
-[:210]% 23
(
-[:310]% 24
)
(
-[:230]% 25
)
-[:150]O% 26
(
-[:90]% -> 8
)
}
};
\coordinate (carbonE1) at (-.25,-.25);
\coordinate (carbonE2) at (-.25,-.4);
\coordinate (carbonE3) at (.4,.1);
\coordinate (carbonE4) at (.55,0);
\coordinate (carbonE5) at (.45,-.7);
\coordinate (carbonE6) at (.35,-.8);
\coordinate (carbonE7) at (-.10,0);
\coordinate (carbonE8) at (.1,.1);
\coordinate (carbonXpt1) at (-.20,-.65);
\coordinate (carbonXpt2) at (.65,-.1);
\fill[black] (carbonE1) circle (1.5pt);
\fill[black] (carbonE2) circle (1.5pt);
\fill[black] (carbonE3) circle (1.5pt);
\fill[black] (carbonE4) circle (1.5pt);
\fill[black] (carbonE5) circle (1.5pt);
\fill[black] (carbonE6) circle (1.5pt);
\fill[black] (carbonE7) circle (1.5pt);
\fill[black] (carbonE8) circle (1.5pt);
\fill[blue] (carbonXpt1) circle (1.5pt);
%\fill[blue] (carbonXpt2) circle (1.5pt);
\node (electron_1) at (-1,-2) {
$\mathrm{e^-_{(solv)}}$
};
\node (electron_2) at (2.3,1) {
$\mathrm{e^-_{(solv)}}$
};
\draw[-{Stealth[left]}, blue, thick]
(electron_1.west) .. controls +(180:1.5cm) and +(180:1.5cm) .. (carbonXpt1.west);
% \draw[-{Stealth[left]}, blue, thick]
% (electron_2.west) .. controls +(90:1.5cm) and +(90:1.5cm) .. (carbonXpt2.west);
\draw[->, >=stealth, red, thick]
(carbonE3) .. controls +(45:.5cm) and +(45:.5cm) .. (carbonE8);
\draw[->, >=stealth, red, thick]
(carbonE1) .. controls +(120:.5cm) and +(120:.5cm) .. (carbonE7);
\draw[-{Stealth[left]}, blue, thick]
(carbonE2) .. controls +(165:.5cm) and +(165:.5cm) .. (carbonXpt1);
\draw[-{Stealth[left]}, blue, thick]
(carbonE4) .. controls +(60:.5cm) and +(60:.5cm) .. (carbonXpt2);
\node (arrow) at (3.3,0) {$\longrightarrow$};
\node (THCA_Int) at (7,0) {
\rotatebox{-30}{%
\chemfig
{
% 1
-% 2
-[:60]% 3
-% 4
-[:60]% 5
-% 6
=_[:60]% 7
(
-[:120]% 8
(
-[:180,,,1]\charge{90=\:,180=\:,270=\:}{O^-}%10
)
=[:60]O% 9
)
-\charge{270={\raisebox{1pt}{\scalebox{3}{\color{blue}$\cdot$}}}}
{\rotatebox{30}{\ensuremath{\mathrm C}}}% 11
(
-[:60,1.3]\charge{90=\:,180=\:,270=\:}{O^-}% 12
)
-[:300]% 13
=^[:240]% 14
(
-[:180]\charge{200={\color{blue}\scalebox{2}{\:}},70={\hspace{5pt}\rotatebox{35}{\ensuremath{-}}}}
{\rotatebox{35}{\ensuremath{\mathrm C}}}%15
-[:120]% -> 6
)
-[:300]O% 16
-% 17
(
-[:260]% 18
)
(
-[:340]% 19
)
>[:60]% 20
-% 21
-[:60]% 22
-[:120]% 23
(
-[:60]% 24
)
=^[:180]% 25
>:[:240]% 26
(
-[:180]% -> 13
)
(
-[:300]% -> 20
)
}}
};
\node (arrow) at (7.6,-2.5) {$\textbf{OR}$};
\node (THCA_Int2) at (7.3,-4.5) {
\rotatebox{-30}{%
\chemfig
{
% 1
-% 2
-[:60]% 3
-% 4
-[:60]% 5
-% 6
=_[:60]% 7
(
-[:120]% 8
(
-[:180,,,1]\charge{90=\:,180=\:,270=\:}{O^-}%10
)
=[:60]O% 9
)
-\charge{200={\color{blue}\scalebox{2}{\:}},
70={\hspace{5pt}\rotatebox{35}{\raisebox{0.2pt}{\rule{6pt}{1.5pt}}}}}
{\rotatebox{30}{\ensuremath{\mathrm C}}}% 11
(
-[:60,1.3]\charge{90=\:,180=\:,270=\:}{O^-}% 12
)
-[:300]% 13
=^[:240]% 14
(
-[:180]\charge{270={\raisebox{1pt}{\scalebox{3}{\color{blue}$\cdot$}}}}
{\rotatebox{35}{\ensuremath{\mathrm C}}}%15
-[:120]% -> 6
)
-[:300]O% 16
-% 17
(
-[:260]% 18
)
(
-[:340]% 19
)
>[:60]% 20
-% 21
-[:60]% 22
-[:120]% 23
(
-[:60]% 24
)
=^[:180]% 25
>:[:240]% 26
(
-[:180]% -> 13
)
(
-[:300]% -> 20
)
}}
};
\end{tikzpicture}
\newpage
\medskip
\noindent
\textbf{The final process is to QUENCH (PROTONATE) the solvent with a proton donor in order to protonate the ring and the remaining anionic oxygen sites. Citric acid is selected as the proton donor because it is a triprotic acid and therefore provides three nominal proton equivalents per molecule. However, I went to tackle this from a saturation/super saturation leveling effect methodology, and will use 1 mol Citric Acid per 4 sites that need to be protonated. The structure of citric acid is:}
\medskip
\noindent
\begin{center}
\chemfig{
HO% 1
-[:60,,2]% 2
(
=[:120]O% 3
)
-% 4
-[:300]% 5
(
-[,,,1]OH% 6
)
(
-[:240]% 7
(
-[:300,,,1]OH% 9
)
=[:180]O% 8
)
-[:60]% 10
-% 11
(
=[:330]O% 12
)
-[:60,,,1]OH% 13
}
\end{center}
\medskip
\noindent
\textbf{Its 3 successive acid dissociations can be represented as:}
\[
\mathrm{
H_3Cit \rightleftharpoons H^+ + H_2Cit^-
}
\]
\[
\mathrm{
H_2Cit^- \rightleftharpoons H^+ + HCit^{2-}
}
\]
\[
\mathrm{
HCit^{2-} \rightleftharpoons H^+ + Cit^{3-}
}
\]
\medskip
\noindent
\textbf{Thus, one mole of citric acid contains three potential proton equivalents. ! equiv. per each of Citric Acvid's corner OH groups (but NOT the center OH.) So:}
\[
1\ \mathrm{mol\ H_3Cit}
\longrightarrow
3\ \mathrm{mol\ H^+\ equivalents}
\]
\medskip
\noindent
\textbf{The intermediate substrate requires four proton accepting sites that must be protonated:}
\medskip
\noindent
\textbf{The intermediate's reactive sites can be classified into four principal locations: (1) $\mathrm{C^-}$, (2) $\mathrm{C^{\bullet}}$, (3) $\mathrm{O^-}$, and (4) $\mathrm{COO^-}$}
\textbf{Recall from the intermediate:}
\begin{center}
\begin{tikzpicture}
\node (THCA_Int) at (7,0) {
\rotatebox{-30}{%
\chemfig
{
% 1
-% 2
-[:60]% 3
-% 4
-[:60]% 5
-% 6
=_[:60]% 7
(
-[:120]% 8
(
-[:180,,,1]\charge{90=\:,180=\:,270=\:}{O^-}%10
)
=[:60]O% 9
)
-\charge{270={\raisebox{1pt}{\scalebox{3}{\color{blue}$\cdot$}}}}
{\rotatebox{30}{\ensuremath{\mathrm C}}}% 11
(
-[:60,1.3]\charge{90=\:,180=\:,270=\:}{O^-}% 12
)
-[:300]% 13
=^[:240]% 14
(
-[:180]\charge{200={\color{blue}\scalebox{2}{\:}},70={\hspace{5pt}\rotatebox{35}{\ensuremath{-}}}}
{\rotatebox{35}{\ensuremath{\mathrm C}}}%15
-[:120]% -> 6
)
-[:300]O% 16
-% 17
(
-[:260]% 18
)
(
-[:340]% 19
)
>[:60]% 20
-% 21
-[:60]% 22
-[:120]% 23
(
-[:60]% 24
)
=^[:180]% 25
>:[:240]% 26
(
-[:180]% -> 13
)
(
-[:300]% -> 20
)
}}
};
\end{tikzpicture}
\end{center}
\medskip
\noindent
\textbf{The carbon-centered radical, $\mathrm{C^{\bullet}}$, will subsequently interact with the remaining e- in the system to form a carbon-centered carbanion, $\mathrm{C^-}$, which can then undergo protonation. These protonations will produce the two new C--H bonds required for formation of the nonconjugated $1,4$-cyclohexadiene ring system}
\medskip
\medskip
\noindent
\textbf{The basic acid--base calculation for any proton-transfer
reaction can be represented as:}
\[
\mathrm{B^- + HA
\rightleftharpoons
HB + A^-}
\]
\[
K=\frac{[\mathrm{HB}][\mathrm{A^-}]}
{[\mathrm{B^-}][\mathrm{HA}]}
\]
\medskip
\noindent
\textbf{where $\mathrm{B^-}$ is the proton acceptor and $\mathrm{HA}$ is
the proton donor; their conjugate acid and base are $\mathrm{BH}$ and
$\mathrm{A^-}$, respectively. The equilibrium constant is then estimated
from the difference in $pK_a$ values:}
\[
\log K
\approx
pK_a(\mathrm{BH})-pK_a(\mathrm{HA})
\]
\[
K
\approx
10^{\,pK_a(\mathrm{BH})-pK_a(\mathrm{HA})}
\]
\medskip
\noindent
\textbf{Defining:}
\[
\Delta pK_a
=
pK_a(\mathrm{BH})-pK_a(\mathrm{HA})
\]
\medskip
\noindent
\textbf{Gives:}
\[
K\approx10^{\Delta pK_a}
\]
\medskip
\noindent
\textbf{and therefore:}
\[
\Delta pK_a>0
\quad\Longrightarrow\quad
K>1.
\]
\medskip
\noindent
\textbf{A large positive $\Delta pK_a$ indicates that proton transfer from $HA$
to $\mathrm{B^-}$ is strongly favored}
\medskip
\noindent
\textbf{The general reaction exact liquid-ammonia $pK_a$ of the THCA-derived cyclohexadienyl species is not available to researchers yet, so an estimate must be used. A useful reference value for a cyclohexadiene-type acid is approximately:}
\[
pK_a\sim45
\]
\medskip
\noindent
\textbf{From a conventional aqueous/organic-acid scale, liquid ammonia substantially changes acidities, and aqueous pK values should not simply be used directly for the ammonia based calculations. Using an approximate liquid-ammonia carbon-acid correlation (from research papers conducting other liquid ammonia experiments,) gives a rough estimate in the low-to-mid $30$ range. For the purposes of this mixture, I used a representative value of:}
\[
pK_a\underset{NH_3}{}(Cyclo-Intermediate)\approx34
\]
\medskip
\noindent
\textbf{Because the reaction occurs in liquid ammonia, the relevant comparison
is between the estimated $pK_a$ of the cyclohexadienyl intermediate in liquid
ammonia and the acidity of the proton donor. The aqueous citric-acid $pK_a$
values are therefore used only as approximate reference values, rather than
being directly converted into liquid-ammonia $pK_a$ values.}
\[
\Delta pK_a
=
34-3.13
=
30.87
\]
\medskip
\noindent
\textbf{Therefore the 2nd CitAcid Proton:}
\[
\Delta pK_a
=
34-4.76
=
29.24
\]
\medskip
\noindent
\textbf{For the third citric-acid proton:}
\[
\Delta pK_a
=
34-6.40
=
27.60
\]
\medskip
\noindent
\textbf{For all three acidic protons of citric acid, the estimated $pK_a$ of the
protonated carbon-centered intermediate is substantially higher than the
corresponding $pK_a$ of citric acid. Therefore, citric acid is the stronger
proton donor in each case, and proton transfer to the carbon-centered anion is
strongly thermodynamically favored. The calculated $\Delta pK_a$ values of
$30.87$, $29.24$, and $27.60$ correspond to equilibrium constants of
approximately $10^{30.87}$, $10^{29.24}$, and $10^{27.60}$, respectively.
Thus, each of the three citric-acid protons is capable of protonating the
carbon-centered anions, with the first proton transfer being the most strongly
favored. The same qualitative conclusion applies to the second carbon-centered site.}
%=============================================================================
%=============================================================================
\medskip
\noindent
\textbf{Protonation mechanism (step 1 of 2) of the Intermediate-Cyclohexadienyl carbanion is shown below (Note: red dashed full hooked arrow represents potential other citric acid sites that could be deprotonated):}
\medskip
\begin{tikzpicture}
\node (THCA_Int) at (0,0) {
\rotatebox{-30}{%
\chemfig
{
% 1
-% 2
-[:60]% 3
-% 4
-[:60]% 5
-% 6
=_[:60]% 7
(
-[:120]% 8
(
-[:180,,,1]\charge{90=\:,180=\:,270=\:}{O^-}%10
)
=[:60]O% 9
)
-\charge{270={\raisebox{1pt}{\scalebox{3}{\color{blue}$\cdot$}}}}
{\rotatebox{30}{\ensuremath{\mathrm C}}}% 11
(
-[:60,1.3]\charge{90=\:,180=\:,270=\:}{O^-}% 12
)
-[:300]% 13
=^[:240]% 14
(
-[:180]\charge{200={\color{blue}\scalebox{2}{\:}},70={\hspace{5pt}\rotatebox{35}{\ensuremath{-}}}}
{\rotatebox{35}{\ensuremath{\mathrm C}}}%15
-[:120]% -> 6
)
-[:300]O% 16
-% 17
(
-[:260]% 18
)
(
-[:340]% 19
)
>[:60]% 20
-% 21
-[:60]% 22
-[:120]% 23
(
-[:60]% 24
)
=^[:180]% 25
>:[:240]% 26
(
-[:180]% -> 13
)
(
-[:300]% -> 20
)
}
}
};
\node at (3,0) {$+$};
\node (CitricAcid) at (5,0) {
\chemfig{
HO% 1
-[:60,,2]% 2
(
=[:120]O% 3
)
-% 4
-[:300]% 5
(
-[,,,1]OH% 6
)
(
-[:240]% 7
(
-[:300,,,1]OH% 9
)
=[:180]O% 8
)
-[:60]% 10
-% 11
(
=[:330]O% 12
)
-[:60,,,1]OH% 13
}
};
\node (arrow) at (8,0) {$\longrightarrow$};
\node (THCA_IntProd) at (12,0) {
\rotatebox{-30}{%
\chemfig
{
% 1
-% 2
-[:60]% 3
-% 4
-[:60]% 5
-% 6
=_[:60]% 7
(
-[:120]% 8
(
-[:180,,,1]\charge{90=\:,180=\:,270=\:}{O^-}%10
)
=[:60]O% 9
)
-\charge{270={\raisebox{1pt}{\scalebox{3}{\color{blue}$\cdot$}}}}
{\rotatebox{30}{\ensuremath{\mathrm C}}}% 11
(
-[:60,1.3]\charge{90=\:,180=\:,270=\:}{O^-}% 12
)
-[:300]% 13
=^[:240]% 14
(
-[:180]%15
-[:120]% -> 6
)
-[:300]O% 16
-% 17
(
-[:260]% 18
)
(
-[:340]% 19
)
>[:60]% 20
-% 21
-[:60]% 22
-[:120]% 23
(
-[:60]% 24
)
=^[:180]% 25
>:[:240]% 26
(
-[:180]% -> 13
)
(
-[:300]% -> 20
)
}}
};
\coordinate (THCAIntC1) at (-.55,-.35);
\coordinate (CitAcidH1) at (3.6,.10);
\coordinate (CitAcidH2) at (6.3,1);
\coordinate (CitAcidH3) at (5.3,-.8);
\draw[->, >=stealth, red, thick]
(THCAIntC1) .. controls +(45:.5cm) and +(45:.5cm) .. (CitAcidH1);
\draw[->, >=stealth, red, thick, dashed]
(THCAIntC1) .. controls +(45:.5cm) and +(45:.5cm) .. (CitAcidH2);
\draw[->, >=stealth, red, thick, dashed]
(THCAIntC1) .. controls +(45:.5cm) and +(45:.5cm) .. (CitAcidH3);
\end{tikzpicture}
\newpage
%=============================================================================
%=============================================================================
\medskip
\noindent
\textbf{Protonation mechanism (step 2 of 2) of the Intermediate-Cyclohexadienyl carbon-radical is shown below (red dashed full hooked arrow represents potential other citric acid sites to become protonated from) and remember back on page 3 that spectating e- present in the system? Here's where it plays its part to form yet another Carbonanion when the e- attacks the carbon radical (Note: red dashed full hooked arrow represents potential other citric acid sites that could be deprotonated):}
\medskip
\begin{tikzpicture}
\node (THCA_IntProd2) at (0,0) {
\rotatebox{-30}{%
\chemfig
{
% 1
-% 2
-[:60]% 3
-% 4
-[:60]% 5
-% 6
=_[:60]% 7
(
-[:120]% 8
(
-[:180,,,1]\charge{90=\:,180=\:,270=\:}{O^-}%10
)
=[:60]O% 9
)
-\charge{270={\raisebox{1pt}{\scalebox{3}{\color{blue}$\cdot$}}}}
{\rotatebox{30}{\ensuremath{\mathrm C}}}% 11
(
-[:60,1.3]\charge{90=\:,180=\:,270=\:}{O^-}% 12
)
-[:300]% 13
=^[:240]% 14
(
-[:180]%15
-[:120]% -> 6
)
-[:300]O% 16
-% 17
(
-[:260]% 18
)
(
-[:340]% 19
)
>[:60]% 20
-% 21
-[:60]% 22
-[:120]% 23
(
-[:60]% 24
)
=^[:180]% 25
>:[:240]% 26
(
-[:180]% -> 13
)
(
-[:300]% -> 20
)
}}
};
\node at (4,0) {$+$};
\node (electron_1) at (6,1) {
$\mathrm{e^-_{(solv)}}$
};
\node (arrow) at (8,0) {$\longrightarrow$};
\node (THCA_IntProd3) at (12,0) {
\rotatebox{-30}{%
\chemfig
{
% 1
-% 2
-[:60]% 3
-% 4
-[:60]% 5
-% 6
=_[:60]% 7
(
-[:120]% 8
(
-[:180,,,1]\charge{90=\:,180=\:,270=\:}{O^-}%10
)
=[:60]O% 9
)
-\charge{200={\color{blue}\scalebox{2}{\:}},
70={\hspace{5pt}\rotatebox{35}{\raisebox{0.2pt}{\rule{6pt}{1.5pt}}}}}
{\rotatebox{30}{\ensuremath{\mathrm C}}}% 11
(
-[:60,1.3]\charge{90=\:,180=\:,270=\:}{O^-}% 12
)
-[:300]% 13
=^[:240]% 14
(
-[:180]%15
-[:120]% -> 6
)
-[:300]O% 16
-% 17
(
-[:260]% 18
)
(
-[:340]% 19
)
>[:60]% 20
-% 21
-[:60]% 22
-[:120]% 23
(
-[:60]% 24
)
=^[:180]% 25
>:[:240]% 26
(
-[:180]% -> 13
)
(
-[:300]% -> 20
)
}}
};
\coordinate (THCAIntC2) at (.55,0);
\draw[-{Stealth[left]}, blue, thick]
(electron_1) .. controls +(60:.5cm) and +(60:.5cm) .. (THCAIntC2);
\end{tikzpicture}
\medskip
\noindent
\textbf{A fesh citric acid molecule (or the other remaining two sites from the previous citric acid,) can also protonate this new formed carbanion (Note: red dashed full hooked arrow represents potential other citric acid sites that could be deprotonated):}
\medskip
\begin{tikzpicture}
\node (THCA_IntProd3) at (0,0) {
\rotatebox{-30}{%
\chemfig
{
% 1
-% 2
-[:60]% 3
-% 4
-[:60]% 5
-% 6
=_[:60]% 7
(
-[:120]% 8
(
-[:180,,,1]\charge{90=\:,180=\:,270=\:}{O^-}%10
)
=[:60]O% 9
)
-\charge{200={\color{blue}\scalebox{2}{\:}},
70={\hspace{5pt}\rotatebox{35}{\raisebox{0.2pt}{\rule{6pt}{1.5pt}}}}}
{\rotatebox{30}{\ensuremath{\mathrm C}}}% 11
(
-[:60,1.3]\charge{90=\:,180=\:,270=\:}{O^-}% 12
)
-[:300]% 13
=^[:240]% 14
(
-[:180]%15
-[:120]% -> 6
)
-[:300]O% 16
-% 17
(
-[:260]% 18
)
(
-[:340]% 19
)
>[:60]% 20
-% 21
-[:60]% 22
-[:120]% 23
(
-[:60]% 24
)
=^[:180]% 25
>:[:240]% 26
(
-[:180]% -> 13
)
(
-[:300]% -> 20
)
}}
};
\node at (3,0) {$+$};
\node (CitricAcid) at (5,0) {
\chemfig{
HO% 1
-[:60,,2]% 2
(
=[:120]O% 3
)
-% 4
-[:300]% 5
(
-[,,,1]OH% 6
)
(
-[:240]% 7
(
-[:300,,,1]OH% 9
)
=[:180]O% 8
)
-[:60]% 10
-% 11
(
=[:330]O% 12
)
-[:60,,,1]OH% 13
}
};
\node (arrow) at (8,0) {$\longrightarrow$};
\node (THCA_IntProd4) at (12,0) {
\rotatebox{-30}{%
\chemfig
{
% 1
-% 2
-[:60]% 3
-% 4
-[:60]% 5
-% 6
=_[:60]% 7
(
-[:120]% 8
(
-[:180,,,1]\charge{90=\:,180=\:,270=\:}{O^-}%10
)
=[:60]O% 9
)
-% 11
(
-[:60,1.3]\charge{90=\:,180=\:,270=\:}{O^-}% 12
)
-[:300]% 13
=^[:240]% 14
(
-[:180]%15
-[:120]% -> 6
)
-[:300]O% 16
-% 17
(
-[:260]% 18
)
(
-[:340]% 19
)
>[:60]% 20
-% 21
-[:60]% 22
-[:120]% 23
(
-[:60]% 24
)
=^[:180]% 25
>:[:240]% 26
(
-[:180]% -> 13
)
(
-[:300]% -> 20
)
}}
};
\coordinate (CarbonXpt3) at (.3,.10);
\coordinate (CitAcidH1) at (3.6,.10);
\coordinate (CitAcidH2) at (6.3,1);
\coordinate (CitAcidH3) at (5.3,-.8);
\draw[->, >=stealth, red, thick]
(CarbonXpt3) .. controls +(45:.5cm) and +(45:.5cm) .. (CitAcidH1);
\draw[->, >=stealth, red, thick, dashed]
(CarbonXpt3) .. controls +(45:.5cm) and +(45:.5cm) .. (CitAcidH2);
\draw[->, >=stealth, red, thick, dashed]
(CarbonXpt3) .. controls +(45:.5cm) and +(45:.5cm) .. (CitAcidH3);
\end{tikzpicture}
%==========================================================================================
\medskip
\noindent
\textbf{For the phenolic oxygen-centered site, conjugate acid is an alcohol/phenol-type species. Liquid-ammonia measurements show that phenolic acidities are substantially different from their aqueous values. Although published liquid-ammonia values vary depending on the experimental method and definition of the apparent acidity. A representative value is:}
\[
pK_a\underset{NH_3}{}(PhOH)\approx6.0
\]
\medskip
\noindent
\textbf{Using the representative value $6.0$ for the first proton of citric acid:}
\[
\Delta pK_a
=
6.0-3.13
=
2.87
\]
\medskip
\noindent
\textbf{and:}
\[
K\sim10^{2.87}\approx7.4\times10^2
\]
\medskip
\noindent
\textbf{For the second citric proton:}
\[
\Delta pK_a
=
6.0-4.76
=
1.24
\]
\medskip
\noindent
\textbf{So:}
\[
K\sim10^{1.24}\approx17
\]
\medskip
\noindent
\textbf{For the third citric proton:}
\[
\Delta pK_a
=
6.0-6.40
=
-0.40
\]
\medskip
\noindent
\textbf{Giving:}
\[
K\sim10^{-0.40}\approx0.40
\]
\medskip
\noindent
\textbf{Therefore, unlike the carbon-centered sites, the oxygen-centered site cannot be described as quantitatively protonated solely by comparison with the third citric-acid $pK_a$. The first and second citric-acid protons are sufficiently acidic to strongly favor protonation of the phenoxide-type site ($K \approx 7.4 \times 10^{2}$ and $K \approx 17$, respectively), whereas transfer of the third proton is near equilibrium ($K \approx 0.4$). Consequently, if fresh citric acid is available with all three acidic protons accessible, protonation of the phenolic oxygen site can occur readily through the first two proton-transfer events. However, after two citric-acid proton sites have been consumed, the remaining third proton is a much weaker proton donor, and protonation of the phenolic oxygen becomes dependent on acid concentration, ion pairing, solvent speciation, and mass-action leveling effects}
%==========================================================================================
%==========================================================================================
\medskip
\noindent
\textbf{The mechanism showing the a lone pair of the phenoxide oxygen of the Intermediate-phenoxide oxygen-centered site attacking and deprotonating a citric acid site (Note: red dashed full hooked arrow represents potential other citric acid sites that could be deprotonated):}
\begin{tikzpicture}
\node (THCA_IntProd4) at (0,0) {
\rotatebox{-30}{%
\chemfig
{
% 1
-% 2
-[:60]% 3
-% 4
-[:60]% 5
-% 6
=_[:60]% 7
(
-[:120]% 8
(
-[:180,,,1]\charge{90=\:,180=\:,270=\:}{O^-}%10
)
=[:60]O% 9
)
-% 11
(
-[:60,1.3]\charge{90=\:,180=\:,270=\:}{O^-}% 12
)
-[:300]% 13
=^[:240]% 14
(
-[:180]%15
-[:120]% -> 6
)
-[:300]O% 16
-% 17
(
-[:260]% 18
)
(
-[:340]% 19
)
>[:60]% 20
-% 21
-[:60]% 22
-[:120]% 23
(
-[:60]% 24
)
=^[:180]% 25
>:[:240]% 26
(
-[:180]% -> 13
)
(
-[:300]% -> 20
)
}}
};
\node at (3,0) {$+$};
\node (CitricAcid) at (5,0) {
\chemfig{
HO% 1
-[:60,,2]% 2
(
=[:120]O% 3
)
-% 4
-[:300]% 5
(
-[,,,1]OH% 6
)
(
-[:240]% 7
(
-[:300,,,1]OH% 9
)
=[:180]O% 8
)
-[:60]% 10
-% 11
(
=[:330]O% 12
)
-[:60,,,1]OH% 13
}
};
\node (arrow) at (8,0) {$\longrightarrow$};
\node (THCA_IntProd4) at (12,0) {
\rotatebox{-30}{%
\chemfig
{
% 1
-% 2
-[:60]% 3
-% 4
-[:60]% 5
-% 6
=_[:60]% 7
(
-[:120]% 8
(
-[:180,,,1]\charge{90=\:,180=\:,270=\:}{O^-}%10
)
=[:60]O% 9
)
-% 11
(
-[:60,1.3]{OH}% 12
)
-[:300]% 13
=^[:240]% 14
(
-[:180]%15
-[:120]% -> 6
)
-[:300]O% 16
-% 17
(
-[:260]% 18
)
(
-[:340]% 19
)
>[:60]% 20
-% 21
-[:60]% 22
-[:120]% 23
(
-[:60]% 24
)
=^[:180]% 25
>:[:240]% 26
(
-[:180]% -> 13
)
(
-[:300]% -> 20
)
}}
};
\coordinate (Phenox) at (1.2,.2);
\coordinate (CitAcidH1) at (3.6,.10);
\coordinate (CitAcidH2) at (6.3,1);
\coordinate (CitAcidH3) at (5.3,-.8);
\draw[->, >=stealth, red, thick]
(Phenox) .. controls +(45:.5cm) and +(45:.5cm) .. (CitAcidH1);
\draw[->, >=stealth, red, thick, dashed]
(Phenox) .. controls +(45:.5cm) and +(45:.5cm) .. (CitAcidH2);
\draw[->, >=stealth, red, thick, dashed]
(Phenox) .. controls +(45:.5cm) and +(45:.5cm) .. (CitAcidH3);
\end{tikzpicture}
%===============================================================================================
\medskip
\noindent
\textbf{Now for the final protonation of the carboxylate of the THCA-intermediate. The corresponding conjugate acid is the carboxylic acid. A precise liquid-ammonia $pK_a$ for the particular substituted carboxylate is not available. Consequently, assigning a single numerical liquid-ammonia $pK_a$ would give a false impression of precision. Liquid-ammonia acid--base chemistry is strongly affected by ion pairing and the relatively low dielectric constant of the solvent. Carboxylic acids that are weak acids in water can behave as substantially stronger acids in liquid ammonia. Therefore, the appropriate conclusion is that carboxylate protonation is thermodynamically accessible, but its quantitative equilibrium cannot be
reliably calculated from an aqueous water based $pK_a$ alone that we find in texts and literature. The carboxylate protonation should consequently be considered together with the total proton-donor activity and the changing ionic composition of the medium. Mechanism shown below:}
\medskip
\begin{tikzpicture}
\node (THCA_IntProd4) at (0,0) {
\rotatebox{-30}{%
\chemfig
{
% 1
-% 2
-[:60]% 3
-% 4
-[:60]% 5
-% 6
=_[:60]% 7
(
-[:120]% 8
(
-[:180,,,1]\charge{90=\:,180=\:,270=\:}{O^-}%10
)
=[:60]O% 9
)
-% 11
(
-[:60,1.3]{OH}% 12
)
-[:300]% 13
=^[:240]% 14
(
-[:180]%15
-[:120]% -> 6
)
-[:300]O% 16
-% 17
(
-[:260]% 18
)
(
-[:340]% 19
)
>[:60]% 20
-% 21
-[:60]% 22
-[:120]% 23
(
-[:60]% 24
)
=^[:180]% 25
>:[:240]% 26
(
-[:180]% -> 13
)
(
-[:300]% -> 20
)
}}
};
\node at (3,0) {$+$};
\node (CitricAcid) at (5,0) {
\chemfig{
HO% 1
-[:60,,2]% 2
(
=[:120]O% 3
)
-% 4
-[:300]% 5
(
-[,,,1]OH% 6
)
(
-[:240]% 7
(
-[:300,,,1]OH% 9
)
=[:180]O% 8
)
-[:60]% 10
-% 11
(
=[:330]O% 12
)
-[:60,,,1]OH% 13
}
};
\node (arrow) at (8,0) {$\longrightarrow$};
\node (THCA_Analog) at (12,0) {
\rotatebox{-30}{%
\chemfig
{
% 1
-% 2
-[:60]% 3
-% 4
-[:60]% 5
-% 6
=_[:60]% 7
(
-[:120]% 8
(
-[:180,,,2]HO% 10
)
=[:60]O% 9
)
-% 11
(
-[:60,,,1]OH% 12
)
-[:300]% 13
=^[:240]% 14
(
-[:180]% 15
-[:120]% -> 6
)
-[:300]O% 16
-% 17
(
-[:260]% 18
)
(
-[:340]% 19
)
>[:60]% 20
-% 21
-[:60]% 22
-[:120]% 23
(
-[:60]% 24
)
=^[:180]% 25
>:[:240]% 26
(
-[:180]% -> 13
)
(
-[:300]% -> 20
)
}}
};
\coordinate (Carbox) at (-.1,1.3);
\coordinate (CitAcidH1) at (3.6,.10);
\coordinate (CitAcidH2) at (6.3,1);
\coordinate (CitAcidH3) at (5.3,-.8);
\draw[->, >=stealth, red, thick]
(Carbox) .. controls +(45:.5cm) and +(45:.5cm) .. (CitAcidH1);
\draw[->, >=stealth, red, thick, dashed]
(Carbox) .. controls +(45:.5cm) and +(45:.5cm) .. (CitAcidH2);
\draw[->, >=stealth, red, thick, dashed]
(Carbox) .. controls +(45:.5cm) and +(45:.5cm) .. (CitAcidH3);
\end{tikzpicture}
%=============================================================================
\medskip
\noindent
\textbf{As the proton donor in this system is specifically citric acid, it contains
three carboxylic acid protons. The relevant proton-transfer equilibrium is:}
\[
\mathrm{Citric\ acid{-}COOH + THCA{-}COO^- \rightleftharpoons Citrate^- + THCA{-}COOH}
\]
\medskip
\noindent
\textbf{The direction of proton transfer is determined by the relative acidities of the citric acid carboxyl group and the corresponding THCA carboxylic acid:}
\[
K=10^{\Delta pK_a}
\]
\medskip
\noindent
\textbf{Where:}
\[
\Delta pK_a =
pK_a(\mathrm{THCA{-}COOH})-
pK_a(\mathrm{Citric\ acid{-}COOH})
\]
\medskip
\noindent
\textbf{A positive $\Delta pK_a$ favors proton transfer from citric acid to the THCA carboxylate. In liquid ammonia, carboxylic acids exhibit enhanced acidity compared with their behavior in water. For the general equilibrium:}
\[
\mathrm{RCOOH + NH_3 \rightleftharpoons RCOO^- + NH_4^+}
\]
\medskip
\noindent
\textbf{The equilibrium relationship is:}
\[
\log K =
pK_a(\mathrm{NH_4^+})-pK_a(\mathrm{RCOOH})
\]
\medskip
\noindent
\textbf{Using:}
\[
pK_a(\mathrm{NH_4^+})\approx10.4
\]
\medskip
\noindent
\textbf{And the observation that carboxylic acids undergo strongly favored
neutralization in liquid ammonia:}
\[
K>10^5
\]
\medskip
\noindent
\textbf{Gives:}
\[
pK_a(\mathrm{RCOOH})<5.4
\]
\medskip
\noindent
\textbf{Therefore, the carboxylic acid groups of citric acid become substantially stronger proton donors in liquid ammonia. The aqueous pKa values are not used as absolute liquid-ammonia values, but they establish the relative order of proton-transfer ability:}
\[
pK_{a1}=3.13
\]
\[
pK_{a2}=4.76
\]
\[
pK_{a3}=6.40
\]
\medskip
\noindent
\textbf{The first citric acid proton is the most acidic proton and provides the most favorable proton transfer to the THCA carboxylate:}
\[
\mathrm{Citric{-}COOH + THCA{-}COO^- \rightarrow Citrate^- + THCA{-}COOH}
\]
\medskip
\noindent
\textbf{This proton-transfer event is strongly favored:}
\[
K\gg1
\]
\medskip
\noindent
\textbf{The second citric acid proton is weaker than the first but remains a carboxylic acid proton. Because liquid ammonia increases the acidity of carboxylic acids, the second proton can also protonate the remaining THCA carboxylate oxygen:}
\[
\mathrm{Citric{-}COOH + THCA{-}COO^- \rightarrow Citrate^- + THCA{-}COOH}
\]
\medskip
\noindent
\textbf{with:}
\[
K>1
\]
\medskip
\noindent
\textbf{The third citric acid proton represents the least favorable proton-transfer event. Although this proton also originates from a carboxylic acid group and is strengthened in liquid ammonia, its higher intrinsic pKa makes it the weakest proton donor of the three. Even when the aqueous pKa values are used
as a relative comparison, the third proton transfer approaches equilibrium rather than proceeding quantitatively. Thus, the effect of liquid ammonia is not to remove the difference between the three citric acid protons, but rather to increase the acidity of all carboxylic acid sites. Consequently, the ability of citric acid to protonate the THCA carboxylate decreases in the order:}
\[
\boxed{H_1>H_2>H_3}
\]
\medskip
\noindent
\textbf{The first two citric acid protons efficiently protonate the two oxygen-centered negative sites of the THCA carboxylate:}
\[
\mathrm{THCA{-}(COO^-)_{2}}
\;\xrightarrow[\text{citric acid}]{}\;
\mathrm{THCA{-}(COOH)(COO^-)}
\;\xrightarrow[\text{citric acid}]{}\;
\mathrm{THCA{-}(COOH)_{2}}
\]
\medskip
\noindent
\textbf{while the third proton remains accessible its K constant point is borderline equlibrium with no strong enough protonation for the 3rd proton of citric acid to protonate the THCA carboxylate group. Yet the first two will. Hence I went a saturation route with citric acid}
\medskip
\noindent
\textbf{Acid Addition, Leveling, and Mass Action and the term ``supersaturation'' is useful as an intuitive description of maintaining a large reservoir of proton donor, but it is more precise to
describe the effect as increased proton-donor activity and mass action.}
\medskip
\noindent
\textbf{Adding an excess of proton donor does not change the intrinsic $K$ of the reaction. Instead, it changes the reaction quotient and drives the equilibrium toward the protonated species. In the present system, the conceptual sequence was therefore:}
\[
\mathrm{
H_3Cit
\longrightarrow
H_2Cit^-
\longrightarrow
HCit^{2-}
\longrightarrow
Cit^{3-}
}
\]
\medskip
\noindent
\textbf{While proton equivalents are distributed among the strongest available bases:}
\[
\mathrm{
NH_2^-,
\qquad
C^-,
\qquad
O^-,
\qquad
COO^-
}
\]
\medskip
\noindent
\textbf{The carbon-centered Birch anions had an enormous estimated proton-transfer driving force and therefore represented the least problematic protonation events. The residual $\mathrm{NH_2^-}$ is also a very strong base and must be included in the proton balance. Continued proton availability drives this
pool toward:}
\[
\mathrm{
NH_2^-
\longrightarrow
NH_3
\longrightarrow
NH_4^+
}
\]
\medskip
\noindent
\textbf{ After the strongest bases have consumed their available proton equivalents, the remaining protonation equilibria involve the oxygen-centered and carboxylate sites. Their quantitative behavior is more sensitive to the actual liquid-ammonia solvent environment, ion pairing, concentration, and
phase equilibria. Thus the overall thermodynamic picture is:}
\[
\begin{array}{c}
\text{Birch-derived reduced substrate}\\[2mm]
\mathrm{C^-\quad C^-\quad O^-\quad COO^-}
\end{array}
\]
\medskip
\noindent
\textbf{Followed by proton transfer toward:}
\[
\begin{array}{c}
\text{neutralized substrate}\\[2mm]
\mathrm{C-H\quad C-H\quad O-H\quad COOH}
\end{array}
\]
\medskip
\noindent
\textbf{While the residual ammonia-derived base is simultaneously driven toward:}
\[
\mathrm{
NH_2^-
\longrightarrow
NH_3
\longrightarrow
NH_4^+
}
\]
\medskip
\noindent
\textbf{ Therefore, the most appropriate interpretation of a large citric-acid excess is not that supersaturation changes the $pK_a$, but that a sufficiently large proton-donor reservoir maintains the acid-side chemical potential while the substrate and residual ammonia/amide pool consume proton
equivalents.}
\medskip
\noindent
\textbf{Conclusion}
\medskip
\noindent
\textbf{The four-site protonation can therefore be divided into two categories. First, the two Birch carbon-centered protonations are overwhelmingly favored according to the analogue-based conjugate-$pK_a$ comparison:}
\[
\Delta pK_a\sim28--31
\]
\medskip
\noindent
\textbf{For the estimated liquid-ammonia carbon-acid scale and the three citric-acid proton donors. Second, protonation of $\mathrm{O^-}$ and $\mathrm{COO^-}$ is more dependent on the actual solvent and ionic environment. Consequently, continued acid availability and the overall mass-action balance are important for driving the system toward the fully protonated state. The residual $\mathrm{NH_2^-}$ pool must also be included in the proton inventory. If $n$ moles of residual $\mathrm{NH_2^-}$ remain per mole of substrate, the nominal proton requirement is:}
\[
4+2n\ \mathrm{H^+\ equivalents}
\]
\medskip
\noindent
\textbf{And the corresponding stoichiometric citric-acid lower bound is:}
\[
n_{\mathrm{Cit}}
=
\frac{4+2n}{3}.
\]
\medskip
\noindent
\textbf{For the present reaction mixture, the minimum citric-acid requirement is therefore determined by the proton equivalents required for the substrate and any remaining lithium amide base. In the absence of residual \(\mathrm{NH_2^-}\):}
\[
n=0
\]
\medskip
\noindent
\textbf{Giving:}
\[
n_{\mathrm{Cit}}
=
\frac{4}{3}
=
1.33
\]
\medskip
\noindent
\textbf{With the citric-acid equivalents per equivalent of substrate, if residual lithium amide remains after the Birch reduction, each equivalent requires two additional proton equivalents:}
\[
n_{\mathrm{Cit}}
=
\frac{4+2n}{3}
\]
\medskip
\noindent
\textbf{Thus, the practical citric-acid requirement is the calculated lower bound plus any additional excess required to account for residual base, ion-pairing effects, and incomplete mixing during quench.}
\medskip
\noindent
\textbf{This provides the appropriate framework for evaluating an excess-citric-acid quench without incorrectly treating the aqueous citric-acid $pK_a$ values as exact liquid-ammonia measurements.}
\medskip
\noindent
\textbf{Following proton transfer from citric acid, the resulting ammonium ions will participate in the evolving ionic equilibrium of the liquid-ammonia medium. Citrate anions generated during proton donation can associate with available ammonium ions to form ammonium citrate-type ion pairs, while the
lithium counterions originating from the Birch reduction conditions may similarly associate with citrate-derived anions to form lithium citrate-type species. The final composition is therefore governed by ion
pairing, solvent coordination, and the relative abundance of lithium, ammonium, and citrate ions rather than by a single isolated salt product}
\medskip
\noindent
\textbf{Consequently, the final reaction mixture should be considered as a dynamic ionic system containing the protonated THCA-derived product together with lithium-containing and ammonium-containing citrate species formed during quench and proton-transfer equilibration}
\newpage
\medskip
\noindent
\begin{center}
\textbf{APPENDIX}
\end{center}
\medskip
\noindent
\textbf{In this appendix I've included an initial book-keeping list of what is needed. Also, I have ran this idea through IBM RXN and included its printout (keeping in mind IBM RXN only provides a prediction when it does not have actual quantifiable lab data. We can see that it predicted a C=O off the Ph-O- whihc we know now wont form a double bond due to the presence of the carbon radical/anion. I still included it for review and analysis}
\medskip
\begin{center}
\includegraphics[
width=\textwidth,
height=0.9\textheight,
keepaspectratio
]{Book1-8.png}
\end{center}
\medskip
\begin{center}
\includegraphics[
width=\textwidth,
height=0.9\textheight,
keepaspectratio
]{Book2-8.png}
\end{center}
\medskip
\begin{center}
\includegraphics[
width=\textwidth,
height=0.9\textheight,
keepaspectratio
]{Book3-8.png}
\end{center}
\medskip
\begin{center}
\includegraphics[
width=\textwidth,
height=0.9\textheight,
keepaspectratio
]{Book4-8.png}
\end{center}
\medskip
\begin{center}
\includegraphics[
width=\textwidth,
height=0.9\textheight,
keepaspectratio
]{Book5-8.png}
\end{center}
\medskip
\begin{center}
\includegraphics[
width=\textwidth,
height=0.9\textheight,
keepaspectratio
]{Book6-8.png}
\end{center}
\medskip
\begin{center}
\includegraphics[
width=\textwidth,
height=0.9\textheight,
keepaspectratio
]{Book7-8.png}
\end{center}
\medskip
\begin{center}
\includegraphics[
width=\textwidth,
height=0.9\textheight,
keepaspectratio
]{Book8-8.png}
\end{center}
\medskip
\begin{center}
\includegraphics[
width=\textwidth,
height=0.9\textheight,
keepaspectratio
]{IBMRXN_Intermediate.png}
\end{center}
\medskip
\begin{center}
\includegraphics[
width=\textwidth,
height=0.9\textheight,
keepaspectratio
]{IBMRXN_Product.png}
\end{center}
\medskip
\begin{center}
\includegraphics[
width=\textwidth,
height=0.9\textheight,
keepaspectratio
]{IBMRXN_Prod_FIN.png}
\end{center}
\medskip
\begin{center}
\includegraphics[
width=\textwidth,
height=0.9\textheight,
keepaspectratio
]{IBMRXN_REAGENTS_PREDICT.png}
\end{center}
\medskip
\begin{center}
\includegraphics[
width=\textwidth,
height=0.9\textheight,
keepaspectratio
]{Precip_Byprods.png}
\end{center}
\end{document}
) and dont recommend anyone to do so due to the fact small molecular changes can have big consequences
