Consider some reversible chemical reaction in a closed system: $$\ce{A <=> B + D} \tag{1}$$ where $\ce{A}$ is some input reagent and $\ce{B}$, $\ce{D}$ are some output reagents. Let's denote forward and backward rates as $k_f$ and $k_b$ correspondingly.

Now, consider the same reaction in the presence of some catalyst $\ce{C}$:

$$\ce{A + C <=> B + D + C} \tag{2}$$

Due thermodynamic constraints the equilibrium state cannot change, which means that forward and backward rates of catalyzed reaction ($c_f$ and $c_b$) must be equal to the rates of uncatalyzed reaction multiplied by the same multiplier, $k$: $$ \begin{matrix} c_f = k \, k_f \\ c_b = k \, k_b \\ \end{matrix} \tag{3} $$

Now, consider the same scenario if the reagents and catalysts are chiral so that each of the reagents has an enantiomer: $$\hat{E}(A) = a, \hat{E}(B) = b, \hat{E}(C) = c, \hat{E}(D) = d \tag{4}$$ where $\hat{E}$ denotes an enantiomer.

Then the first reaction splits into two: $$ \begin{matrix} \ce{A <=> B + D} \\ \ce{a <=> b + d} \\ \end{matrix} \tag{5} $$ with the same forward and backward rates $k_f$ and $k_b$ and catalytic version splits into 4:

$$ \begin{matrix} \ce{A + C <=> B + D + C} \\ \ce{a + c <=> b + d + c} \\ \end{matrix} \tag{6} $$

and $$ \begin{matrix} \ce{A + c <=> B + D + c} \\ \ce{a + C <=> b + d + C} \\ \end{matrix} \tag{7} $$

where (6) is just the same as the catalytic reaction (2) above but now expressed for both enantiomers and (7) accounts for the fact that chiral catalysts are not 100% enantioselective: if chiral reaction is catalyzed by a chiral catalyst, then it is also catalyzed by enantiomer of that catalyst (though with different rate coefficients).

To consider thermodynamic equilibrium in such a system we must also take inter-conversion of enantiomers (racemization) into account: the concentration of both enantiomers must be equal in equilibrium for each pair of enantiomers. This leads to the following constraints: $$ \begin{matrix} c_f + c^e_f = k \, k_f \\ c_b + c^e_b = k \, k_b \\ \end{matrix} \tag{8} $$

where $c_f$, $c_b$ are forward and backward rate coefficients of reactions (6) and $c^e_f$, $c^e_b$ are forward and backward rate coefficients of reactions (7).

So, thermodynamics imposes two constraints on four parameters. If we consider many catalysts for reaction (5), then all of them will obey equation (8) with some value of $k$.

The question is: are there any other constraints, which can be imposed on coefficients of reactions (6) and (7)?

  • $\begingroup$ Why bother with species D, it just makes things more complicated. Also if you do not add reactions $A\leftrightharpoons a$ and similarly for the other species, you effectively have two separate reactions, one for species $A$ and another for $a$. $\endgroup$ – porphyrin Dec 17 '18 at 9:26
  • $\begingroup$ The considerations without D species and without reactions like $\ce{A <=> a}$ results in wrong thermodynamic considerations and, subsequently, in erroneous conclusion that chiral catalyst would result in the same multiplier for forward and backward rates of chiral reaction, thus not changing even local equilibrium state. But it does! So, both D species and racemization are required :( $\endgroup$ – Konstantin Konstantinov Dec 17 '18 at 9:57
  • $\begingroup$ As it happens I disagree with your conclusion, $A\leftrightharpoons a$ etc is the racemisation step. $\endgroup$ – porphyrin Dec 19 '18 at 8:29
  • $\begingroup$ @porphyrin Great! That's why I am here. If we put a racemase and just one of the species (L or D), then what will we observe after a while? Will it be a racemic mixture of L + D regardless where we start OR will it be just mostly one of the species? Of course, we will need enough cofactor and maybe even some inflow of energy. But, ... what will be the outcome? I can't find the answer on the web :( This is the key underlying question. $\endgroup$ – Konstantin Konstantinov Dec 19 '18 at 11:40

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