# 12.4.4: The conjugate base mechanism

$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

The substitution reaction of acidic octahedral complexes (with ligands that can donate a proton) can be catalyzed in the presence of hydroxide ion ($$\ce{OH^-}$$. Through an initial acid-base reaction, the metal complex can be deprotonated to its conjugate base. The deprotonation increases electron density on the metal center, facilitating the loss of the leaving group ligand, especially when the leaving ligand is trans to the deprotonation site. This mechanism is referred to as the conjugate-base mechanism or the $${\mathrm{S}}_{\mathrm{N}}\mathrm{1CB}$$ (Substitution, Nucleophilic, first-order in the Conjugate Base mechanism) mechanism. Although it involves a dissociative rate-limiting step, the rate law is consistent with the second-order kinetics because the rate also depends on the initial reaction of the base with the metal complex. In this mechanism, the first step is formation of the conjugate base of the metal complex by deprotonation of one of its ligands. This is a case where numerous lines of evidence are necessary to establish the mechanism.

Cobalt pentammine complexes, $$\ce{Co(NH3)5X^{n+}}$$, have been studied extensively and are found to react through the conjugate-base mechanism. The way in which the actual mechanism was determined, and how other possible mechanism were ruled out are described below.

## Evidence under acidic aqueous conditions

In acidic aqueous solutions, the reaction

$\ce{ [Co(NH3)_5X]^{2+} + Y^{-} -> [Co(NH3)_5Y]^{2+} + X^{-}} \nonumber$

was found to proceed through an aquo complex intermediate, $$\ce{[Co(NH3)_5{OH_2}]^{3+}}$$, which is also the dominant product despite the identity of the incoming ligand. In other words, to form the product of the reaction above, there are two substitution steps necessary (shown below).

\begin{align*} \ce{[Co(NH3)_5X]^{2+} + H2O} &\rightleftharpoons \ce{[Co(NH3)_5(OH2)]^{3+} + X^{-}} \tag{step 1, form aquo intermediate}\\[4pt] \ce{[Co(NH3)5(OH2)]^{3+} + Y^{-}} &\rightleftharpoons \ce{[Co(NH3)5Y]^{2+} + H2O } \tag{step 2, form final product} \end{align*}

The first step in the reaction is the breaking of a $$\ce{Co \bond{-} X}$$ bond and the formation of a $$\ce{Co \bond{-} OH2}$$ bond (step 1). Subsequently, $$\ce{Y^{-}}$$ can replace the aquo group (step 2).

In aqueous solution, water is always present at a much higher concentration than the various possible entering groups $$\ce{Y^{-}}$$, so it is reasonable that the aquo complex should be favored in the competition to form the new bond to $$\ce{Co(III)}$$. Nevertheless, the strength of the $$\ce{Co \bond{-} Y}$$ bond should depend on the nucleophilicity of $$\ce{Y^{-}}$$ in these substitution reactions. Thus, the amount of $$\ce{[Co(NH3)_5Y]^{2+}}$$ product should increase with increasing nucleophilicity of $$\ce{Y^{-}}$$. The fact that the aquo complex is the predominant reaction product despite the identity of $$\ce{Y^{-}}$$ strongly suggests that the the $$\ce{Co \bond{-} X}$$ bond breaking is more important in the reaction outcome than formation of the new bond to the incoming $$\ce{Y^{-}}$$.

This evidence indicates dissociative character in the reaction. But the nature of the substitution mechanism is still unclear. For example, is the original $$\ce{Co \bond{-}X}$$ bond completely broken before the new $$\ce{Co \bond{-} OH2}$$ bond has begun to form so that $$\ce{Co(NH3)_5^{3+}}$$ is a true intermediate? Does the mechanism involve a transitory intermediate where bonds are breaking and forming in a concerted fashion?

## Evidence under basic aqueous conditions

The evidence of reaction mechanism under basic conditions is seemingly contradictory to what is described above. When the entering group, $$Y^{-}$$, is the hydroxide ion, the reaction is

$\ce{ Co(NH3)_5X^{n+} + OH^{-} \to Co(NH3)_5OH^{2+} + X^{q-}} \nonumber$

Substitution by hydroxide ion is kinetically faster than by the aquo ligand in acidic solutions, and the rate is law found to be second-order:

$\frac{d\left[\ce{Co(NH3)_5OH^{2+}}\right]}{dt}=k\ \left[\ce{Co(NH3)_5X^{n+}}\right]\left[\ce{OH^{-}}\right] \nonumber$

This rate law is consistent with an associative reaction involving nucleophilic attack by the hydroxide ion at the cobalt center. In an associative interchange mechanism, the $$\ce{Co \bond{-} OH}$$ bond formation occurs simultaneously with breaking of the $$\ce{Co \bond{-} X}$$ bond. However, just as in the case above, the hydroxo complex dominates even in the presence of $$\ce{Y^{-}}$$. Thus, the hydroxide ion seems to be a uniquely effective nucleophile toward cobalt(III). However, nucleophilic displacements have been investigated on many other electrophile, and in general, hydroxide is not a particularly effective nucleophile toward other electrophilic centers. So, assignment of an associative mechanism to this reaction is reasonable only if we can explain why hydroxide is uniquely reactive in this case and not in others.

## The conjugate-base mechanism

An alternative mechanism, referred to as the conjugate-base mechanism or the $${\mathrm{S}}_{\mathrm{N}}\mathrm{1CB}$$ (Substitution, Nucleophilic, first-order in the Conjugate Base mechanism) mechanism, is also consistent with the second-order rate law. In this mechanism, the first step is formation of the conjugate base of the metal complex by deprotonation of one of its ligands. An important requirement of this mechanism is the presence of a metal-bound ligand with acidic protons (including amine, ammine, or aquo ligands). Under basic conditions the mecahnism is catalyzed by hydroxide ion.

In the case of $$\ce{[Co(NH3)_5{OH_2}]^{3+}}$$ reacting under basic conditions, the hydroxide removes a proton from one of the ammine ligands to give a six-coordinate conjugate base intermediate (step 1 below). This conjugate base intermediate contains an amido ($$\ce{NH^{-}_2}$$) ligand which is a $$\pi$$-donor ligand, allowing facilitated dissociation of one of the other ligands (particulary the ligand in the trans position). This intermediate loses the leaving group $$X^{-}$$ in the rate determining step to form a five-coordinate intermediate, $$\ce{Co(NH3)4NH^{2+}}$$ (step 2 below). This five-coordinate intermediate quickly picks up a water molecule from the bulk solution to give the aquo complex (step 3 below). In a series of proton transfers to (step 4) and from (step 5) the aqueous solvent, the aquo complex rearranges to the final product. The mechanism is shown below for $$\ce{[Co(NH3)_5{OH_2}]^{3+}}$$ reacting with $$\ce{Cl^{-}}$$:

\begin{align*} \ce{[Co(NH3)_5{Cl}^{2+}+OH]^{-}} &\rightleftharpoons \ce{[Co(NH3)_4(NH2){Cl}]^{+} + H_2O} \tag{step 1, conjugate base}\\[4pt] \ce{[Co(NH3)_4(NH2){Cl}]^{+}} &\rightleftharpoons \ce{[Co(NH3)_4{(NH2)}]^{2+} + {Cl}^{-}} \tag{step 2, dissociation} \\[4pt] \ce{Co(NH3)_4{(NH2)}^{2+} + H_2O} &\rightleftharpoons \ce{Co(NH3)_4{(NH2)(OH2)}^{2+}} \tag{step 3, association} \\[4pt] \ce{Co(NH3)_4{(NH2)(OH2)}^{2+} + OH^{-}} &\rightleftharpoons \ce{Co(NH3)_4{(NH2)OH}^{+} + H_2O} \tag{step 4, proton transfer} \\[4pt] \ce{Co(NH3)_4{(NH2)OH}^{+} + H_2O} &\rightleftharpoons \ce{Co(NH3)_5{OH}^{2+} + OH^{-}} \tag{step 5, proton transfer} \end{align*}

The evidence in favor of the $${\mathrm{S}}_{\mathrm{N}}\mathrm{1CB}$$ mechanism is persuasive. It requires that the ammine protons be acidic, so that they can undergo the acid–base reaction in the first step. That this reaction occurs is demonstrated by proton-exchange experiments. In basic $$\ce{D_2O}$$, the ammine protons undergo $$\ce{H-D}$$ exchange according to

$\ce{[Co(NH3)5Cl]^{2+} + D2O \rightleftharpoons [Co(ND3)5Cl]^{2+} + HDO} \nonumber$

The ammine protons are also necessary; the reaction does not occur for similar compounds, like $$\ce{Co(2,2^'-bipyridine)_2(O_2CCH_3)^{+}_2}$$, which lack acidic protons on the nitrogen atoms that are bound to cobalt (i.e., there are no $$\ce{H-N-Co}$$ moieties).

The evidence that $$\ce{Co(NH3)_4{(NH2)}^{2+}}$$ is an intermediate is also persuasive. When the base hydrolysis reaction is carried out in the presence of other possible entering groups, $$Y^{p-}$$, the rate at which $$\ce{[Co(NH3)_5X]^{n+}}$$ is consumed is unchanged, but the product is a mixture of $$\ce{[Co(NH3)_5{OH}]^{2+}}$$ and $$\ce{[Co(NH3)5Y]^{n+}}$$. If this experiment is done with a variety of leaving groups, $$X^{q-}$$, the proportions of $$\ce{[Co(NH3)_5{OH}^{2+}]}$$ and $$\ce{[Co(NH3)5Y]^{n+}}$$ are constant despite which leaving group exists in the reactant complex. These observations are consistent with the hypothesis that all reactants, $$\ce{[Co(NH3)_5X]^{n+}}$$, give the same intermediate, $$\ce{[Co(NH3)_4{(NH2)}]^{2+}}$$. The product distribution is always the same, because it is always the same species undergoing the product-forming reaction.

This page titled 12.4.4: The conjugate base mechanism is shared under a not declared license and was authored, remixed, and/or curated by Kathryn Haas.