Question 6.FP.6: The following is a schematic mechanism for the bromination o......
The following is a schematic mechanism for the bromination of an organic compound RH:
{\mathrm{Br}}_{2}+{\mathrm{M}}\ {\overset{k_{1}}{\longrightarrow}}\ 2\,{\mathrm{Br}}^{•}+{\mathrm{M}}
\mathrm{Br}^{•}+\mathrm{RH}\stackrel{k_{2}}{\longrightarrow}\mathrm{R}^{•}+\mathrm{HBr}
\mathrm{R}^{•}+\mathrm{B}\mathrm{r}_{2}\stackrel{k_{3}}{\longrightarrow}\mathrm{RBr}+\mathrm{Br}^{•}
2\,\mathrm{Br}^{•}+\,{\bf M}\stackrel{k_{4}}{\longrightarrow}\,\mathrm{Br}_{2}+{\bf M}
Derive the steady state rate for the production of HBr for this mechanism. If the inhibition step
\mathrm{R}^{•}+\mathrm{HBr}{\stackrel{k_{-2}}{\longrightarrow}}\mathrm{Br}^{•}+\mathrm{RH}
were included how would this affect the steady state rate expression?
Derive this rate expression.
Reaction without inhibition:
+\frac{\mathrm{d}[\mathrm{Br}^{•}]}{\mathrm{d}t}= 2\,k_{1}\left[\mathrm{Br}_{2}\right]\left[\mathrm{M}\right]-k_{2}[\mathrm{Br}^{\mathrm{•}}]\left[\mathrm{RH}\right]+k_{3}[\mathrm{R}^{\mathrm{•}}]\left[\mathrm{Br}\right]_{2} -\,2\,k_{4}[\mathrm{Br}^{•}]^{2}[{\bf M}] = 0 (I)
+\frac{\mathrm{d}[\mathbf{R}^{\bullet}]}{\mathrm{d}t}=k_{2}[\mathbf{Br}^{\bullet}][\mathbf{RH}]-k_{3}[\mathbf{R}^{\bullet}][\mathbf{Br}_{2}]=0 (II)
Add (I) and (II):
2\,k_{1}[\mathrm{Br}_{2}][\mathrm{M}]=2\,k_{4}[\mathrm{Br}^{\mathrm{•}}]^{2}[\mathrm{M}]
\left[\mathbf{Br}^{•}\right]=\left({\frac{k_{1}}{k_{4}}}\right)^{1/2}\left[\mathbf{Br}_{2}\right]^{1/2} (III)
+\frac{\mathrm{d}[\mathrm{HBr}]}{\mathrm{d}t}=k_{2}[\mathrm{Br}^{\mathrm{{•}}}][\mathrm{RH}]
=k_{2}{\bigg(}{\frac{k_{1}}{k_{4}}}{\bigg)}^{1/2}[\mathbf{Br}_{2}]^{1/2}[\mathbf{RH}] (IV)
Note in this mechanism that Equation (II) implies that ‘the rates of the two propagation steps are equal’.
Reaction with inhibition:
+\frac{\mathrm{d}[\mathrm{Br}^{•}]}{\mathrm{d}t}=2\,k_{1}[\mathrm{Br}_{2}][\mathrm{M}]-k_{2}[\mathrm{Br}^{•}][\mathrm{RH}]+k_{-2}[\mathrm{R}^{•}][\mathrm{HBr}]
+\,k_{3}[{\bf R}^{•}][{\bf B}{\bf r}_{2}]-2\,k_{4}[{\bf B r}^{•}]^{2}[{\bf M}]=0 (V)
+\frac{\mathrm{d}[\mathbf{R}^{\bullet}]}{\mathrm{d}t}=k_{2}[\mathbf{Br}^{\bullet}][\mathbf{R}\mathbf{H}]-k_{-2}[\mathbf{R}^{\bullet}][\mathbf{H}\mathbf{B}\mathbf{r}]-k_{3}[\mathbf{R}^{\bullet}][\mathbf{B}\mathbf{r_{2}}]=0 (VI)
Add (V) and (VI):
2\,k_{1}[\mathrm{Br}_{2}][\mathrm{M}]=2\,k_{4}[\mathrm{Br}^{\mathrm{•}}]^{2}[\mathrm{M}]
\left[\mathbf{Br}^{•}\right]=\left({\frac{k_{1}}{k_{4}}}\right)^{1/2}\left[\mathbf{Br}_{2}\right]^{1/2} (VII)
Note: this equation is identical to the corresponding equation for the mechanism without inhibition. This is as it should be since addition of the steady state equations should generate the algebraic statement that ‘the total rate of formation of the chain carriers should be balanced by the total rate of removal of the chain carriers’.
This should hold whether there is inhibition or not.
However, equation (VI) differs from equation (II). Here the rate term for the inhibition step is included, so that the rates of the two propagation steps are no longer equal. The long chains approximation in this form cannot be used for the inhibition mechanism, though it is valid for the reaction without inhibition.
+\frac{\mathrm{d}[{\mathrm{HBr}}]}{\mathrm{d}t}=k_{2}[{\mathrm{Br}}^{\mathrm{•}}][{\mathrm{RH}}]-k_{-2}[{\mathrm{R}}^{\mathrm{•}}][{\mathrm{HBr}}]
=k_{2}{\left({\frac{k_{1}}{k_{4}}}\right)}^{1/2}[\mathbf{B}\mathbf{r}_{2}]^{1/2}[\mathbf{R}\mathbf{H}]-k_{-2}[\mathbf{R}^{•}][\mathbf{H}\mathbf{B}\mathbf{r}]
From equation (VI),
k_{2}\left(\frac{k_{1}}{k_{4}}\right)^{1/2}[\mathrm{Br}_{2}]^{1/2}[\mathrm{RH}]=[\mathrm{R}^{•}]\{k_{-2}[\mathrm{HBT}]+k_{3}[\mathrm{Br}_{2}]\}
[\mathbf{R}^{•}]=k_{2}\left(\frac{k_{1}}{k_{4}}\right)^{1/2}\left(\frac{\left[\mathbf{B}\mathbf{r}_{2}\right]^{1/2}[\mathbf{RH}]}{k_{-2}[\mathbf{HBr}]+k_{3}[\mathbf{Br}_{2}]}\right)
+\frac{{\mathrm{d}}[{\mathrm{HBr}}]}{\mathrm{d}t}=k_{2}{\bigg(}{\frac{k_{1}}{k_{4}}}{\bigg)}^{1/2}[{\mathrm{Br}}_{2}]^{1/2}[{\mathrm{RH}}]
-\,k_{-2}[{\mathrm{HBr}}]k_{2}\left({\frac{k_{1}}{k_{4}}}\right)^{1/2}\left({\frac{[{\mathrm{Br}}_{2}]^{1/2}[{\mathrm{RH}}]}{k_{-2}[{\mathrm{HBr}}]+k_{3}[{\mathrm{Br}}_{2}]}}\right)
=k_{2}{\left({\frac{k_{1}}{k_{4}}}\right)}^{1/2}[\mathbf{Br}_{2}]^{1/2}[\mathbf{RH}]
-\,k_{-2}\,k_{2}\left(\frac{k_{1}}{k_{4}}\right)^{1/2}\left(\frac{[\mathrm{Br}_{2}]^{1/2}[\mathrm{RH}][\mathrm{HBr}]}{k_{-2}[\mathrm{HBr}]+k_{3}[\mathrm{Br}_{2}]}\right)
=k_{2}\left(\frac{k_{1}}{k_{4}}\right)^{1/2}[{\bf B r}_{2}]^{1/2}[\mathrm{RH}]\left\{1-k_{-2}\left[\frac{1}{k_{-2}+k_{3}\frac{[{\mathrm{Br}}_{3}]}{[{\mathrm{HBr}}]}}\right]\right\}
Note: the presence of the inhibition step dramatically complicates the rate expression.