The mechanism for a reaction is a set of elementary reactions that detail the bond–making and bond–breaking steps leading from reactants to products.
There is often more than one plausible mechanism (pathway) for a reaction.
Elementary reactions occur on a single collision between the reacting species.
There is often more than one plausible mechanism (pathway) for a reaction.
For example, consider the reaction:
CH3Br + OH–
CH3OH + :Br–
two step mechanism
step 1: slow
CH3Br+CH3 + :Br – (C–Br bond breaks)
step 2: fast
H3C+ + :OH– CH3–OH (C–O bond forms)
single step mechanism
CH3Br + :OH– CH3OH + :Br–
(simultaneous C–Br bond breaking and C–O bond formation)
Multistep mechanisms involve intermediates like +CH3. These are formed in one step of the mechanism and consumed in a subsequent step.
It may be possible to distinguish between plausible mechanisms by comparing the experimental rate law with the overall rate law that the mechanism would predict. If they are the same, that mechanism could be operating.
The rate of an elementary reaction depends on the concentrations of the reactant species. Therefore the predicted rate law for any step in a proposed mechanism has the product of its reactant concentrations raised to their coefficient in the balanced equation and then multiplied by k.
For Step 1 of the two-step mechanism:
rate = k[CH3Br]
For the single-step mechanism:
rate = k[CH3Br][OH–]
The slow step determines the rate of the overall reaction and is referred to as rate-determining. If the rate-determining step is the first step in a reaction mechanism, the rate law predicted for the overall reaction will be the same as the rate-law for that step.
CH3Br + OH–
CH3OH + :Br–two step mechanism
step 1: slow
CH3Br
+CH3 + :Br – (C–Br bond breaks)step 2: fast
H3C+ + :OH–
CH3–OH (C–O bond forms) single step mechanism
CH3Br + :OH–
CH3OH + :Br–(simultaneous C–Br bond breaking and C–O bond formation)
Multistep mechanisms involve intermediates like +CH3. These are formed in one step of the mechanism and consumed in a subsequent step.
Intermediates, like catalysts, do not appear in the overall equation for the reaction. Catalysts are different to intermediates in that they are a reactant in first step and a product in a subsequent step.
It may be possible to distinguish between plausible mechanisms by comparing the experimental rate law with the overall rate law that the mechanism would predict. If they are the same, that mechanism could be operating.
The rate of an elementary reaction depends on the concentrations of the reactant species. Therefore the predicted rate law for any step in a proposed mechanism has the product of its reactant concentrations raised to their coefficient in the balanced equation and then multiplied by k.
For Step 1 of the two-step mechanism:
rate = k[CH3Br]
For the single-step mechanism:
rate = k[CH3Br][OH–]
The slow step determines the rate of the overall reaction and is referred to as rate-determining. If the rate-determining step is the first step in a reaction mechanism, the rate law predicted for the overall reaction will be the same as the rate-law for that step.