Question

What is a chemical reaction mechanism?

Answer 1

It's basically what's happening in a reaction in great detail. What we write out for an overall reaction does not include each step of the mechanism, but is a result of the mechanism itself.

Knowing the mechanism often helps us predict how the products of new reactions might turn out.

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In this answer, we will examine only complex, and no elementary reactions. I will also assume that you know about rate constants, rate laws, rates of reaction, reaction order, and other basic facets of kinetics.

A basic mechanism has these main components:

• A catalysis, if desired.
• Reactants and Products
• Intermediates (zero or more)

Past the High School level, you should additionally know about:

• Transition States (one or more)
• Electron movement
• Bond breaking/making

Mechanisms can range from a few steps (`1~5`) to many steps (`5~10`). I don't think I've seen more than that for a single reaction.

HIGH SCHOOL EXAMPLE

A two-step example with catalysis is:

`O_3(g) + stackrel("Catalyst")overbrace(Cl(g)) stackrel(k_1" ")(=>) stackrel("Intermediate")overbrace(ClO(g)) + O_2(g)`
`underbrace(ClO(g)) + O(g) stackrel(k_2" ")(=>) O_2(g) + underbrace(Cl(g))`
`stackrel("Intermediate")""" "" "" "" "" "" "" "stackrel("Catalyst")""`
`"-----------------------------------------"`
`stackrel("Overall reaction")overbrace(underbrace(O_3(g) + O(g)) stackrel(k_"obs"" ")(->) underbrace(2O_2(g)))`
`" "stackrel("Reactant(s)")""" "" "" "stackrel("Product(s)")""`

where `k_i` is the rate constant for step `i`, and `k_"obs"` is the rate constant for the overall reaction.

I used `=>` for elementary steps (which occur as-written) and `->` for complex steps. Here, I've identified the catalyst and intermediate.

• A catalyst speeds up the rate of reaction. It is placed into the reaction, and is regenerated at the end.
• An intermediate was created in the middle of a reaction, but does not stick around at the end.

Note that steps 1 and 2 were added to give the overall reaction, and intermediates and catalysts cancel.

This is something appropriate for the High School level, and you wouldn't need to identify transition states for this kind of reaction.

If you wanted to write a rate law for this kind of reaction, you would do so for the overall reaction, but you may have trouble identifying the true value of `k_"obs"`, as it is a combination of `k_1` and `k_2` in some way.

`r(t) = k_"obs"["O"_3]^m["O"]^n`

Furthermore, if you wanted to know `k_"obs"` in terms of `k_1` and `k_2`, the reactant orders `m` and `n` are unknown without further work, as the rate law must include ONLY reactants, and no intermediates or catalysts.

This would generally require complicated substitutions, or particular assumptions.

POST-HIGH SCHOOL EXAMPLE

An example without catalysis but with well-known electron movement is this one-step arrow-pushing mechanism, called a concerted mechanism:

• Reactants: `"C"_2"H"_4` and `"CH"_3"COOOH"`
• Products: `"Epoxide"` and `"CH"_3"COOH"`
• Transition states: `1`
• Intermediates: `0`

You can see electrons moving according to the curved arrows; old bonds have been broken and new ones have been made.

Transition states are the point when important bonds are made and broken at each step, and they occur for any reaction, but only if there are `n+1` transition states are there `n` intermediates.