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CH334

Organic Chemistry

Reaction Mechanisms
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I'd like us to consider initially a complex reaction:  free-radical bromination of ethylbenzene (Problem 4-46g).

Bromination of Ethylbenzene

We'd like to know what product, or products, will form; more generally, we'd like to know why one or some products form but others do not.  There are thousands or maybe even millions of similar transformations we might consider doing; few of them have been tried yet, and so we'd like a predictive model that can tell us the likely outcome.  Understanding aspects of the reaction mechanism will allow us to do this.

A mechanism is a specific set of bond-breaking and bond-forming processes that lead from reactants to products.  All chemical reactions involve breaking at least one bond and forming at least one new bond; usually more than one may be be involved.  So we have to propose a logical order for these reorganizations to occur that encompasses known systematic behavior of a wide array of organic and other molecules.  The outcome should consistently predict the observed outcome.

In this case, a simple place to start is to ask:  what bonds are broken?
  • We have to break a Br-Br bond.
  • We have to break a C-H bond in every possible outcome.  There are 6 different C-H bonds that we could break.
Now, what new bonds are formed:
  • In all cases, we get a Br-H bond.
  • In all cases we form a new C-Br bond, though we are making 6 different bonds in the 6 different possible products.
→We have to understand some chemistry to go from here. Bromine (Br2) has a weak Br-Br bond which can be broken either thermally (adding enough heat to provide the 192 kJ/mol needed to break the bond) or photolytically: adding light that has enough energy to break the bond. I'll express this for thermal initiation:

Br2 → 2 Br·

The equilibrium for this lies far to the left; at 80°C this works out to 1 part in 1014 that gets converted to radicals. But remember that even a millimole has 6 x 1023 particles, so in a millimole of bromine at 80°C there will be millions of bromine atoms present.
The bromine atoms that form have unpaired electrons: any such specied is called a free radical. Because of the absence of an octet free radicals are highly reactive. It can engage in an atom-transfer reaction, pulling an electron out of a covalent bond and taking one of the atoms with it. Hydrogens are easy to get to in ethylbenzene, but they will differ in how easy they are to pull off. We can judge this by the different bond-dissociation enthalpies:

Bond Dissociation Enthalpies

We can then predict selectivity based on the cost of pulling off a hydrogen atom, less the enthalpic gain of forming H-Br (368 kJ/mol):

CH3: +410 kJ/mol - 368 kJ/mol = +42 kJ/mol
CH2: +356 kJ/mol - 368 kJ/mol = -12 kJ/mol
Aromatic C-H: +473 kJ/mol - 368 kJ/mol = +105 kJ/mol

The only reaction that is enthalpically favored is abstraction of the CH2 hydrogen! So we predict that this will be favored.

Now, just because the initial reaction is uphill does not mean it cannot occur at all; if we get back more energy then we'd be OK. let's figure out where to go next. We have, at this stage:
  • Broken Br-Br bond;
  • Broken a C-H bond (and figured out how to predict which one);
  • formed H-Br bond.
We need to form the C-Br bond. The organic free radical is just like the Br atom: highly reactive, electron deficient, seeking to complete its octet via atom abstraction. It could pull another H atom from a second molecule of ethylbenzene, but at best this is a neutral reaction that doesn't change anything. Or we could interact with Br-Br, and abstract bromine:

Bromine atom abstraction

The energetics are the energy of the new bond, less the energy of the broken bond:

C-Br = 293 kJ/mol
Br-Br = 192 kJ/mol

Net enthalpy change: -101 kJ/mol.

When we combine this, we get the following:

Substitute CH3 hydrogen: Net ΔH°= +42 kJ/mol -101 kJ/mol = -59 kJ/mol. This is possible.
Substitute CH2 hydrogen: Net ΔH° = -12 kJ/mol - 101 kJ/mol = -113 kJ/mol. This is the best reaction energetically.
Substitute an aromatic C-H: Net ΔH° = +105 kJ/mol - 101 kJ/mol = +4 kJ/mol; this is energetically uphill and will not happen.

Note also: we generate the same species (a Br atom) that was used to start the process: this can react with anothe ethylbenzene molecule and so forth. It is very unlikely that it will recombine with another radical; remember only one part in 1014 of the system is present as a free radical.

The observed outcome was reported in 1937 to give exclusively (1-bromoethyl)benzene, the CH2 substitution product, although photolytic initiation was used rather than thermal initiation.