MORISSON   Organic Chemistry

MORISSON Organic Chemistry

DisciplinaQuímica Orgânica I15.009 materiais276.565 seguidores
Pré-visualização50 páginas
difference is this: since halogenation is a chain reaction, dissocia-
tion of each molecule of halogen gives rise ultimately to many molecules of methyl
halide; hence, even though dissociation is very slow, the overall reaction can be
fast. The attack of iodine atoms on methane, however, is a chain-carrying step and
if it is slow the entire reaction must be slow; under these circumstances chain-
terminating steps (e.g., union of two iodine atoms) become so important that
effectively there is no chain.
2.21 Structure of the methyl radical, sp2 Hybridization
We have spent a good part of this chapter discussing the formation and
reactions of the methyl free radical CH 3 - . Just what is this molecule like? What
is its shape? How are the electrons distributed and, in particular, where is the odd
electron ?
These are important questions, for the answers apply not only to this simple
radical but to any free radical, however complicated, that we shall encounter.
The shape, naturally, underlies the three-dimensional chemistry the stereo-
chemistry of free radicals. The location of the odd electron is intimately involved
with the stabilization of free radicals by substituent groups.
As we did when we "made" methane (Sec. 1.11), let us start with the elec-
tronic configuration of carbon,
Is 2s 2p
c O O O
and, to provide more than two unpaired electrons for bonding, promote a 2s
electron to the empty 2p orbital:
Is 2s 2p ,
* One electron promoted:
O O O O four unpaired electrons
Like boron in boron trifluoride (Sec. 1.10), carbon here is bonded to three
other atoms. Hybridization of the 2s orbital and two of the p orbitals provides the
is 2s 2p
| | sp
\S Sp 2 In
c O O O O
necessary orbitals: three strongly directed sp
2 orbitals which, as we saw before,
lie in a plane that includes the carbon nucleus, and are directed to the corners of
an equilateral triangle.
If we arrange the carbon and three hydrogens of a methyl radical to permit
maximum overlap of orbitals, we obtain the structure shown in Fig. 2.9a. It is
Figure 2.9. Methyl radical, (0) Only a bonds shown, (b) Odd electron in
p orbital above and below plane of a bonds.
flat, with the carbon atom at the center of a triangle and the three hydrogen atoms
at the corners. Every bond angle is 120.
Now where is the odd electron? In forming the sp2 orbitals, the carbon atom
has used only two of its three p orbitals. The remaining p orbital consists of two
equal lobes, one lying above and the other lying below the plane of the three sp2
orbitals (Fig. 2.9b); it is occupied by the odd electron.
This is not the only conceivable electronic configuration for the methyl radi-
cal: an alternative treatment would lead to a pyramidal molecule like that of am-
monia, except that the fourth sp 3 orbital contains the odd electron instead of an
electron pair (Sec. 1.12). Quantum mechanical calculations do not offer a clear-
cut decision between the two configurations. Spectroscopic studies indicate that
the methyl radical is actually flat, or nearly so. Carbon is trigonal, or not far from
it; the odd electron occupies a/? orbital, or at least an orbital with much/? charac-
Compare the shapes of three molecules in which the central atom is bonded to three
other atoms: (a) boron trifluoride, with no unshared electrons, trigonal; (b) ammonia,
with an unshared pair, tetrahedral; and (c) the methyl radical, with a single unshared
electron, trigonal or intermediate between trigonal and tetrahedral.
There is stereochemical evidence (for example, Sec. 7.10) that most other free
radicals are either flat or, if pyramidal, undergo rapid inversion like that of the
ammonia molecule (Sec. 1.12).
Problem 2.3 Besides free radicals, we shall encounter two other kinds of reac-
tive particles, carbonium ions (positive charge on carbon) and carbanions (nega-
tive charge on carbon). Suggest an electronic configuration, and from this predict
the shape, of the methyl cation, CH 3 + ; of the methyl anion, CH3 :~.
2.22 Transition state
Clearly, the concept of act is to be our key to the understanding of chemical
reactivity. To make it useful, we need a further concept: transition state,
A chemical reaction is presumably a continuous process involving a gradual
transition from reactants to products. It has been found extremely helpful, how-
ever, to consider the arrangement of atoms at an intermediate stage of reaction as
though it were an actual molecule. This intermediate structure is called the tran-
sition state; its energy content corresponds to the top of the energy hill (Fig. 2.10).
Transition state
Progress of reaction >
Figure 2.10. Potential energy changes during progress of reaction : transi-
tion state at top of energy hump.
The reaction sequence is now:
reactants > transition state products
Just as A// is the difference in energy content between reactants and products, so
act is the difference in energy content between reactants and transition state.
The transition state concept is useful for this reason: we can analyze the struc-
ture of the transition state very much as though it were a molecule, and attempt
to estimate its stability. Any factor that stabilizes the transition state relative to
the reactants tends to lower the energy of activation; that is to say, any factor that
lowers the top of the energy hill more than it lowers the reactant valley reduces the
net height we must climb during reaction. Transition state stability will be the
basis whether explicit or implicit of almost every discussion of reactivity in
this book.
But the transition state is only a fleeting arrangement of atoms which, by its
very nature lying at the top of an energy hill cannot be isolated and examined.
How can we possibly know anything about its structure? Well, let us take as an
example the transition state for the abstraction of hydrogen from methane by a
halogen atom, and see where a little thinking will lead us.
To start with, we can certainly say this: the carbon-hydrogen bond is stretched
but not entirely broken, and the hydrogen-halogen bond has started to form but
is not yet complete. This condition could be represented as
H f H
| 18- 8-
H-C-H + -X > H-C--H-X
A I A .
Reactants Transition state
H-C- + H-X
where the dashed lines indicate partly broken or partly formed bonds.
Now, what can we say about the shape of the methyl group in this transition
state? In the reactant, where methyl holds the hydrogen, carbon is tetrahedral
(spMiybridized); in the product, where methyl has lost the hydrogen, carbon is
trigonal (,sp2-hybridized). In the transition state, where the carbon-hydrogen
bond is partly broken, hybridization of carbon is somewhere between sp3 and sp2 .
The methyl group is partly but not completely flattened; bond angles are greater
than 109.5 but less than 120.
+ H X
Reactant Transition state Product
Tetrahedral Becoming trigonal Trigonal
Finally, where is the odd electron? It is on chlorine in the reactants, on the
methyl group in the products, and divided between the two in the transition state.
(Each atom's share is represented by the symbol 8-.) The methyl group partly
carries the odd electron it will have in the product, and to this extent has taken on
some of the character of the free radical it will become.
Thus, in a straightforward way, we have drawn a picture of the transition state
that shows the bond-making and bond-breaking, the spatial arrangement of the
atoms, and the distribution of the electrons.
(This particular transition state is intermediate between reactants and products
not only in the time sequence but also in structure.