Roberts, Caserio - Basic Principles of Organic Chemistry (2nd edition, 1977)

b. Classify each of the following as chlral, achlral, asymmetric, dlssymmetr~cor symmetric a cup, a sh~rt,a b~cycle,a tennis racket, an automobile, a penny, a pair of scissors, a flat splral (4), and a conical spring (5) lndlcate any arnb~gu~tlesthat may be rnvolved

5-36 The structures of some biochemically interesting compounds are shown below. Mark the chiral carbons in each and calculate by the 2" rule how many stereoisomers might be expected. Explain why only one pair of enantiomers is known for camphor (ball-and-stick models will be very helpful here). How many different stereoisomers would you expect actually could be prepared of the quinine and codeine structures?

BONDING IN

ORGANIC MOLECULES. ATOMIC-ORBITAL MODELS

In previous chapters, we have shown how you can use ball-and-stick models to predict the general arrangements in space of organic molecules. The sticks correspond to chemical bonds, which we represent in structural formulas as lines, or in Lewis struc'tures as pairs of dots denoting shared pairs of electrons. Remembering that electrons and nuclei are charged particles, and that it is electrical forces of attraction and repulsion between the electrons and nuclei that determine the bonding, perhaps we should be surprised that such simple mechanical models provide so much useful information. What we will try to do in this chapter is to show you how the modern electronic theory of chemical bonding provides strong support for the use of ball-and-stick models for many organic molecules, and also where it indicates that the models need to be modified or cannot properly represent the structural arrangement.

There are several qualitative approaches to bonding in polyatomic molecules, but we shall discuss here the most widely used and currently popular approach. This approach involves setting up appropriate atomic orbitals for the atoms and considering that each bond arises from the attractive electrical forces of two or more nuclei for a pair of electrons in overlapping atomic orbitals, with each orbital on a different atom. The geometry of the bonds is assumed to be determined by the geometry of the orbitals and by the repulsive forces between the electrons. In the course of showing how this approach

can be applied, we shall discuss ways of formulating bonding and geometries for several important kinds of organic compounds. Finally, we will show you some of the results currently being obtained by sophisticated quantummechanical calculations, which provide strong support for our qualitative formulations.

6-1 HYDROGENLIKE ATOMIC ORBITALS

With the modern concept of a hydrogen atom we do not visualize the orbital electron traversing a simple planetary orbit. Rather, we speak of an atomic orbital, in which there is only a probability of finding the electron in a particular volume a given distance and direction from the nucleus. The boundaries of such an orbital are not distinct because there always remains a finite, even if small, probability of finding the electron relatively far from the nucleus.

There are several discrete atomic orbitals available to the electron of a hydrogen atom. These orbitals differ in energy, size, and shape, and exact mathematical descriptions for each are possible. Following is a qualitative description of the nature of some of the hydrogen atomic orbitals.

The most stable or ground state of a hydrogen atom is designated 1s.l In the 1s state the electron is, on the average, closest to the nucleus (i.e., it is the state with the smallest atomic orbital). The Is orbital is spherically symmetrical. This means that the probability of finding the electron at a given distance r from the nucleus is independent of the direction from the nucleus. We shall represent the 1s orbital as a sphere centered on the nucleus with a radius such that the probability of finding the electron within the boundary surface is high (0.80 to 0.95); see Figure 6-1. This may seem arbitrary, but an orbital representation that would have a probability of 1 for finding the electron within the boundary surface would have an infinite radius. The reason is that there is a finite, even if small, probability of finding the electron at any given distance from the nucleus. The boundary surfaces we choose turn out to have sizes consistent with the distances between the nuclei of bonded atoms.

The 2s orbital is very much like the 1s orbital except that it is larger and therefore more diffuse, and it has a higher energy. For principal quantum number 2, there are also three orbitals of equal energies called 2p orbitals, which have different geometry than the s orbitals. These are shown in Figure

'The index number refers to the principal quantum number and corresponds to the " K shell" designation often used for the electron of the normal hydrogen atom. The principal quantum number 2 corresponds to the L shell, 3 to the M shell, and so on. The notation s (also p, d, f to come later) has been carried over from the early days of atomic spectroscopy and was derived from descriptions of spectroscopic lines as "sharp," "principal," "diffuse," and "fundamental," which once were used to identify transitions from particular atomic states.

6 Bonding in Organic Molecules. Atomic-Orbital Models

6-2, in which we see that the respective axes passing through the tangent spheres of the three p orbitals lie at right angles to one another. The p orbitals are not spherically symmetrical.

The 3s and 3 p states are similar to the 2s and 2p states but are of higher energy. The 3d, 4d, 4f, . - . ,orbitals have still higher energies and quite different geometries; they are not important for bonding in most organic substances, at least for carbon compounds with hydrogen and elements in the first main row (Li--Ne) of the periodic table. The sequence of orbital energies is shown in Figure 6-3.

Figure 6-2 The shapes and orientations of the three 2p orbitals of a hydrogen atom. Notice that p orbitals have two lobes, one on each side of the nucleus.

6-1 Hydrogenlike Atomic Orbitals

I ., energy

Figure 6-3 Schematic diagram of the relative energies of the hydrogen like atomic orbitals

The famous Pauli exclusion principle states that no more than two electrons can occupy a given orbital and then only if they differ with respect to a property of electrons called electron spin. An electron can have only one of

resent two unpaired electrons, which may not go into a single orbital.

If we assume that all atomic nuclei have orbitals like those of the hydrogen atom, we can see how atoms more complex than hydrogen can be built up by adding electrons to the orbitals in accord with the Pauli principle. The lowest-energy states will be those in which the electrons are added to the lowest-energy orbitals. For example, the electronic configuration of the lowestenergy state of a carbon atom is shown in Figure 6-4, which also shows the relative energies of the Is through 4p atomic orbitals. The orbitals of lowest energy are populated with the proper number of electrons to balance the nuclear charge of +6 for carbon and to preserve the Pauli condition of no more than two paired electrons per orbital. However, the two highest-energy electrons are put into different 2p orbitals with unpaired spins in accordance with Hund's rule. The rationale of Hund's rule depends on the fact that electrons repel each other and the degree of repulsion becomes greater as the electrons come closer together. Now, suppose there are two electrons that can go into two different orbitals of the same energy (so-called degenerate orbitals). Hund's rule tells us that the repulsion energy between these electrons will be less if they have unpaired spins ( T ). Why is this so? Because if they have unpaired

6 Bonding in Organic Molecules. Atomic-Orbital Models

Figure 6-4 D~agramshowing the most stable electron~cconf~gurat~on, (Is)2(2s)~2px)'(2p,)',of a carbon atom Add~tionof more electrons In accord w ~ t hHund'srule glves the electronic conf~gurat~onof the other atoms indicated by the atomlc symbols

spins they cannot be in the same orbital at the same time. Therefore they will not be able to approach each other as closely as they would if they could be in the same orbital at the same time. For this reason the electronic configuration

is expected to be more stable than the configuration

if both orbitals have the same energy.

States such as the one shown in Figure 6-4 for carbon are built up through the following steps. Helium has two paired electrons in the 1s orbital; its configuration can be written as (1s ) ~the, superscript outside the parentheses denoting two paired electrons in the Is orbital. For lithium, we expect Li ( l s ) 2 ( 2 s ) 1to be the ground state, in which the Is electrons must be paired according to the exclusion principle. Continuing in this way, we can derive the electronic configurations for the elements in the first three rows of the periodic table, as shown in Table 6-1. These configurations conform to the

6-2 Bond Formation Using Atom~cOrbitals

Table 6-1

Electronrc Configurations of Ground States of Atoms

principle that an s orbital can accommodate a maximum of two paired electrons and a set of three p orbitals a maximum of six paired electrons. The first electronic configuration should conform to Hund's rule, as shown by the example of carbon in Figure 6-4.

Exercise 6-1 Write the ground-state configuration for a helium atom with two un- paired electrons (H e T T ) that is in accord with the Pauli principle. Give your reasoning.

6-2 BOND FORMATION USING ATOMIC ORBITALS

In writing the conventional Lewis structures for molecules, we assume that a covalent chemical bond between two atoms involves sharing of a pair of electrons, one from each atom. Figure 6-5 shows how atomic orbitals can be considered to be used in bond formation. Here, we postulate that a single bond is formed by the pulling together of two atomic nuclei by attractive forces exerted by the nuclei for the two paired electrons in overlapping atomic orbitals.

Because two atomic orbitals can hold a maximum of four electrons, it is reasonable to ask why it is that two rather than one, three, or four electrons normally are involved in a bond. The answer is that two overlapping atomic

6 Bonding in Organic Molecules. Atomic-Orbital Models

overlap \

orbitals can be considered to combine to give one low-energy bonding molecular orbital and one high-energy antibonding molecular orbital (see the top part of Figure 6-6(a).ZOrbitals that overlap as shown in Figure 6-6(a) are said to overlap in the sigma manner? and the bonding orbital is called a sigma orbital (a); the antibonding orbital is called a a*orbital (read "sigma star"). Two paired electrons suffice to fill the a orbital. Any additional electrons must go into the high-energy a*orbital and contribute not to bonding but to repulsion between the atoms.

The hydrogen molecule-ion, Hz@,can be regarded as having one electron in a a orbital. It has been studied in the vapor state by spectroscopic means and found to have a dissociation energy to H@and H. of 61 kcal mole-' compared to the 104.2 kcal mole-' bond energy for Hz. Several possible combinations of two hydrogen orbitals and from one to four electrons are shown in Figure 6-6(b).

similar to Figure 6-6(b). The ion He," has been detected spectroscopically; suggest a reason why this ion is more stable than Hzo.

Exercise 6-3 Write electronic configurations, as in Figure 6-6, for three different excited states of Hz,all of which agree with the Pauli principle. Arrange them in order of expected stability. Show your reasoning.

2More about the difference between bonding and antibonding orbitals is given in Section 21-2. For now we will say that the property of orbitals that leads to bonding or antibonding is a property analogous to phase. An in-phase combination of two orbitals is bonding, and an out-ofphase combination is antibonding.

3The designation sigma (cr) denotes that orbital overlap and electron density are greatest along the internuclear axis.

6-3 Electron Repulsion and Bond Angles Orb~taiHybridization

6-3 ELECTRON REPULSION AND BOND ANGLES.

ORBITAL HYBRIDIZATION

In predicting bond angles in small molecules, we find we can do a great deal with the simple idea that unlike charges produce attractive forces while like charges produce repulsive forces. We will have electron-nuclear attractions, electron-electron repulsions, and nucleus-nucleus repulsions. Let us first consider the case of a molecule with just two electron-pair bonds, as might be expected to be formed by combination of beryllium and hydrogen to give beryllium hydride, H:Be:H. The problem will be how to formulate the bonds and how to predict what the H-Be-H angle, 8, will be:

If we proceed as we did with the H-H bond, we might try to formulate bond formation in BeH, by bringing two hydrogen atoms in the ( 1 s ) l state up to beryllium in the ( 1s) ( 2 s ) ground state (Table 6-1).But there is a problem -