LECTURE 2 STRUCTURE AND PROPERTIES OF ORGANIC MOLECULES 1. Atomic wave functions and orbitals. LCAO. The important thing to know is that atomic orbitals are represented by wave functions, and they have signs (have nothing to do with charges, rather mathematical). If value of function and its amplitude are 0 at certain point or region, then this is a node (or nodal region). s-orbitals are spherical and have no nodal planes. p-orbitals consist of two lobes and do have nodal planes. On either side of the nodal plane signs are opposite. Atomic orbitals can and do combine to form either molecular or hybrid orbitals, depending on the origin of the starting set. Show and discuss rules. 2. Molecular orbitals. The important thing here to stress is the bonding region, that between the nuclei. If the electron density is increased there due to the combination of atomic orbitals, then this leads to a decrease of energy and positive bonding process. Reason the electrons keep the two nuclei close, because of favorable +/- interactions, while screening the unfavorable +/+ interaction of the nuclei themselves. There is always an optimal distance at which these interactions are most favorable, it would be called the bond length.
The simplest case of AO interaction leading to MO is formation of hydrogen molecule. Draw and discuss the formation of bonding (constructive overlap) and antibonding (destructive overlap) orbitals. The bonding orbital has most of the el. density in the region between the nuclei very favorable; σ-bond cylindrical or axial symmetry. Now show and discuss the whole orbital scheme for hydrogen. Other orbitals can also form σ-bonds, provided they are oriented in the appropriate fashion, i.e. along the internuclear axis. Draw and discuss the cases of two p-orbitals, then one s- and one p-orbital. Bonding is also possible even if not along the internuclear axis. Two p-orbitals perpendicular to the internuclear axis. Overlap above and below the nodal plane (containing the nuclei). Still forming a bonding (π) and antibonding (π*) orbitals. This is the π-bond. Weaker, because of weaker overlap, since it is not right between the nuclei, but at an angle. 3. Hybridization and molecular geometry. One should expect, based on the geometry of s- and p-orbitals, that a molecule such as methane should have bond angles of 90 o. And more so, un-equivalent C-H bonds. Neither is true. For methane, 109.5 o, equal C-H bonds, ethylene 120 o, acetylene 180 o. These angular relationship, in agreement with the VSEPR theory, would correspond to a maximum separation (minimum interaction) of 4, 3 and 2 el. pairs respectively. Another way to explain the geometry (and equivalency of C-H bonds) is needed. This is
HYBRIDIZATION. Atomic orbitals from the same atom can combine to form a new set (same number) of orbitals, with the proper spatial orientation and equivalent to each other, so as to explain the observed facts. A/ sp 3 -hybridization combine one s- and three p-orbitals. Leads to the formation of four sp 3 -orbitals, at a 109.5 o angle. Draw the scheme for formation, discuss, tetrahedral molecular geometry. Example CH 4, in detail. Ethane, in detail. Bond length of a single C C bond: 1.53Å. Other molecules NH 3, NH + 4, H 2 O; both bonds and lone pairs count for hybridization. B/ sp 2 -hybridization now combine one s- and two p-orbitals. Leads to the formation of three sp 2 -orbitals, in a plane, at a 120 o angle to each other. Planar molecular geometry. What is left is a non-hybridized p-orbital, perpendicular to the plane. Example ethylene, in detail. The formation of the double bond. Bond length of the double C = C bond: 1.32 Å. Other molecules formaldehyde, BH 3 (tell them to consider the latter themselves). C/ sp-hybridization combine one s- and one p-orbital. Forms two sp-hybrid orbitals, in a linear arrangement. Linear geometry. Two non-hybridized p-orbitals left, perpendicular to the hybrid orbitals and to each other. Example acetylene, in detail. The formation of the triple bond. Bond length of the triple C = C bond: 1.18 Å. Other molecules HCN, BeH 2 (Tell them to consider the latter themselves).
Now if there is time, offer a more complex example CH 3 CN or CH 3 CHO, or allene. How do we determine the hybridization of a particular center: Count σ-bonds + lone pairs = number of necessary hybrid orbitals, thus determine hybridization. What happens when there is resonance? Always consider the lowest possible hybridization fitting any of the important resonance structures. The textbook has the example of CH 2 NH + 2. Consider: CH 2 =CHO -. 4. Drawing of three-dimensional structures. It is done by the use of dashed and bold lines or wedges, to denote atoms or group above or below the plane of paper. Regular lines are used to show atoms or groups in the plane. 5. Molecular structure and bond rotation. One of the most important properties of the σ-bond is its cylindrical symmetry, which makes the overlap and bond absolutely independent of possible rotation and angle change. This is the reason for the free rotation in molecules such as ethane. Each variety is known as a conformation. To the contrary, double or triple bonds have π-components, for which the rotation would lead to an immediate decrease of overlap and essential breakage of the π-bond. This is very unfavorable, and because of it, rotation around double or triple bonds requires a lot of energy, it is not free. The supposed different rotation forms are
now not interconvertible, and are different compounds. Typical example: cis- and trans-2-butene. 6. Isomers. A/ Constitutional (or structural) same molecular formula, different structure. Dependant on the number of non-hydrogen atoms. Increase with increasing number. Example 1-butene vs. 2-butene. B/ Stereoisomers same molecular formula, same structural formula, but different spatial orientation. Example cis-2-butene vs. trans-2- butene. 7. Dipole Moments. As we mentioned earlier, because of electronegativity differences, covalent bonds between different elements tend to polarize and form dipoles, characterized by dipole moments. The polarity and diploe moment usually increase with increasing electronegativity difference: ethane, methylamine, methanol and chloromethane. The dipole moment is the product of the charge x the distance between the charged centers: µ = δ x d Dipole moments are measured in debyes, with 1 debye (D) = 3.34 x 10-30 coulomb.meters. So if a proton and an electron are separated by 1Å, then the dipole moment is: µ = (1.60 x10-19 ) x 10-10 = 1.60 x 10-29 coulomb.meter. Or in debyes: µ = 1.60 x 10-29 /3.34 x 10-30 = 4.8 D
Which is the standard dipole moment for 1e charges at 1Å distance. Should the distance or charges be different, then we simply multiply the above number by the corresponding values: µ = 4.8 x δ(charge as a fraction of 1e charge) x d (in angstroms) The obvious advantage of this formula is the possibility to work with charges in fractions of the electron charge and distance in angstroms. The above formula s value is in the possibility to go backwards, if we know µ, and calculate the charges on the atoms. Thus the C=O bond has µ = 2.4 D, and the distance is 1.21A. This means that each of the atoms has a charge of 2.4/4.8x1.21 = 0.4 electron charges, which is positive at carbon and negative at oxygen. Aside from bond dipole moments, there are also molecular dipole moments, which is the dipole moment of the whole molecule. It is a good indicator of the polarity of the molecule, but caution must be exercised, because the moilecular dipole moment is a vector sum of the individual bond dipole moments. With the appropriate geometry, such vector sum could cancel out to 0. Thus: formaldehyde (2.3 D) and carbon dioxide (0.0 D). Or chloromethane (1.9 D), chloroform (1.0 D) and carbon tetrachloride (0.0 D). Lone pairs have to be taken into account when one tries to explain the magnitude of dipole moments. They are negative charges and a corresponding positive charge at the corresponding nucleus is
associated with it, thus creating additional dipoles and dipole moments. This is the reason why molecules with lone pairs, such as acetonitrile and water, have even larger than expected dipole moments. 8. Intermolecular Attraction and Repulsion Forces. Molecules are generally attracted to each other when they are at larger distances and this continues until they get so close that interfere with each other s atomic radii, at which point a strong repulsion force pulls them apart. These forces of attraction act to various degree and are of varying nature in different molecules, but they can nevertheless be grouped into several general types. A/ Dipole dipole forces. If the molecules have permanent dipole moments, then each molecule is an entity with two oppositely charged poles. The +/- attraction is stabilizing and generally prevalent, as molecules tend to orient in such a fashion. This stabilization then must be overcome in terms of extra energy, in order to pull such molecules apart. This reflects on higher boiling points for such materials. B/ The London Dispersion Forces. Materials with very small or non-existing permanent dipoles also tent to attract each other, albeit with much weaker forces, known as London forces, part in general of the van der Waals forces. Their appearance and exertion can be viewed as follows: Even in non-polar molecules, because of the fact that electron distribution changes constantly, some fleeting, short-
lived dipoles appear, due to a momentary un-even distribution of the electron density. Such temporary dipole immediately causes polarization in the neighboring molecules, and for a short time a cluster of such temporary dipoles is built, with them oriented favorably in the +/- fashion. In the next moment this arrangement id destroyed, but another one appears, etc. The overall effect is a weak attraction. Example with carbon tetrachloride. These forces are greatly dependent on the molecular size and shape, since they rely on close intermolecular contacts. The larger the molecule, the larger the attraction. If molecules are isomers with the same # of carbon atoms, then shape is important. Linear are with more attraction, branched with less: n-pentane (bp = 36), isopentane (28) and neopentane (10). The more branched, the more spherical, the less surface for interaction. C/ Hydrogen bond. It is a particularly strong dipole-dipole attraction, between molecules, containing hydrogen atom connected to oxygen nitrogen or fluorine. And since H F bonds are not known in organic chemistry, so only OH and NH materials. The reason for its appearance is the high electronegativity of the element connected to oxygen. This causes an extreme polarization of the element hydrogen bond, and electron depletion at the hydrogen. Such highly positive hydrogen will now tend to interact with the lone pairs of the element (O or N) of a neighboring molecule, leading to a relatively strong attachment. It must be mentioned that, regardless of
its name, the hydrogen bond is far weaker than true bonds (thus O H ---O interaction is ~ 5 kcal/mol, while O H bond is ~ 100 kcal/mol). Nevertheless, it is cumulative and brings about a tremendous change of melting/boiling point, vaporization properties, etc. In the sequence H 2 S, H 2 Se, H 2 Te the bp increases, but the bp of water, instead of being the lowest, is actually 160 o C higher. Also tremendous change in boiling points of alcohols, amines, in comparison with isomers, which do not contain OH or NH functionalities. The forces between molecules are also responsible for the solubility properties of materials. Generally like dissolves like. Four possible combinations: 1) Polar solute in polar solvent. Such as salts in water, or HCl in water, or alcohol and water. Needs to break energy of both solute and solvent, which is significant. But upon solvation energy is gained, also positive entropy. 2) Polar solute in a non-polar solvent. Such as salt in gasoline. No dissolution!! Reason: The eventual interaction of the solvent with the solute molecules (solvation) is too weak and not sufficient to break up the strong interaction between the molecules or ions of the solute. 3) Non-polar solute in a polar solvent. Oil in water. No dissolution. Reason: The eventual solvation of the solute is too weak to balance the energy needed to overcome the interactions between the solvent molecules. 4) Non-polar solute in a nonpolar solvent. Paraffin in gasoline. Dissolves. Reason: Both are nonpolar
materials with weak interactions within themselves and between each other. So not much energy is required to break starting interactions, not much is gained upon solvation. But entropy change is positive and this drives the reaction.