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Basic Principles for Introduction to Organic Chemistry

Building Bridges to Knowledge

What is Organic Chemistry?

Organic Chemistry is the study of compounds containing carbon atoms. Organic compounds are somewhat unique in that they form molecules where carbon atoms bond together. This is not a phenomenon exclusive to carbon, but carbon does it better than any other element. Boron and silicon are two examples of elements that can also form covalent bonds with themselves, i.e., boron to boron

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and silicon to silicon.

Carbon has a unique position on the Periodic Chart, because it has the ability to expand its octet and bond with itself. No other element on the periodic chart can compete with carbon’s ability to bond with itself; therefore, the carbon atom can be facetiously pictured as dominating the Periodic Chart. The perception that carbon dominates the Periodic Chart is an indication of the important role it plays in biology and chemistry. Carbon is singularly different, because it has an expanded octet, i.e., it has the ability to bond ad infinitum to other carbon atoms. In addition, carbon atoms have the ability to bond with other elements such as hydrogen, nitrogen, oxygen, sulfur, phosphorus, etc. Carbon’s ability to bond to other carbon atoms and other elements is extremely important in designing and representing the existence of millions of organic compounds with thousands of new organic

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compounds synthesized every year. The remaining elements on the Periodic Chart form more than 300,000 inorganic compounds; consequently, the number of carbon containing compounds* far exceeds the number of inorganic compounds. Up to the middle of the nineteenth century, scientists believed that organic compounds came from living organisms. The theory that organic compounds originated from living organisms is known as the Vital Force Theory. The one thing that all organic compounds have in common is the element carbon. The science of organic chemistry didn’t evolve until around the first quarter of the nineteenth century. Since that time, organic chemistry has grown exponentially spanning three overlapping periods. The first period in the development of organic chemistry, as a separate and distinct discipline, is the reproducibility of chemical transformations independent of supernatural influences. Such observations resulted in the abandonment of the Vital Force Theory. The Vital Force Theory’s demise began in 1828 when Frederick Wöhler stumbled upon the formation of urea in his attempts to prepare ammonium cyanate from various inorganic compounds. Frederick Wöhler was a Heidelberg physician and a member of the Royal Swedish Academy of Sciences turned chemist. He taught chemistry at three colleges (Polytechnic School in Berlin, the Higher Polytechnic School in Kassel, and the University of Göttingen). Wöhler wanted to prepare ammonium cyanate using several different starting materials. He attempted to prepare ammonium cyanate from cyanic acid and ammonia (equation 1); however, ammonium cyanate is an unstable intermediate, and it decomposed to urea, an organic compound. He attempted to prepare ammonium cyanate from silver cyanate and ammonium chloride (equation 2), but the end product was urea, an organic compound. He discovered that heating lead (II) cyanate and ammonia resulted in the formation of urea (equation 3), and that heating mercury (II) cyanate and ammonia resulted in the formation of urea (equation 4). Wöhler’s multiple attempts to prepare ammonium cyanate resulted in the formation of urea, because the desired ammonium cyanate was too unstable to be isolated. This amazed Wöhler, because the thinking in his day was that organic compounds must come from

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living substances and not from an inorganic source. Wöhler’s observations appeared to be “divine intervention,” because his experimental results brought serious doubts to the contemporary scientific community concerning the invariability of the Vital Force Theory. (1)

(2)

5

(3)

(4)

Wöhler wrote to his teacher and mentor, "I cannot, so to say, hold my chemical water and must tell you that I can make urea without

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thereby needing to have kidneys, or anyhow, an animal, be it human or dog". Wöhler’s assertion that he was able to prepare urea from a purely inorganic source was not fully accepted by some of his contemporaries; therefore, it took several years before the eminent death of the Vital Force Theory. The synthesis of urea from inorganic substances made Frederick Wöhler the father of organic chemistry. By 1850, the “Vital Force Theory” was dead. This was primarily due to the work of Adolph Wilhelm Hermann Kolbe (1818-1884). Kolbe synthesized acetic acid, an organic compound, starting with the inorganic compounds carbon disulfide (CS2) and chlorine gas (Cl2). Kolbe’s approach to the preparation of acetic acid from carbon disulfide and chlorine gas made it difficult to question the viability of starting exclusively with inorganic precursors to synthesize organic compounds. There were four steps in Kolbe’s synthesis that resulted in irrevocably rejecting the Vital Force Theory. The synthesis of carbon tetrachloride from carbon disulfide and chlorine gas; the synthesis of tetrachlorethylene from carbon tetrachloride, the synthesis of trichloroacetic acid from tetrachloroethylene; and the synthesis of acetic acid from trichloroacetic acid.

(1) The initial step in the synthesis was the conversion of carbon disulfide to carbon tetrachloride. The reaction scheme involved reacting carbon disulfide with chlorine to produce carbon tetrachloride and disulfur dichloride.

(2) The second step was the pyrolysis of carbon tetrachloride to

tetrachloroethylene.

tetrachloroethylene

(3) The third step was the aqueous chlorination of tetrachloroethylene to produce trichloracetic acid.

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(4) The final step of the synthesis was the electrolysis of

trichloroacetic acid in the presence of Zn to form acetic acid, an organic compound.

Kolbe’s synthesis of acetic acid from carbon disulfide and chlorine was the breakthrough needed for the demise of the Vital Force Theory. It ushered in the birth of organic chemistry. By the middle of the nineteenth century, organic chemistry emerged as a viable discipline. Understanding mechanisms of organic reactions may dispel perceived mysteries associated with the reactions of organic molecules. Also, an understanding of the mechanisms of organic chemistry reactions gives insight into the stoichiometry of reactions, and it rationalizes the formation of the products. After obtaining a clearer understanding of the mechanisms of organic reactions, students may want to return to Kolbe’s synthesis to obtain further insight into the mechanisms of these reactions. The second period in the development of organic chemistry came in 1859 with the birth of structural theory. Berzelius (Wöhler’s mentor) and Wöhler were curious, from a structural perspective, about the details associated with the conversion of ammonium cyanate into urea. An understanding of molecular interactions of organic as well

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as inorganic molecules will rationalize the processes by which ammonium cyanate can easily be converted to urea. Modern theories in chemistry, including Valence Shell Electron Repulsion (VSEPR) Theory and Molecular Orbital Theory (MOT), provide insight into an understanding of Wöhler’s experiment. Structural theory was independently proposed by August Kekulé, from Germany; Archibald Couper, from Scotland; and Alexander Butlerov, from Russia. The results of their work, the nature of the covalent bond, will be discussed later in this Chapter. The changes that occur with the chemical bonds in Wöhler’s experiment are indeed dramatic.

The symbol Δ represents heat. When ammonium cyanate is transformed to urea, the chemical bonds in the ammonium cyanate change. Changes in the orbital transformations of ammonium cyanate can be illustrated by the following diagrams.

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After reading the chapter, return to the above illustrations, and see if you can understand the details of the bond changes that occur when ammonium cyanate is transformed into urea. The information presented in this chapter as well as subsequence chapters explains the nature of the hybridization around the carbon atom of the cyanate anion and the carbon atom of the carbonyl group of urea. At the molecular level, a number of changes occur when ammonium cyanate transforms into urea including hybridization about the atoms, and changes in the bond angles. The hybridization about the cyanate carbon atom changes from a 2sp hybridized atomic orbital to a 2sp2 hybridized atomic orbital in urea. Also, the atomic orbital of the nitrogen atom in the cyanate changes from 2p to 2sp3 in urea. The hybridization of the nitrogen of the ammonium ion is 2sp3 , and in urea both nitrogen atoms are 2sp3 hybridized. The bond angle in cyanate changes from 180o to 120o in urea. The bond angle in the ammonium ion changes from 109.5o to 120o in urea. Finally, an ionic compound, ammonium cyanate, transforms to a polar covalent compound, urea. The third period of development of organic chemistry involves the use of modern instrumentation for separating, analyzing, and identifying organic compounds. These modern methods of analyses, e.g., mass spectrometry, infrared spectrophotometry, nuclear magnetic

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resonance spectrometry, and UV-visible spectrophotometry will be discussed in a future paper. Most organic compounds are synthesized from a variety of inorganic substances, e.g., carbonates or cyanides and primarily from other organic compounds. There are two large reservoirs for obtaining organic compounds-petroleum and coal. There are more than a million organic compounds known today compared to just thousands of inorganic compounds. Organic compounds outnumber inorganic compounds, because of the expanded octet of carbon atoms, i.e., the ability of carbon atoms to bond with other carbon atoms. Structural Theory takes into consideration the number of atoms comprising the molecule, the order of attachment of the atoms, the electrons holding the atoms together, and the shapes and sizes of the molecules. Structural Theory follows the Valence Shell Electron Repulsion Theory (VSEPR) proposed by Ronald James Gillespie and Sir Ronald Sydney Nyholm. Before 1926, the important aspects of the nature of the chemical bond included the ionic bond proposed by Walter Kossel in 1916 and the covalent bond described by G. N. Lewis. Kossel and Lewis based their bonding theories on the belief:

(1) That the positive charged nucleus is surrounded by electrons (2) That electrons exist in shells with a designated maxima (3) That stability is reached when the outer shell is full (4) That ionic and covalent bonds are consequences of the ability

of atoms to obtain the stable electron configuration of noble gases

Ionic bonds involve electron transfers. For example, lithium fluoride, LiF, forms an ionic bond where an electron in the outer shell of Li is transferred to the outer shell of fluorine. The driving force for this transfer results from the lithium ion exhibiting an electron configuration that is isoelectronic (same electron configuration) with the electron configuration of the noble gas helium and the fluoride ion exhibiting an electron configuration that is isoelectronic with the electron configuration of the noble gas neon. Figure 1.1 illustrates the loss of an electron from the outer energy level of lithium to make a lithium ion, a cation isoelectronic with He, and the gain of an

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electron to the outer energy level of fluorine to make a fluoride ion, an anion that is isoelectronic with Ne.

Li → Li+ + e-

F + e- → F- Figure 1.1 the formation of the ionic bond The electrostatic interaction of lithium fluoride can be represented as Li+ F-. In reality, lithium fluoride exists as a body-centered cubic lattice crystalline structure. There are twelve lithium ions on edge and one lithium ion in the body of the cubic lattice structure. Each cation on edge is divided by 4; therefore, there are three lithium ions represented on the edges of the cubic lattice structure, and the one lithium cation in the center making a total of four lithium ions in the unit cell of lithium fluoride. There are eight fluoride ions on the corners of the unit cell, and six fluoride ions in the face of the unit cell. The fluoride ions in the face are divided by 2, and the fluoride ions on the corners are divided by 8; therefore, there are three fluoride ions in the face, and one fluoride ion on the corners of the unit cell. That would make a total of 4 fluoride ions in a unit cell of lithium fluoride. The ratio of lithium ions and fluoride ions in a unit cell of lithium fluoride is 4 to 4 or 1:1; therefore, the formula of lithium fluoride would

.

.. e

--

Li3 Li3

.

.

+

.

. . e- .

.F F9 9..

.

..

.....

..

.._

+

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be LiF. Consequently, lithium fluoride has a mass based on formula units, where the unit cell contains four lithium cation and four fluoride anion, but 4:4 is reduced to 1:1; therefore, the mass of the formula unit would be based on a formula of LiF for lithium fluoride. Li+ has a stable electronic configuration that is isoelectronic with He, 1s2. F- has a stable electronic configuration that is isoelectronic with Ne, 1s2 2s2 2p6. LiF is formed by the transfer of an electron from Li to F to form LiF, an ionic compound. Covalent bonds are formed from the sharing of electrons. For example, the hydrogen molecule, H2, is formed from the sharing of an electron of one hydrogen atom with an electron from another hydrogen atom. When this sharing process occurs, both hydrogen atoms can complete their energy shells; therefore, gaining the stable configuration of He, 1s2. The same consideration could be given to the formation of F2. In the formation of F2, the outer shell electrons would reach the stable outer shell configuration of Ne, 1s2 2s2 2p6, eight (8) electrons, generally referred to as a stable octet. Figure 1.2 represents covalent bonding in H2 and F2

Figure 1.2 covalent bonding in hydrogen and fluorine Figure 1.3 represents the covalent bonding in CF4 where the fluorine and carbon atoms share electrons to satisfy the octet rule resulting in a stable octet around the carbon atom and a stable octet around the fluorine atom. An important consideration in this illustration is that the electron pairs of the four covalent bonds around carbon tetrachloride are not equally shared with the carbon atom and the four fluorine atoms. The bonding electrons are closer to the fluorine atoms than the carbon atom, because fluorine is more electronegative than carbon. Each fluorine atom would exhibit a partial negative charge, and the carbon atom would exhibit a partial positive four charge.

H. + Hx H .x H

F FF:..... + x

x

xxx

xx

F....... x xxx x

x

x

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Figure 1.3 represents the covalent bonding in carbon tetrafluoride In 1926, Erwin Schrödinger used quantum mechanics to develop models that describe the motion of electrons and the energies associated with electrons in energy levels about the nucleus. These models suggest that electrons behave as waves; therefore, they are referred to as wave equations. These equations cannot be solved in an exact manner, but give approximate solutions called wave functions and eigenvalues (energy values). The lower the energy value, the more nearly correct the wave function. The mathematical equation leading to the wave functions in its most simplistic form is Hψ = Eψ where H is the Hamiltonian operator, ψ is the wave function or eigenfunction, and E is the eigenvalue or energy value. The Hamiltonian is a complex mathematical expression that operates on a wave function that describes the behavior of the electron. The Hamiltonian operates on the wave function in a manner that regenerates the wave function and generates energy values called eigenvalues. When the Hamiltonian operates on the wave function to regenerate the wave function and produces an energy value, the mathematics is referred to as quantum mechanics and involves differential equations. A simple example of such behavior can be illustrated by considering an electron traveling a distance x’ along the x-axis only. If the electron behaves like a wave, its wave function could be expressed as

where ψ represents the wave function and λ represents the wavelength of the electron. Hψ = Eψ

'2 x = sin

πλ

Ψ

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For simplicity of mathematical calculations, let’s assume that the Hamiltonian operator in this one-electron scenario is

Therefore,

This mathematical operation is familiar to anyone who has taken a calculus course. The operation is not too complicated and can be easily understood, and the solution can be expressed by equation 1.1 Equation 1.1

where the wave function is,

and the eigenvalue or energy value is

or

2

2'x∂∂

2 '

2

2 x sin

'xπλ

∂∂

2 '2 2 x- sin

π πλ λ

⎛ ⎞⎜ ⎟⎝ ⎠

'2 xsin

πλ

22- πλ

⎛ ⎞⎜ ⎟⎝ ⎠

2

2

4πλ

−

16

The Hamiltonian operator is

the eigenvalue or wave function is

and the eigenvalue or energy value is

The mathematics may not be familiar to anyone who has not had a course in differential calculus or differential equations. If so, simply understand that the equations that describe electron behavior are referred to as Quantum Mechanics. Quantum Mechanics suggest that electrons exhibit the properties of waves, and the wave behavior of electrons as they move around atomic nuclei. The solutions to these wave equations are described in the section on “Atomic Orbitals. Atomic Orbitals Solutions to the Schrödinger equation, Hψ = Eψ, (using spherical coordinates instead of Cartesian coordinates) for an electron in a quantized atomic orbital leads to three quantum numbers. The three quantum numbers are the principle quantum number, n, the energy level in which the electron resides; the azimuthal or secondary quantum number, l, the shape that describes the probability of finding the electron in a volume element; and the magnetic quantum number,

∂2

∂x2 sin 2π xλ

⎡⎣⎢

⎤⎦⎥

⎛⎝⎜

⎞⎠⎟

= - 4π 2

λ sin 2π x

λ⎡⎣⎢

⎤⎦⎥

2

2'x∂∂

'2 xsin

πλ

2

2

4πλ

−

17

ml, the orientation of the volume element describing the probability of finding the electron in three dimensions. These solutions lead to the wave behavior of electrons in three dimensions. The wave equations cannot simultaneously give the exact position of the electron at any particular moment and how fast the electron is moving, an application of the Heisenberg Uncertainty Principle. Also, no specific pathway about the nucleus can be plotted by the wave equation. Atomic or molecular orbitals map the region of space that defines the probability of finding the electron within a volume element. Consequently, atomic or molecular orbitals may be defined as probability densities distributions for finding the electron or electrons within some volume element. As indicated earlier, the shape of the orbital depends upon the energy of the electron. The cloud that describes the probability of finding the electron may be visualized as a blurred photograph of a rapidly moving electron. The electron cloud density is not uniform, but densest in the region where the probability of finding the electron is highest, i.e., where the average negative charge is the greatest. Figure 1.4 is the result of solutions of the Schrödinger equation for the lowest eigenvalue, the 1s atomic orbital. Figure 1.5 is the result of solutions of the Schrödinger equation for the next higher energy atomic orbital, the 2s atomic orbital. Higher potential energy than the 2s atomic orbital is the three degenerate (equal in energy) 2p atomic orbitals. Figure 1.6 represents solutions of the Schrödinger equation for the 2p atomic orbitals. The three degenerate 2p atomic orbitals have three orientations along the Cartesian coordinate axes. These degenerate energy atomic orbitals are the 2px, 2py and 2pz atomic orbitals. They are given the magnetic quantum numbers of -1, 0, and +1.

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Figure 1.4, the 1s atomic orbital

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Figure 1.5, the 2s atomic orbital with one node

-1 0 +1

x

y

z

x

y

zz

x

y

2 px2 p

2 py

z

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Figure 1.6,the 2 p atomic orbitals Each 2p atomic orbital is dumbbell shaped. *The chemistry of carbon compounds is referred to as organic chemistry. Electron Configuration An atomic orbital can have a maximum of two electrons, but the two electrons must have opposite spins (Pauli Exclusion Principle). The spin quantum number is not a solution to the Schrödinger equation, but an outcome of the Stern-Gerlach Experiment. The Stern-Gerlach Experiment demonstrated that electrons have two spin orientations (Figures 1.7 and 1.8). In the Stern-Gerlach Experiment, a beam of hydrogen atoms is passed through a magnetic field. The beam of hydrogen atoms is split into two beans suggesting that the electrons in the atoms have magnetic properties that are associated with opposite spins.

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Figure 1.7 The Stern-Gerlach Experiment Exhibiting the two spins for electrons Figure 1.8 shows that that one beam consist of hydrogen electrons with a spin of +½ and the other with a spin of -½

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Figure 1.8 Two Spins for the Hydrogen Electrons The hydrogen electrons in each beam act as magnets because of their spins. The orientation-of-spins are different, i.e., one hydrogen electron spins clockwise and the other hydrogen electron spins counterclockwise. The spinning electrons generate small magnetic fields. The electron spin is susceptible to quantum restrictions, i.e., ms = +½ or ms = -½ Two electrons can occupy a single orbital, but the electrons must have opposite spins. http://www.upscale.utoronto.ca/GeneralInterest/Harrison/SternGerlach/SternGerlach.html gives a summary of the Stern-Gerlach experiment on spin. There are four quantum numbers, including the spin quantum number. No two electrons can have the same set of four quantum numbers in a single orbital. This is a statement of the Pauli Exclusion

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Principle. If two electrons reside in the same orbital, then one electron would have a spin of +½ and the other would have a spin of -½ since the principle quantum number “n,” the azimuthal or secondary quantum number “l,” and the magnetic quantum number “ml” would be the same. Also, Hund’s Rule states that one electron must be in each atomic orbital of a set of degenerate atomic orbitals before two electrons can occupy any of the degenerate atomic orbitals. Electron Configuration and Orbital Diagrams As indicated previously, solutions to the Schrödinger equation give the probability of finding the electron in a volume of space. These volumes are the s, p, d and f atomic orbitals. There are three degenerate “p” atomic orbitals; five degenerate “d” atomic orbitals; and seven degenerate “f” atomic orbitals. “d” Orbitals can split in the presence of specific ligands (Crystal Feld Theory). “f” Atomic orbitals do not exhibit splitting in the presence of specific ligands. Each atomic orbital can hold two electrons, and the two electrons must have opposite spins. The total number of electrons in three degenerate “p” atomic orbitals is six electrons. The d atomic orbitals can hold a maximum of ten electrons, and the f atomic orbitals can hold a maximum of fourteen electrons. Most organic compounds use 2s and 2p atomic orbitals to form molecular bonds; however, higher atomic orbitals are used when Cl, Br, I, S, P, and transition metals are included in molecular formulas. Subshells consist of the different atomic orbitals within a specified volume of space. For example, when n = 2, the total number of electrons that are possible in energy level 2 would be 2n2 where n equals the energy level in question. Eight electrons are possible for n = 2, i.e., 2(2)2 = 8. When n = 2, two electrons can reside in the 2s atomic orbital and six electrons can reside in three degenerate 2p atomic orbitals. Table 1.1 indicates the maximum number of electrons that can exist in each subshell.

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Subshell l Number of Orbitals*

Maximum Number of Electrons

s 0 1 2 P 1 3 6 d 2 5 10 f 3 7 14 Table 1.1 the maximum number of electrons available for each subshell. Subshells were described in the section entitled “pictorial solutions to the Schrödinger equation. They are probability densities of finding electrons about the nucleus of an atom. Each atomic orbital can hold a maximum of two electrons, but the two electrons must have opposite spins. This constraint can be applied to writing the electron configuration of Li. Lithium has three electrons in atomic orbitals surrounding the nucleus. The first energy level, n=1, can only hold two electrons [2(1)2 = 2]; therefore, there is only one subshell, the 1s atomic orbital. The second energy level has two subshells, the 2s and 2p subshells. The third electron of lithium must reside in the second energy level in the 2s atomic orbital (lower in potential energy than the 2p atomic orbital). The electron configuration for Li would be 1s22s1. Figure 1.9 is an orbital diagram that represents the potential energy relationships of the atomic orbital electrons in the lithium atom.

Figure 1.9 the electron configuration of Li

Li 1s22s13

1s

2s

potential energy

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The following arrangement of electrons in an atomic orbital would be forbidden (a violation of the Pauli Exclusion Principle).

Another important rule is Hund’s rule that was proposed by Friedrich Hermann Hund (1896-1997). As indicated earlier, Hund’s rule states that for a set of degenerate atomic orbitals, i.e., orbitals with the same energy, each degenerate atomic orbital must have one electron in it before a second electron can be placed in the atomic orbital. When a second electron is placed in the atomic orbital, it must have an opposite spin. For example, there are three 2p atomic orbitals with the same energy in the 2p subshell. If three electrons are in the 2p subshell, than there must be one electron in the 2px atomic orbital, one electron in the 2py atomic orbital, and one electron in the 2pz atomic orbital before a second electron of opposite spin can be added to the set of degenerate atomic orbitals. Nitrogen has seven electrons in atomic orbitals surrounding the nucleus. The first energy level, n=1, can only hold two electrons [2(1)2 = 2]; therefore, there is only one subshell consisting of the 1s atomic orbital. The second energy level, n=2, can hold eight electrons [[2(2)2 = 8]. Since two of the seven nitrogen electrons reside in the 1s atomic orbital, the remaining five electrons must reside at energy level 2. Two of these electrons would be in the 2s atomic orbital, and the remaining three would be in the 2p atomic orbitals. As indicated earlier, the Pauli Exclusion Principle, proposed by physicist Wolfgang Pauli, states that no two electrons can have the same set of four quantum numbers. Since two electrons within the same orbital can have the same values for n, l, and ml ,the two electrons in the orbital must have opposite spins, i.e., ms = +½ and ms = -½. The electron configuration for nitrogen is 1s22s22px

1 2py1 2pz

1.

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Using Hund’s rule and the Pauli Exclusion, Figure 1.10 is an orbital diagram that represents the potential energy relationships of the orbital electrons in the nitrogen atom.

Figure 1.10 electron configuration of nitrogen obeying Hund’s rule Table 1.2 represents the configuration for hydrogen using four quantum numbers (three from the Schrödinger equation and one from the experiments of Stern and Gerlach). electron N l ml ms 1 1 0 0 -½ Table 1.2 the four quantum numbers for hydrogen Table 1.3 represents the electron configuration for helium (two electrons about the nucleus). Each electron is described using four quantum numbers (three from the Schrödinger equation and one from the experiments of Stern and Gerlach). Helium is a noble gas and is inert, i.e., it does not combine with other elements to form compounds.

1s2 2s2 2px1 2py

12pz1

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electron N l ml ms 1 1 0 0 -½ 2 1 0 0 +½ Table 1.3: the two electrons have identical numbers for n, l, and ml, but different spin quantum numbers. Table 1.4 represents the configuration for lithium (three electrons about the nucleus). Each electron is described using four quantum numbers (three from the Schrödinger equation and one from the experiments of Stern and Gerlach). electron N l ml ms 1 1 0 0 -½ 2 1 0 0 +½ 3 2 0 0 -½ Table 1.4 the quantum numbers for the three lithium electrons Table 1.5 represents the configuration for beryllium (four electrons about the nucleus). Each electron is described using four quantum numbers (three from the Schrödinger equation and one from the experiments of Stern and Gerlach). electron N l ml ms 1 1 0 0 -½ 2 1 0 0 +½ 3 2 0 0 -½ 4 2 0 0 +½ Table 1.5 the quantum numbers for the four electrons Table 1.6 represents the configuration for boron (five electrons about the nucleus). Each electron is described using four quantum numbers (three from the Schrödinger equation and one from the experiments of Stern and Gerlach).

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electron N l ml ms 1 1 0 0 -½ 2 1 0 0 +½ 3 2 0 0 -½ 4 2 0 0 +½ 5 2 1 -1 -½ Table 1.6: the quantum numbers for the five B electrons and the electron configuration for B is

Table 1.7 represents the configuration for carbon (six electrons about the nucleus). Each electron is described using four quantum numbers (three from the Schrödinger equation and one from the experiments of Stern and Gerlach). electron N l ml ms 1 1 0 0 -½ 2 1 0 0 +½ 3 2 0 0 -½ 4 2 0 0 +½ 5 2 1 -1 -½ 6 2 1 0 -½ Table 1.7: the quantum numbers for the six carbon electrons and the electron configuration for C is

Table 1.8 represents the configuration for nitrogen (seven electrons about the nucleus). Each electron is described using four quantum numbers (three from the Schrödinger equation and one from the experiments of Stern and Gerlach).

2 2 1 0 01 2 2 2 2x y zs s p p p

2 2 1 1 01 2 2 2 2x y zs s p p p

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electron N l ml ms 1 1 0 0 -½ 2 1 0 0 +½ 3 2 0 0 -½ 4 2 0 0 +½ 5 2 1 -1 -½ 6 2 1 0 -½ 7 2 1 1 -½ Table 1.8: The quantum numbers for the seven electrons for N and the electron configuration for N is

Table 1.9 represents the configuration for oxygen (eight electrons about the nucleus). Each electron is described using four quantum numbers (three from the Schrödinger equation and one from the experiments of Stern and Gerlach). electron N l ml ms 1 1 0 0 -½ 2 1 0 0 +½ 3 2 0 0 -½ 4 2 0 0 +½ 5 2 1 -1 -½ 6 2 1 0 -½ 7 2 1 1 -½ 8 2 1 -1 +½ Table 1.9: The electron configuration for O is

Table 1.10 represents the configuration for fluorine (nine electrons about the nucleus). Each electron is described using four quantum numbers (three from the Schrödinger equation and one from the experiments of Stern and Gerlach).

2 2 1 1 11 2 2 2 2x y zs s p p p

1s2 2s2 2px2 2py

12pz1

30

electron N l ml ms 1 1 0 0 -½ 2 1 0 0 +½ 3 2 0 0 -½ 4 2 0 0 +½ 5 2 1 -1 -½ 6 2 1 0 -½ 7 2 1 1 -½ 8 2 1 -1 +½ 9 2 1 0 +½ Table 1.10: The quantum numbers for the nine electrons of F and the electron configuration for F is

Table 1.11 represents the configuration for neon (ten electrons about the nucleus). Each electron is described using four quantum numbers (three from the Schrödinger equation and one from the experiments of Stern and Gerlach). Neon is a noble gas and will not react with other elements to form compounds. electron N l ml ms 1 1 0 0 -½ 2 1 0 0 +½ 3 2 0 0 -½ 4 2 0 0 +½ 5 2 1 -1 -½ 6 2 1 0 -½ 7 2 1 1 -½ 8 2 1 -1 +½ 9 2 1 0 +½ 10 2 1 1 +½ Table 1.11: The quantum numbers for the ten electrons for Ne and the electron configuration for Ne is

2 2 2 2 11 2 2 2 2x y zs s p p p

2 2 2 2 21 2 2 2 2x y zs s p p p

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The above discussion represents the Aufbau Principle, i.e., the building of the electron configuration of the elements as they are listed in the periodic chart. Molecular Orbitals Molecular orbitals result from the linear combination of atomic orbitals. This occurs when the electrons of two atoms form a chemical bond. The electron pair or pairs in the molecular orbital are located in proximity to the two nuclei forming the chemical bond. The shape of the molecular orbital is related to the shapes of the atomic orbitals in the component atoms. Following are descriptions of molecular bonds in methane, ammonia, water, carbon-carbon double bonds, and carbon-carbon triple bonds. Methane, CH4 To understand the bonding that occurs in the covalent bonds between C and H, the electron configurations of both carbon and hydrogen should be revisited. The electron configurations of carbon and hydrogen are: C 1s2 2s2 2p2 H 1s1 In order to understand the nature of the chemical bonds in methane, an electron in the 2s atomic orbital of carbon must be promoted to a 2p atomic orbital of carbon. The promotion of an electron from the 2s atomic orbital of C to a 2p atomic orbital of carbon is an energy consuming process, and can be illustrated by Diagram 1.1.

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Diagram 1.1 The 2p atomic orbital that contains the promoted electron linearly combines with the remaining two 2p atomic orbitals of carbon (each containing an electron) and the 2s atomic orbital that contains one electron to form four degenerate (equal in energy) 2sp3 hybridized atomic orbitals. Hybridization consumes energy. The resulting hybridized atomic orbitals are lower in energy than the 2p atomic orbitals from which they were formed and higher in energy than the 2s atomic orbital from which they were formed. Each 2sp3 atomic orbital contains one electron (Diagram 1.2).

Diagram 1.2

1s

2s

2px 2pz2py 2py 2pz2px

2s

1s

1s

2sp3 2sp3 2sp3 2sp3

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Each C-H covalent bond in methane is formed from the linear combination of a1s atomic orbital of H with a 2sp3 hybridized atomic orbital of carbon to form a sigma bonding molecular orbital and a sigma anti-bonding molecular orbital. Two electrons, with opposite spins, reside in the carbon-hydrogen bonding molecular orbital. Diagram 1.3 describes the bonding molecular orbital and the anti-bonding molecular orbital in each C-H sigma bond of methane. Bond formation releases energy. The release of energy in the formation of bonds more than compensates for the energy required to promote an electron from the “s” atomic orbital to the “p” atomic orbital and the energy required for hybridization. When two atomic orbitals linearly combine to form two molecular orbitals, the molecular orbital with the lower potential energy is the bonding molecular orbital. The molecular orbital with the higher potential energy orbital is the anti-bonding molecular orbital. The two electrons reside in the bonding molecular orbital, and they have opposite spins.

Diagram 1.3 Each bonding molecular orbital between carbon and hydrogen is a single covalent bond (referred to as a sigma, σ bond). The σ bond is formed from the linear combination of a 2sp3 hybridized atomic orbital

2sp3

1s

σ(2sp 3 + 1s )

σ(2sp 3 - 1s )*

2sp3

1s

σ(2sp 3 + 1s )

σ(2sp 3 - 1s )*

34

of carbon with a 1s atomic orbital of hydrogen. The resulting bonding molecular orbital is designated as σ(2sp3 +1s). The four σ bonds of methane are illustrated in Diagram 1.4, and the energy required to break a C-H bond is 427 kJ/mol.

Diagram 1.4, a tetrahedral molecular with a H-C-H bond angle equal to 109.5o Ammonia, NH3 To understand the bonding that occurs between N and H, the electron configurations of both nitrogen and hydrogen should be revisited. The electron configurations of nitrogen and hydrogen are: N 1s2 2s2 2p3 H 1s1 In the case of nitrogen, there is no need to promote an electron in the 2s orbital to a 2p orbital. The only energy requirement would be the need to hybridize the three 2p atomic orbitals each containing one electron and the lower energy 2s orbital containing two electrons to form four degenerate 2sp3 hybridized orbitals as illustrated in Diagram 1.5.

C

H H

H

H

σ(2sp3

+ 1s)

+ 1s)

σ(2sp3+ 1s)

σ(2sp3+ 1s)σ(2sp3

35

Diagram 1.5 The four 2sp3 hybridized atomic orbitals have lower potential energy than the 2p atomic orbitals from which they were formed, and higher potential energy than the 2s atomic orbital from which they were formed. The three hybridized 2sp3 atomic orbitals have one electron each and one 2sp3 hybridized atomic orbital has two electrons. The three N-H covalent bonds are formed from the linear combination of a 1s atomic orbital of H with a 2sp3 hybridized atomic orbital of nitrogen to form a bonding molecular orbital and an anti-bonding molecular orbital. Diagram 1.6 illustrates the bonding molecular orbital and the anti-bonding molecular orbital.

Diagram 1.6

2py 2pz2px

2s

1s

2sp32sp32sp32sp3

1s

2sp3

1s

σ(2sp 3 + 1s )

σ(2sp 3 - 1s )*

2sp3

1s

σ(2sp 3 + 1s )

σ(2sp 3 - 1s )*

36

Each bond between nitrogen and hydrogen is a single covalent bond (referred to as a sigma, σ bond). The σ bond is formed from the linear combination of a 2sp3 hybridized atomic orbital of nitrogen with a 1s atomic orbital from hydrogen. The resulting bonding molecular orbital is designated as σ(2sp3 +1s). The N-H bond length is 1.01 Å, and the energy required to break the N-H bond is 431kJ/mol. The lone pair of electrons (the two electrons that occupied the 2sp3 hybridized atomic orbital) reside in a nonbonding 2sp3 molecular orbital. Diagram 1.7 illustrates the three σ bonds and the lone pair of electrons in the nonbonding 2sp3 molecular orbitals of ammonia.

Diagram 1.7 NH3 is has a trigonal pyramidal structure, a special case of a tetrahedral structure, with a H-N-H bond angle that is approximately 107o Water, H2O To understand the bonding that occurs in the covalent bonds between O and H, let’s revisit the electron configurations of both oxygen and hydrogen. The electron configuration of O and H are: O 1s2 2s2 2p4 H 1s1

N

H HH

..

non-bonding 2sp 3 molecular orbital

σ(2sp3 + 1s)

+ 1s)σ(2sp3

σ(2sp3 + 1s)

37

In the case of oxygen, there is no need to promote an electron in the 2s atomic orbital to a 2p atomic orbital. The only energy requirement would be the hybridization of the three 2p atomic orbitals. Two “2p” atomic orbitals contain one electron each and one “2p” atomic orbital contains two electrons. The lower energy “2s” orbital contains two electrons. These four atomic orbitals hybridize to form four degenerate 2sp3 atomic orbitals as illustrated in Diagram 1.8.

Diagram 1.8 The four 2sp3 hybridized atomic orbitals have lower potential energy than the three “2p” atomic orbitals from which they were formed, and higher potential energy than the 2s atomic orbital from which they were formed. Two of the “2sp3“ hybridized atomic orbitals have one electron each and each of the two “2sp3” hybridized atomic orbitals have two electrons (with opposite spins). Each of the two O-H covalent bonds are formed from the linear combination of a 1s atomic orbital of H with a 2sp3 hybridized atomic orbital of oxygen to form a bonding and an anti-bonding molecular orbital. The O-H bonds in water are described by Diagram 1.9.

1s

2sp3 2sp3 2sp3 2sp3

1s

2s

2px 2pz2py

38

Diagram 1.9 Each bond between oxygen and hydrogen is a single covalent bond (referred to as a sigma, σ bond). The σ bond is formed from the linear combination of a 2sp3 hybridized atomic orbital of oxygen and a 1s atomic orbital of hydrogen. The resulting bonding molecular orbital is designated as σ(2sp3 +1s). The O-H bond length is 0.96 Å, and the energy required to break the O-H bond is 494kJ/mol. Four electrons would reside in the two nonbonding 2sp3 molecular orbitals (each 2sp3 molecular orbital of oxygen would contain two electrons with opposite spins). The result is described in Diagram 1.10.

Diagram 1.10, a bent molecule with an H-O-H bond angle equal to 105o

2sp3

1s

σ(2sp 3 + 1s )

σ(2sp 3 - 1s )*

2sp3

1s

σ(2sp 3 + 1s )

σ(2sp 3 - 1s )*

O

HH

..

non-bonding 2sp 3 molecular orbital

+ 1s)σ(2sp3

σ(2sp3 + 1s)..

39

The C-C Double Bond The simplest molecule with a carbon-carbon double bond is ethylene (ethylene), C2H4. The structural formula for C2H4 is (following the octet rule)

A sigma, σ, and a pi, π, bond comprise the double bond. In order to understand the nature of the double bond, let’s revisit the electron configuration of carbon. This configuration is 1s2 2s2 2p2. An electron in the 2s atomic orbital of carbon can be promoted to the 2p atomic orbital. Diagram 1.1 (revisited) initiates this process, i.e., an illustration of the promotion process.

Diagram 1.1 The 2p atomic orbital that contains the promoted electron linearly combines with the 2s atomic orbital that contains one electron to form three degenerate 2sp2 hybridized atomic orbitals. Diagram 1.11 is an illustration of this process.

C C

H

H

H

H

1s

2s

2px 2pz2py 2py 2pz2px

2s

1s

40

Diagram 1.11 A 2sp2 hybridized atomic orbital of carbon linearly combines with a 1s atomic orbital of hydrogen to form a C-H sigma bond. Four of these C-H bonds are formed in the ethylene molecule. Diagram 1.12 describes each C-H single bond.

Diagram 1.12 The third 2sp2 hybridized atomic orbital linearly combines with the 2sp2 hybridized atomic orbital of the second carbon to form a (2sp2 + 2sp2) σ bond. The remaining electron in the 2p orbital of carbon linearly combines with the electron in the 2p orbital of the second carbon to form a π bond. The (2sp2 + 2sp2) σ and the (2p+2p) π molecular orbitals comprise the C-C double bond. Diagram 1.13

1s

2sp2 2sp2 2sp2

2p

1s

2sp2

σ

σ

*

(2sp2 + 1s)

(2sp2 - 1s)

41

illustrates these linear combinations of atomic orbitals that form the molecular orbitals of the C-C double bond.

Diagram 1.13 The overall picture of the ethylene molecule is illustrated in Diagram 1.14

42

Diagram 1.14 The pi bond can be described as the vertical overlap of two 2p atomic orbitals, and the sigma bond can be described as the horizontal overlap of two 2sp2 hybridized atomic orbitals. Together, the pi and sigma bonds form the C-C double bond. Each C-H single bond can be described as the horizontal overlap of a 1s atomic orbital of hydrogen with a 2sp2 atomic orbital of carbon. The H-C-H bond angle is approximately120o, and the H-C-C bond angle is approximately120o. The Triple Bond The simplest molecule with a triple bond is acetylene (ethyne), C2H2. Following the octet rule, the structure for C2H2 would be:

A sigma, σ, and two pi, π, bonds comprise the triple bond. To understand the nature of the triple bond, let’s revisit the electron configuration of each of the two carbons that comprise the acetylene molecule. The electron configuration of carbon is 1s2 2s2 2p2. In forming the triple bond, an electron in the 2s atomic orbital of carbon is promoted to a 2p atomic orbital of carbon. Diagram 1.1 (revisited) initiates this process, i.e., an illustration of the promotion process.

C CH

H H

H

σ(2sp2 + 2sp2)

π(2p +2p)

C CH H

43

Diagram 1.1 A 2p atomic orbital of carbon containing one electron linearly combines with the 2s atomic orbital to form two degenerate 2sp hybridized atomic orbitals. Diagram 1.15 illustrates this process.

Diagram 1.15 A 2sp hybridized atomic orbital of carbon linearly combines with a 1s atomic orbital of hydrogen to form a C-H sigma bond. Two of these C-H bonds are formed in the acetylene molecule. Diagram 1.16 describes each C-H single bond.

1s

2s

2px 2pz2py 2py 2pz2px

2s

1s

2p

2sp2sp

1s

2p

44

Diagram 1.16 The second 2sp hybridized atomic orbital linearly combines with the 2sp hybridized atomic orbital of the second carbon to form a (2sp + 2sp) σ bond. The remaining two “2p” atomic orbitals of carbon linearly combine with the remaining two “2p” atomic orbitals of the second carbon to form two π bonds. Diagram 1.17 is an illustration of this process.

1s

2sp

σ

σ

*

(2sp + 1s)

(2sp - 1s)

45

Diagram 1.17 The overall representation of the acetylene molecule is illustrated in Diagram 1.18.

46

Diagram 1.18 Two pi bonds are formed from the vertical overlap of four p orbitals, and the sigma bond is the horizontal overlap of two 2sp hybridized atomic orbitals of one carbon atom with another carbon atom. Together, the two pi bonds and the sigma bond comprise the C-C triple bond. Each C-H single bond can be described as the horizontal overlap of a 1s atomic orbital of hydrogen with a 2sp atomic orbital of carbon. The H-C-C bond angle is 180o. Physical Parameters of Molecules Intermolecular and Intramolecular forces Intermolecular forces are the forces of attraction between molecules. Intramolecular forces are the forces of attraction within a molecule. The two types of intramolecular forces encountered in organic systems are repulsive forces and attractive forces. Repulsive forces exist when electrons remain as far apart as possible because they have the same charge or the same spin. Attractive forces exist when the atomic nuclei attract electrons due to their opposite charges. Intermolecular forces include molecular association, London or Van der Waal force, dipole-dipole force, induce dipole force, and ion-dipole interactions. Molecular association or hydrogen bonding results in molecules possessing hydrogen atoms that are covalently bonded to highly

C

C

H

H

47

electronegative atoms such as oxygen, nitrogen, and fluorine. Hydrogen bonding/molecular association is a strong electrostatic interaction. In fact, hydrogen bonding is stronger than Van der Waal forces or dipole-dipole interaction. London or Van der Waal force results from the induced instantaneous polarization caused by electron movement within molecules. The force occurs when electrons in adjacent molecules begin to disperse from one another when they approach each other. The electron dispersions within proximity molecules produce instantaneous dipoles that result in molecular attractions. Dipole–dipole interactions are forces of attraction between polar molecules resulting from their charge separations. The positive regions of molecules are attracted toward the negative regions of adjacent molecules, and the negative regions of molecules are attracted toward the positive regions of adjacent molecules. Induced dipole forces result from the polarizable attractive interaction between the permanent dipole of a molecule with an induced dipole of another molecule. Ion-dipole forces are between ions and molecules. They are analogous to dipole-dipole and induced-dipole interactions. The ion-dipole force involves the attractive interaction between an ion and a polar molecule. For example, a positively charged ion can interact with the negative region of a polar molecule. Bond Polarity Bond polarity is a property of the covalent bond where the two nuclei don’t share the covalently bonded electrons equally. The electron cloud is denser about one atom more than the other; therefore, one end of the bond is relatively negative and the other is positive. The following examples illustrate bond polarity and the symbols that represent bond polarity:

48

Properties Associated with Bond Polarity Bond polarity occurs when there is a difference in electronegativity between atoms. The greater the difference in electronegativity, the more polar the bond. Fluorine is the most electronegative element. Electronegativity is a measure of the ability of an atom in a molecule to draw bonding electrons to itself. Linus Pauling (1901-1994, awarded two Nobel prizes- one in chemistry in 1954 and the other in peace in 1963) suggested electronegativity values from bond energies. However, to make this work, he arbitrarily assigned an electronegativity value of 4.0 to the most electronegative element, fluorine (F). Therefore, the electronegativity values of the other elements were calculated using 4.0 as the highest possible value for an atom. The symbol for electronegativity is χ, and χ can be calculated using equation 1.2. Equation 1.2

Where χA is the more electronegative element; therefore, in Equation 1.2, χA must have a greater value than χB, i.e., the more electronegative atom has the value of χA and the less electronegative atom has the value of χB . Examples 1.1 and 1.2 are calculations of the electronegativity values for hydrogen and bromine using Pauling’s equation

H Fδ δ+ - O

H H

..

..

δ δ

δ

+ +

-2

NH

HH

..

δδ

δ

δ -3

++

+

2 2

12

A B A B12

mol - = 0.102 - ( )( )

kJχ χ ΑΒΔΗ ΔΗ ΔΗ

49

Example 1.1 Calculate the electronegativity of hydrogen using the bond energies from Table 1.12 . The electronegativity of F, the most electronegative atom, is arbitrarily selected to be 4.0.

Example 1.2 Calculate the electronegativity of bromine using the value for the electronegativity of hydrogen calculated in the previous problem and the bond energies from Table 1.12. Using equation 1.2, the electronegativity of bromine can be calculated in the following manner:

Compare this value to the electronegativity value for Br given in Chart 1.1. As indicated previously, electronegativity values are used to determine the extent of polarity in a bond. The polarity between two elements increases with increase in the electronegativity difference between the elements. For example, the electronegativity of hydrogen is 2.2 (Chart 1.1), and the

2 2

12

F H HF F H12

mol - = 0.102 - ( )( )kJ

χ χ ΔΗ ΔΗ ΔΗ

12

F H 12

mol - = 0.102 565 kJ/mol - (155 kJ/mol)(432 kJ/mol)kJ

χ χ

H = 1.78 - 4.0χ−H = 1.78 - 4.0 = -2.2χ−

H = 2.2χ

12

Br 12

mol - 2.2 = 0.102 362 kJ/mol - (432 kJ/mol)(190 kJ/mol)

kJχ

Br - 2.2 = 0.89χ

Br = 0.89 + 2.2 = 3.1χ

50

electronegativity of chlorine is 3.2; therefore, for HCl, the electronegativity difference would be 3.2-2.2 = 1.0. The electronegativity of fluorine is 4.0; therefore, for HF, the electronegativity difference would be 4.0-2.2 = 1.8. Consequently, HF would be a more polar molecule than HCl. H C N O S F Cl Br I H 432 C 411 346 N 386 305 167 O 459 358 201 142 S 363 272 226 F 565 485 283 190 284 155 Cl 428 327 313 218 255 249 240 Br 362 285 201 217 249 216 190 I 295 213 201 278 208 175 149 Table 1.12 single bond energies in kilojoules per mole

51

H

2.2

He

Li

1.0

Be

1.6

B

2.0

C

2.6

N

3.0

O

3.4

F

4.0

Ne

Na

0.9

Mg

1.3

Al

1.6

Si

1.9

P

2.2

S

2.6

Cl

3.2

Ar

K

0.8

Ca

1.0

Sc

1.4

Ti

1.5

V

1.6

Cr

1.7

Mn

1.6

Fe

1.8

Co

1.9

Ni

1.9

Cu

1.9

Zn

1.7

Gd

1.8

Ge

2.0

As

2.2

Se

2.6

Br

3.0

Kr

Rb

0.8

Sr

1.0

Y

1.2

Zr

1.3

Nb

1.6

Mo

2.2

Tc

2.1

Ru

2.2

Rh

2.3

Pd

2.2

Ag

1.9

Cd

1.7

In

1.9

Sn

2.0

Sb

2.1

Te

2.1

I

2.7

Xe

2.6

Cs

0.8

Ba

0.9

La

1.1

Hf

1.3

Ta

1.5

W

1.7

Re

1.9

Os

2.2

Ir

2.2

Pt

2.2

Au

2.4

Hg

1.9

Tl

1.6

Pb

1.8

Bi

1.9

Po

2.0

At

2.2

Rn

Fr

0.7

Ra

0.9

Ac

1.1

Chart 1.1: Pauling Electronegativity values of elements Electronegativity values of the periodic elements increase from left to right within a period, and the electronegative values of elements decrease from top to bottom within a family. The electronegativity of Br is 3.0, and the electronegativity difference between H and Br is 0.9. The electronegativity of I is 2.7, and the electronegativity difference between H and I is 0.5. It is apparent that these data indicate the HF is more polar than HCl; HCl is more polar than HBr; and HBr is more polar than HI. Polarity of Molecules A molecule is polar when its center of negative charge does not coincide with its center of positive charge. Polar bonds in the

52

molecule exhibit dipoles when two equal and opposite charges are separated in space. Under these circumstances the molecule exhibits a dipole moment, and the dipole is represented as

The arrow points from the positive end of the molecule to the negative end of the molecule. The dipole moment of the molecule is represented by µ, where µ, can be defined by equation 1.3. Equation 1.3 µ (in Debye units) = e (esu) x d (in angstroms, Å) where “e” is the charge in electrostatic units (esu) and “d” is the internuclear distance in angstroms 1 debye = 10-10 esu-Å H2 , O2, N2 , and Cl2 have zero dipole moments. These molecules are nonpolar because the two atoms have the same electronegativity; therefore, e = 0 and µ = 0. There is no simple calculation for the dipole moment; however, if the partial charge, e, is given and the distance in Å between the nuclei, then the calculations for µ becomes quite simple. For example, the partial charge on the fluoride ion in HF is 1.84 x 10-10 esu. Therefore the dipole moment for HF would be µ (in Debye units) = e (esu) x d (in angstroms, Å) µ = 1.84 x 10-10 esu x 0.92 Å = 1.7 x 10-10 esu- Å = 1.7 D The dipole moment depends on the polarity of its individual bonds and the direction of the bonds; however, calculating the dipole moment becomes more complicated when more than two atoms are involved. Carbon tetrachloride (III) is an example of a molecule that exhibits a zero dipole moment, and chloroform (IV) is a molecule that has a dipole moment. Remember that the dipole moment of a molecule is the vector sum of the bonds dipole moments (this is not easily calculated).

53

As illustrated in the diagram, the vector sum in carbon tetrachloride is zero; therefore the net dipole moment is zero.

The vector sum is not zero; therefore, the molecule has a net dipole moment. The net dipole moment of ammonia (V) is 1.46 D

C

Cl

ClCl

Cl

µ = 0

III

C

Cl

µ

HH

H

IV

= 1.86 D

54

NF3 (VI) has a smaller dipole moment than NH3 because the differences in the directions of the bond dipoles.

Useful physical parameters of organic molecules include the dipole moment, index of refraction, the melting point, the boiling point, the

µ

HH

H

= 1.46 D

V

N..

55

solubility, and the density. These physical parameters provide information about the compound that may be helpful in identifying its structure. Physical parameters can be used to determine methods for separating organic compounds from mixtures. Sometimes the isolation and purification methods are more complicated than the method of synthesis. Resonance Resonance is the phenomenon in which the actual electron structure is a hybrid between two or more electron structures. The ozone molecule exhibits resonance. Using VSEPR, ozone would exhibit the following two equivalent structures.

VII

VIII The actual structure is not VII or VIII, but a hybrid between VII and VIII similar to IX

IX

O

O

O

..

: .. ...... -

+

- ........:

..

OO

O+

+

.....

..:

..

OO

O.. . ...

δδ --

56

The following can represent resonance in ozone, where the double- headed arrow represents resonance.

The Lewis structure for benzene is:

or

An abbreviated representation of these two structures is

And resonance in benzene is

O

O

O:..

..

:: .. -

+ +

-..: :

..

.. :O

O

O

C C

C

CC

C

H H

H

HH

H

C C

C

CC

C

H H

H

HH

H

57

Acid and Bases An Arrhenius acid is a substance that ionizes in aqueous solution to produce hydronium ions, H3O+. An Arrhenius base is a substance that ionizes in aqueous solution to produce hydroxide ions. A Brønsted-Lowry acid is a substance that is a proton donor, and a Brønsted-Lowry base is a proton acceptor. For example, sulfuric acid, H2SO4, ionizes in H2O to produce hydronium ions, H3O+.

Also, sulfuric acid, H2SO4, is the Brønsted-Lowry acid, and water is the Brønsted-Lowry base. The greater the concentration of hydronium ions, [H3O+], in aqueous solution, the more acidic the solution. Hydrochloric acid, HCl, reacts with ammonia, NH3 , to form the ammonium ion and the chloride ion.

Hydrochloric acid, HCl, is the Brønsted-Lowry acid, and ammonium, NH3, is the Brønsted-Lowry base. The strength of the acid depends upon its tendency to give up protons and provide hydronium ions in solution. A strong base dissociates 100% in aqueous solution to form hydroxide ions, and a strong acid dissociates 100% in aqueous solution to produce hydronium ions.

-2 4 (aq) 2 (l) 3 (aq) 4 (aq)H SO + H O H O + HSO+→

-(aq) 3 (aq) 4 (aq) (aq)HCl + NH NH + Cl+→

58

Sulfuric acid, a strong acid, dissociates in aqueous solution to produce hydronium ions, and sodium hydroxide, a strong base dissociates in aqueous solution to produce hydroxide ions. A weak cationic acid like the ammonium ion in aqueous ammonium chloride, NH4Cl (aq), would react with a strong base like sodium hydroxide, NaOH, to form water and a weak base.

The weak cationic acid, the ammonium ion, functions as a relatively strong acid since its conjugate base, ammonia, is weak. NH4

+ + H2O ⥄ H3O+ + NH3 Ka = 1.8 x 10-5 A Lewis acid is an electron pair acceptor, and a Lewis base is an electron pair donor. An example of a Lewis acid-base reaction is

Boron trifluoride is the Lewis acid, and ammonia is the Lewis base. Ammonia has a lone pair of electrons that it can donate to the boron trifluoride. The strength of acids and bases can be determined by examining their Ka values, and the strength of bases can be ascertained by examining their Kb values. For example, if a general organic acid, indicated below, dissociates in aqueous (water) to form hydronium ions, H3O+, then the Ka, the equilibrium constant, may be used to determine the strength of the acid.

+ -4 2 3NH + OH H O + NH

stronger stronger weaker weakeracid base acid base

⎯⎯→←⎯

59

Equation 1.4 can be used to calculate the concentration of the hydronium ion. Equation 1.4

The larger the Ka values, the stronger the acid. Ka is related to the pKa by equation 1.5; therefore, the larger the Ka value, the smaller the value of the pKa, and the stronger the acid. It follows, therefore, that the larger the value of pKa, the weaker the acid. Equation 1.5

pH is defined as the negative logarithm to the base 10 (common logarithms) of the hydronium ion concentration. Equation 1.6 gives the relationship between pH and the hydronium ion concentration. The numerical value for pH decreases with increase acid strength. Equation 1.6 can be used to calculate pH of acids. Equation 1.6

The same analogy can be used to determine the base strength of a solution. For example, a general amine (weak bases) with the formula RR’R”N: where the R groups represent small molecular mass alkyl groups or hydrogen atoms, can react with water to form hydroxide ions in aqueous solution.

C

OH

O

+ H2O

O

C

O- H3O+ +R R

- +3

a2

[RCOO ] [H O ]K = [RCO H]

a apK = -log K

3pH = -log [H O ]+

60

Equation 1.7 can be used to determine the equilibrium constant for the above Arrhenius base reaction in aqueous solution. Equation 1.7

The negative common logarithm of Kb would be

and the relationship between the pOH and pH is given by equation 1.8. Equation 1.8

N :

R'

R"

R + H2O R

R"

R'

:N H+

+ -OH

+' '' -

b[RR R N H][ OH]K =

[RR'R''N:]

b bpK = -log K

pH + pOH= 14

61

Problems Basic Principles for Introduction to Organic Chemistry

1. After reading this paper, write a comprehensive definition of

organic chemistry?

2. Write one or two paragraphs giving a brief historical summary of organic chemistry as a comprehensive science.

3. What is the vital force theory, and how did it have an impact on the field of organic chemistry as a viable scientific discipline?

4. What impact, if any, does the following equation play in the development of organic chemistry?

5. Describe Adolph Kolbe’s contribution to organic chemistry.

6. What are the three periods of development for organic chemistry?

7. Use VSEPR and MOT to show a more detail analyses for the conversion of ethene to ethane.

8. Describe the nature of the chemical bond in CaCl2.

9. Describe the nature of the C-I chemical bond in CI4.

10. Describe the nature of the C-I chemical bond from a quantum mechanical perspective.

C C

H H

H H

+

HH

HH

CCH2 H H

62

11. Given two electrons within a 2s atomic orbital, which of the following figures best represent these two electrons. Give a rationale for your answer. What experimental evidence would give rise to the correct answer?

(a)

(b)

(c)

ee

e e

e e

63

12. Which of the following is in agreement with both the Pauli Exclusion Principle and Hund’s rule of Maximum Multiplicity? Give a rationale for your choice.

(a)

(b)

(c)

(c)

(d)

13. Which of the diagrams in problem 12 is the best representation of Hund’s rule?

14. Use VSEPR theory to describe the bonding in CH4. Also, include bond angels in your explanation.

2p2p 2pzyx

x y z2p 2p 2p

2p 2p 2pzyx

2p 2p 2pzyx

2p 2p 2pzyx

64

15. Use VSEPR theory to describe the bonding in ammonia. Also, include bond angels in your explanation.

16. Use VSEPR theory to describe the bonding in water. Also, include bond angels in your explanation.

17. Use VSEPR theory to describe the bonding in propene. Also, include bond angels in your explanation.

18. Use VSEPR theory to describe the bonding in the following alkyne. Also, include bond angels in your explanation.

19. Which of the following molecules would have the greatest bond polarity?

H

H

HC

H

NH

H

H

..

H H

..

..O

H

H

C

H

C

C

H

H

H

H

H

C

H

CC H

65

(a) HF (b) HCl (c) HBr (d) HI

20. Which of the following molecules would be the strongest acid?

(a)

pKa = 25 (b)

pKa = 36 (c)

pKa = 42

21. What is the hybridization about the carbon atom that contains the acidic hydrogen atom in 20 (a)?

22. What is the hybridization about the carbon atom that contains the acidic hydrogen atom in 20 (b)?

23. What is the hybridization about the carbon atom that contains the acidic hydrogen atom in 20 (c)?

24. Can a relationship be made between acidity and percent s-

character of the carbon atom containing the acidic hydrogen in hydrocarbons?

25. Is there a relationship between electronegativity and bond dissociation energy? If so, explain such a relation.

26. What is meant by the following statement: “This molecule exhibits a dipole moment?”

CH3C C H

CH3C C

H

H

H

CH3CH2CH2CH3

66

27. Which of the following molecules would have a higher dipole moment?

28. Calculate the electronegativity of Br if the electronegativity of hydrogen is 2.2.

29. Calculate the dipole moment of H-Br. (Research the necessary mathematical parameters you would need to make that calculation.)

30. Does trichloroamine, NCl3, have a dipole moment? If so, how would its value compare to the value of the dipole moment for trifluoroamine, NF3?

31.Write resonance structures for the sulfate anion.

32. Can resonance exist in sodium propanoate (sodium propionate)? if so, write resonance structures for the propanoate anion.

33. The following table lists the Ka of four acids.

C

BrBr

Br

Br

BrBr

Br

C

H

..

_

_

:

::....

....:

O

OOS

CH3CH2C

O

O

:

::

..

.. - Na+

67

Acid Name Ka [H3O+] pH CH3COOH acetic acid 1.77x10-5 ClCH2COOH chloroacetic acid 1.36x10-3 Cl2CHCOOH dichloroacetic

acid 5.50x10-2

Cl3CCOOH trichloroacetic acid

0.23

(a) Calculate the hydronium ion concentration, [H3O+], and the pH of 0.100 M solution of each acid.

(a) On the basis of your calculations, list acetic acid, chloroacetic acid, dichloroacetic acid, and trichloracetic acid in order of their increasing acidities.

34. Is aluminum chloride an acid or a base? (a) What definition would define it as an acid or a base. (b) Write an acid-base equation between aluminum chloride and

chlorine gas. (c) Identify the conjugate base. (d) Identify the conjugate acid.

35. In the following reaction:

an incipient positive charge resides on the carbon atom that has a double bond connected to the oxygen atom.

(a) Why is there an incipient positive charge on that particular carbon atom and not on the others?

(b) In this reaction, what chemical entity is the base? Give a rationale for your answer.

(c) What chemical entity is the acid? Give a rationale for your answer.

(d) What species is the conjugate base? (e) What species is the conjugate acid?

CH3CH2C

O :..

Cl

+ CH3OH....

..:O

CH3CH2C

OCH3....

+ HCl

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36. Write the equilibrium constant expression for reactions (a) and (b)

(a)

(b)

37. Which acid in problem 36 is a stronger acid? Give a rationale for your answer.

38. n-Propylamine is soluble in water and can function as a base in water.

(a) Write an equilibrium expression showing the basicity of n-

propylamine. (b) Calculate the pH of a 0.100 M solution of n-propylamine if

the pKb of the base is 3.3.

39. Write an equation to show how diethyl malonate can react with sodium ethoxide, Na+ -OCH2CH3, in an acid base reaction.

C

OH

O

+ H2O

O

C

O- H3O+ +

pKa = 4.19

C

OH

O

+ H2O

O

C

O- H3O+ +

pKa = 3.48NO2 NO2

CH3CH2CH2NH2 + H2O(l) CH3CH2CH2NH3(aq)+ + -OH(aq)

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diethyl malonate

C

C

C

HH

O

O

OCH2CH3

CH3CH2O