Chapter 1

Introduction—Molecular Structure and Reactivity

Abstract

To understand the way atoms bond to one another and then transform into new structures it is important to have a common set of rules and definitions. Herein, we describe key aspects of bonding, equilibrium, and driving forces that influence the structure and allow us to model and predict what are otherwise very complicated events. Many of the concepts presented in this chapter are a review for the student who has already taken the basic Organic series and the General Chemistry sequence, with a few new concepts and further explanations on some specific topics. This chapter helps create a common language and concepts that will be encountered in the book.

Keywords

Acidity; Aromaticity; Basicity; Bond; Charge; Dipole; Electronegativity; Electrophile; Formal charge; Hybridization; Lewis structure; Nucleophile; Octet; Resonance; Tautomers Reaction mechanisms offer us insights into how molecules react, enable us to manipulate the course of known reactions, aid us in predicting the course of known reactions using new substrates, and help us to develop new reactions and reagents. In order to understand and write reaction mechanisms, it is essential to have a detailed knowledge of the structures of the molecules involved and to be able to notate these structures unambiguously. In this chapter, we present a review of the fundamental principles relating to molecular structure and of the ways to convey structural information. A crucial aspect of structure from the mechanistic viewpoint is the distribution of electrons, so this chapter outlines how to analyze and depict electron distributions. Mastering the material in this chapter will provide you with the tools you need to propose reasonable mechanisms and to convey these mechanisms clearly to others.

1. How to Write Lewis Structures and Calculate Formal Charges

The ability to construct Lewis structures is fundamental to writing or understanding organic reaction mechanisms. It is particularly important because lone pairs of electrons frequently are crucial to the mechanism but often are omitted from structures appearing in the chemical literature. There are two methods commonly used to show Lewis structures. One shows all electrons as dots. The other shows all bonds (two shared electrons) as lines and all unshared electrons as dots.

A. Determining the Number of Bonds

HINT 1.1 To facilitate the drawing of Lewis structures, estimate the number of bonds. For a stable structure with an even number of electrons, the number of bonds is given by the equation:

ElectronDemand?ElectronSupply)/2=NumberofBonds

  The electron demand is two for hydrogen and eight for all other atoms usually considered in organic chemistry. (The tendency of most atoms to acquire eight valence electrons is known as the octet rule.) For elements in group IIIA (e.g., B, Al, Ga), the electron demand is six. Other exceptions are noted, as they arise, in examples and problems. For neutral molecules, the contribution of each atom to the electron supply is the number of valence electrons of the neutral atom. (This is the same as the group number of the element when the periodic table is divided into eight groups.) For ions, the electron supply is decreased by one for each positive charge of a cation and is increased by one for each negative charge of an anion. Use the estimated number of bonds to draw the number of two-electron bonds in your structure. This may involve drawing a number of double and triple bonds (see the following section).

B. Determining the Number of Rings and/or ? Bonds (Degree of Unsaturation)

The total number of rings and/or ? bonds can be calculated from the molecular formula, bearing in mind that in an acyclic saturated hydrocarbon the number of hydrogens is 2n + 2, where n is the number of carbon atoms. Each time a ring or ? bond is formed, there will be two fewer hydrogens needed to complete the structure. HINT 1.2 On the basis of the molecular formula, the degree of unsaturation for a hydrocarbon is calculated as (2m + 2 ? n)/2, where m is the number of carbons and n is the number of hydrogens. The number calculated is the number of rings and/or ? bonds. For molecules containing heteroatoms, the degree of unsaturation can be calculated as follows: Nitrogen: For each nitrogen atom, subtract 1 from n. Halogens: For each halogen atom, add 1 to n. Oxygen: Use the formula for hydrocarbons. This method cannot be used for molecules in which there are atoms like sulfur and phosphorus whose valence shell can expand beyond eight. EXAMPLE 1.1 CALCULATE THE NUMBER OF RINGS AND/OR ? BONDS CORRESPONDING TO EACH OF THE FOLLOWING MOLECULAR FORMULAS a. C2H2Cl2Br2 There are a total of four halogen atoms. Using the formula (2m + 2 ? n)/2, we calculate the degree of unsaturation to be (2(2) + 2 ? (2 + 4))/2 = 0. b. C2H3N There is one nitrogen atom, so the degree of unsaturation is (2(2) + 2 ? (3?1)) = 2.

C. Drawing the Lewis Structure

Start by drawing the skeleton of the molecule, using the correct number of rings or ? bonds, and then attach hydrogen atoms to satisfy the remaining valences. For organic molecules, the carbon skeleton frequently is given in an abbreviated form. Once the atoms and bonds have been placed, add lone pairs of electrons to give each atom a total of eight valence electrons. When this process is complete, there should be two electrons for hydrogen; six for B, Al, or Ga; and eight for all other atoms. The total number of valence electrons for each element in the final representation of a molecule is obtained by counting each electron around the element as one electron, even if the electron is shared with another atom. (This should not be confused with counting electrons for charges or formal charges; see Section 1.D.) The number of valence electrons around each atom equals the electron demand. Thus, when the number of valence electrons around each element equals the electron demand, the number of bonds will be as calculated in Hint 1.1. Atoms of higher atomic number can expand the valence shell to more than eight electrons. These atoms include sulfur, phosphorus, and the halogens (except fluorine). HINT 1.3 When drawing Lewis structures, make use of the following common structural features. 1. Hydrogen is always on the periphery because it forms only one covalent bond. 2. Carbon, nitrogen, and oxygen exhibit characteristic bonding patterns. In the examples that follow, the R groups may be hydrogen, alkyl, or aryl groups, or any combination of these. These substituents do not change the bonding pattern depicted. (a) Carbon in neutral molecules usually has four bonds. The four bonds may all be ? bonds, or they may be various combinations of ? and ? bonds (i.e., double and triple bonds).
    There are exceptions to the rule that carbon has four bonds. These include CO, isonitriles (RNC), and carbenes (neutral carbon species with six valence electrons; see Chapter 4). (b) Carbon with a single positive or negative charge has three bonds.
(c) Neutral nitrogen, with the exception of nitrenes (see Chapter 4), has three bonds and a lone pair.
(d) Positively charged nitrogen has four bonds and a positive charge; exceptions are nitrenium ions (see Chapter 4).
(e) Negatively charged nitrogen has two bonds and two lone pairs of electrons.
(f) Neutral oxygen has two bonds and two lone pairs of electrons.
(g) Oxygen–oxygen bonds are uncommon; they are present only in peroxides, hydroperoxides, and diacyl peroxides (see Chapter 5). The formula, RCO2R, implies the following structure:
(h) Positive oxygen usually has three bonds and a lone pair of electrons; exceptions are the very unstable oxenium ions, which contain a single bond to oxygen and two lone pairs of electrons.
3. Sometimes a phosphorus or sulfur atom in a molecule is depicted with 10...