Organic Chemistry Text Book (CHEM 3401 and 3402)

Structural Formulas
It is necessary to draw structural formulas for organic compounds because in most cases a molecular formula does not uniquely represent a single compound. Different compounds having the same molecular formula are called isomers, and the prevalence of organic isomers reflects the extraordinary versatility of carbon in forming strong bonds to itself and to other elements.
When the group of atoms that make up the molecules of different isomers are bonded together in fundamentally different ways, we refer to such compounds as constitutional isomers. There are seven constitutional isomers of C₄H₁₀O, and structural formulas for these are drawn in the following table. These formulas represent all known and possible C₄H₁₀O compounds, and display a common structural feature. There are no double or triple bonds and no rings in any of these structures.. Note that each of the carbon atoms is bonded to four other atoms, and is saturated with bonding partners.

Structural Formulas for C₄H₁₀O Isomers

Kekulé Formula	Condensed Formula	Shorthand Formula

Simplification of structural formulas may be achieved without any loss of the information they convey. In condensed structural formulas the bonds to each carbon are omitted, but each distinct structural unit (group) is written with subscript numbers designating multiple substituents, including the hydrogens. Shorthand (line) formulas omit the symbols for carbon and hydrogen entirely. Each straight line segment represents a bond, the ends and intersections of the lines are carbon atoms, and the correct number of hydrogens is calculated from the tetravalency of carbon. Non-bonding valence shell electrons are omitted in these formulas.
Developing the ability to visualize a three-dimensional structure from two-dimensional formulas requires practice, and in most cases the aid of molecular models. As noted earlier, many kinds of model kits are available to students and professional chemists, and the beginning student is encouraged to obtain one. 

Distinguishing Carbon Atoms
When discussing structural formulas, it is often useful to distinguish different groups of carbon atoms by their structural characteristics. A primary carbon (1º) is one that is bonded to no more than one other carbon atom. A secondary carbon (2º) is bonded to two other carbon atoms, and tertiary (3º) and quaternary (4º) carbon atoms are bonded respectively to three and four other carbons. The three C₅H₁₂ isomers shown below illustrate these terms.

Structural differences may occur within these four groups, depending on the molecular constitution. In the formula on the right all four 1º-carbons are structurally equivalent (remember the tetrahedral configuration of tetravalent carbon); however the central formula has two equivalent 1º-carbons (bonded to the 3º carbon on the left end) and a single, structurally different 1º-carbon (bonded to the 2º-carbon) at the right end. Similarly, the left-most formula has two structurally equivalent 2º-carbons (next to the ends of the chain), and a structurally different 2º-carbon in the middle of the chain. A consideration of molecular symmetry helps to distinguish structurally equivalent from nonequivalent atoms and groups. The ability to distinguish structural differences of this kind is an essential part of mastering organic chemistry. It will come with practice and experience.

Although structural formulas are essential to the unique description of organic compounds, it is interesting and instructive to evaluate the information that may be obtained from a molecular formula alone. Three useful rules may be listed:

1. The number of hydrogen atoms that can be bonded to a given number of carbon atoms is limited by the valence of carbon. For compounds of carbon and hydrogen (hydrocarbons) the maximum number of hydrogen atoms that can be bonded to n carbons is 2n + 2 (n is an integer). In the case of methane, CH₄, n=1 & 2n + 2 = 4. The origin of this formula is evident by considering a hydrocarbon made up of a chain of carbon atoms. Here the middle carbons will each have two hydrogens and the two end carbons have three hydrogens each. Thus, a six-carbon chain (n = 6) may be written H-(CH₂)₆-H, and the total hydrogen count is (2 x 6) + 2 = 14. The presence of oxygen (valence = 2) does not change this relationship, so the previously described C₄H₁₀O isomers follow the rule, n=4 & 2n + 2 = 10. Halogen atoms (valence = 1) should be counted equivalent to hydrogen, as illustrated by C₃H₅Cl₃, n = 3 & 2n + 2 = 8 = (5 + 3). If nitrogen is present, each nitrogen atom (valence = 3) increases the maximum number of hydrogens by one. 

   Some Plausible Molecular Formulas	C7H16O3, C9H18, C15H28O3, C6H16N2
   Some Impossible Molecular Formulas	C8H20O6, C23H50, C5H10Cl4, C4H12NO

2. For stable organic compounds the total number of odd-valenced atoms is even. Thus, when even-valenced atoms such as carbon and oxygen are bonded together in any number and in any manner, the number of remaining unoccupied bonding sites must be even. If these sites are occupied by univalent atoms such as H, F, Cl, etc. their total number will necessarily be even. Nitrogen is also an odd-valenced atom (3), and if it occupies a bonding site on carbon it adds two additional bonding sites, thus maintaining the even/odd parity.

   Some Plausible Molecular Formulas	C4H4Cl2, C5H9OBr, C5H11NO2, C12H18N2FCl
   Some Impossible Molecular Formulas	C5H9O2, C4H5ClBr, C6H11N2O, C10H18NCl2

3. The number of hydrogen atoms in stable compounds of carbon, hydrogen & oxygen reflects the number of double bonds and rings in their structural formulas. Consider a hydrocarbon with a molecular structure consisting of a simple chain of four carbon atoms, CH₃CH₂CH₂CH₃. The molecular formula is C₄H₁₀ (the maximum number of bonded hydrogens by the 2n + 2 rule). If the four carbon atoms form a ring, two hydrogens must be lost. Similarly, the introduction of a double bond entails the loss of two hydrogens, and a triple bond the loss of four hydrogens. 

    

    

    

   From the above discussion and examples it should be clear that the molecular formula of a hydrocarbon (C_nH_m) provides information about the number of rings and/or double bonds that must be present in its structural formula. By rule #2 m must be an even number, so if m < (2n + 2) the difference is also an even number that reflects any rings and double bonds. A triple bond is counted as two double bonds. 

   The presence of one or more nitrogen atoms or halogen substituents requires a modified analysis. The above formula may be extended to such compounds by a few simple principles:

4. The presence of oxygen does not alter the relationship.
5. All halogens present in the molecular formula must be replaced by hydrogen.
6. Each nitrogen in the formula must be replaced by a CH moiety.