Oxidation-reduction in organic chemistry

Energy diagram comparing heats of combustion of isomeric CgHig alkanes. 

The small differences in stability between branched and unbranched alkanes result from an interplay between attractive and repulsive forces within a molecule (intramolecular forces). These forces are nucleus-nucleus repulsions, electron-electron repulsions, and nucleus-electron attractions, the same set of fundamental forces we met when talking about chemical bonding (Section 2.2) and van der Waals forces between molecules (Section 2.17). When the energy associated with these interactions is calculated for all of the nuclei and electrons within a molecule, it is found that the attractive forces increase more than the repulsive forces as the structure becomes more compact. Sometimes, though, two atoms in a molecule are held too closely together. WeTl explore the consequences of that in Chapter 3. 

Without consulting Table 2.3, arrange the following compounds in order of decreasing heat of combustion pentane, 2-methylbutane, 2,2-dimethylpropane, hexane. 

As we have just seen, the reaction of alkanes with oxygen to give carbon dioxide and water is called combustion, A more fundamental classification of reaction types places it in the oxidation-reduction category. To understand why, let s review some principles of oxidation-reduction, beginning with the oxidation number (also known as oxidation state). 

There are a variety of methods for calculating oxidation numbers. In compounds that contain a single carbon, such as methane (CH4) and carbon dioxide (CO2), the oxidation number of carbon can be calculated from the molecular formula. For neutral molecules the algebraic sum of all the oxidation numbers must equal zero. Assuming, as is customary, that the oxidation state of hydrogen is +1, the oxidation state of carbon in CH4 then is calculated to be —4. Similarly, assuming an oxidation state of 

The carbon in methane has the lowest oxidation number ( 4) of any of the compounds in Table 2.6. Methane contains carbon in its most reduced form. Carbon dioxide and carbonic acid have the highest oxidation numbers (+4) for carbon, corresponding to its most oxidized state. When methane or any alkane undergoes combustion to form carbon dioxide, carbon is oxidized and oxygen is reduced. 

Oxidation of carbon corresponds to an increase in the number of bonds between carbon and oxygen or to a decrease in the number of carbon-hydrogen bonds. Conversely, reduction corresponds to an increase in the number of carbon-hydrogen bonds or to a decrease in the number of carbon-oxygen bonds. From Table 2.6 it can be seen that each successive increase in oxidation state increases the number of bonds between carbon and oxygen and decreases the number of carbon-hydrogen bonds. Methane has four C—H bonds and no C—O bonds carbon dioxide has four C—O bonds and no C—H bonds. 

Among the various classes of hydrocarbons, alkanes contain carbon in its most reduced state, and alkynes contain carbon in its most oxidized state. 

Increasing oxidation state of carbon (decreasing hydrogen content) 

TABLE 2.6 Oxidation Number of Carbon in One-Carbon Compounds  

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Chemical Properties Combustion of Alkanes 80 Thermochemistry 83 Oxidation-Reduction in Organic Chemistry 83 Summary 86 Problems 90... 

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