Degree of Unsaturation

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Degree of Unsaturation How to determine the number of rings and multiple bonds in a compound from its molecular formula No hydrocarbon can contain a greater number of hydrogens than fits the formula CnH2n+2 The number of hydrogens is always even. All such compounds are acyclic CH3CH2CH2CH3 They may be straight (normal) chains as in n-butane (C4H10) or CH3CH(CH3)CH3 they may be branched as in isobutane (C4H10) If two hydrogen atoms on non-adjacent carbons of n-butane (C4H10) are removed then a ring, cyclobutane (C4H8), is formed with the loss of H2 This process is not necessarily a chemical reaction but rather a conceptual device. If two hydrogen atoms on adjacent carbons of n-butane (C4H10) are removed then a double bond is formed with the loss of H2. In this case, the alkene, 1-butene (C4H8), is formed. A compound with the molecular formula C4H8 is either an acyclic alkene (olefin) or cycloalkane. How to determine the Degree of Unsaturation (DU) of this compound? C4H10 - C4H8 = H2 divided by 2 = 1 DU most saturated C4 compound Try the following formulas: DU Example C6H6 4 benzene C7H10 3 norbornene C10H8 7 naphthalene How is the Degree of Unsaturation of a hydrocarbon containing halogen, or other monovalent atom, determined? Every halogen in a hydrocarbon replaces a hydrogen. Replace each halogen with hydrogen and then compare this hydrocarbon with the most saturated hydrocarbon. The alkyl halide C8H5BrCl2 becomes C8H8 C8H18 - C8H8 = H10 /2 = 5 DU One example of an alkyl halide C8H5BrCl2 Br Cl 1 DU - C6 H H 3 DU 1 DU - C8 H 5 DU H Cl C8 H5BrCl2 H How is the Degree of Unsaturation of a hydrocarbon containing oxygen, or other divalent atom, determined? Ignore divalent atoms! C4H8O C4H8 1 DU 1 DU 2,4-Dichlorophenoxyacetic acid carbonyl - 1 DU benzene ring - 4 DU 2,4-Dichlorophenoxyacetic acid C8H6Cl2O3 C8H6Cl2 C8H8 (drop O3) (- Cl2 + H2) C8H18 - C8H8 = H10 / 2 = 5 DU How is the Degree of Unsaturation of a hydrocarbon containing nitrogen, or other trivalent atom, determined? Substitute CH for every N. (- N + CH) pyridine benzene C5H5N C6H6 C6H14 - C6H6 = H8 / 2 = 4 DU Cocaine - C17H21NO4 C17H21NO4 C17H21N C18H22 C18H38 - C18H22 = H16 / 2 = 8 DU 1 DU 1 DU (5-membered ring) 4 DU 1 DU 1 DU (6-membered ring) Degree of Unsaturation The End F. E. Ziegler 2003.

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