Calculating the Degrees of Unsaturation From a Compound s Molecular Formula

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1 Calculating the Degrees of Unsaturation From a Compound s Molecular Formula Alkanes have the molecular formula C n. Alkanes are saturated hydrocarbons because each member of the family has the maximum number of hydrogen atoms per carbon atom in its molecular formula. Alkenes and cycloalkanes both have the molecular formula C n H 2. These are examples of unsaturated hydrocarbons. The presence of a double bond or a ring in a compound's structure causes there to be two fewer hydrogen atoms in the molecule than there are in an alkane with the same number of carbon atoms. C n H 2 = C n H 2 alkane cycloalkane or formula alkene formula The total number of π-bonds and rings in the structure of a compound is called the degree of unsaturation of the compound (sometimes called the index of hydrogen deficiency). Every π-bond and every ring in a compound s structure results in one degree of unsaturation. degree of unsaturation = Σ (π-bonds + rings) There are three steps in calculating the degree of unsaturation of a compound from its molecular formula: Step 1: Step 2: Use the formula C n to determine the number of hydrogen atoms that a saturated hydrocarbon with the same number of carbon atoms as the compound in question would have. Determine the number of hydrogen atom equivalents * in the molecular formula of the compound in question. * Not all organic compounds are hydrocarbons. The presence of atoms other than carbon and hydrogen in a compound s molecular formula modifies the determination of the number of hydrogen atoms in this step. The rules for handling a molecular formula that contains certain heteroatoms are given later in this handout. Step 3: Subtract the number of hydrogen atom equivalents in the compound s molecular formula (Step 2) from the number of hydrogens atoms in the corresponding saturated hydrocarbon (Step 1) and then divide this difference by two. This gives the degree of unsaturation of the compound. degree of unsaturation = [(# of hydrogens from step 1) (# of hydrogens from step 2)] / 2

2 The calculated degree of unsaturation tells you the total number of π-bonds and rings in a compound s structure. It does not tell you the actual number of rings or the actual number of π-bonds present in the compound s structure. For example, a compound with the molecular formula C 5 has two degrees of unsaturation. You cannot tell from this information whether these two degrees of unsaturation arise from the presence of two π-bonds, or from one π-bond and one ring, or from two rings in the compound s structure. Each of the following C 5 isomers has two degrees of unsaturation. Determining exactly how many of the calculated degrees of unsaturation in a compound s structure are from π-bonds and how many of the degrees of unsaturation are from rings takes additional analysis, usually involving the results of a π-bond addition reaction. For example, a compound with the molecular formula C 8 H 12 has three degrees of unsaturation. If this compound reacts with H 2 /Pt to give a product with the molecular formula C 8 H 16, then you know that there are two π-bonds in the compound s structure (since two equivalents of hydrogen add to the compound in the hydrogenation reaction). The difference between the number of π-bonds in the compound and the calculated degree of unsaturation for the compound tells you the number of rings in the compound s structure. degree of unsaturation # of π-bonds = number of rings Since, in this example, there are three degrees of unsaturation from the molecular formula, and two of these degrees of unsaturation are from π-bonds, the remaining degree of unsaturation must be from the presence of a ring in the compound s structure. Handling Heteroatoms in a Molecular Formula The degree of unsaturation can also be calculated for compounds that contain atoms other than just carbons and hydrogens. We will look at how to deal with three different heteroatoms found in many organic compounds: halogens, oxygen, and nitrogen. Halogens Halogen atoms, like hydrogen atoms, are monovalent. A halogen atom can be thought of as replacing a hydrogen atom in a compound s. The structures of ethene, C 2, and bromoethene, C 2 H 3 Br, are shown below. Each compound has one degree of unsaturation; the halogen atom in the second compound replaces one of the hydrogen atoms in the first structure.

3 To determine the degree of unsaturation of an organic compound with halogens in its molecular formula, add the number of halogen atoms to the number of hydrogen atoms in the compound s molecular formula, and then compare that number of hydrogen atom equivalents to the number of hydrogen atoms in a corresponding saturated hydrocarbon. H 3 Br 3 is treated as BrClF 2 is treated as H 12 in the calculation (1 degree of unsaturation) Oxygen Oxygen atoms are the easiest heteroatom to deal with in the degree of unsaturation calculation. Oxygen is divalent and can be considered tos simply insert into a single bond in the structure. Any number of oxygen atoms can be added to a molecule without changing the number of hydrogen atom equivalents in the molecular formula. Ethane (C 2 ) has zero degrees of unsaturation; the oxygen containing compounds dimethyl ether and ethanol (both C 2 O), and ethylene glycol (C 2 ) also have zero degrees of unsaturation. The additional oxygen atom(s) are inserted into single bonds in the compounds structures. So, whenever a molecular formula contains oxygen atoms, ignore the oxygen atoms in the molecular formula when determining the degree of unsaturation for the compound. is treated as in the calculation (3 degrees of unsaturation) is treated as H 10

4 Nitrogen Nitrogen is trivalent, and its presence in a compound effectively adds a hydrogen to the compound s molecular formula. For example, ethane (C 2 ) has zero degrees of unsaturation; the nitrogen containing compounds ethylamine and dimethylamine (both C 2 H 7 N) have one more hydrogen atom in their molecular formulas than ethane, but they still have zero degrees of unsaturation. When a molecular formula contains nitrogen atoms, subtract the number of nitrogen atoms in the formula from the number of hydrogen atoms in the formula, and then compare that number of hydrogen atom equivalents to the number of hydrogen atoms in the formula of the corresponding saturated hydrocarbon. H 5 N is treated as (3 degrees of unsaturation) C 5 H 11 N 3 is treated as C 5 (2 degrees of unsaturation) By using the rules for these heteroatoms in conjunction with the three steps outlined initially for the hydrocarbons, you can easily determine the degree of unsaturation for many organic compounds directly from the compound's molecular formula. A few examples are: is treated as H 10 ClF 2 N 3 is treated as (3 degrees of unsaturation) C 7 BrN 3 is treated as C 7 (5 degrees of unsaturation) C 29 4 Br 3 Cl 5 N 6 O 4 is treated as C 29 6 (7 degrees of unsaturation) ===========================================================================

5 I prefer to do the degree of unsaturation determination in the stepwise fashion outlined above. There is a more general approach that uses the valencies of the atoms in the structure, as opposed to the identity of the atoms, to determine the degree of unsaturation for a molecule. In this approach, the calculation is: degree of unsaturation = n + 1 [(m t)/2] where n is the number of tetravalent atoms (carbon) in the formula, m is the number of monovalent atoms (hydrogen and halogen) in the formula, and t is the number of trivalent atoms (nitrogen and phosphorus) in the formula. (divalent atoms (oxygen and sulfur) are ignored in this calculation also) Using this equation and an example from above, for. n = 6, t = 0, and m = 10; the degree of unsaturation is [(10-0)/2] = 2 degrees of unsaturation Another example from above, for C 7 BrN 3, n = 7, t = 3, and m = 9; the degree of unsaturation is [(9-3)/2] = 5 degrees of unsaturation It makes no difference which method you chose to use, they both give you the same answer in the end.