Molecular Formula Determination

Transcription

1 Molecular Formula Determination Classical Approach Qualitative elemental analysis Quantitative elemental analysis Determination of empirical formula Molecular weight determination Molecular formula determination Modern Approach High Resolution Mass Spectroscopy (HRMS) Precise masses for substances of molecular mass 44 amu Compound Exact Mass (amu) CO N 2 O C 2 H 4 O C 3 H

2 Knowledge of Molecular Formula Great deal of information about the structure of an unknown substance can be learned from molecular formula. Alkane C n H 2n +2 Cycloalkane or alkene C n H 2n Alkyne C n H 2n 2 3

3 Molecular formula with non carbon or non hydrogen elements Applicable to an open-chain, saturated hydrocarbon Formula containing Group V elements (N, P, As, Sb, Bi) One additional hydrogen atom must be added to the molecular formula for each Group V element present C 2 H 6 C 2 H 7 N C 2 H 8 N 2 C 2 H 9 N 3 Formula containing Group VI elements (O, S, Se, Te) No change in the number of hydrogens is required C 2 H 6 C 2 H 6 O C 2 H 6 O 2 C 2 H 6 O 3 Formula containing Group VII elements (F, Cl, Br, I) One additional hydrogen atom must be subtracted from the molecular formula for each Group V element present C 2 H 6 C 2 H 5 F C 2 H 4 F 2 C 2 H 3 F 3 4

4 Index of Hydrogen Deficiency (Unsaturation Index) Number of bonds and/or rings a molecule contains The formula of the unknown substance is compared with the formula of the corresponding acyclic, saturated compound. The differences in the numbers of hydrogens between these formulas, when divided by 2, gives the index of hydrogen deficiency. 5

5 Information from hydrogen deficiency index Compound with an Index of one must have one double bond or one ring, but not both. Consult IR spectrum to find the presence of a double bond. If no double bond is present, the compound is cyclic and saturated. Compound with an index of two could have a triple bond, or it could have two double bonds, or two rings, or one of each. Benzene contains one ring and three double bonds and thus an index of four. Any substance with an index of four or more may contain a benzenoid ring: a substance with an index less than four cannot contain such a ring. 6

6 Index of Hydrogen Deficiency - Example 1 Molecular Formula C 7 H 14 O 2 Determine the formula for the saturated, acyclic hydrocarbon containing the same number of carbon atoms as the unknown substance C n H 2n+2, where n = 7. Calculated formula C 7 H 16 Correct this formula for the nonhydrocarbon elements present in the unknown. Correction for oxygens (no change in the number of hydrogens) C 7 H 16 O 2 Compare this formula with the molecular formula of the unknown. Determine the number of hydrogens by which the two formula differ. Compare C 7 H 16 O 2 with C 7 H 14 O 2 7

7 Index of Hydrogen Deficiency - Example 1 The index of hydrogen deficiency equals one. There must be one ring or one double bond in the unknown substance. IR analysis indicates the presence of carbonyl (C=O) group. Further analysis identifies the unknown as isopentyl acetate. O H 3 C O CH 3 CH 3 8

8 The rule of thirteen A method for generating possible molecular formulas for a given molecular mass An example Unknown Substance Molecular Wt. 94 Generate a base formula, which contains only carbon and hydrogen. Divide the molecular wt. By 13. M/13 = n + r/13 The base formula becomes C n H n+r In the example, 94/13 = 7 + 3/13, n = 7 and r = 3 The base formula = C 7 H 10 9

9 The rule of thirteen and index of hydrogen deficiency The index of hydrogen deficiency (U) is determined by applying the relationship U = (n r + 2) / 2 In the example, U = ( ) / 2 = 3 The unknown substance could have three rings or multiple bonds. CH 3 10

10 The rule of thirteen and index of hydrogen deficiency Compound with same molecular mass but with one oxygen atom. Base Formula = C 7 H 10 U = 3 Add: +O Subtract: -CH4 Change in U: U = 1 New Formula = C 6 H 6 O New index of hydrogen deficiency: U = 4 OH 11

11 Additional molecular formulas Molecular mass 94 amu C 5 H 2 O 2 U = 5 C 5 H 2 S U = 5 C 6 H 8 N U = 3 ½ CH 3 Br U = 0 Impossible combinations U = 3 ½ U = < 0 If enough hydrogens are not present, we can subtract 1 carbon and add 12 hydrogens. If the value of U > 7, we can add 1 carbon and subtract 12 hydrogens. 12

12 Nitrogen Rule When the number of nitrogen atoms present in the molecule is odd, the molecular mass will be an odd number When the number of nitrogen atoms present in the molecule is even or zero, the molecular mass will be an even number. For example, ethylamine, C 2 H 5 NH 2 has one nitrogen atom, and its mass is an odd number (45). Ethylenediamine, H 2 N-CH 2 -CH 2 -NH 2, has two nitrogen atoms, and its mass is an even number (60) 13