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CHEM 430 – Structural Analysis of Organic Compounds

Spring 2014. CHEM 430 – Structural Analysis of Organic Compounds. 1.1 Introduction - Organic Structural Analysis. Attributes necessary for success: Chemical common sense - What structures are possible? General knowledge of the principles of organic chemistry

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CHEM 430 – Structural Analysis of Organic Compounds

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  1. Spring 2014 CHEM 430 – Structural Analysis of Organic Compounds

  2. 1.1 Introduction - Organic Structural Analysis Attributes necessary for success: Chemical common sense - What structures are possible? General knowledge of the principles of organic chemistry Develop your own organized and systematic approach Understand how to interpret data from each spectral method Resist using negative or unreliable data to make positive conclusions about a structure Maintain intellectual composure when conflicting data sets seem to be in hand – draw on past experience Simultaneously utilize all spectral data to look for multiple pieces of data to support individual conclusions Reevaluate conclusions – self critique – peer review

  3. 1.2 Steps in Establishing a Final Structure Most ideal starting point is the determination of molecular formula

  4. 1.2 Steps in Establishing a Final Structure Methods for determining molecular formula Classic approach involves three steps with a “cold” unknown: • Qualitative elemental analysis Whatelements are present in the unknown? • Quantitative elemental analysis What ratio of each element is in the unknown? • Molecular mass determination By combining the relative amount of each element to give a determined mass an exact formula can be obtained Or the determination of a high-resolution mass spectrum (HRMS)!!!

  5. 1.2 Steps in Establishing a Final Structure Methods for determining molecular formula Qualitative elemental analysis • We can assume that almost all organic compounds contain C and H • If in doubt – burn in presence of O2 • If CO2 is detected there must be carbon • If H2O is detected there must be hydrogen • Even in 2011, there is no routine qualitative test for the presence of oxygen! • N, S and X (= F, Cl, Br, I) are qualitatively determined by the dreaded Na-fusion test (in sophomore organic lab texts) • When an organic compound containing these elements is “fused” with molten Na metal, reductive decomposition affords the corresponding salts: R-N, R-S, R-X NaCN, Na2S, NaX + decomposed R • Wet chemical qual tests for these ions infers the presence of each

  6. 1.2 Steps in Establishing a Final Structure Methods for determining molecular formula Quantitative elemental analysis • Commercial laboratories perform these analyses on a routine basis for research and industry: • Midwest Microlabs • Intertek • Galbraith Laboratories • A small (10-50 mg) sample of the pure material is submitted • Carbon and hydrogen are determined by quantitative analysis of CO2 and H2O production from the pyrolysis of the sample in an O2chamber • Remember your earliest general and organic chemistry: CxHy + O2 (excess)  X CO2 + Y/2 H2O

  7. 1.2 Steps in Establishing a Final Structure Methods for determining molecular formula Quantitative elemental analysis • This analysis is required for the publication of any new compound in the literature • % Found values must be within 0.4% of the theoretical value Typical form sent with sample:

  8. 1.2 Steps in Establishing a Final Structure Methods for determining molecular formula Quantitative elemental analysis The weight of one mole of a substance in conjunction with the empirical formula will give the overall molecular formula Old methods – these methods are based on colligative properties, you may have done them in a general chemistry laboratory course: Vapor Density Method – material is vaporized and weighed, and after converting the volume to standard temperature and pressure, the exact fraction of a mole the sample represents would be known (PV = n RT) CryoscopicMethod (freezing point depression) – the material is dissolved in a solvent of a known freezing point, the amount of deviation of the freezing point of the resulting solution from the pure solvent gives the molal concentration of the solute (DT = kf m, where m is moles per Kg of solvent) Vapor Pressure Osmometry (related to boiling point elevation) – the change in vapor pressure of a pure solvent by adding a known weight of solute gives the molal concentration of the solute Titration of an unknown acid – titration of a known weight of the unknown with standardized base gives the molecular weight

  9. 1.2 Steps in Establishing a Final Structure Methods for determining molecular formula Quantitative elemental analysis New method – HRMS – High Resolution Mass Spectrometry: • Mass spectroscopy works on the principle of “punching” an electron out of a sample molecule and determining the mass of the ion fragments produced (which we will study in detail later) • The largest fragment is one that differs in mass from the original molecule by the negligible weight of a single e- • From our study of molecular formulas, we will see only a limited subset of possible formulas gives a crude molecular mass • If we take the isotopic weights of each naturally occurring element into account (i.e. C is 12.011 not 12; H is 1.0079 not 1, O is 15.9994 not 16) the “subset” of possible combinations falls dramatically • HRMS determines the molecular ion to a precision of 6-8 significant figures

  10. 1.2 Steps in Establishing a Final Structure Methods for determining molecular formula Quantitative elemental analysis HRMS – High Resolution Mass Spectrometry • From tables of combinations of formula masses with the natural isotopic weights of each element, it is possible to find an exact molecular formula from HRMS • Example: HRMS gives you a molecular ion of 98.0372 • From a table of mass 98 data: C3H4N3O 98.0355 C3H6N4 98.0594 C4H4NO2 98.0242 C4H6N2O 98.0480 C4H8N3 98.0719 C5H6O2 98.0368  gives us the exact formula C5H8NO 98.0606 C5H10N2 98.0845 C6H10O 98.0732 C7H14 98.1096

  11. 1.2 Steps in Establishing a Final Structure Revisit - Attributes necessary for success: • Chemical common sense - What structures are possible? • General knowledge of the principles of organic chemistry • Develop your own organized and systematic approach Let’s take these three attributes and utilize them to their fullest extent! Now we have a molecular formula, what can we do with it? • Index of hydrogen deficiency (HDI, IHD, U, etc.) • Rule of thirteen

  12. 1.2 Steps in Establishing a Final Structure A molecular formula provides the jigsaw puzzle • All the pieces must be used • The number of pieces determines the size of the picture • Pieces can only fit together in a logical pattern • Pieces can only be used once • Each piece can only connect to other pieces in a predetermined way

  13. 1.2 Steps in Establishing a Final Structure Hydrogen Deficiency Index - HDI Organic molecules exist as discrete sets of covalent bonds based on the valence of the elements that comprise them i.e. hydrogen is monovalent, oxygen divalent and carbon tetravalent… • If a molecular formula is known: • Functional groups can be implied or ruled out Obvious, but often overlooked tool • The number of times valence rules for elements are violated is implied most commonly for carbon, which is called the index of unsaturation or hydrogen deficiency index (HDI), • Less commonly for elements such as oxygen and nitrogen that may be involved in acid-base chemistry (i.e. nitrogen has a valence of 4 in ammonium salts)

  14. 1.2 Steps in Establishing a Final Structure Hydrogen Deficiency Index - HDI For simple straight chain or branched hydrocarbons, there is always a certain ratio of hydrogen to carbon necessary to make the entire structure saturated: We say these molecules are not hydrogen deficient, and set the index at 0

  15. 1.2 Steps in Establishing a Final Structure Hydrogen Deficiency Index - HDI If a double bond is present in the structure with the same number of carbons, the requisite amount of hydrogen required to saturate the chain is reduced by H2 We say these molecules are hydrogen deficient, and the index increases by one for each double bond added, for the first structure, the index is 1, for the second the index is 2

  16. 1.2 Steps in Establishing a Final Structure Hydrogen Deficiency Index - HDI A ring closure, like a double bond requires the “sacrifice” of H2from the formula, increasing the index by 1 for each closed ring in the compound Triple bonds act as two double bonds increasing the index by two for each one in a molecule Hence, the first structure has an index of 1, the second an index of 2

  17. 1.2 Steps in Establishing a Final Structure Hydrogen Deficiency Index - HDI Nitrogen (for the azaphobics) is usually assumed to be trivalent - obviously ammonium salts and nitro compounds violate this. When working “cold”, assume nitrogen is trivalent, and therefore, for every nitrogen in a structure, one less hydrogen is needed to “fill” its valence requirement than carbon

  18. 1.2 Steps in Establishing a Final Structure Hydrogen Deficiency Index - HDI Halogens (normally monovalent) merely replace hydrogen in a like-indexed formula Oxygen, with a valence of two will have no bearing on the HDI! Higher elements, found commonly in biologically interesting organic compounds, such as sulfur and phosphorus exist in almost equal populations in the various valences they are capable of and are typically not considered by this method directly

  19. 1.2 Steps in Establishing a Final Structure Hydrogen Deficiency Index - HDI Furthermore, it is tedious to go through a structural analysis to get the index of unsaturation. It can be algebraically expressed by combining the effects of each of the common elements. Thus, given a molecular formula, CxHyNzOthe HDI becomes: HDI = x - y/2 + z/2 + 1 Remember to count halogens as hydrogen and to omit oxygen Your text gives an algebraic variation of this equation: U = C + 1 – ½ (X – N) Where X = number of hydrogens and halogens

  20. 1.2 Steps in Establishing a Final Structure Hydrogen Deficiency Index - HDI Examples: C4H6 HDI = 4 – 6/2 + 0/2 + 1 = 4 – 3 + 1 = 2 This compound could contain 2 double bonds (db) OR 1 triple bond (tb) OR 2 rings (r) OR 1 ring and 1 double bond There are eight working structures that fit this formula:

  21. 1.2 Steps in Establishing a Final Structure Hydrogen Deficiency Index - HDI More Examples: C7H12 C5H5N C14H10 C60 C6H12O6 C6H8O

  22. 1.2 Steps in Establishing a Final Structure Hydrogen Deficiency Index - HDI More Examples (only one representative structure is shown): C7H12 C5H5N C14H10 C60 C6H12O6 C6H8O

  23. 1.2 Steps in Establishing a Final Structure Rule of Thirteen For the formula “connoisseur” there is another algebraic treatment of low resolution molecular mass that can lead to possible molecular formulas When a molecular mass, M, is known, a base formula can be generated from the following equation: M = n + r 13 13 the base formula becomes: CnHn+ r For this formula, the HDI can be calculated from the following formula: HDI = ( n – r + 2 ) 2

  24. 1.2 Steps in Establishing a Final Structure Rule of Thirteen The result of this equation is only a first estimate that can be further refined. Example: We determine by MS the molecular mass is 98 98 / 13 = n + r / 13 98  13 = 7 remainder 7 = 7 + 7 / 13 Base formula = C7 H7+ 7 = C7H14 (n = 7, r = 7) and HDI (U) = (7 – 7 + 2)/2 = 1 Remember, this is only the first of several possible formulas that give a molecular mass of 98!

  25. 1.2 Steps in Establishing a Final Structure Rule of Thirteen To explore further possible formulas we can substitute other elements, provided the appropriate adjustment must be made: For example, if we wish to consider that the base formula also includes oxygen, with atomic mass 16, one carbon (12) and four hydrogens (4 x 1) must be removed from the hypothetical formula to give the same MW Likewise, an adjustment to hydrogen deficiency must be made. Important: Not all formulas generated by the rule of thirteen are possible! HDI is the test for ‘correctness’: • If HDI is an integer, the possible formula is valid • If HDI is a fraction, the possible formula is invalid

  26. 1.2 Steps in Establishing a Final Structure Rule of Thirteen To explore further possible formulas we can substitute other elements, provided the appropriate adjustment must be made:

  27. 1.2 Steps in Establishing a Final Structure Rule of Thirteen Going back to our previous example: MW = 98 has a first possible formula of C7H14, HDI = 1 The unknown may be one of the isomeric heptenes: Or it may be a saturated carbocycle: With the knowledge of molecular mass, we can whittle the infinity of possible organic compounds to two families of closely related isomers

  28. 1.2 Steps in Establishing a Final Structure Rule of Thirteen Remember, off of the base formula we can begin to substitute other elements into the formula to come up with other possibilities that give a molecular mass of 98: If we now assume the unknown has a single oxygen: Base formula: C7H14 Add oxygen: C7H14O (mol. mass now 114) Subtract CH4: C6H10O (mol. mass now 98) HDI correction: 1 + 1 = 2 You can check HDI by earlier formula We can formulate compounds C6H10O with HDI = 2:

  29. 1.2 Steps in Establishing a Final Structure Rule of Thirteen We can add other elements (or multiples of elements) exhaustively: Observe how for a low molecular mass the inference of big elements greatly simplifies the number of possible structures: Base formula: C7H14 Add Nitrogen: C6H12N (sub. CH2) HDI: 1.5 Probably an incorrect formula, it is unlikely this compound has nitrogen Base formula: C7H14 Add Sulfur: C5H6S (sub. C2H8) HDI: 3 Very few possibilities with only 6 hydrogens and an HDI of 3 Base formula: C7H14 Add Bromine: CH7Br (sub. C6H7) HDI: -2 Impossible structure

  30. 1.2 Steps in Establishing a Final Structure HDI HDI calc. HDI calc. Molecular Formula Molecular Mass Functional group inference Rule of 13 HRMS & Legacy methods • Qualitative elemental analysis • Quantitative elemental analysis • Determination of Molecular Formula

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