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

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Presentation on theme: "CHEM 430 – Structural Analysis of Organic Compounds Spring 2014 1."— Presentation transcript:

1 CHEM 430 – Structural Analysis of Organic Compounds Spring

2 1.1 Introduction - Organic Structural Analysis Attributes necessary for success: 1.Chemical common sense - What structures are possible? 2.General knowledge of the principles of organic chemistry 3.Develop your own organized and systematic approach 4.Understand how to interpret data from each spectral method 5.Resist using negative or unreliable data to make positive conclusions about a structure 6.Maintain intellectual composure when conflicting data sets seem to be in hand – draw on past experience 7.Simultaneously utilize all spectral data to look for multiple pieces of data to support individual conclusions 8.Reevaluate conclusions – self critique – peer review 2

3 1.2 Steps in Establishing a Final Structure 3 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: 1.Qualitative elemental analysis What elements are present in the unknown? 2.Quantitative elemental analysis What ratio of each element is in the unknown? 3.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)!!! 4

5 1.2 Steps in Establishing a Final Structure Methods for determining molecular formula 1.Qualitative elemental analysis We can assume that almost all organic compounds contain C and H ▫ If in doubt – burn in presence of O 2 ▫ If CO 2 is detected there must be carbon ▫ If H 2 O 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, Na 2 S, NaX + decomposed R Wet chemical qual tests for these ions infers the presence of each 5

6 1.2 Steps in Establishing a Final Structure Methods for determining molecular formula 2.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 CO 2 and H 2 O production from the pyrolysis of the sample in an O 2 chamber Remember your earliest general and organic chemistry: C x H y + O 2 (excess)  X CO 2 + Y/2 H 2 O 6

7 1.2 Steps in Establishing a Final Structure Methods for determining molecular formula 2.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: 7

8 1.2 Steps in Establishing a Final Structure Methods for determining molecular formula 3.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) Cryoscopic Method (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 (  T = k f 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 8

9 1.2 Steps in Establishing a Final Structure Methods for determining molecular formula 3.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 not 12; H is not 1, O is not 16) the “subset” of possible combinations falls dramatically ▫ HRMS determines the molecular ion to a precision of 6-8 significant figures 9

10 1.2 Steps in Establishing a Final Structure Methods for determining molecular formula 3.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 ▫ From a table of mass 98 data: C 3 H 4 N 3 O C 3 H 6 N C 4 H 4 NO C 4 H 6 N2O C 4 H 8 N C 5 H 6 O  gives us the exact formula C 5 H 8 NO C 5 H 10 N C 6 H 10 O C 7 H

11 1.2 Steps in Establishing a Final Structure 11 Revisit - Attributes necessary for success: 1.Chemical common sense - What structures are possible? 2.General knowledge of the principles of organic chemistry 3.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 12

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) 13

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 14

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 H 2 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 15

16 1.2 Steps in Establishing a Final Structure Hydrogen Deficiency Index - HDI A ring closure, like a double bond requires the “sacrifice” of H 2 from 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 16

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 17

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 18

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, C x H y N z O the 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 19

20 1.2 Steps in Establishing a Final Structure Hydrogen Deficiency Index - HDI Examples: C 4 H 6 HDI = 4 – 6/2 + 0/2 + 1 = 4 – = 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: 20

21 1.2 Steps in Establishing a Final Structure Hydrogen Deficiency Index - HDI More Examples: 21 C 7 H 12 C5H5NC5H5NC 14 H 10 C 60 C 6 H 12 O 6 C6H8OC6H8O

22 1.2 Steps in Establishing a Final Structure Hydrogen Deficiency Index - HDI More Examples (only one representative structure is shown): 22 C 7 H 12 C5H5NC5H5NC 14 H 10 C 60 C 6 H 12 O 6 C6H8OC6H8O

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 the base formula becomes: C n H n + r For this formula, the HDI can be calculated from the following formula: HDI = ( n – r + 2 ) 2 23

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 / 13 = n + r /  13 = 7 remainder 7 = / 13 Base formula = C 7 H = C 7 H 14 ( 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! 24

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 25

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: 26 Element addedSubtract:  HDI (  U in text) Element addedSubtract:  HDI (  U in text) CH ClC 2 H 11 3 H 12 C-7 79 BrC6H7C6H7 -3 OCH 4 1FCH 7 2 NCH 2 1/2SiC2H4C2H4 1 SC2H8C2H8 2PC2H7C2H7 2 IC 9 H 19 0

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 C 7 H 14, 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 27

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:C 7 H 14 Add oxygen:C 7 H 14 O(mol. mass now 114) Subtract CH 4 :C 6 H 10 O(mol. mass now 98) HDI correction:1 + 1 = 2 You can check HDI by earlier formula We can formulate compounds C 6 H 10 O with HDI = 2: 28

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: C 7 H 14 Add Nitrogen: C 6 H 12 N (sub. CH 2 ) HDI: 1.5 Probably an incorrect formula, it is unlikely this compound has nitrogen Base formula: C 7 H 14 Add Sulfur: C 5 H 6 S (sub. C 2 H 8 ) HDI: 3 Very few possibilities with only 6 hydrogens and an HDI of 3 Base formula: C 7 H 14 Add Bromine: CH 7 Br (sub. C 6 H 7 ) HDI: -2 Impossible structure 29

30 1.2 Steps in Establishing a Final Structure 30 Molecular Formula Molecular Formula HDI Functional group inference Functional group inference Molecular Mass Molecular Mass Rule of 13 HDI calc. HRMS & Legacy methods HRMS & Legacy methods 1.Qualitative elemental analysis 2.Quantitative elemental analysis 3.Determination of Molecular Formula 1.Qualitative elemental analysis 2.Quantitative elemental analysis 3.Determination of Molecular Formula


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