Structural characterization begins with a purity check!

Transcription

1 Structural characterization begins with a purity check! Refractive Index Melting Point Elemental Analysis (EA) Thin Layer Chromatography (TLC) High Performance Liquid Chromatography (HPLC) Course: Analytical Chemistry, R. Letcher Gas Chromatography (GC) only for volatile compounds Course: Analytical Chemistry, R. Letcher Other chromatographic or electrophoretic methods

2 Purity of commercially available reagents and solvents Most reagents come in purities between 95 % and 99 % (by weight or volume). The price of a chemical can increase significantly if high purity is requested. Example (Aldrich prices): 500 g of PBr % = $ % = $ % = $119 A real chemical/material can not be pure and the preparation of highly purified chemicals is tedious and consumes time. Consequently, it is crucial to decide what degree of purity is required for a specific experiment. It is important to know what impurities contaminate the sample to predict whether they will interfere with a certain reaction or with a physical measurement, such as a spectroscopic investigations. These details are provided by the manufacturer or distributor.

3 Refractive index (n) of liquids/solvents The purity of a liquid (solvent) is usually determined by GC and/or its refractive index n. The refractive index (n) is the ratio of the velocity of light in a vacuum to the velocity in some medium. Refractive indices generally increase with the atomic number of the constituent atoms. A high density or high atomic number elements usually results in high refractive indices. Abbe Refractometer Medium refractive index of a vacuum refractive index of air refractive index of ice refractive index of a water (20 ºC) Refractive Index 1.0 (exactly) Digital Laboratory Refractometer refractive index of glass iodine crystal

4 Refraction and the refractive index When light enters a transparent medium of different refractive index, n, it is refracted (Snell s Law): n = sinθ 1 / sinθ 2 (angle of incident & refraction) sinθ 1 / sinθ 2 = n 2 / n 1 The velocity of a light wave changes when light enters a transparent medium of different refractive index but not the frequency: velocity = νλ; n = c / v = λ vac / λ subs

5 Molecular Theory of Refractive Indices Lorentz local field for an isotropic medium: E loc = [(ε + 2) / 3] E where ε is the mean permittivity Using ε = n 2 derived from the Maxwell s Equations, the Lorenz-Lorentz expression relates the refractive index to the mean molecular polarizability: n 2 1 / n = N α / 3ε 0 where N is the number density (d N A / M ), α the mean molecular polarizability, and ε 0 = As/Vm n e2 1 / n = N α / 3ε 0 n o2 1 / n = N α / 3ε 0 with n 2 = 1/3 (n e2 + 2n o2 )

6 Melting Points of Solids Freezing Point Melting Point The melting point of a substance is sensitive to its composition (freezing point depression T f = -K f m). Measuring the melting point of a chemical is and easy and powerful method for determining its identity and purity, if reference values are available. The melting point (freezing point) of a substance is defined as the temperature at which the solid and liquid form of the substance can co-exist.

7 MELTING POINTS are measured with a melting point apparatus. The sample is first heated at several K a minute until the solid is molten. This crude melting temperature is now determined more accurately by using a slow heating rate (2 K / min). The temperature at which the edges of a solid start to melt is defined as on-set temperature and the final melting temperature is reached when all the solid has become a liquid. The exact melting temperature is in between these two temperatures and is reached when both phases, the solid and the liquid, are present at the same time. Determination of a mixed melting point The melting point of an unknown compound is determined to 99.5 ºC and a reference table (e.g. from the Handbook of Chemistry and Physics) suggests 8 possible compounds. If we assume malic acid might be the unknown compound, we first determine the m.p. of the pure malic acid (about ºC ). Both compounds are then mixed and run again. The melting point does not change if both compounds are identical.

8 Thin Layer Chromatography (TLC) A small amount of the compound under investigation, or its solution in a volatile solvent, is spotted onto a substrate that is coated with an active stationary phase such as silica gel. One side of the substrate, the spotted side, is then exposed to a solvent that runs up the plate due to capillary forces. Compounds run with the solvent but might be slowed down by interactions with the stationary phase. The ratio between the distance one spot runs and the distance the solvent front runs is given as R f values (retention factor). Optimum R f values are between 0.1 and 0.5 (1.0 = solvent front). Good separations are obtained only if the right combination of solvent and stationary phase has been chosen. A polar stationary phase (silica gel) is available in the labs and the least polar organic solvent that runs your compounds should be chosen. Generally, polar solvents for used for non-polar stationary phases and vice versa. Three common stationary phases are silica, aluminum oxide, and reverse phases such as RP-18. A solvent must also dissolve your sample well enough. Streaking spots appear if the solubility of your compounds is not high enough.

9 Elemental Analysis (EA) Elemental analysis on carbon, hydrogen, and nitrogen is the oldest investigation performed to characterize and/or prove the elemental composition of an organic or inorganic sample. In fact, it used to be the only available routine method before modern spectroscopy was established in the 1950s. The majority of organic compounds only contain the elements C, H, N, and O and the latter is seldom determined separately. Our departmental analyser gives weight percentages of C, H, N done by combustion in O 2 and gas chromatographic analysis of CO 2, H 2 O and NO 2. As an option S can be measured as SO 2 and oxygen as CO. Other elements such as Cl, Br, and I must be determined by other means. Perkin Elmer 2400 Series II CHNS/O Analyzer

10 Measuring Principle The sample under test is weighed in using a tin capsule. The required amount is 2 to 3 mg of organic material and can hardly exceed 10 mg, if inorganic matter with little carbon content is investigated. After folding the capsule (looking rather like wrapped tin foil) the sample is placed in the autosampler. The tin capsule enclosing the sample falls into the reactor chamber where excess oxygen is introduced before. At about 990 C the material is "mineralized". Formation of carbonmonoxide is probable at this temperature even under these conditions of excess oxygen. The complete oxidation is reached at a tungsten trioxide catalyst which is passed by the gaseous reaction products. The resulting mixture should thus consist of CO 2, H 2 O und NO x. But also some excess O 2 passes the catalyst. The product gas mixture flows through a silica tube packed with copper granules. In this zone, held at about 500 C, remaining oxygen is bound and nitric/nitrous oxides are reduced to N 2. The leaving gas stream includes the analytically important species CO 2, H 2 O und N 2. Eventually included SO 2 or hydrohalogenides are absorbed at appropriate traps.

11 High purity helium (Quality 5.0) is used as carrier gas. Finally the gas mixture is brought to a defined pressure/volume state and is passed to a gas chromatographic system. Separation of the species is done by so called zone chromatography. In this technique a staircase type signal is obtained and the step height is proportional to the substance amount in the mixture. Blank values are taken from empty tin capsules Calibration is done by elemental analysis of standard substances supplied by the instrument's manufacturer for this purpose. Working Range Mineralization and detection covers every species of the analyte elements in the sample. Beside purely organic samples various metal organic compunds and even inorganic samples as carbides and nitrides have successfully been characterized. The detection limit for carbon and nitrogen at sample amounts of 2 to 3 mg was found to be at about 0.05 w-% (500 ppm) in what case the uncertainty stays at about 0.02 w-%. According to the apparatus's supplier the instrument's uncertainty in the medium range stays below 0.3 w-% as required by journals to prove the expected composition. With very carbon rich samples we found this tolerance was slightly exceeded (still within 0.5 w-%).

12 Problems and Interferences The weighing of oily or fluid substances is impossible using the thin walled tin capsules. For this purpose alumina pans with a lid are available. These pans are tightly closed by cold welding to prevent loss of sample by spillage and evaporation. As the blank value for nitrogen is dramatically increased be the enclosed volume of air the determination limit for N in liquid samples is increased to 0,1 to 0,2 w-%. With highly viscous or even glassy materials elemental analysis is even impossible with the above method. It has long been known that phosphorus can interfere in the mineralization of organic material. In literature the formation of glassy P 2 O 5 xh 2 O yc has been described. Elemental analysis of phosphorus containing compounds can thus suffer from systematic deviations in the determined carbon content exceeding the tolerance limit of 0,3 w-%. This effect can be controlled by the addition of vanadium pentoxide (V 2 O 5 ). Fluorine is mineralized to form HF which reacts at the wall of the silica tubes which form the main part of the reaction zone. The gaseous products, such as SiF 4 and relatives, can cause systematic errors which rarely become significant with respect to the 0,3 w-% tolerance. The mineralization of metal containing samples can also be affected by interferences. By modification of the method most of these can be compensated for.

13 Example Calculations Percentage Composition C x H y O z (9.83 mg) + excess O 2 gives CO 2 (23.26 mg) + y/2 H 2 O (9.52 mg) Calculations: mg of CO 2 / mg/mmol = mmoles of CO 2 mmoles of CO 2 = mmoles of C in original sample ( mmoles of C)(12.01mg/mmol C) = 6.35 mg of C in original sample 9.52 mg of H 2 O / mg/mmol = mmoles of H 2 O mmoles of H 2 O = 1/2 mmoles of H in original sample (0.528 mmoles of H) (2) (1.008mg/mmol H) = 1.06 mg of H in original sample

14 Weight Percentage Composition C x H y O z (9.83 mg) + excess O 2 x CO 2 (23.26 mg) + y/2 H 2 O (9.52 mg) %C = 6.35 mg/9.83 mg x 100 = 64.6% %H = 1.06 mg/9.83 mg x 100 = 10.8% %O = 100-( ) = 24.6% Calculation of Empirical Formula assume for example a 100g sample 64.6% of C: 64.6 g/12.01 g/mol = 5.38 moles of C 10.8% of H: 10.8 g/1.008 g/mol = 10.7 moles of H 24.6% of O: 24.6 g/16.0 g/mol = 1.54 moles of O Thus: C5.38 H10.7 O1.54 converting to simplest ratio: C5.38/1.54; H10.7/1.54; O1.54/1.54 C 3.50 H 7.00 O 1.00 = C 7 H 14 O 2 What are theoretically possible structures?