“Structure Elucidation”-Comprehensive Spectral Interpretation Lab 2 “Structure Elucidation”-Comprehensive Spectral Interpretation

Objectives Suggest empirical formulae from given elemental analysis data and molecular weight of molecules. Identify unknown compounds by combined spectral interpretation of NMR, IR, and MS spectra, and with respect to their empirical formulae.

Steps for verifying the structure of an unknown compound 1- CHN (molecular formula) 2- IHD (degree of un saturation) 3- FT-IR (functional groups) 4-1H-NMR (carbon skeleton) 5- MS (M.Wt.)

1- Elemental analysis - CHN : How to Get the Empirical Formula of an Unknown -Qualitative -Quantitative CHNX analysis Most Common: CHN analysis, is accomplished by combustion analysis. In this technique, a sample is burned in an excess of oxygen—> carbon dioxide, water, and nitric oxide. The weights of these combustion products can be used to calculate the composition of the unknown sample.

Molecular weight of an unknown can be measured directly by mass-spectrometry EXAMPLE: practical elemental analysis data of an unknown are found to be: C: 42.7% ,H: 4.92% , N: 31.1%, others: 21.3%   the molecular weight of the same unknown was found to be: 225.2 now, the number of atoms in the molecule are calculated as: nC* 12.011 / 225.2 = 0.427 => nC = 8 C-atoms   nH * 1.0079 / 225.2 = 0.0492 => nH = 11 H-atoms nN * 14.007 / 225.2 = 0.311 => nN = 5 N-atoms empirical formula : C8H11N5 ???  molecular weight : 177.21. difference of 47.99..WHY?? “other” atoms present in the molecule. So the remaining part could be e.g. 3 O-atoms (3 * 15.999 = 47.997), this has to be verified by interpretation of the charts.

Aromatic Ring+ 1 double bond 2- Calculate IHD IHD = [(2C + 2) – X + N – H ] / 2 Example: C7H6O IHD= [(2*7+2)-0+0-6]/2= 5 Aromatic Ring+ 1 double bond

3- How to Interpret an IR Spectrum Is a carbonyl group present? 1820-1660 cm-1.  strongest in the spectrum and of medium width.   If C=O is present, ACIDS O-H present? broad band near 3500-2500 cm-1 ESTERSC-O present? strong band near 1300-1000 cm-1 AMIDESN-H present? medium band near 3500 cm-1 ANHYDRIDES two C=O at 1810 / 1760 cm-1 ALDEHYDE is aldehyde C-H present? two weak bands near 2850 / 2750 cm-1 (Fermi resonance) KETONES if none of the previous five match -

Regions where various common types of bonds absorb (Base values) It is recommended at first to establish the “broad patterns” given below. Then as a second step a “typical absorption value” can be memorized for each of the functional groups in this pattern. Fingerprint Region C=O =C-H str. O= C-H str. acid chlorides anhydrides esters ketones aldehydes carboxylic acids amides C-H oop bend Aromatic compound CC CO CN Str. C C C N OH NH Bond str. Olefinic aromatic C-H str. Aldehyde Fermi resonance C=C C=N OH CH bend aliphatic C=C aromatic N-H bend 4000 3300 3000 2800 2500 2000 1400 1000 600 Frequency (cm-1)

Other functional groups Bond Base Value Strength / Shape Comments C=O 1715 (typical) 1600-1800 s, "finger" exact position depends on the environment of carbonyl O—H alcohol 3300-3500 s, brd broad due to H bonding O—H acid 2500-3500 s, very brd   N—H m can tell primary from secondary C-H just to the right of 3000 s Aliphatic, sp3 just to the left of 3000 Olefinic or aromatic, sp2 3300 Acetylenic, sp 2750-2850 w, Fork like “Fermi resonance” of aldehydic group C—O 1100-1300 also check for OH and C=O C=C 1650 w alkene  m-s aromatic  Alkene w due to low polarity  Aromatic usually in pairs C≡C 2150 w, sharp characteristic C≡N 2250 m, sharp

4- How to Interpret an 1H-NMR Spectrum The number of signals present in the spectrum. The chemical shift [ppm] of the signals. The signal intensity = AUC (area under the curve) = integration of each signal. The multiplicity of each signal.

Identify the compound from its molecular formula and its 1H-NMR spectrum. Propyl benzene

Remember: protons of OH, NH and COOH are often “exchangeable”, this means they can easily be replaced by 2H (deuterium), which will NOT give a signal under common NMR settings.

5- How to Interpret a Mass Spectrum The signal with the greatest m/z value (on the x-axis) usually is the molecular ion peak. If electron impact ionization was used in the mass detector, the m/z value is at the same time the molecular weight of the analyte. The signal with the biggest intensity in the spectrum is the base peak. It reflects the most stable fragment of the analyte. Check for eye-catching isotope peaks of Chlorine (n and n+2, relative intensity 75% and 25%) and Bromine (n and n+2, relative intensity 50% and 50%) Generally, all the peaks in a mass spectrum should be assigned to fragments of the analyte. These fragments can arise from direct fragmentation, or from re-arrangement including eliminations from the molecular ion or from other fragments.

Mass spectrum Base peak (highest relative abundance) Fragments (daughter peaks) Molecular ion peak (parent peak)