1. IHD IR Spectroscopy Mass Spectroscopy NMR

2. 11.3: Analytical techniques can be used to determine the structure of a compound, analyze the composition of a substance, or determine the purity of a compound. Spectroscopic techniques are used in the structural identification of organic and inorganic compounds. 20.1: Although spectroscopic characterization techniques form the backbone of structural identification of compounds, typically no one techniques results in a full structural identification of a molecule

3. Analytical techniques Qualitative analysis: detection of presence but not quantity of substance in mixture Quantitative analysis: measurement of the quantity of a particular substance in a mixture Structural analysis: description of how atoms are arranged in molecular structures

4. Index of Hydrogen Deficiency Learning outcomes Understand how the number of double bonds/rings can be worked out from molecular formula

5. Index of Hydrogen Deficiency Alkanes have maximum number of hydrogen atoms For every two hydrogen atoms fewer than in alkane with same number of carbon atoms, there is one double bond or ring present. (double bond equivalent) The number of double bond equivalents sometimes called degree of unsaturation or the Index of hydrogen deficiency.

6. How the index of hydrogen deficiency works. 1 A double bond and ring each counts as one IHD. 2 A triple bond counts as two IHD Hydrocarbons (C x H y ): IHD=1/2[2C+2-H] (where x = number of carbon, Y= number of hydrogen) Example C 4 H 8 ½(8+2-8)= 1

7. Compounds Containing Elements Other than C and H O and S atoms do not affect the IHD. Halogens (F, Cl, Br, I) are treated like H atoms (CH 2 Cl 2 has the same IHD as CH4). For each N, add one to the number of C and one to the number H (CH 5 N is treated as C 2 H 6. CH 4 N 2 O is treated as C 3 H 6 by adding 2 to # of C and 2 to # of H).

8. Calculate IHD for C 3 H 5 N will be treated as C 4 H 6 IHD = 1/2(4x2 + 2- (6) = 4/2=2 Possible structures

9. Calculate IHD for C 6 H9Cl will be treated like C 6 H 10 IHD=1/2[2C+2-H] IHD =1/2[12+2-10]=4/2=2

10. Practice problems CH3CHCHCH2CHCH2 CH3C ≡ CCOCH3 IHD = 3 IHD = 2 IHD = 5 IHD = 1 IHD = 3

11. Learning outcomes Understanding Mass spectrometry (MS), proton nuclear magnetic resonance spectroscopy (1H NMR) and infrared spectroscopy (IR) are techniques that can be used to help identify compounds and to determine their structure.

12. Learning outcomes Deduction of information about the structural features of a compound from percentage composition data, MS, 1H NMR or IR.

14. Mass spectrometer Picture Source http://chemistry.umeche.maine.edu/CHY251/Ch13-Overhead4.html

15. Using a mass spectrum to find relative formula mass The formation of molecular ions When the vaporized organic sample passes into the ionization chamber of a mass spectrometer, it is bombarded by a stream of electrons. These electrons have a high enough energy to knock an electron off an organic molecule to form a positive ion. This ion is called the molecular ion.

16. The molecular ion is often given the symbol M + or M - the dot in this second version represents the fact that somewhere in the ion there will be a single unpaired electron. That's one half of what was originally a pair of electrons - the other half is the electron which was removed in the ionization process. The molecular ions tend to be unstable and some of them break into smaller fragments. These fragments produce the familiar stick diagram. Fragmentation is irrelevant at this stage.

17. Using the molecular ion to find the relative formula mass In the mass spectrum, the heaviest ion (the one with the greatest m/z value) is likely to be the molecular ion. For example, in the mass spectrum of pentane, the heaviest ion has an m/z value of 72.

18. Analyzing mass spectrometer data C 3 H 8

19. MS data of N,N-diethylmethylamine

20. WHAT IS AN INFRA-RED SPECTRUM? If a range of infra-red frequencies shine at an organic sample, some of the frequencies get absorbed by the compound. A detector attached to the other end of spectrum measured frequencies absorbed or transmitted. How much of a particular frequency gets through the compound is measured as percentage transmittance.

21. What an infra-red spectrum looks like A graph is produced showing how the percentage transmittance varies with the frequency of the infra-red radiation.

22. Analyzing IR spectrum IR spectra are not particularly easy to analyse, nor do they give definitive information about structure. There are however, two different stages in an analysis. 1 Identification of absorptions 2 Fingerprinting The first stage involves looking for characteristic absorptions and attempting to ascribe them to specific structural features. The second stage is usually carried out after a series of analyses leads to a possible conclusion.

23. IDENTIFICATION OF PARTICULAR BONDS IN A MOLECULE INFRA RED SPECTRA - USES The presence of bonds such as O-H and C=O within a molecule can be confirmed because they have characteristic peaks in identifiable parts of the spectrum. IDENTIFICATION OF COMPOUNDS BY DIRECT COMPARISON OF SPECTRA The only way to completely identify a compound using IR is to compare its spectrum with a known sample. The part of the spectrum known as the ‘Fingerprint Region’ is unique to each compound.

24. Infra-red spectra are complex due to the many vibrations in each molecule. Total characterisation of a substance based only on its IR spectrum is almost impossible unless one has computerised data handling facilities for comparison of the obtained spectrum with one in memory. However, the technique is useful when used in conjunction with other methods such as nuclear magnetic resonance (nmr) spectroscopy and mass spectroscopy. Peak position depends on bond strength masses of the atoms joined by the bond strong bonds and light atomsabsorb at lower wavenumbers weak bonds and heavy atoms absorb at high wavenumbers INFRA RED SPECTRA - INTERPRETATION

25. Vertical axisAbsorbance the stronger the absorbance the larger the peak Horizontal axisFrequencywavenumber (waves per centimetre) / cm -1 Wavelengthmicrons (m); 1 micron = 1000 nanometres INFRA RED SPECTRA - INTERPRETATION

26. FINGERPRINT REGION organic molecules have a lot of C-C and C-H bonds within their structure spectra obtained will have peaks in the 1400 cm -1 to 800 cm -1 range this is referred to as the “fingerprint” region the pattern obtained is characteristic of a particular compound the frequency of any absorption is also affected by adjoining atoms or groups.

27. IR SPECTRUM OF A CARBONYL COMPOUND carbonyl compounds show a sharp, strong absorption between 1700 and 1760 cm -1 this is due to the presence of the C=O bond

28. IR SPECTRUM OF AN ALCOHOL alcohols show a broad absorption between 3200 and 3600 cm -1 this is due to the presence of the O-H bond

29. IR SPECTRUM OF A CARBOXYLIC ACID carboxylic acids show a broad absorption between 3200 and 3600 cm -1 this is due to the presence of the O-H bond they also show a strong absorption around 1700 cm -1 this is due to the presence of the C=O bond

30. IR SPECTRUM OF AN ESTER esters show a strong absorption between 1750 cm -1 and 1730 cm -1 this is due to the presence of the C=O bond

31. WHAT IS IT! O-H STRETCH C=O STRETCH O-H STRETCH C=O STRETCH AND ALCOHOL ALDEHYDE CARBOXYLIC ACID One can tell the difference between alcohols, aldehydes and carboxylic acids by comparison of their spectra.

32. O-HC=OC-O N-H Aromatic C-CC-H C=CC-C alkanes CNCN C-Cl CHARACTERISTIC FREQUENCIES

33. BondClass of compoundRange / cm -1 Intensity C-HAlkane2965 - 2850strong C-CAlkane1200 - 700weak C=CAlkene1680 - 1620variable C=OKetone1725 - 1705strong Aldehyde1740 - 1720strong Carboxylic acid1725 - 1700strong Ester1750 - 1730strong Amide1700 - 1630strong C-OAlcohol, ester, acid, ether1300 - 1000strong O-HAlcohol (monomer)3650 - 3590variable, sharp Alcohol (H-bonded)3420 - 3200strong, broad Carboxylic acid (H-bonded)3300 - 3250variable, broad N-HAmine, Amide3500 (approx)medium C  NNitrile 2260 - 2240medium C-XChloride800 - 600strong Bromide600 - 500strong Iodide500 (approx)strong CHARACTERISTIC ABSORPTION FREQUENCIES

34. Nuclear Magnetic Resonance

35. NMR Nuclear magnetic resonance relies on the magnetic field produced by a spinning nucleus containing an odd number of nucleons (protons or neutrons). In the presence of an external magnetic field the nucleus can exhibit more than one spin state and can move between these states by the absorption of electromagnetic radiation of a specific frequency (energy). The energy absorbed can be detected and from this information about the location (environment) of the nucleus can be deduced.

37. https://www.youtube.com/watch?v=k0eR8YqcA8c https://www.youtube.com/watch?v=OrvAUDgVoT8&f eature=iv&src_vid=k0eR8YqcA8c&annotation_id=ann otation_877622 https://www.youtube.com/watch?v=OrvAUDgVoT8&f eature=iv&src_vid=k0eR8YqcA8c&annotation_id=ann otation_877622

38. NMR is probably the most useful tool in the organic chemists arsenal for structural determination. As organisms are mainly water (containing H atoms with an odd number of nucleons), NMR has developed into an invaluable medical diagnostics tool, called an MRI (magnetic resonance instrument) scan.

39. Nuclear magnetic resonance This tells us the number of hydrogen atoms in different environments within the molecule. As hydrogen is present in (almost) all organic compounds this technique is very useful. The pattern produced by the hydrogen atoms is often split into finer structure, that also gives information about the number of hydrogen adjacent to the absorbing atoms.

40. Nuclear energy levels Nuclei with odd numbers of nucleons (protons and neutrons) can have different energy levels. To help us differentiate these energy levels we say that they have 'spin'. A hydrogen nucleus can have two different 'spin' states. These are designated as spin = +1/2 and spin = - 1/2. For a hydrogen nucleus to change spin states it must absorb energy (promotion) or lose energy (relaxation). Animation

41. Low resolution NMR A low resolution spectrum looks much simpler because it can't distinguish between the individual peaks in the various groups of peaks The numbers against the peaks represent the relative areas under each peak. That information is extremely important in interpreting the spectra

42. Interpreting L R NMR spectrum Different sizes of peak give valuable information Are underneath a peak is proportional to number of hydrogen atoms in that environment. Area underneath peaks can be worked out by integration trace. The vertical heights of the steps in Integration trace are proportional to the number of hydrogen atoms in each envirnoment. Three peaks in the spectrum corresponds to different chemical environments of H atoms

43. Interpreting L R NMR spectrum The Chemical Shift gives information about the environment of protons( Hydrogen atoms). The protons in different chemical environment give different chemical shift Detail about chemical shift will be covered in HL syllabus. Chemical Shift: The Horizontal scale on NMR, is given by the symbol δ has a unit parts per million ppm.

44. NMR Spectrum of Pentan-3-one Symmetrical molecule Two peaks show two different chemical environment Heights of peaks as ratio of 2;3 in integration trace, show four H atom in one environment and 6 in other.

45. Identify number of different chemical environments and ratio of H atoms in each environment 3 2 3

46. Chemical Shift ( HL only) The horizontal scale on a nuclear magnetic resonance spectrum is called chemical shift. The symbol for chemical shift is δ. It is measured as parts per million The Chemical Shift gives information about the environment of protons( Hydrogen atoms). The protons in different chemical environment give different chemical shift Chemical shift are measured relative to TMS Chemical shift for TMS is Zero

47. Tetramethylsilane is the standard All Hs are the same = 1 signal Si Has lower EN than Carbon Si absorbs in a different part of the spectrum than C when bonded to H Si(CH 3 ) 4 Has low boiling point Is chemically inert (non- reactive) Is soluble in most organic solvents

48. Chemical Shift values relative to TMS Values are given in data book let page 26, table 27

49. Using Chemical Shift

50. High-resolution 1 H NMR Can show the difference in the spins of nuclei Right: Hi-res 1 H NMR Below: 1 H NMR

51. What a low resolution NMR spectrum tells you Low Resolution High Resolution Remember: The number of peaks tells you the number of different environments the hydrogen atoms are in. The ratio of the areas under the peaks tells you the ratio of the numbers of hydrogen atoms in each of these environments. The chemical shifts give you important information about the sort of environment the hydrogen atoms are in. In a high resolution spectrum, the low resolution spectrum are split into clusters of peaks. 1 peak a singlet 2 peaks in the cluster a doublet 3 peaks in the cluster a triplet 4 peaks in the cluster a quartet The amount of splitting of the peaks gives you important extra information.

52. Interpreting a high resolution spectrum The n+1 rule The amount of splitting tells you about the number of hydrogens attached to the carbon atom or atoms next door to the one you are currently interested in. The number of sub-peaks in a cluster is one more than the number of hydrogens attached to the next door carbon(s). Singlet next door to carbon with no hydrogens attached doubletnext door to a CH group tripletnext door to a CH2 group quartetnext door to a CH3 group

53. Multiplicity (Splitting) NEIGHBOR hydrogens = number of peaks -1

54. Multiplicity (Splitting) 3 peaks (triplet) 2 neighbor H’s (probably CH 2 ) 6 peaks (hextet) 5 H’s (CH 3 & CH 2 ) 3 peaks (triplet) 2 neighbor H’s (CH 2 )

56. Practice Example 11.6 page 540