1. WM4 Instrumental analysis

2. The 3 key instrumental techniques How do we know that salicylic acid contains – OH and –COOH groups? Mass spectroscopy (m.s.). Infrared (i.r.) spectroscopy. Nuclear magnetic resonance (n.m.r.) spectroscopy.

3. Making use of infrared spectroscopy Any unidentified, new substance has its i.r. spectrum recorded.

4. 6.4 Infrared spectroscopy Energy in molecules is quantised. In i.r., molecules are exposed to radiation between 10 14 Hz – 10 13 Hz (wavelengths 2.5µm -15 µm). Remember: c = λ v Bonds vibrate and stretch (pull apart then push together again) as they absorb energy.

5. Vibrational changes of CO 2 Asymmetric stretch Symmetric stretch bending

6. Infrared spectra: signals = stretches An IR spectrum hangs down from a baseline (100% transmittance = no absorbtion). The signals (look like ‘icicles’) on an IR spectra correspond to bonds absorbing a packet of energy and vibrating more. The –OH and –CO bonds in salicylic acid absorb energy at specific wavelengths (λ)/µm and so wavenumbers (1/ λ)/ cm -1.

7. Generally, particular bonds can be matched to a particular absorption region. BondLocationWavenumber/cm -1 Intensity C-HAlkanes2850 -2950M-S C=CAlkenes1620 – 1680M ArenesSev peaks 1450-1650Variable C Alkynes2100 – 2260M C=OAldehydes1720 – 1740S C-OAlcohols, ethers phenols 1050 - 1300S C-Ffluoroalkanes1000 – 1400S O-HAlcohols3600 – 3640S N-H1 o amines3300 - 3500M-S

8. Interpreting spectra: 2-ethylbut-1-ene ?

9. Interpreting spectra: propanone

10. Regions in the IR spectrum where typical absorptions occur Absorption range/cm -1 Bonds responsibleExamples 4000 – 2500Single bonds to hydrogenO-H, C-H, N-H, 2500 – 2000Triple bondsC C, N N 2000 – 1500Double bondsC=C, C=O Below 1500 (fingerprint region) Various (not used to ID functional groups) C-O, C-X (halogen) Label these regions on the previous two examples; sketch the structures and link them to the main signals. Do this for the following examples, too.

11. Examples of infrared spectra Butane Strong absorption at 2970 cm -1 characteristic of C-H stretching in aliphatic compounds. No indication of any functional groups.

12. Examples of infrared spectra Methylbenzene 2 types of C-H absorption ~3000 cm-1 (above = C- H on benzene; below = C-H on methyl group). No indication of any functional groups. Absorption pattern ~700 cm -1 is typical of a benzene ring with a substituted group.

13. Examples of infrared spectra Benzoic acid A sharp absorption at 3580 cm -1 is due to O-H bond. A strong absorption at 1760 cm -1 is the C=O group. Position of C-H absorption suggests an aromatic compound.

14. Summary of IR spectroscopy An IR spectrum measures the extent to which electromagnetic radiation is transmitted through a sample of substance. Frequency ranges absorbed give clues about functional groups which are present. IR spectrum of salicylic acid gives evidence of C=O and –OH groups.

15. Evidence from nuclear magnetic resonance (n.m.r.) spectroscopy. This technique helps to determine structure, as it investigates the different environments in which (hydrogen) nuclei are situated. The n.m.r. spectrum for salicylic acid shows signals for the different environments of the 6 protons: One proton in a –COOH environment. One proton in a phenolic –OH environment. Four protons attached to a benzene ring.

16. n.m.r. spectrum for salicylic acid 12 11 Absorption

17. The evidence so far…. A combination of i.r. and n.m.r. spectroscopy shows that salicylic acid has an –OH group and –COOH group both attached to a benzene ring; we can now rename it HYDROXYBENZOIC ACID. However, it could be one of 3 possible isomers: 2-hydroxybenzoic acid, 3- hydroxybenzoic acid and 4-hydroxybenzoic acid. Mass spectroscopy can determine which isomer we have.

18. The mass spectrum of salicylic acid Signals correspond to positively charged ions formed from the parent compound, and fragment ions. Parent or molecular ion The fragmentation pattern is characteristic of a particular compound…the fragment at 120 can only come from 2- hydroxybenzoic acid…can you see why? mass

19. Fragmentation Positive ions in a mass spectrometer can break down into ‘building blocks’. Example: 2-ethoxybutane. CH 3 -CH 2 -CH-O-CH 2 CH 3 CH 3 (M=102) + CH 3 -CH 2 -CH-O-CH 2 CH 3 (M=87) ? + HO-CH-CH 3 (M=45) Loss and rearrangement of CH 3 CH=CHCH 3 + CH 3 -CH 2 -CH (M=87) CH 3 ? Loss of CH 3 -CH 2 ? ? ? + Loss of CH 3 CH 3 -CH 2 (M=29) +

20. Positively charged fragments form. Mass difference suggests functional groups. For each fragmentation, one product has a positive charge: M + A + + B A + B + The most stable ion usually forms. Mass Difference Group that is suggested 15CH 3 17OH 28C=O or C 2 H 4 29C2H5C2H5 43COCH 3 45COOH 77C6H5C6H5

21. Isotope peaks : heights are in the same ratio of abundance for particular elements. Pairs of peaks correspond to isotopes of 35 Cl and 37 Cl in the ratio of 75%:25% ie. 3:1. Highlight these. mass

22. Now it’s over to you! Do activity WM4: use accurate M r values, isotope peaks and a database to lead you to the formula of salicylic acid. It shows you how chemists use fragmentation patterns to deduce or confirm a molecular structure. Do assignments 1 and 2 C.S. p110-111 Do ‘Problems for 6.5’ on mass spectrometry, C.I., p145-146.