1. Demonstrate understanding of spectroscopic data in chemistry Chemistry A.S. 3.2 91388 3 internal credits

2. Spectroscopy Spectroscopy is the study of how EMR interacts with matter. High wavelength = low energy We will look at 3 spectroscopic techniques: IR = infrared spectroscopy NMR = nuclear magnetic resonance spectroscopy Mass spectroscopy

3. Infrared spectroscopy When IR radiation interacts with a molecule covalent bonds will bend, stretch and twist (vibrate) If the vibrations result in a change in bond dipole then the molecule will absorb energy at that particular wavelength. A plot of energy absorbed Vs wavenumber* (cm -1 ) will show absorption spectra characteristic of certain functional groups. *Wavenumber = 1/λ

4. Interpreting IR Spectrum Each trough represents the narrow range of frequencies where each functional group vibrates and absorbs energy. Sharp trough around 1700cm -1 indicates C=O C-H bond causes a trough around 3000cm -1 The presence of O-H or N-H bonds cause a broad trough around 3500cm -1 Why do you think this value is higher than that of C-H?

5. Identify the sample Use your knowledge to identify this 3 carbon compound. It has a molecular mass of 60. When justifying your choice explain how you used the info to select your molecule and eliminate others. It has a distinct absorption at 2900cm -1 indicating C-H bonds and also an absorption at 3400cm -1 indicating O-H or N-H. There is no absorption at 1700cm -1 therefore there are no C=O groups. Without a C=O group the compound can’t be an amide, aldehyde, ketone or carboxylic acid therefore it must be an amine or alcohol. Propan-1-ol has a mass of 60 propan-1-amine doesn’t.

6. 13 C NMR Spectroscopy The chemical shift is affected by: – The electronegativity of elements e.g. F will draw electrons away from C – The type of bonding e.g. double bonds draw electrons away from C – chemical environment e.g. the shielding of each C atom is dependent on the total of all the shielding effects acting on it. Powerful magnets are used to influence the spin of nuclei Electrons around an atom will shield the nucleus from the magnetic field Chemical shift ( δ ) is a measure, in parts per million (ppm), of how much the nucleus is affected. i.e. greater ppm = less shielded

7. Chemical environment C bonded to the same groups will have the same shielding and are considered to be the same environment The number of chemical environments translates to the number of peaks in a NMR spectrum and therefore the minimum number of carbons in the molecule How many peaks would show in pentane? Draw it and justify your answer. C1 and C5 are bonded to 3Hs and a butyl group so they represent 1 environment C2 and C4 are bonded to a methyl and propyl group so they a second environment C3 is bonded to 2 ethyl groups so it is in a third environment

8. Interpreting NMR spectrum The number of peaks = number of environments C=O (aldehyde, ketone) appears at 180-220ppm C=O (carboxylic acid, ester, acyl chloride, amide) appears at 160-185ppm C=C appears at 100-150ppm (if only 1 line appears the molecule is symmetrical about the C=C bond) C-O bond in Esters 60-80 C- high EN element or C-C=O appears at 30-60ppm CH 3 appears at 0 - 30 ppm Ignore any signal at 75ppm this is the solvent

9. Identifying a sample Draw the structures of the following isomers and then identify which produced this spectrum. Justify your answer. a) 2,2 dimethylpropanal b) Pentanal c) 3-methylbutan -2-one d) 1-penten -3 -ol a has 3 environments and c has 4 so the spectrum is b or d. The peak at 220ppm indicates a carbonyl group so it must be b.

10. Mass spectrometry The sample is vaporised and ionised so that the mass can be determined. These conditions often cause the sample to fragment, giving a series of peaks. The mass of the molecular ion [M + ] is always furthest right The base peak is always the highest peak. If the molecular ion has an odd numbered mass it must have an odd number of N atoms in the sample = Nitrogen rule If a sample has a molecular ion of 31 does it contain Nitrogen? The sample is an amine what is it called? Note: this is the isotope peak don’t confuse it with the molecular ion

11. Halogen isotope analysis The isotopes of Cl,Br and I often show up on analysis and can be identified because… Cl has isotopes 35 and 37 in a 3:1 ratio so if you see a molecular ion with a second peak [M+2] 1/3 the size 2 m/z higher it is due to the presence of Cl Br 79 and 81 appear in a 1:1 ratio I has a large mass of 127 so it often appears separated from other fragments. If this happens the base peak will be 127 less than the molecular ion. I also often partners with H + to form a HI peak at 128.

12. Fragmentation analysis When energy is supplied to the sample it can break into fragments which give an indication of functional groups. If there is a peak 1, 15, 17 or 18 less than the molecular ion it indicates… Fragment massFragment lostIndicates the presence of 1HydrogenCOOH, CHO or NH 15MethylCH 3 17HydroxylCOOH or OH 18waterOH usually alcohol

13. Comparing data Use the different spectroscopic data to identify the unknown sample. Mass spectroscopy will give you info on the mass and the presence of N, Cl, Br, I NMR will give you info about the structure and the presence of carbonyl or C=C IR will give you info about the functional groups C=O, OH/NH