Bringing it all together Elemental analysis C 66.63% H 11.18% O 22.19% 66.63/12 1. Work out empirical formula 11.18/1 22.19/16 C H O =5.5525 =11.18 =1.386875 =4 =8 =1 2. Using molecular ion peak m/z value from mass spectrum calculate Molecular formula molecular ion peak m/z value= 144 Mr empirical formula C4H8O = 72 If Mr molecular formula 144 then compound is C8H16O2 3. Use IR spectra to identify main bonds/functional group C8H16O2 could be an ester, carboxylic acid or combination of alcohol and carbonyl. Look for IR spectra for C=O and O-H bonds 4 3 2 1 5 δ ppm 2 2 3 9 C=O C-H There is a C=O but no O-H absorptions, so must be an ester. 4. Use NMR spectra to give details of carbon chain 4 peaks – only 4 different environments. singlet of area 9 At δ =0.9 Means 3 CH3 groups C CH3 H3C CH3 Peak at δ 4 shows H–C–O Area 2 suggests CH2 Quartet means next to a CH3 O C H H H Peak at δ 2.2 shows H–C=O Area 2 suggests CH2 Singlet means adjacent to C with no hydrogens O C H H C O H Peak at δ 1.2 shows R-CH3 Area 3 means CH3 Triplet means next to a CH2 CH3 Put all together to give final structure
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6.3.2 Spectroscopy
Combined techniques (h) deduction of the structures of organic compounds from different analytical data including: (i) elemental analysis (see also 2.1.3c) (ii) mass spectra (iii) IR spectra. M3.1 Limited to functional groups encountered in this specification. Learners will not be expected to interpret mass spectra of organic halogen compounds. HSW3,5,6 Analysis and interpretation of different analytical data.