No single technique proves a structure. In exam questions you combine evidence from several:• Elemental analysis (percentage composition) gives the empirical formula.
• Mass spectrometry: the molecular ion peak, M+, gives the relative molecular mass, and so the molecular formula. Fragment peaks show parts of the molecule.
• Infrared spectroscopy identifies functional groups: C=O 1630-1820 cm-1, broad O-H of a carboxylic acid 2500-3300 cm-1, O-H of an alcohol 3200-3600 cm-1.
• 13C NMR gives the number of different carbon environments and their types.
• 1H NMR gives the number of proton environments, their types (δ), the ratio of protons (peak areas) and the neighbouring protons (splitting).Worked example: compound X contains C, H and O only.
1. Empirical formula C2H4O (M = 44). M+ peak at m/z = 88, so the molecular formula is C4H8O2.
2. IR: strong peak at 1740 cm-1 (C=O), no broad absorption at 2500-3300 cm-1, so not a carboxylic acid: likely an ester.
3. 1H NMR: singlet at 2.0 ppm (3H), quartet at 4.1 ppm (2H), triplet at 1.2 ppm (3H). The triplet and quartet show an ethyl group; the quartet at 4.1 ppm means the CH2 is bonded to O. The singlet is a CH3 next to C=O.X is ethyl ethanoate, CH3COOCH2CH3. Always check the final structure against every piece of data.
• Mass spectrometry: the molecular ion peak, M+, gives the relative molecular mass, and so the molecular formula. Fragment peaks show parts of the molecule.
• Infrared spectroscopy identifies functional groups: C=O 1630-1820 cm-1, broad O-H of a carboxylic acid 2500-3300 cm-1, O-H of an alcohol 3200-3600 cm-1.
• 13C NMR gives the number of different carbon environments and their types.
• 1H NMR gives the number of proton environments, their types (δ), the ratio of protons (peak areas) and the neighbouring protons (splitting).Worked example: compound X contains C, H and O only.
1. Empirical formula C2H4O (M = 44). M+ peak at m/z = 88, so the molecular formula is C4H8O2.
2. IR: strong peak at 1740 cm-1 (C=O), no broad absorption at 2500-3300 cm-1, so not a carboxylic acid: likely an ester.
3. 1H NMR: singlet at 2.0 ppm (3H), quartet at 4.1 ppm (2H), triplet at 1.2 ppm (3H). The triplet and quartet show an ethyl group; the quartet at 4.1 ppm means the CH2 is bonded to O. The singlet is a CH3 next to C=O.X is ethyl ethanoate, CH3COOCH2CH3. Always check the final structure against every piece of data.