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Deducing Structures from Combined Data

No single technique identifies an unknown compound on its own. Combine the evidence step by step:
1. Molecular formula: from percentage composition (empirical formula) and the molecular ion peak, M+, in the mass spectrum, which gives the relative molecular mass. Fragment peaks give clues to parts of the structure.
2. Infrared: functional groups, e.g. C=O at about 1700 to 1750 cm-1, broad O-H in alcohols around 3200 to 3550 cm-1, very broad O-H in carboxylic acids 2500 to 3300 cm-1.
3. 13C NMR: number of carbon environments.
4. 1H NMR: H environments, chemical shift, integration ratio and splitting.
5. Propose a structure and check it fits every piece of data.Worked example: compound X, C4H8O2, has M+ at m/z = 88. IR: strong absorption at 1740 cm-1, no broad O-H. 13C NMR: 4 peaks. 1H NMR: δ 4.1 (2H, quartet), δ 2.0 (3H, singlet), δ 1.3 (3H, triplet).
• C=O but no O-H: an ester
• quartet (2H) + triplet (3H): an ethyl group; δ 4.1 means this CH2 is bonded to an oxygen atom
• singlet (3H) at δ 2.0: a CH3 next to C=O with no H on the adjacent carbon
X is ethyl ethanoate, CH3COOCH2CH3, which has 4 carbon environments.Its isomer methyl propanoate, CH3CH2COOCH3, is ruled out: its CH3 singlet is bonded to O (about δ 3.7) and its quartet is next to C=O (about δ 2.3).

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Specification: Edexcel Chemistry (9CH0), A-level, Topic 19A: Mass Spectrometry and Topic 19B: Nuclear magnetic resonance (NMR). Topic: Modern Analytical Techniques.

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