No single technique proves the structure of an unknown compound. Combining evidence from different techniques narrows the options until only one structure fits.What each technique tells you
• Elemental analysis (percentage composition) gives the empirical formula.
• The mass spectrum gives Mr from the molecular ion peak, so the molecular formula; fragment peaks show parts of the structure.
• The IR spectrum shows which functional groups are present, such as C=O, O-H and C-O.Strategy
1. Find the molecular formula (empirical formula and M+).
2. Use IR to identify the functional groups.
3. Draw all structures that fit.
4. Use fragment peaks to decide between them.Worked example: compound Y contains only C, H and O and has one C=O group. Its molecular ion peak is at m/z 58.
• C3H6O has Mr = 36 + 6 + 16 = 58, so the molecular formula is C3H6O.
• IR: a strong trough at 1715 cm−1 (C=O) and no broad O-H trough, so Y is an aldehyde or a ketone.
• Possible structures: propanal, CH3CH2CHO, or propanone, CH3COCH3.
• The mass spectrum has a large fragment peak at m/z 43 (CH3CO+) and a peak at 15 (CH3+), with no large peak at 29. Propanal would give a large peak at m/z 29 from C2H5+.
So Y is propanone.
• Elemental analysis (percentage composition) gives the empirical formula.
• The mass spectrum gives Mr from the molecular ion peak, so the molecular formula; fragment peaks show parts of the structure.
• The IR spectrum shows which functional groups are present, such as C=O, O-H and C-O.Strategy
1. Find the molecular formula (empirical formula and M+).
2. Use IR to identify the functional groups.
3. Draw all structures that fit.
4. Use fragment peaks to decide between them.Worked example: compound Y contains only C, H and O and has one C=O group. Its molecular ion peak is at m/z 58.
• C3H6O has Mr = 36 + 6 + 16 = 58, so the molecular formula is C3H6O.
• IR: a strong trough at 1715 cm−1 (C=O) and no broad O-H trough, so Y is an aldehyde or a ketone.
• Possible structures: propanal, CH3CH2CHO, or propanone, CH3COCH3.
• The mass spectrum has a large fragment peak at m/z 43 (CH3CO+) and a peak at 15 (CH3+), with no large peak at 29. Propanal would give a large peak at m/z 29 from C2H5+.
So Y is propanone.