3.2.12 (HL)—Combined data analysis
- Syllabus
- First assessment 2025
- Objective
- 3.2.12
- Level
- HL
Combining Organic Evidence
HL only| Evidence | Constraint or question |
|---|---|
| Molecular formula | atom totals and degree of unsaturation/rings-plus-π-bonds |
| Molecular ion | relative molecular mass consistent with the formula |
| MS fragments | which candidate substructures can produce the supplied m/z ions? |
| IR | which functional groups are required or excluded? |
| ¹H NMR | do signal count, shift, integration and splitting all fit? |
| Final candidate | does one connectivity satisfy every constraint, and what ambiguity remains? |
Write the specific peak, range, ratio or fragment beside each inference; “the spectrum looks like it” is not evidence.
Use each technique as an independent constraint, then reject any candidate that contradicts one of the supplied data sets.
Use an elimination workflow: molecular mass and formula limit atom totals, IR requires or excludes functional groups, and NMR fixes hydrogen environments and neighbours. Write each constraint beside a candidate and reject it immediately when one spectrum conflicts; agreement with a single striking peak is not enough.
Determining Structure from Combined Data
HL onlyAssessment in practice
Representative question
Question 1
[Maximum number: 11]
The mass spectrum, infrared spectrum and details of the 1HNMR spectrum of compound X are given below.
Mass spectrum:
Infrared spectrum:
1 HNMR spectrum:
| Peak with splitting | Integration trace (area under peak) |
|---|---|
| Singlet | 1 |
| Singlet | 6 |
| Triplet | 3 |
| Quartet | 2 |
Analyse these three spectra and, using relevant information, deduce the identity of the compound.
Mass spectrum:
Infrared spectrum:
1 HNMR spectrum:
Identity of X :
Mass spectrum:
molecular ion peak at 88/M+=88 shows molecular formula is C5H12O;
absorption at 73 due to (M−CH3)+/X contains a methyl group as peak at M-15 / OWTTE;
absorption at 59 due to (M−C2H5)+/X contains an ethyl group as peak at M-29;
Penalise once only if + charge omitted.
Marking guidance:
Accept that X contains a CHO group due to M-29 but in fact it cannot as there are too many hydrogen atoms in the compound for it to be an aldehyde.
Infrared spectrum:
peak in range at 3200−3600 cm−1 shows it contains an OH group / OWTTE;
(sharp) peaks just below 3000 cm−1/in range 2850−3100 cm−1 due to C-H absorptions;
lack of peak at approximately 1700 cm−1 shows it does not contain C=O;
absorption between 1050 and 1410 cm−1 due to C-O;
Allow "due to alcohol" instead of due to C-O.
Accept "absorption between 1050 and 1410 cm−1 due to ether or ester" although
it cannot be either as there is only one O atom and it has been identified as bonded to H.
fingerprint region specific to compound but needs to be compared with library / OWTTE;
1 HNMR spectrum:
(12 protons are in) four different chemical environments (in the ratio 1:2:6:3);
singlet (with integration trace of 1) due to OH proton;
singlet (with integration trace of 6) suggests (two CH3 ) groups attached to a carbon atom with no Hs attached to it;
quartet (with integration trace of 2) due to CH2 next to CH3;
triplet (with integration trace of 3) due to CH3 next to CH2;
Reference must be made to the association of the splitting pattern (singlet, triplet etc.) to the specific carbon fragments.
( X is) 2-methylbutan-2-ol/ CH3CH2C(CH3)2OH;
No ECF throughout 2(b).
Organic Analysis Summary
Retrieve the route: translate formulae, identify functional groups and series, name and classify isomers, then use mass, IR, and NMR evidence together to determine structure.
Check connectivity, functional-group evidence, formula/mass constraint, shifts and integration, splitting neighbours, and agreement across every technique.