5.6.1Unit 5. Transition Metals and Organic Nitrogen Chemistry

Deducing Organic Structures — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions

5 min read 5 parts Lesson 38 of 43 Unit quiz

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Section 1 of 5A Systematic Strategy

Start from the molecular formula, which comes from the empirical formula and the molecular ion peak at the highest m/z in the mass spectrum.

Calculate the degree of unsaturation or compare with the saturated formula CₙH₂ₙ₊₂ to see whether rings or double bonds are present.

Use infrared absorptions to list the functional groups present, such as O–H, C=O or N–H.

Use chemical tests to confirm or separate possibilities, for example distinguishing an aldehyde from a ketone.

Use NMR chemical shifts, integration and splitting to place groups and assemble the carbon skeleton.

Finally, check that every piece of data, including fragment peaks, fits your proposed structure.

Each technique answers a different question, so use them together.

Mass spectrum

The molecular ion gives the relative molecular mass, and fragment peaks differ by lost groups such as 15 for CH₃ or 29 for C₂H₅.

Infrared

A strong absorption near 1700 cm⁻¹ shows C=O, and a broad band between 2500 and 3300 cm⁻¹ points to a carboxylic acid O–H.

¹³C NMR

The number of peaks equals the number of distinct carbon environments, which reveals symmetry in the molecule.

¹H NMR

Peak area gives the ratio of hydrogens, and the n + 1 rule uses splitting to show how many neighbouring hydrogens are present.

Worked Reasoning Pattern

Identify the class

Decide whether the compound is an alcohol, carbonyl, acid, ester, amine or arene using IR bands and test results.

Count environments

Match the number of ¹³C peaks to the carbon framework, remembering that identical environments give a single peak.

Use splitting

A triplet next to a quartet strongly suggests an ethyl group, CH₃CH₂–, in the structure.

Check fragments

A peak at m/z 43 may suggest CH₃CO⁺ or C₃H₇⁺, so confirm with the other data before deciding.

Question

A compound has formula C₃H₆O, shows a strong IR band at 1710 cm⁻¹, gives an orange precipitate with 2,4-DNPH and no silver mirror.
Identify it.

C₃H₆O has one degree of unsaturation compared with C₃H₈O, so there is one C=O or C=C or ring.
The IR band at 1710 cm⁻¹ shows a C=O group, and the 2,4-DNPH result confirms an aldehyde or ketone.
No silver mirror rules out an aldehyde, so the compound is a ketone.
The only ketone with three carbons is propanone, CH₃COCH₃, which has just one ¹H NMR peak and two ¹³C peaks.

Answer Propanone, CH₃COCH₃

Molecular ion The ion formed by loss of one electron from the whole molecule, M⁺.
Chemical shift The position of an NMR peak relative to tetramethylsilane, in ppm.
Integration The relative area of an NMR peak, proportional to the number of hydrogens.
Splitting The division of an NMR peak into a pattern caused by hydrogens on adjacent carbons.

Using only one technique and ignoring contradictions with the other data.
Counting the number of peaks as the number of carbons rather than carbon environments.
Forgetting that OH and NH peaks in ¹H NMR may be broad and do not always split.
Assigning the molecular ion to the tallest peak instead of the peak with highest m/z.

Last updated: 28 October 2026