Determining structures from analytical dataEdexcel International A Level Chemistry: Revision notes

The empirical formula is the simplest whole-number ratio of atoms. For percentages, divide each by Ar, then divide by the smallest. For combustion analysis, convert CO₂ to carbon (n(C) = n(CO₂)) and H₂O to hydrogen (n(H) = 2 × n(H₂O)), and find oxygen by difference.

Worked example: 54.5% C, 9.1% H, 36.4% O. Moles: C 4.54, H 9.1, O 2.275. Divide by 2.275 to get 2 : 4 : 1, so C₂H₄O (Mr 44). A molecular ion at m/z 88 is twice this, so the molecular formula is C₄H₈O₂.

Characteristic reactions identify groups:

• Alkene: bromine water is decolourised.
• Aldehyde or ketone (C=O): 2,4-DNPH gives an orange precipitate; only aldehydes give a silver mirror with Tollens' reagent.
• Primary/secondary alcohol or aldehyde: acidified dichromate(VI) turns orange to green; tertiary alcohols give no change.
• Carboxylic acid: effervesces with sodium carbonate.

Bonds absorb infrared radiation at characteristic wavenumbers (cm⁻¹). Typical values (check the data booklet):

• C=O: 1680–1750 (strong)
• O–H in alcohols: 3200–3550 (broad)
• O–H in carboxylic acids: 2500–3300 (very broad)
• N–H: 3300–3500
• C–H: 2850–3100

The presence or absence of a peak is evidence. An absorption near 1700 with no O–H points to an aldehyde, ketone or ester.

The molecular ion peak (M⁺) gives the relative molecular mass. Fragmentation gives peaks for positive fragments, e.g. CH₃⁺ 15, C₂H₅⁺ 29, CH₃CO⁺ 43, C₃H₇⁺ 43. The tallest peak is the base peak. A compound with one bromine atom shows M and M+2 peaks of equal height (⁷⁹Br and ⁸¹Br); one chlorine atom gives M and M+2 in a ratio of about 3 : 1. Subtracting fragment masses shows what was lost, e.g. M – 15 shows loss of CH₃.

Each different carbon environment gives one peak, so the number of peaks shows how many types of carbon there are; symmetry gives fewer peaks than carbons. Typical shifts (ppm): alkyl C 5–40, C–O 50–90, C=C 110–150, C=O 160–220. For example, propanone has two peaks (the two CH₃ carbons are equivalent), while 1-bromopropane has three.

In proton NMR the chemical shift shows the environment, the integration (relative area) shows the number of hydrogens, and the splitting follows the n + 1 rule: n hydrogens on adjacent carbons give n + 1 lines (singlet, doublet, triplet, quartet, septet). Hydrogens on the same carbon or on equivalent carbons do not split each other. Typical δ: CH₃–C 0.9, CH₃–C=O 2.1, O–CH₃ 3.7, R–COOH 10–12.

To solve a structure: find the formula; use IR and chemical tests for the functional group; use mass spectrum fragments; then NMR to assemble the carbon skeleton. Check that every piece of data fits.