Identifying Organic Compounds Using Data

No single analytical technique gives you the complete structure of an unknown organic compound. Mass spectrometry tells you about mass, infrared (IR) spectroscopy tells you about bonds, NMR spectroscopy tells you about the environments of atoms, and elemental analysis tells you about the ratio of elements present. In A Level Chemistry, questions on identifying organic compounds using data expect you to read several pieces of evidence together and narrow down the possible structures until only one fits every piece of data.

This topic pulls together ideas you may already have met when studying specific functional groups, such as aldehydes and ketones and carboxylic acids, acyl chlorides and esters. Here the focus is on the strategy for combining that knowledge with real spectroscopic and analytical data.

Combustion analysis gives the percentage by mass of carbon, hydrogen and (by difference) oxygen or other elements in a compound. To find the empirical formula, convert each percentage to moles by dividing by the relative atomic mass, then divide through by the smallest number of moles to get a whole-number ratio.

For example, a compound containing 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass gives moles of \( \dfrac{40.0}{12} = 3.33 \), \( \dfrac{6.7}{1} = 6.7 \) and \( \dfrac{53.3}{16} = 3.33 \). Dividing by the smallest value, 3.33, gives a ratio of C : H : O = 1 : 2 : 1, so the empirical formula is \( CH_2O \).

The empirical formula only gives a ratio. To find the molecular formula you need the relative molecular mass, which is where mass spectrometry comes in.

The molecular ion peak, \( M^{+} \), appears at the highest mass-to-charge ratio in the spectrum (ignoring any small isotope peaks) and its \( m/z \) value equals the relative molecular mass of the compound. Comparing this value with the mass of the empirical formula tells you how many empirical formula units make up one molecule, using \( n = \dfrac{M_r}{M_{ef}} \) written without units inside the ratio, so for the example above with \( M_r = 60 \) and an empirical formula mass of 30, \( n = 2 \), giving a molecular formula of \( C_2H_4O_2 \).

Fragment peaks below the molecular ion are just as useful. Each fragment forms when a bond breaks and a piece of the molecule is lost as a neutral fragment, leaving a smaller positive ion. Common mass losses include 15 for loss of \( CH_3 \), 17 for loss of \( OH \), 29 for loss of \( CHO \) or \( C_2H_5 \), and 45 for loss of \( COOH \).

An infrared spectrum shows which bonds are present by the wavenumber at which they absorb radiation. You are not expected to memorise every value precisely, but you should recognise the key ranges and match them to functional groups.

Bond	Wavenumber range (cm-1)	Typical functional group
O-H (carboxylic acid)	2500 to 3300, broad	Carboxylic acids
O-H (alcohol)	3230 to 3550, broad	Alcohols, phenols
N-H	3300 to 3500	Amines, amides
C=O	1680 to 1750	Aldehydes, ketones, acids, esters, amides
C-H	2850 to 3100	Almost all organic compounds

A broad O-H absorption around 2500 to 3300 cm-1 together with a sharp C=O absorption near 1710 cm-1 is a strong indicator of a carboxylic acid, distinguishing it from an alcohol or an aldehyde. The same C=O absorption without the broad O-H peak points instead towards an aldehyde or ketone, covered in more depth in aldehydes and ketones: properties and reactions.

\( ^{13}C \) NMR shows the number of chemically distinct carbon environments in a molecule, with each peak's chemical shift, measured in ppm, indicating the type of carbon (for example, a shift around 170 to 220 ppm suggests a carbonyl carbon).

\( ^{1}H \) NMR gives extra detail: the number of peaks shows how many distinct hydrogen environments exist, the chemical shift of each peak suggests what that hydrogen is attached to, the integration (relative peak area) gives the ratio of hydrogens in each environment, and the splitting pattern (the n + 1 rule) shows how many hydrogens are on neighbouring carbons. A peak split into a triplet, for instance, has two hydrogens on an adjacent carbon, while a quartet indicates three neighbouring hydrogens.

Exchangeable protons, such as the O-H of an alcohol or acid and the N-H of an amine, often appear as broad singlets and can shift or disappear when the sample is shaken with \( D_2O \), which helps you tell them apart from other hydrogens. This is particularly useful when distinguishing amine or amide protons, discussed further in amines, amides and amino acids.

An unknown compound gives the following data: combustion analysis shows 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass; the mass spectrum shows \( M^{+} \) at \( m/z = 60 \) with fragment peaks at 45 and 43; the infrared spectrum shows a broad absorption between 2500 and 3300 cm-1 and a sharp peak at 1710 cm-1; the \( ^{1}H \) NMR spectrum shows two peaks, a singlet at about 11.5 ppm (relative area 1) and a singlet at about 2.1 ppm (relative area 3).

Working through the data: the combustion analysis gives an empirical formula of \( CH_2O \) (mass 30). Since \( M_r = 60 \), the molecular formula is \( C_2H_4O_2 \). The IR spectrum's broad O-H absorption plus a C=O near 1710 cm-1 points strongly to a carboxylic acid. The loss of 17 (60 to 43) matches loss of \( OH \), and the loss of 15 (60 to 45) matches loss of \( CH_3 \), consistent with a \( CH_3COOH \) fragmentation pattern. Finally, the \( ^{1}H \) NMR shows only two environments in a 1 : 3 ratio, exactly as expected for the single acidic proton and the three equivalent methyl protons of ethanoic acid, \( CH_3COOH \). Every data set agrees on the same structure, which is how you confirm an identification with confidence rather than relying on one technique alone.

Work through the evidence in a logical order rather than trying to guess the structure straight away. Use elemental analysis or the molecular ion to fix the molecular formula first, use IR data to identify likely functional groups, use \( ^{13}C \) and \( ^{1}H \) NMR to confirm the number and type of atomic environments and rule out isomers, and use mass spectrometry fragment peaks as a final check that the proposed structure would break apart in a way that matches the spectrum. If a structure fits the molecular formula but does not explain every peak or absorption given, it is not yet the correct answer, so keep testing alternative isomers, including any possible optical isomers, against the full data set.

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