Notes · ChemistryUK · A-Levels

Organic analysis

Determining the structure of an unknown organic compound combines chemical tests with instrumental methods. This chapter covers the test-tube reactions that identify functional groups, the use of mass spectrometry to find the relative molecular mass and deduce fragments, and infrared spectroscopy to identify bonds and functional groups - and how to combine these lines of evidence to determine a structure.

4 sections~12 min reading time3 competenciesLevel Standard 3 · Advanced 1

T·131313 / 18

Exam profile

AO1 · Recall the functional-group tests and the basis of mass and infrared spectroscopyAO2 · Deduce Mr and fragments from a mass spectrum and functional groups from an infrared spectrumAO3 · Combine chemical tests and spectra to determine the structure of an unknown compound

Operators:describeidentifydeduceinterpretdetermine

basic level

AS-Level requires the functional-group tests, the molecular-ion peak in mass spectrometry and the identification of bonds by infrared spectroscopy.

higher level

The full A-Level combines these with NMR and chromatography in the later analysis topic to determine complete structures of unknown compounds.

Depth

Text

Contents · 4 sections

Organic analysis

Test-tube reactions for functional groups#

StandardAQA 7405 3.3.6.1DfE GCE Chemistry - identification of functional groups

Functional-group tests

Fig. 1Each test gives a characteristic observation that identifies a functional group.

Simple test-tube reactions identify the functional group present in an organic compound by a characteristic observation. An alkene (C=C) decolourises bromine water from orange to colourless, by electrophilic addition across the double bond - the standard test for unsaturation. A halogenoalkane, when warmed with aqueous silver nitrate (in ethanol), slowly gives a precipitate of the silver halide (white, cream or yellow for chloride, bromide or iodide), because the halide ion is released by hydrolysis.

A carboxylic acid (-COOH) is the only common organic functional group acidic enough to react with a carbonate or hydrogencarbonate, releasing carbon dioxide gas (effervescence) that turns limewater milky. This distinguishes carboxylic acids from phenols and alcohols, which are too weakly acidic to release

\text{CO}_2

from a carbonate. It is a quick, decisive test for the acid group.

Alcohols and aldehydes are distinguished by oxidation and by specific carbonyl tests. Acidified potassium dichromate(VI) is turned from orange to green by primary and secondary alcohols and by aldehydes, but not by tertiary alcohols or ketones. Aldehydes, being readily oxidised, give a positive result with Tollens' reagent (a silver mirror) and with Fehling's solution (a brick-red precipitate), whereas ketones do not - so these two tests distinguish an aldehyde from a ketone.

The order in which tests are done matters when several functional groups may be present, and the observations must be described precisely (the colour before and after, any gas or precipitate). A complete answer names the reagent, states the observation for a positive result, and identifies the functional group - for example, 'add bromine water; if it decolourises from orange to colourless, a C=C double bond (alkene) is present'.

Worked example

Distinguishing an aldehyde from a ketone

Describe a chemical test to distinguish propanal (an aldehyde) from propanone (a ketone).

Choose the reagent
Use Tollens' reagent (ammoniacal silver nitrate) and warm gently.
Aldehyde result
Propanal, an aldehyde, is oxidised and reduces the silver ions to a silver mirror on the tube.
Ketone result
Propanone, a ketone, is not oxidised, so no silver mirror forms.

Result: Propanal gives a silver mirror with Tollens' reagent; propanone does not, distinguishing the aldehyde from the ketone.

Typical mistakes

Saying an alcohol releases CO2 from a carbonate - only the more acidic carboxylic acid does.
Giving a vague observation ('it changes colour') instead of the specific before-and-after colours.

Active revision

Describe chemical tests you would use to show that a compound is a carboxylic acid rather than an alcohol, giving the reagents and observations.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content for the sciences (Department for Education) · AQA A-level Chemistry 7405 specification (AQA)

Mass spectrometry of organic molecules#

StandardAQA 7405 3.3.7.1DfE GCE Chemistry - mass spectrometry

Mass spectrum of propanone

Fig. 2The M+ peak at m/z 58 gives Mr; the base peak at 43 is the loss of CH3 (15), forming the CH3CO+ acylium ion.

In the mass spectrum of an organic molecule, the peak at the highest mass/charge ratio (ignoring small isotope peaks) is the molecular ion peak,

\text{M}^+

, formed when the whole molecule loses one electron. Its

m/z

value gives the relative molecular mass of the compound directly, which is often the single most useful piece of information for identifying an unknown - it constrains the molecular formula.

During electron-impact ionisation the molecular ion often breaks apart into smaller fragments, giving a pattern of peaks at lower

m/z

values (fragmentation). Each fragment peak corresponds to a positively charged piece of the molecule, and the difference in mass between the molecular ion and a fragment tells you what neutral piece was lost. Common losses are

15

(a methyl group,

\text{CH}_3

17

(hydroxyl,

\text{OH}

28

(carbon monoxide,

\text{CO}

) and

29

(an ethyl group

\text{C}_2\text{H}_5

or a

\text{CHO}

group).

The fragment ions themselves are diagnostic. A strong peak at

m/z = 43

often indicates an acylium ion (

\text{CH}_3\text{CO}^+

, from a methyl ketone or ethanoyl group) or a propyl group; a peak at

29

suggests

\text{CHO}^+

(an aldehyde) or

\text{C}_2\text{H}_5^+

; a peak at

15

\text{CH}_3^+

. The most intense peak (the base peak) corresponds to the most stable fragment ion, which is why methyl ketones show a strong peak at

43

The high-resolution mass spectrometer measures

m/z

to several decimal places, which lets you distinguish molecular formulae of the same nominal mass. For example,

\text{C}_3\text{H}_8

and

\text{C}_2\text{H}_4\text{O}

both have a nominal mass of

44

, but their accurate masses (

44.0626

and

44.0262

) differ, so an accurate molecular-ion mass fixes the molecular formula. Interpreting a mass spectrum means reading

\text{M}^+

for the

M_r

, then using the mass losses and fragment ions to work towards the structure.

Worked example

Interpreting a mass spectrum

A compound has a molecular ion at m/z 46 and a peak at m/z 31. Deduce the Mr, the fragment lost and a possible structural feature.

Relative molecular mass
The molecular ion at m/z 46 gives Mr = 46.
Mass lost
46 - 31 = 15, the mass of a methyl group CH3.
Fragment retained
The m/z 31 fragment is CH2OH+ (or CH3O+), suggesting an -OH group; consistent with ethanol (Mr 46) losing CH3 to leave CH2OH+.

Result: Mr = 46; loss of 15 (CH3) leaves a CH2OH+ fragment at 31, consistent with ethanol.

Typical mistakes

Taking a fragment peak, rather than the highest-mass molecular-ion peak, as Mr.
Confusing the mass lost (a neutral fragment) with the mass of the observed ion.

Active revision

The mass spectrum of a ketone has its molecular ion at $m/z = 72$ and a strong peak at $m/z = 43$ . Deduce the mass lost and identify a likely structural feature.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

Infrared spectroscopy and the greenhouse effect#

StandardAQA 7405 3.3.8.1DfE GCE Chemistry - infrared spectroscopy

Characteristic infrared absorptions

Fig. 3Each bond absorbs at a characteristic wavenumber; the axis runs from high to low wavenumber, as in a real IR spectrum.

Infrared (IR) spectroscopy identifies the bonds, and hence the functional groups, in a molecule. Bonds vibrate (stretch and bend) at frequencies that lie in the infrared region, and each type of bond absorbs infrared radiation of a characteristic frequency, measured as a wavenumber in

\text{cm}^{-1}

. An IR spectrum is a plot of the percentage of radiation transmitted against wavenumber; where a bond absorbs, less is transmitted, so the spectrum shows a dip (an absorption 'peak' pointing downwards). The wavenumber axis is drawn decreasing from left to right by convention.

Characteristic absorptions let you identify functional groups from a data sheet. A broad absorption around

3200

3550\ \text{cm}^{-1}

indicates the O-H of an alcohol; a very broad one around

2500

3300\ \text{cm}^{-1}

the O-H of a carboxylic acid; a strong absorption around

1680

1750\ \text{cm}^{-1}

the C=O of a carbonyl (aldehyde, ketone, acid or ester); and absorptions around

3300\ \text{cm}^{-1}

(sharp) the N-H of an amine. Matching the observed absorptions to the data sheet identifies which groups are present.

The region below about

1500\ \text{cm}^{-1}

is called the fingerprint region. It contains a complex pattern of absorptions unique to each molecule, like a fingerprint, and is used to confirm the identity of a compound by matching its spectrum against a database of known spectra, rather than to identify individual bonds. Confirming an identity by the fingerprint region is a routine analytical use.

Infrared absorption also underlies the greenhouse effect. Greenhouse gases such as carbon dioxide, methane and water vapour absorb infrared radiation emitted by the Earth's surface, because the vibrations of their bonds are in the infrared, and then re-emit it, warming the atmosphere. This is a direct application of the same bond-vibration idea: bonds that absorb infrared (particularly C=O and O-H bonds) make a gas an effective absorber of the Earth's outgoing radiation, linking molecular spectroscopy to climate.

Worked example

Identifying a compound class from IR

An IR spectrum shows a strong absorption at about 1740 cm^-1 and no broad O-H absorption. What functional group is present, and what is ruled out?

Interpret the 1740 absorption
A strong absorption near 1700-1750 cm^-1 indicates a C=O (carbonyl) bond.
Interpret the absence of O-H
The absence of a broad O-H absorption rules out an alcohol and a carboxylic acid.
Conclude
A carbonyl with no O-H suggests an aldehyde or a ketone (or an ester).

Result: A C=O group is present (1740 cm^-1) with no O-H, so the compound is likely an aldehyde, ketone or ester rather than an alcohol or acid.

Typical mistakes

Confusing the broad carboxylic-acid O-H (2500-3300) with the narrower alcohol O-H (3200-3550).
Forgetting that the IR axis is wavenumber, decreasing from left to right, and that absorptions point downwards.

Active revision

An IR spectrum shows a strong absorption at $1710\ \text{cm}^{-1}$ and a broad one from $2500$ to $3300\ \text{cm}^{-1}$ . Identify the two functional-group features and the likely class of compound.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

Combining evidence to determine a structure#

AdvancedAQA 7405 3.3.6.1AQA 7405 3.3.7.1AQA 7405 3.3.8.1DfE GCE Chemistry - structure determination

A decision key for functional groups

Fig. 4Working through the tests in turn identifies the functional group, which is then combined with Mr and formula.

Determining an unknown structure is a matter of combining independent lines of evidence, each of which narrows the possibilities. The molecular formula (from combustion analysis or high-resolution mass spectrometry) tells you what atoms are present and lets you calculate the degree of unsaturation. The mass spectrum gives the relative molecular mass and, through its fragments, clues to how the molecule is built. Chemical tests and the IR spectrum identify the functional groups.

A systematic approach works best. Start from the molecular formula and

M_r

, then use the IR spectrum to identify functional groups (a C=O, an O-H, and so on), then use chemical tests to confirm or refine them (is the C=O an aldehyde or a ketone?), and finally use the mass-spectrum fragments to work out how the pieces are joined. Each result should be consistent with the others; a good answer explains what each piece of evidence shows.

A decision key or flow of tests is a useful way to organise the reasoning. For instance: does the compound release carbon dioxide from a carbonate (carboxylic acid)? Does it decolourise bromine water (alkene)? Does it turn acidified dichromate green (a primary/secondary alcohol or an aldehyde)? Does it give a silver mirror (aldehyde)? Following the branches in turn identifies the functional group, and combining that with the

M_r

and formula pins down the structure.

The skill assessed is bringing several results together into a single, consistent conclusion, and justifying it. When you determine a structure, quote each piece of evidence and what it shows - for example, 'the IR shows a C=O at

1715\ \text{cm}^{-1}

; the compound gives a silver mirror, so it is an aldehyde;

M_r = 44

from the mass spectrum, so it is ethanal,

\text{CH}_3\text{CHO}

'. Every deduction should be supported by the data, never asserted.

Worked example

Determining a structure from combined evidence

A compound has Mr = 60 (mass spectrum), gives effervescence with sodium carbonate, and its IR shows a broad absorption 2500-3300 cm^-1 and a strong peak at 1715 cm^-1. Deduce its structure.

Chemical test
Effervescence with a carbonate shows a carboxylic acid (-COOH) group.
IR evidence
The broad 2500-3300 absorption is the acid O-H and the 1715 peak the C=O - consistent with a carboxylic acid.
Fit to Mr
A carboxylic acid of Mr 60 is ethanoic acid, CH3COOH (Mr = 24 + 3 + 32 + 1 = 60).

Result: The evidence (carbonate test, IR, Mr 60) all point to ethanoic acid, CH3COOH.

Typical mistakes

Asserting a structure without linking it to the evidence, or ignoring evidence that contradicts a guess.
Using only one technique when the question provides several lines of evidence to be combined.

Active revision

A compound of $M_r = 60$ gives effervescence with sodium carbonate and shows a broad IR absorption from $2500$ to $3300\ \text{cm}^{-1}$ and a peak at $1715\ \text{cm}^{-1}$ . Deduce its structure.

Active recall

Recall the key points — then reveal.

Sources: AQA A-level Chemistry 7405 specification (AQA)

References & sources

Sources

Department for Education

GCE AS and A level subject content for the sciences

AQA

AQA A-level Chemistry 7405 specification

Organic analysis

4 sections~12 min reading time3 competenciesLevel Standard 3 · Advanced 1

Organic analysis

Key points

Distinguishing an aldehyde from a ketone

Key points

Interpreting a mass spectrum

Key points

Identifying a compound class from IR

Key points

Determining a structure from combined evidence

Sources

Organic analysis

Key points

Distinguishing an aldehyde from a ketone

Key points

Interpreting a mass spectrum

Key points

Identifying a compound class from IR

Key points

Determining a structure from combined evidence

Sources