Test-tube reactions for functional groups, and high-resolution mass spectrometry

Four reagents, four functional groups

Each of four reagents reacts with one functional group and shows it with a colour change or a gas. Cyclohexene, propanal, propan-1-ol and ethanoic acid are all colourless liquids, and the four reagents tell them apart in a few minutes of test-tube work.

The test for an alkene is bromine water, which is orange. Shaken with an alkene it turns colourless, because bromine adds across the C=C double bond and the dibromo product is colourless. An alkane gives no change. Write 'orange to colourless' or 'decolourised'. 'Clear' is not the same word: a solution can be clear and still be orange.

C₆H₁₀(l) + Br₂(aq) → C₆H₁₀Br₂(l)Cyclohexene decolourises bromine water by electrophilic addition.

Two reagents test for an aldehyde. Tollens' reagent is a colourless solution of the complex ion [Ag(NH₃)₂]⁺. Warmed with an aldehyde in a water bath, it gives a silver mirror on the inside of the tube. The aldehyde is oxidised to a carboxylic acid and silver(I) is reduced to silver metal. Fehling's solution is a deep blue solution of copper(II) ions, kept in solution in alkali by a complexing agent. Warmed with an aldehyde, it gives a brick-red precipitate of copper(I) oxide, Cu₂O. A ketone does neither: Tollens' stays colourless and Fehling's stays blue.

[Ag(NH₃)₂]⁺(aq) + e⁻ → Ag(s) + 2NH₃(aq)Silver(I) is reduced; the electron comes from the aldehyde.

The difference between the two carbonyls comes from what is attached to the C=O carbon. An aldehyde has a hydrogen there, and oxidation replaces that C–H with C–OH, giving a carboxylic acid. A ketone has two carbons there, and oxidising it would mean breaking a C–C bond, which these mild oxidising agents cannot do.

Primary and secondary alcohols are shown by potassium dichromate(VI) acidified with dilute sulfuric acid, which is orange. Warmed with a primary or a secondary alcohol, it turns green, as Cr₂O₇²⁻ is reduced to Cr³⁺. A tertiary alcohol has no hydrogen on the carbon bearing the OH and is not oxidised, so the mixture stays orange. Aldehydes are oxidised too, so they also turn it green.

CH₃CH₂CH₂OH(l) + [O] → CH₃CH₂CHO(l) + H₂O(l)A primary alcohol oxidised to an aldehyde; [O] stands for the oxidising agent.

A carboxylic acid is shown by solid sodium hydrogencarbonate, or sodium carbonate solution, with which it gives effervescence. The gas is carbon dioxide and turns limewater milky. Alcohols and aldehydes are far too weakly acidic to react. Write 'effervescence' or 'bubbles'; 'a gas is produced' is a conclusion, since nobody sees a gas.

CH₃COOH(aq) + NaHCO₃(s) → CH₃COONa(aq) + H₂O(l) + CO₂(g)

Functional group

The atom or group of atoms in a molecule that is responsible for its characteristic chemical reactions.

Tollens' reagent

An aqueous solution of [Ag(NH₃)₂]⁺ that is reduced to silver by an aldehyde, giving a silver mirror, but does not react with a ketone.

Fehling's solution

An alkaline solution of a copper(II) complex that is reduced by an aldehyde to a brick-red precipitate of copper(I) oxide.

Working through an unknown

One reagent can respond to two groups, so the order in which the results are read matters. Acidified dichromate turns green with a primary alcohol, a secondary alcohol and an aldehyde. A green result means 'one of those three', and only the Tollens' or Fehling's result says whether an aldehyde is present.

Describe the drawing

A decision tree of four test-tube tests running down the left. Sodium hydrogencarbonate first: effervescence that turns limewater milky means a carboxylic acid. Otherwise bromine water: orange to colourless shows a C=C, but the branch rejoins the tree because other groups may still be present. Then Tollens' reagent or Fehling's: a silver mirror or brick-red precipitate means an aldehyde. Last, acidified potassium dichromate(VI): orange to green means a primary or secondary alcohol, and staying orange leaves a tertiary alcohol, a ketone or none of these groups. A note warns that an aldehyde would also turn dichromate green, which is why that test comes after Tollens'.

Dichromate alone cannot tell a primary alcohol from a secondary one, because both turn it green. Oxidise the alcohol with dichromate in distillation apparatus and test the distillate with Tollens' reagent. A primary alcohol gives an aldehyde, which distils off and forms a silver mirror. A secondary alcohol gives a ketone, which does not.

High-resolution mass spectrometry

Nitrogen, carbon monoxide and ethene are three colourless gases with the same relative molecular mass to the nearest whole number, 28. A low-resolution mass spectrometer shows each of them as a molecular ion at m/z 28 and cannot tell them apart. A high-resolution instrument measures m/z to four decimal places, and at that precision the three are different.

They differ because atomic masses are not whole numbers. Carbon-12 is exactly 12 by definition, but every other isotope is slightly off a whole number, because the mass of a nucleus is not simply the sum of the masses of its protons and neutrons. The precise masses used are for the most abundant isotope of each element, not the relative atomic masses of the periodic table.

Adding these gives 2 × 14.0031 = 28.0062 for N₂, 12.0000 + 15.9949 = 27.9949 for CO and 2 × 12.0000 + 4 × 1.0078 = 28.0312 for C₂H₄. A molecular ion measured at 28.0313 can only be ethene.

Hydrogen and nitrogen sit slightly above a whole number and oxygen slightly below, so a molecule rich in hydrogen is a little heavier than its nominal mass and one rich in oxygen a little lighter. Formulae with the same nominal mass usually differ by 0.01 or more, and a high-resolution instrument measures to within about 0.001, so the match is unambiguous. The technique gives a molecular formula, not a structure: propanal and propanone have the same formula and the same precise mass, and only a test or a spectrum separates them.

Describe the drawing

A number line of m/z from 27.98 to 28.04 with three tall lines on it: CO at 27.9949, N₂ at 28.0062 and C₂H₄ at 28.0312, the second gap more than twice the first. Above it a small low-resolution axis from 26 to 30 shows a single line at 28 carrying all three formulae, with dashed lines opening that one line out into the high-resolution axis below.

Molecular ion

The ion formed when a molecule loses one electron, M⁺; its m/z gives the relative molecular mass.

Precise relative molecular mass

The relative molecular mass calculated from the precise masses of the most abundant isotopes, given to four decimal places.

In an electron-impact spectrometer, a fast electron knocks an electron out of the molecule, giving the molecular ion. Its peak is the one with the highest m/z, ignoring a very small peak one unit beyond it from molecules containing carbon-13, and its m/z is the relative molecular mass. If the sample was ionised by electrospray instead, the peak is MH⁺, one proton heavier, and 1.0073 is subtracted: the proton, not the hydrogen atom, because the added H⁺ brings no electron with it.

M(g) + e⁻ → M⁺(g) + 2e⁻

Other boards: 2,4-DNP, triiodomethane, phenols and fragmentation

OCR assesses the four tests of required practical 6 in PAG 7, Edexcel within its core practicals, and Cambridge in its practical papers, with the same reagents and the same observations.

AQA's required practical stops at the four tests in the table. OCR, Edexcel and Cambridge also examine a test for the C=O group itself; Edexcel and Cambridge add a test for a methyl group next to it, and OCR a way of telling a phenol from a carboxylic acid.

2,4-Dinitrophenylhydrazine, 2,4-DNP or Brady's reagent, is used dissolved in methanol and sulfuric acid. A few drops of any aldehyde or ketone give a yellow or orange precipitate at room temperature. The NH₂ group of the reagent joins to the C=O carbon and a molecule of water is lost, giving a solid 2,4-dinitrophenylhydrazone. The C=O of a carboxylic acid or an ester does not react, so the test picks out aldehydes and ketones together, and Tollens' reagent then says which of the two is present.

CH₃COCH₃(l) + C₆H₃(NO₂)₂NHNH₂ → (CH₃)₂C=NNHC₆H₃(NO₂)₂(s) + H₂O(l)Propanone and 2,4-DNP condense, losing water. The reagent is in methanol and acid, so it carries no state symbol.

The precipitate also names the compound. Filtered off, recrystallised from ethanol and dried, the derivative has a sharp melting point, and each aldehyde and ketone gives a derivative with its own. The measured melting point is compared with a data table. This works where boiling points fail: propanal and propanone boil only 7 °C apart, but their derivatives melt about 30 °C apart. OCR sets both uses of the reagent, the test and the melting point; Edexcel and Cambridge set the test.

Edexcel and Cambridge add the triiodomethane test, also called the iodoform test. The unknown is warmed with iodine solution and aqueous sodium hydroxide. A pale yellow precipitate of CHI₃, with an antiseptic smell, shows a CH₃CO group: ethanal and every methyl ketone give it. Propanone gives the precipitate and propanal does not, so the test separates those two isomers. Alcohols with a CH₃CH(OH) group, such as ethanol and butan-2-ol, give it too, because the alkaline iodine first oxidises them to a CH₃CO compound.

CH₃COCH₃(aq) + 3I₂(aq) + 4NaOH(aq) → CHI₃(s) + CH₃COONa(aq) + 3NaI(aq) + 3H₂O(l)The three hydrogens of one CH₃ group are replaced by iodine, then the C–C bond to the carbonyl carbon breaks.

OCR adds phenols. Phenol, C₆H₅OH, is a weak acid: it reacts with sodium hydroxide solution to form sodium phenoxide, but it is too weak an acid to release carbon dioxide from a carbonate or hydrogencarbonate. A compound that reacts with sodium hydroxide but gives no effervescence with sodium hydrogencarbonate is a phenol; a carboxylic acid does both.

C₆H₅OH(s) + NaOH(aq) → C₆H₅ONa(aq) + H₂O(l)

2,4-Dinitrophenylhydrazine

Brady's reagent, which gives a yellow or orange precipitate with an aldehyde or a ketone; the melting point of the purified precipitate identifies the carbonyl compound.

Triiodomethane test

Warming with iodine and aqueous sodium hydroxide; a pale yellow precipitate of CHI₃ shows a CH₃CO group or a CH₃CH(OH) group.

AQA asks for the precise-mass calculation and not for fragment patterns, which OCR, Edexcel and Cambridge all examine. The molecular ion has excess energy and some break apart. A fragmentation gives one positive ion and one neutral radical, and only the ion is accelerated and detected.

CH₃COCH₂CH₃⁺(g) → CH₃CO⁺(g) + •CH₂CH₃(g)Butanone losing an ethyl radical gives the ion at m/z 43.

Describe the drawing

A line mass spectrum of butanone with m/z from 10 to 80 along the bottom and relative abundance up the side. The tallest line, the base peak at 43, is CH₃CO⁺. The molecular ion M⁺ at 72 reaches about a quarter of its height, with a tiny M+1 line at 73. Smaller lines at 57, 29 and 15 are labelled CH₃CH₂CO⁺, CH₃CH₂⁺ and CH₃⁺; lines at 27 and 42 are unlabelled. A skeletal structure of butanone shows dashed cuts either side of the C=O carbon giving 43 and 57.

Read a fragment as a difference. From m/z 72, a peak at 57 is a loss of 15, a CH₃ group, leaving CH₃CH₂CO⁺; a peak at 43 is a loss of 29, an ethyl group, leaving CH₃CO⁺. The peaks at 29 and 15 are CH₃CH₂⁺ and CH₃⁺ themselves. Butanal, an isomer, has no CH₃CO group, yet it also gives a peak at 43, from CH₃CH₂CH₂⁺. Two different ions can share a nominal m/z, so a fragment is assigned from the structure, not the structure from one fragment.

Fragmentation

The breaking of a molecular ion into a positive ion and a neutral radical, of which only the ion is detected.

Base peak

The tallest peak in a mass spectrum, given a relative abundance of 100.

M+1 peak

The small peak one m/z unit above the molecular ion, from molecules containing one carbon-13 atom; its height relative to M gives the number of carbon atoms.

Cambridge also reads the two small peaks just beyond the molecular ion.

The M+1 peak comes from molecules that contain one carbon-13 atom. About 1.1% of carbon atoms are carbon-13, so each carbon in a molecule has a 1.1% chance of being one. A molecule with n carbon atoms gives an M+1 peak about n × 1.1 per cent of the height of the M peak, and turning that round counts the carbons.

n = (100 × height of M+1 peak) ÷ (1.1 × height of M peak)

The M+2 peak shows chlorine or bromine. Chlorine is about 75% ³⁵Cl and 25% ³⁷Cl, so a compound with one chlorine atom gives two molecular ion peaks two units apart in the height ratio 3:1: chloroethane at m/z 64 and 66. Bromine is ⁷⁹Br and ⁸¹Br in nearly equal amounts, so one bromine atom gives two peaks of equal height: bromoethane at m/z 108 and 110. Two chlorine atoms give three peaks, at M, M+2 and M+4, in the ratio 9:6:1.