Organic analysis

In a nutshell

Organic analysis is how a chemist works out the identity of an unknown organic compound, using simple test-tube reactions and two instrumental techniques.

Three tools do the work:

• Test-tube reactions show which functional group is present (alkene, alcohol, aldehyde or carboxylic acid).
• Mass spectrometry gives the relative molecular mass, and at high resolution the molecular formula.
• Infrared (IR) spectroscopy identifies the bonds present, confirms a compound by its fingerprint, and explains the greenhouse effect.

Assumed knowledge: Alkenes, Alcohols, Atomic structure.

Core content

Identifying functional groups by test-tube reactions

A functional group is found by adding a reagent and watching for a specific colour change, precipitate or gas. The identification rests on the observation, so you must learn the reagent, the conditions and exactly what you would see.

Here is the full set for this subtopic:

Two warnings that decide the mark here:

• Acidified potassium dichromate(VI) turns from orange to green with a primary or secondary alcohol and with an aldehyde. It cannot tell those two apart.
• Tollens' and Fehling's react with aldehydes only, never with ketones. This is what separates an aldehyde from a ketone.

Required practical 6: tests for alkene, alcohol, aldehyde and carboxylic acid

This required practical is about using the reactions above to distinguish and identify unknown organic liquids. Each test in detail:

Alkene: bromine water

• Add bromine water and shake.
• The C=C double bond undergoes electrophilic addition, so the bromine is used up.
• Observation: the orange bromine water is decolourised (goes colourless).

$CH_2{=}CH_2 + Br_2 \rightarrow CH_2BrCH_2Br$

A compound with no C=C leaves the bromine water orange.

Alcohol: acidified potassium dichromate(VI)

• Warm the compound with potassium dichromate(VI) that has been acidified with dilute sulfuric acid.
• A primary or secondary alcohol is oxidised, so the orange dichromate(VI) ion (Cr₂O₇^2-) is reduced to the green chromium(III) ion (Cr³⁺).
• Observation: orange to green.
• A tertiary alcohol is not oxidised, so it stays orange. That is how the test distinguishes tertiary alcohols.

Aldehyde: Tollens' reagent (and Fehling's solution)

• Tollens' reagent is made by adding aqueous ammonia to silver nitrate solution; the active species is the complex ion [Ag(NH₃)₂]⁺.
• Warm gently. An aldehyde is oxidised to a carboxylic acid, and the silver(I) is reduced to silver metal:

$Ag^+ + e^- \rightarrow Ag$

• Observation: a silver mirror coats the inside of the tube.
• Fehling's solution contains blue copper(II) ions. An aldehyde reduces them to a red precipitate of copper(I) oxide (Cu₂O). Observation: blue to red precipitate.
• A ketone gives no change with either reagent.

Carboxylic acid: a named carbonate

• Add a named carbonate (such as sodium carbonate, Na₂CO₃) or a hydrogencarbonate.
• Observation: effervescence as carbon dioxide is released. Bubbling the gas through limewater (which turns milky) confirms it is CO₂.

$2CH_3COOH + Na_2CO_3 \rightarrow 2CH_3COONa + H_2O + CO_2$

Carboxylic acids are the only common organic group acidic enough to release CO₂ from a carbonate, so this is a reliable test for the group.

How to tackle a "distinguish these compounds" question. Do not assume you already know which compound is which. Add the first reagent to every compound, record which react and which do not, then choose a second test to separate the ones that are left. Work from the results, not from a guess.

Mass spectrometry: the molecular ion and M_r

In a mass spectrometer a molecule loses one electron to form a positive ion called the molecular ion:

$M \rightarrow M^{+} + e^{-}$

The molecular ion has effectively the same mass as the whole molecule, so it appears at the highest m/z value in the spectrum. Because its charge is +1, that m/z value is equal to the relative molecular mass (M_r) of the compound.

In the spectrum above the peak at the highest m/z (46) is the molecular ion, so the compound has M_r = 46.

High-resolution mass spectrometry: finding the molecular formula

Different molecular formulae can share the same M_r to the nearest whole number. For example C₃H₈, CO₂ and N₂O all have a whole-number M_r of about 44.

A high-resolution mass spectrometer measures mass to several decimal places. Because each isotope has its own precise atomic mass, each formula has a different precise M_r, so the exact value identifies the formula.

The precise atomic masses are given to you in the question. A typical set is:

The mass of ¹²C is exactly 12.00000 because carbon-12 is the standard against which all other masses are defined.

Still don't get it? · why precise masses separate compounds with the same Mr

Think of three people who all weigh "about 70 kg" on bathroom scales. On those scales you cannot tell them apart. Put them on a laboratory balance that reads to the nearest gram and you get 69.812 kg, 70.244 kg and 70.009 kg: now each is unique.

High-resolution mass spectrometry is that laboratory balance. The reason the exact values differ is that each atom's true mass is not a whole number. A hydrogen atom is 1.00794, not 1, and an oxygen atom is 15.99491, not 16. When you add up the atoms in a formula, those tiny decimals push the total in a direction that is different for every combination of atoms, even when the whole-number total is the same.

Exam version: compounds with the same M_r to one decimal place can still be told apart because high-resolution mass spectrometry measures the precise molecular mass, and each molecular formula has its own precise M_r.

Infrared spectroscopy: bonds absorb at characteristic wavenumbers

Covalent bonds vibrate (stretch and bend). A bond absorbs infrared radiation when the radiation's energy matches the bond's natural vibration, and each type of bond does this at a characteristic wavenumber (measured in cm^-1).

An IR spectrum plots transmittance against wavenumber, so an absorption shows as a dip (trough) pointing down. You are given a data sheet of ranges in the exam; the key ones are:

The single most important exam habit: quote both the wavenumber range and the bond responsible. "An absorption at 1700 cm^-1" scores nothing on its own; "an absorption at 1680 to 1750 cm^-1 due to C=O" earns the mark.

Watch the two O-H absorptions, which examiners test constantly:

• O-H in an alcohol: a broad trough at 3230 to 3550 cm^-1.
• O-H in a carboxylic acid: a very broad trough spread over 2500 to 3000 cm^-1.

Here is an alcohol (ethanol) beside a carboxylic acid (ethanoic acid):

The acid's O-H is lower and far broader than the alcohol's, and the acid has the strong C=O trough at about 1710 cm^-1 that the alcohol lacks. Reading those two features apart is exactly what the exam rewards.

The fingerprint region and identifying impurities

The region below about 1500 cm^-1 is the fingerprint region. It is a complicated pattern of many absorptions that is unique to each individual compound, like a fingerprint.

You do not try to assign every peak in it. Instead a computer compares the whole spectrum against a database of known spectra and looks for an exact match to identify the compound.

Still don't get it? · the fingerprint region

Above 1500 cm^-1, an IR spectrum is like reading someone's job title: it tells you the big features (an O-H here, a C=O there) but several different people could share the same title. Two compounds with the same functional groups can look identical in that upper region.

The fingerprint region is like an actual fingerprint. It is far too detailed to read off bond by bond, but no two different compounds produce the same one. So you do not decode it; you match it. A computer lays your spectrum over a library of known spectra and finds the one it matches exactly.

Exam version: to identify a specific molecule, especially to tell apart two compounds with the same functional groups, you compare the fingerprint region with the spectra of known compounds in a database and look for an exact match.

IR also spots impurities. A pure compound's spectrum matches the reference exactly, so extra (rogue) absorptions that should not be there reveal an impurity.

Infrared absorption and global warming

The same bond vibrations that give IR spectra also happen in the atmosphere. Bonds in carbon dioxide, methane and water vapour absorb infrared radiation, and this is the cause of the greenhouse effect.

The chain of reasoning:

1. Short-wavelength radiation from the Sun passes through the atmosphere and warms the Earth's surface.
2. The warm Earth re-emits energy as longer-wavelength infrared radiation.
3. Bonds in CO₂, CH₄ and water vapour absorb this infrared radiation instead of letting it escape to space.
4. The energy is passed on to other molecules by collisions, so the atmosphere is warmed.

A gas is a more effective greenhouse gas if its bonds absorb more of the Earth's infrared radiation.

Worked examples

Example 1 (calculation): using high-resolution mass spectrometry to find a molecular formula.

An unknown compound has three candidate formulae, all with a whole-number M_r of 60: C₂H₄O₂, C₃H₈O and CH₄N₂O. High-resolution mass spectrometry gives the precise M_r as 60.0216. Which formula is it?

Use the precise atomic masses H = 1.00794, C = 12.00000, N = 14.00307, O = 15.99491. Work out the precise M_r of each candidate.

$C_2H_4O_2 = 2(12.00000) + 4(1.00794) + 2(15.99491)$ $= 24.00000 + 4.03176 + 31.98982 = 60.02158$

$C_3H_8O = 3(12.00000) + 8(1.00794) + 1(15.99491)$ $= 36.00000 + 8.06352 + 15.99491 = 60.05843$

$CH_4N_2O = 1(12.00000) + 4(1.00794) + 2(14.00307) + 1(15.99491)$ $= 12.00000 + 4.03176 + 28.00614 + 15.99491 = 60.03281$

Compare each with the measured 60.0216. The match is C₂H₄O₂ (60.0216 to 4 decimal places).

Where marks go: multiply each atom's precise mass by how many of that atom there are (do not round the atomic masses part-way), then compare all your values with the measured mass. State the formula whose precise M_r matches; do not stop at "they are close".

Example 2 (model answer): identifying a compound from its IR spectrum.

An IR spectrum shows a strong absorption at 1715 cm^-1 and a very broad absorption spread from 2500 to 3000 cm^-1, but no absorption in the range 3230 to 3550 cm^-1. What functional group is present?

1. The absorption at 1680 to 1750 cm^-1 is due to a C=O bond.
2. The broad absorption at 2500 to 3000 cm^-1 is due to the O-H bond of a carboxylic acid.
3. There is no O-H (alcohol) absorption at 3230 to 3550 cm^-1, so it is not an alcohol.
4. C=O together with the acid O-H means the compound is a carboxylic acid.

Where marks go: each bond assignment needs both the wavenumber range and the bond, so each of the first two points is one mark. Naming the carboxylic acid O-H range (2500 to 3000) rather than the alcohol range is the discriminating step.

Example 3 (model answer): distinguishing two compounds by a test-tube reaction.

You have propanal (an aldehyde) and propanone (a ketone). Give one test that distinguishes them.

1. Add Tollens' reagent to a sample of each and warm gently.
2. Propanal (the aldehyde) is oxidised, forming a silver mirror on the tube.
3. Propanone (the ketone) is not oxidised, so there is no change.

A positive result (silver mirror) for the aldehyde and no change for the ketone is the complete answer. Fehling's solution (blue to red precipitate with the aldehyde only) works equally well.

Common exam mistakes

• Quoting a wavenumber without the bond, or a bond without the wavenumber range. The mark needs both ("an absorption at 2500 to 3000 cm^-1 due to O-H in a carboxylic acid").
• Confusing the two O-H absorptions: O-H (alcohol) is 3230 to 3550 cm^-1, O-H (carboxylic acid) is 2500 to 3000 cm^-1. Mixing them up loses the mark.
• Giving the alkene test as a test for the absence of something, or getting the colour change backwards. The positive result is bromine water going orange to colourless.
• With potassium dichromate: forgetting the word acidified, giving the wrong oxidation state (it must be dichromate**(VI)**), or giving an incomplete colour change. The full change is orange to green.
• Thinking acidified dichromate(VI) identifies an aldehyde on its own. It turns green with a primary or secondary alcohol as well, so it cannot separate an aldehyde from those alcohols.
• Suggesting Tollens' or Fehling's for a ketone. They react with aldehydes only. Also, the positive observation is a silver mirror (not "a silver precipitate") and a red precipitate with Fehling's.
• Writing just "carbonate" for the carboxylic-acid test. Name it fully (sodium carbonate / Na₂CO₃), and state the gas is CO₂.
• Writing "nothing" or "no observation" when asked for an observation. A negative result still needs a positive statement of what you see, or you score zero.
• In high-resolution mass spectrometry, not realising that ¹²C is defined as exactly 12.00000, or failing to explain that two compounds share the same M_r to one decimal place but have different precise M_r values.
• For two compounds with the same functional groups, forgetting that you identify them by matching the fingerprint region to a known spectrum, not by the functional-group region.

Key definitions

• Molecular ion - the positive ion formed when a molecule loses one electron ($M \rightarrow M^{+} + e^{-}$). It gives the peak at the highest m/z, which equals the M_r of the compound.
• High-resolution mass spectrometry - mass spectrometry that measures mass precisely enough to use precise atomic masses and the precise molecular mass to determine the molecular formula of a compound.
• Characteristic absorption - the fixed wavenumber range at which a particular type of bond absorbs infrared radiation.
• Fingerprint region - the region of an IR spectrum below about 1500 cm^-1, unique to each compound, used to identify a compound by comparison with the spectra of known compounds.
• Greenhouse gas - a gas (such as CO₂, CH₄ or water vapour) whose bonds absorb infrared radiation emitted by the Earth, warming the atmosphere.

Specification

• I can identify the functional groups alkene, alcohol, aldehyde and carboxylic acid using the test-tube reactions in the specification, giving the reagent and the positive observation.
• I can carry out Required practical 6: tests for alcohol, aldehyde, alkene and carboxylic acid.
• I can state that the molecular ion gives a peak at the highest m/z equal to the M_r of the compound.
• I can use precise atomic masses and the precise molecular mass to determine the molecular formula of a compound.
• I can use infrared spectra and the Chemistry Data Sheet to identify particular bonds, and therefore functional groups, by quoting the wavenumber range and the bond.
• I can use infrared spectra to identify impurities.
• I can explain how fingerprinting allows a molecule to be identified by comparison of spectra.
• I can describe the link between absorption of infrared radiation by bonds in CO₂, methane and water vapour and global warming.

Related notes

• Alkenes
• Alcohols
• Halogenoalkanes
• Atomic structure
• Introduction to organic chemistry
• Leads to: Aldehydes and ketones (3.3.8), Nuclear magnetic resonance spectroscopy (3.3.15) and Chromatography (3.3.16), all A-level only.