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Structure 3.2 · Stereoisomerism

1Stereoisomers 3.2.7 HL only Strong

The syllabus statement: Stereoisomers have the same constitution (atom identities, connectivities and bond multiplicities) but different spatial arrangements of atoms. Same atoms, same connections, different arrangement in space — which is what separates them from the structural isomers of Part A, where the connections themselves differ. This outcome appeared in all six sessions in the collection behind these notes, and carries more evidence than any other in Structure 3.

Structural isomers (Part A) differ in which atoms are joined to which. Stereoisomers have the same constitution — the same atoms, the same connections and the same bond multiplicities — but a different arrangement of those atoms in space. Two kinds are in the syllabus: cis–trans isomers, which arise from restricted rotation, and enantiomers (optical isomers), which arise from a chiral carbon atom. Stereoisomers cannot be interconverted without breaking bonds; they are different compounds, not different poses of one molecule.

Cis–trans isomerism

Rotation about a carbon–carbon single bond is free at room temperature, because the σ bond is symmetrical about the internuclear axis. Rotation about a C=C double bond is restricted: the π bond is formed by sideways overlap of p orbitals above and below the plane of the molecule, and twisting one carbon relative to the other would break that overlap (Structure 2.2). The groups attached to each carbon of a double bond are therefore locked in position.

If each carbon of the C=C carries two different groups, the molecule can exist in two forms. In the cis isomer the two like groups (or the two non-hydrogen groups) are on the same side of the double bond; in the trans isomer they are on opposite sides. But-2-ene, CH₃CH=CHCH₃, exists as cis-but-2-ene and trans-but-2-ene. But-1-ene, CH₂=CHCH₂CH₃, does not show cis–trans isomerism, because C1 carries two identical hydrogen atoms: swapping them produces the same molecule.

A ring does the same job as a double bond. In cyclopropanes and cyclobutanes (the C₃ and C₄ cycloalkanes named in the syllabus) the ring prevents rotation, so two substituents on different ring carbons can lie on the same face of the ring (cis) or on opposite faces (trans): cis- and trans-1,2-dimethylcyclopropane are different compounds.

Cis and trans isomers are different substances with different physical properties. The polar C–Cl bonds of cis-1,2-dichloroethene reinforce each other, so the molecule is polar and boils at 60 °C; in the trans isomer they cancel, the molecule is non-polar and it boils at 48 °C. The trans isomer, being more symmetrical, often packs better into a solid and has the higher melting point. Their chemical properties can also differ when the two groups are close enough to interact: cis-butenedioic acid forms a cyclic anhydride on heating, but the trans isomer cannot because its carboxyl groups are too far apart.

Chirality and enantiomers

A carbon atom bonded to four different atoms or groups is a chiral carbon (a stereocentre, or asymmetric carbon). Its four groups sit at the corners of a tetrahedron, and they can be arranged in two ways that are mirror images of each other. The two arrangements cannot be superimposed, however they are rotated — in the same way that a left hand cannot be superimposed on a right hand. The two stereoisomers are called enantiomers. A molecule that is not superimposable on its mirror image is described as chiral.

Enantiomers have identical physical properties (melting point, boiling point, density, solubility, IR and NMR spectra) and identical chemical properties towards non-chiral reagents. They differ in two ways:

• Optical activity. Each enantiomer rotates the plane of plane-polarized light — by the same angle but in opposite directions. The rotation is measured with a polarimeter: light passes through a polarizing filter, then through the sample, and an analysing filter is turned until the light intensity is at its maximum again; the angle turned is the observed rotation.
• Behaviour in chiral environments. Enantiomers react differently with other chiral molecules, including enzymes and receptors in the body. This is why many drugs are made as a single enantiomer: one form may be active and the other inactive or harmful.

An equimolar mixture of two enantiomers is a racemic mixture (racemate). It is optically inactive, because the rotations produced by the two enantiomers cancel exactly.

Exam focus · what the published papers show

Structure 3.2 · Spectroscopic identification

2Mass spectrometry of organic compounds 3.2.8 HL only Repeated

The syllabus statement: Mass spectrometry (MS) of organic compounds can cause fragmentation of molecules. The skill: deduce information about the structural features of a compound from specific MS fragmentation patterns, and include reference to the molecular ion. The fragment data are in the data booklet, so nothing here is memory work — it is reading.

In a mass spectrometer, vaporized molecules are bombarded with high-energy electrons. An electron is knocked out of a molecule, forming the molecular ion, M⁺:

M(g) + e⁻ → M⁺(g) + 2e⁻

The molecular ion carries a great deal of excess energy, and many molecular ions break apart — fragment — before they reach the detector. Each fragmentation splits M⁺ into a positive ion and a neutral fragment (a radical or molecule):

M⁺ → X⁺ + Y^•

Only charged species are accelerated, deflected and detected, so only X⁺ gives a peak; the neutral fragment Y is invisible and is identified from the difference between the m/z values. The spectrum plots relative abundance against mass-to-charge ratio, m/z; since almost all ions carry a single positive charge, m/z equals the mass of the ion.

• The molecular ion peak is the peak at the highest m/z (ignoring small isotope peaks): it gives the relative molecular mass.
• The tallest peak, the base peak, is the most abundant ion — usually a particularly stable fragment. It is not necessarily the molecular ion.
• Common losses and fragments help to reveal structure. Fragment data are in the data booklet; the masses below follow from atomic masses.

Exam focus · what the published papers show

3Infrared spectroscopy 3.2.9 HL only Repeated

The syllabus statement: Infrared (IR) spectra can be used to identify the type of bond present in a molecule. The skill: interpret the functional group region of an IR spectrum, using a table of characteristic frequencies (wavenumber/cm–1), and include reference to the absorption of IR radiation by greenhouse gases. The table is in the data booklet.

Covalent bonds are not rigid: they vibrate, stretching and bending continuously. Each bond vibrates at a natural frequency that depends on the masses of the two atoms and the strength of the bond. When infrared radiation of exactly that frequency falls on the molecule, the bond absorbs it and vibrates with greater amplitude. Stronger bonds and lighter atoms vibrate at higher frequencies: C≡C absorbs at a higher wavenumber than C=C, which absorbs higher than C–C, and bonds to hydrogen (O–H, N–H, C–H) absorb at high wavenumbers.

A vibration absorbs infrared radiation only if it changes the dipole moment of the molecule. The stretching of the polar C=O bond changes the dipole strongly and gives an intense absorption; the symmetric stretch of CO₂ produces no change in dipole and is IR-inactive, but its asymmetric stretch and its bending vibration are IR-active. Molecules such as N₂ and O₂, with no dipole in any vibration, do not absorb IR at all.

This connects infrared spectroscopy to the greenhouse effect. The Earth’s surface emits infrared radiation. Gases whose molecules absorb IR — water vapour, carbon dioxide, methane, nitrous oxide — absorb some of this radiation and re-emit it in all directions, including back towards the surface, warming the lower atmosphere. Nitrogen and oxygen, the main components of air, are not greenhouse gases because their vibrations do not change a dipole moment.

An IR spectrum plots transmittance (percentage of radiation transmitted) against wavenumber (1/λ, in cm⁻¹, proportional to frequency), with wavenumber decreasing from left to right; absorptions appear as downward troughs. The region above about 1500 cm⁻¹, the functional group region, contains the characteristic absorptions used for identification.

Exam focus · what the published papers show

Past-paper practice · Practice set V · Mass spectrometry and infrared spectroscopy HL only

Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination.

V1IB · November 2016 · SL Paper 1 · Q29 · [1]

What is always correct about the molecular ion, M⁺, in a mass spectrum of a compound?

A. The M⁺ ion peak has the smallest m/z ratio in the mass spectrum.B. The m/z ratio of the M⁺ ion peak gives the relative molecular mass of the molecule.C. The M⁺ ion is the most stable fragment formed during electron bombardment.D. The M⁺ ion peak has the greatest intensity in the mass spectrum.

V2IB · May 2019 · HL Paper 1 · TZ2 · Q40 · [1]

Which technique involves breaking covalent bonds when carried out on an organic compound?

A. infrared spectroscopyB. nuclear magnetic resonance spectroscopyC. X-ray crystallographyD. mass spectrometry

V3IB · May 2019 · HL Paper 1 · TZ1 · Q40 · [1]

Which can be identified using infrared (IR) spectroscopy?

A. functional groupsB. molar massC. 3-D configurationD. bond angle

V4IB · May 2021 · SL Paper 1 · TZ2 · Q28 · [1]

Which spectra would show the difference between propan-2-ol, CH₃CH(OH)CH₃, and propanal, CH₃CH₂CHO?

I. mass    II. infrared    III. ¹H NMR

A. I and II onlyB. I and III onlyC. II and III onlyD. I, II and III

V5IB · May 2021 · HL Paper 2 · TZ2 · Q4(g)(i)–(ii) · [3]

Combustion analysis of an unknown organic compound indicated that it contained only carbon, hydrogen and oxygen.

For the peaks at m/z 58 and m/z 43, deduce two features of this molecule that can be obtained from the mass spectrum.

V6IB · November 2023 · HL Paper 2 · TZ1 · Q3(c)(i) · [2]

Methanoic acid can be converted into methyl methanoate, HCOOCH₃. State one similarity and one difference you would expect in the infrared (IR) spectra of methanoic acid and methyl methanoate in the region of 1500–3500 cm⁻¹. Use the data booklet. [2]

Solutions and mark-scheme guidance · Set V

V1 B

M⁺ is the whole molecule minus one electron, so its m/z (charge 1+) equals the relative molecular mass. It is the highest m/z peak (ignoring isotope peaks), not the smallest, and it is often neither the most stable fragment nor the tallest peak.

V2 D

Electron bombardment in mass spectrometry breaks bonds to give fragments. IR only makes bonds vibrate; NMR changes nuclear spin states; X-ray crystallography diffracts X-rays.

V3 A

Each bond type absorbs IR at characteristic wavenumbers, so IR identifies bonds and hence functional groups.

V4 D

They are isomers (M = 60 for both), but their fragmentation patterns differ (MS); propan-2-ol shows a broad O–H band and propanal a C=O band (IR); and their proton environments differ (propan-2-ol three, propanal three with different shifts and splitting, including the aldehyde H near 9–10 ppm) (NMR).

V5 [3]

m/z 58: the molar / relative molecular mass is 58 (the molecular ion) ✔. m/z 43: loss of a methyl / CH₃ fragment, OR a COCH₃⁺ fragment ✔ (missing charges not penalized on that paper; C₂H₃O⁺ and C₃H₇⁺ also accepted).
(ii) Absorption A (just above 1700 cm⁻¹): C=O ✔ (“carbonyl” accepted).

V6 [2]

Similarity: absorption at 1700–1750 cm⁻¹ by C=O, OR at 2850–3090 cm⁻¹ by C–H ✔. Difference: methanoic acid has the broad absorption at 2500–3000 cm⁻¹ from the O–H of the carboxyl group, absent for methyl methanoate ✔. The scheme does not accept a bond without its wavenumber or a reference to absorption; does not accept C–O (outside the stated region); does not accept “hydroxide”; and does not accept 3200–3600 cm⁻¹, which is the alcohol O–H range.

4¹H NMR: signals, chemical shifts and integration 3.2.10 HL only Strong

The syllabus statement: Proton nuclear magnetic resonance spectroscopy (1H NMR) gives information on the different chemical environments of hydrogen atoms in a molecule. The skill: interpret 1H NMR spectra to deduce the structures of organic molecules from the number of signals, the chemical shifts, and the relative areas under signals (integration traces) — three separate pieces of information, and questions name which one they want.

The nucleus of a hydrogen atom, a proton, has a property called spin that makes it behave like a tiny magnet. In a strong external magnetic field its magnetic moment can align with the field (lower energy) or against it (higher energy). The small energy gap between the two states corresponds to radio-frequency radiation. In a nuclear magnetic resonance spectrometer, the sample is placed in a strong magnetic field and irradiated with radio waves; protons absorb (resonate) when the frequency matches their energy gap.

The exact frequency at which a proton resonates depends on its chemical environment. Electrons around a proton circulate in the applied field and create a small opposing field that shields the nucleus. A proton near an electronegative atom (O, halogen) or a C=O group has less electron density around it: it is deshielded and resonates at a higher chemical shift, δ.

Chemical shifts are measured in parts per million (ppm) relative to a reference compound, tetramethylsilane (TMS), Si(CH₃)₄, which is assigned δ = 0. TMS is chosen because all twelve of its protons are in the same environment, giving one strong signal; silicon is less electronegative than carbon, so its protons are highly shielded and appear upfield of almost all organic protons; and it is inert and volatile, so it does not react with the sample and is easily removed.

Exam focus · what the published papers show

5¹H NMR: splitting patterns 3.2.11 HL only Repeated

The syllabus statement: Individual signals can be split into clusters of peaks. The skill: interpret 1H NMR spectra from splitting patterns showing singlets, doublets, triplets and quartets to deduce greater structural detail. Four patterns, and one rule.

In a high-resolution spectrum many signals appear as clusters of closely spaced peaks. The splitting (spin–spin coupling) arises because the tiny magnetic fields of protons on the adjacent carbon atom either add to or subtract from the applied field experienced by the protons being observed. With n equivalent neighbouring protons there are n + 1 possible combinations of their spins, so the signal is split into n + 1 peaks.

Three conventions apply. Equivalent protons do not split each other (the six protons of propanone give a singlet). The O–H proton of an alcohol usually appears as a singlet and does not split neighbouring signals, because it exchanges rapidly between molecules. And an ethyl group, CH₃CH₂–, attached to an atom with no hydrogens always shows the characteristic pair of a triplet (CH₃, split by 2 H) and a quartet (CH₂, split by 3 H).

Exam focus · what the published papers show

6Combining the techniques 3.2.12 HL only Repeated

The syllabus statement is one line: Data from different techniques are often combined in structural analysis. The skill: interpret a variety of data, including analytical spectra, to determine the structure of a molecule. There is no new chemistry here — only the habit of using everything you were given.

No single technique identifies an unknown compound on its own, because each answers a different question. Structure determination therefore combines evidence, and a good answer uses every piece of data provided and checks the proposed structure against all of it. Combustion analysis or percentage composition gives the empirical formula (Structure 1.4); mass spectrometry gives the relative molecular mass and so the molecular formula; the index of hydrogen deficiency is not required, but counting double bonds and rings from the formula helps; infrared shows the functional groups present; and ¹H NMR reveals the carbon–hydrogen framework.

Exam focus · what the published papers show

Past-paper practice · Practice set W · ¹H NMR and combined techniques HL only

Attempt these before opening the solutions below. References give the session, level, paper and question number of the original examination.

W1IB · May 2018 · SL Paper 1 · TZ1 · Q30 · [1]

What is the ratio of the areas of the signals in the ¹H NMR spectrum of pentan-3-ol?

A. 6:4:1:1B. 6:2:2:2C. 5:5:1:1D. 3:3:2:2:1:1

W2IB · November 2018 · SL Paper 1 · Q29 · [1]

What is the ratio of areas under each signal in the ¹H NMR spectrum of 2-methylbutane?

A. 6 : 1 : 2 : 3B. 3 : 3 : 1 : 5C. 6 : 1 : 5D. 3 : 3 : 1 : 2 : 3

W3IB · May 2021 · HL Paper 1 · TZ2 · Q40 · [1]

What information can be deduced from the splitting pattern of ¹H NMR signals?

A. total number of hydrogen atoms in a compoundB. number of hydrogen atoms on adjacent atom(s)C. functional group on which hydrogen atoms are locatedD. number of hydrogen atoms in a particular chemical environment

W4IB · May 2023 · SL Paper 1 · TZ2 · Q30 · [1]

Which compound will have only one ¹H NMR signal and show a carbonyl group in the IR spectrum?

A. CH₃CHOB. CH₃COOHC. CH₃OCH₃D. CH₃COCH₃

W5IB · November 2020 · HL Paper 1 · Q40 · [1]

A. but-3-en-2-ol, CH₂=CHCH(OH)CH₃B. butanal, CH₃CH₂CH₂CHOC. butanone, CH₃COCH₂CH₃D. but-3-en-1-ol, CH₂=CHCH₂CH₂OH

W6IB · May 2021 · HL Paper 2 · TZ2 · Q4(g)(iii) · [2]

This continues V5. Deduce the identity of the unknown compound using the previous information, the ¹H NMR spectrum and the data booklet. [2]

W7IB · November 2022 · HL Paper 2 · Q6(g) · [4]

(i) Deduce the number of signals and the ratio of areas under the signals in the ¹H NMR spectrum of 2-bromobutane. [2]

W8IB · November 2020 · HL Paper 2 · Q1(d)(v) · [3]

Ethoxyethane (diethyl ether), CH₃CH₂OCH₂CH₃, can be used as a solvent. Deduce the number of signals and chemical shifts with splitting patterns in the ¹H NMR spectrum of ethoxyethane. Use the data booklet. [3]

W9IB · May 2021 · SL Paper 2 · TZ1 · Q5(b) · [1]

Justify why ethene has only a single signal in its ¹H NMR spectrum. [1]

Solutions and mark-scheme guidance · Set W

W1 A

CH₃CH₂CH(OH)CH₂CH₃ is symmetrical: two equivalent CH₃ (6 H), two equivalent CH₂ (4 H), one CH (1 H), one OH (1 H).

W2 A

(CH₃)₂CHCH₂CH₃: two equivalent CH₃ on C2 (6), the CH (1), the CH₂ (2), the terminal CH₃ (3).

W3 B

A signal is split into n + 1 peaks by n hydrogens on the adjacent atom(s). D is what integration gives.

W4 D

Propanone has C=O and six equivalent hydrogens. Ethanal and ethanoic acid have two environments; methoxymethane has one environment but no carbonyl.

W5 B

The signal near 9.7 ppm is an aldehyde H, ruling out C and the alcohols; four signals (CHO, CH₂, CH₂, CH₃) with the upfield triplet and multiplets fit butanal.

W6 [2]

One signal indicates one hydrogen environment / a symmetrical structure, OR the shift near 2.2 ppm indicates H on a carbon next to a carbonyl ✔; compound: propanone, CH₃COCH₃ ✔. This is consistent with M = 58 and CH₃CO⁺ at 43 from V5, and the C=O absorption.

W7 [4]

(i) Number of signals: 4 ✔; ratio of areas 3:1:2:3 (any order) ✔.
(ii) Circle A (CH₃ next to CHBr): doublet ✔; circle B (CH₃ next to CH₂): triplet ✔.

W8 [3]

2 signals ✔; 0.9–1.0 ppm AND triplet ✔; 3.3–3.7 ppm AND quartet ✔ (any values within the ranges; [1] for two correct shifts or two correct splitting patterns only).

W9 [1]

All hydrogen atoms / protons are in the same chemical environment ✔ (“all H equivalent” and “symmetrical” accepted).

Review · Structure 3.2, additional higher level

7Misconceptions, the examiner’s view, and the question types

Six question types cover this whole document, and each has a fixed opening move.

8Quick check

9Summary and knowledge organiser

Essential knowledge

• Stereoisomers: same constitution, different spatial arrangement. Cis–trans needs restricted rotation (C=C or ring) and two different groups on each carbon.
• A chiral carbon has four different groups; its two mirror-image forms are enantiomers. They rotate plane-polarized light equally in opposite directions; a racemic mixture is optically inactive; they behave differently in chiral environments. Draw with wedges and dashes.
• MS: M⁺ (highest m/z) gives M_r; fragments X⁺ are detected; losses (15, 17, 29, 31, 45) reveal groups.
• IR: bonds absorb at characteristic wavenumbers if the vibration changes the dipole; C=O about 1700–1750, O–H acid 2500–3000 (broad), O–H alcohol 3200–3600; IR absorption by CO₂, H₂O and CH₄ is the greenhouse effect.
• ¹H NMR: number of signals = environments; δ relative to TMS = type of environment; integration = ratio; splitting n + 1 = neighbours.
• Combine: formula → MS → IR → NMR → check against all data.

Knowledge organiser