Home / H3 Chemistry 9813 / H3 Spectroscopic Techniques (9813) H3 Spectroscopic Techniques (9813) Learn H3 molecular orbitals, UV/visible, infrared, NMR and mass spectrometry through 39 ordered lessons with diagrams and focused practice.
Learn how quantised molecular energy produces the evidence seen in UV/visible, infrared, NMR and mass spectra—and how to combine that evidence when a structure is unfamiliar. Every technique follows the same chain: molecular structure → interaction with radiation → spectrum evidence → combined identification.
Prerequisites
Atomic orbitals, sigma and pi bonding, electron configurations and molecular shape.
Organic functional groups, structural formulae and intermolecular forces.
Moles, concentration, logarithms, proportional reasoning and graph interpretation.
Lessons (Recommended Order)
Start with the basic principles: the molecular-orbital and energy-level lessons explain why each later technique absorbs where it does. Then work through the four techniques in order. For revision, go straight to the lesson for the statement you need.
Common Exam Traps
Naming a peak without linking its position, shape, area or pattern to a structural feature.
Treating UV/visible, IR, NMR and mass spectra as interchangeable evidence.
Using one spectral feature to claim a unique structure when alternatives still fit.
Learning goals Distinguish atomic and molecular orbitals Classify bonding, antibonding and nonbonding orbitals Distinguish molecular orbitals with sigma and pi symmetry Explain discrete molecular-orbital energy levels Apply LCAO to homonuclear diatomic molecules Apply LCAO to benzene and linear polyenes Construct and interpret diatomic MO diagrams, including HOMO and LUMO Construct and interpret pi-MO diagrams for benzene and polyenes Relate electromagnetic radiation, photons and E = hf Compare electronic, vibrational, rotational and nuclear energy quantisation Explain photon absorption and emission as energy-level transitions Relate electronic transitions and chromophores to UV/visible absorption Predict UV/visible absorption from a chromophore Explain conjugation, energy gaps and longer-wavelength absorption Use the Beer–Lambert law in concentration calculations Plan quantitative analysis using UV/visible spectroscopy Describe stretching vibrations Describe bending vibrations Predict IR absorption count and vibrations for simple molecules Identify characteristic functional-group IR absorptions Suggest structures from IR spectra Predict characteristic IR absorptions from structure Relate polyatomic-gas IR absorption to the greenhouse effect Explain nuclear spin Explain energy absorption in NMR Interpret chemical shift Account for deuterated solvents and labile protons Determine proton equivalence and signal count Use peak integration to count protons Interpret first-order spin–spin splitting and multiplicity Use the delta scale and TMS reference Explain electronegativity and inductive effects on shielding Explain anisotropic effects on chemical shift Explain hydrogen-bonding effects on chemical shift Explain mass-spectrometric ionisation and fragmentation Interpret mass-to-charge ratio Identify a molecular-ion peak Interpret M+1, M+2 and M+4 isotope patterns Suggest major fragment ions without rearrangement Syllabus statements covered understand basic molecular orbital (MO) theory, involving: — atomic and molecular orbitals understand basic molecular orbital (MO) theory, involving: — bonding, anti-bonding and non-bonding orbitals understand basic molecular orbital (MO) theory, involving: — molecular orbitals with σ and π symmetry understand that molecular orbitals represent discrete electronic energy levels in molecules (see also 1.1 (e)(ii)) apply linear combination of atomic orbitals (LCAO) principles to obtain the shape and relative energies of molecular orbitals in the following: — simple homonuclear diatomic molecules such as H2, O2, and F2 apply linear combination of atomic orbitals (LCAO) principles to obtain the shape and relative energies of molecular orbitals in the following: — benzene and linear polyenes (molecular orbitals of π symmetry only) [quantitative treatment of LCAO is not required] construct and interpret molecular orbital diagrams, and identify the highest occupied molecular orbital (HOMO) and lowest unoccupied molecular orbital (LUMO) for the following: — simple homonuclear diatomic molecules such as H2, O2, and F2 construct and interpret molecular orbital diagrams, and identify the highest occupied molecular orbital (HOMO) and lowest unoccupied molecular orbital (LUMO) for the following: — benzene and linear polyenes (molecular orbitals of π symmetry only) [knowledge of orbital mixing between orbitals of the same symmetry is not required] understand the following in relation to the fundamental principles of spectroscopy: — properties of electromagnetic radiation - the electromagnetic spectrum (with range of wavelengths for different types of radiation used in spectroscopy) - the photon as a discrete packet (quantum) of electromagnetic energy - the relationship between wavelength, frequency and speed of light, including the use of the equation, E = hf understand the following in relation to the fundamental principles of spectroscopy: — the quantisation of energy in relation to - electronic, vibrational and rotational energy levels - nuclear energy levels in applied magnetic field understand the following in relation to the fundamental principles of spectroscopy: — energy level transitions associated with the absorption and emission of photons with energy matching the energy gap explain that ultraviolet/visible absorption in organic molecules requires electronic transitions (σ→σ*, n→σ*,π→π*, n→π* transitions; forbidden and allowed transitions) between energy levels in chromophores which contain a double or triple bond, a delocalised system, or a lone pair of electrons [detailed knowledge of instrumentation is not required] predict whether a given organic molecule will absorb in the ultraviolet/visible region by identifying the chromophore explain qualitatively how increasing conjugation in an organic molecule decreases the gap between energy levels and hence shifts the absorption towards longer wavelength use the Beer–Lambert law, absorbance = lg(Io / I) = εcl, where ε is taken merely as a constant characteristic of the substance concerned, to calculate the concentration of a given species (either organic or inorganic) in solution apply ultraviolet/visible spectroscopy to quantitative analysis of a given species (either organic or inorganic) in solution explain the origin of IR spectroscopy in simple molecules in terms of: — stretching vibrations explain the origin of IR spectroscopy in simple molecules in terms of: — bending vibrations [detailed knowledge of instrumentation is not required] predict the number of IR absorptions for a given simple molecule (e.g. CO2 or SO2), and identify the molecular vibrations which give rise to them identify characteristic IR absorptions in the IR spectrum of a compound which may contain different functional groups [absorptions of common functional groups will be provided in the Data Booklet] suggest structures for a compound from its IR spectrum predict the characteristic IR absorptions that will be present in the IR spectrum of a compound, given its structure describe qualitatively, in terms of their IR absorption, the role of polyatomic gases (e.g. CO2, H2O, CHF3) in the greenhouse effect outline the basic principles of NMR with reference to: — nuclear spin outline the basic principles of NMR with reference to: — the process of absorption of energy [quantitative calculations of transitional energy are not required; detailed knowledge of instrumentation is not required] understand the following features and use them in the interpretation and prediction of 1H NMR spectra: — chemical shift understand the following features and use them in the interpretation and prediction of 1H NMR spectra: — deuterated solvents in the identification of labile protons understand the following features and use them in the interpretation and prediction of 1H NMR spectra: — the number of 1H NMR signals: equivalent and non-equivalent protons understand the following features and use them in the interpretation and prediction of 1H NMR spectra: — peak area (integration) and proton counting understand the following features and use them in the interpretation and prediction of 1H NMR spectra: — spin-spin splitting: first order spin-spin coupling; multiplicity explain the use of the δ scale with tetramethylsilane (TMS) as the reference explain the factors affecting chemical shift: — electronegativity: inductive effect of substituents, including shielding and deshielding effects explain the factors affecting chemical shift: — anisotropic effects explain the factors affecting chemical shift: — hydrogen bonding outline the basic principles of mass spectrometry, with reference to: — ionisation and fragmentation outline the basic principles of mass spectrometry, with reference to: — mass/charge ratio, m/z [detailed knowledge of instrumentation is not required] understand the following features and use them in the interpretation and prediction of mass spectra: — molecular ion peak understand the following features and use them in the interpretation and prediction of mass spectra: — isotopic abundances including the use of (M+1) peak caused by 13C and (M+2) and (M+4) peaks for the identification of halogen compounds understand the following features and use them in the interpretation and prediction of mass spectra: — major fragment ions [fragment ions obtained from rearrangements are not included]
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The 1s orbitals of two hydrogen atoms are combined by LCAO. Which statement about the result is correct?
Two atomic orbitals give two molecular orbitals: a lower-energy σ orbital with electron density built up between the nuclei, and a higher-energy σ* orbital with a node between them. Each molecular orbital belongs to the molecule as a whole, not to one nucleus. Two atomic orbitals give one molecular orbital, which holds both electrons of the H–H bond. The molecular orbital is only the region where the two 1s orbitals overlap; outside that region the electrons stay in unchanged atomic orbitals. The out-of-phase combination is lower in energy than the in-phase combination, so electrons enter the σ* orbital first. A molecular orbital in a diatomic molecule is rotated about the internuclear axis. Which test distinguishes σ from π symmetry?
A σ orbital is unchanged by rotation through any angle about the internuclear axis; a π orbital has a nodal plane containing that axis and changes sign on rotation through 180°. A σ orbital has no nodes anywhere; a π orbital has at least one node. A σ orbital comes only from s orbitals and a π orbital only from p orbitals. A σ orbital is always bonding and a π orbital is always antibonding. The valence configuration of O₂ is (σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)². Which statement is correct?
The bond order is (8 − 4)/2 = 2, and the two π*2p electrons occupy the two degenerate π* orbitals singly, so O₂ is paramagnetic; the HOMO is π*2p. The bond order is (8 − 4)/2 = 2, and the two π*2p electrons pair in one π* orbital, so O₂ is diamagnetic. The bond order is (10 − 4)/2 = 3, because the π2p electrons count twice. The HOMO is π2p, because π2p lies above π*2p in the diagram. Benzene is treated by LCAO using one 2p orbital perpendicular to the ring plane on each carbon atom. Which statement about the π system is correct?
Six 2p orbitals give six π molecular orbitals — three bonding and three antibonding — and the six π electrons fill the three bonding orbitals, the lowest of which has no nodal plane perpendicular to the ring. Six 2p orbitals give three π molecular orbitals, one for each localised C=C double bond. Six 2p orbitals give six π molecular orbitals, and the six π electrons occupy one electron per orbital. The lowest π orbital has one nodal plane perpendicular to the ring, because every molecular orbital must have a node. Why does an isolated molecule absorb only certain photon energies rather than absorbing across a continuous range?
Its electrons are confined to the molecule, so the allowed molecular-orbital energies form a discrete set; a photon is absorbed only when its energy matches the gap between two allowed levels. Photons of other energies are absorbed but immediately re-emitted, so they are not detected. Only photons in the visible region carry enough energy to be absorbed by a molecule. Molecular orbital energies are continuous, but only the strongest photons can excite an electron. A mercury lamp emits ultraviolet radiation of wavelength 254 nm. Using h = 6.63 × 10⁻³⁴ J s and c = 3.00 × 10⁸ m s⁻¹, what is the energy of one such photon?
7.83 × 10⁻¹⁹ J 7.83 × 10⁻²⁸ J 2.61 × 10⁻²⁷ J 4.71 × 10⁵ J
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In an MO description of water, one occupied molecular orbital is almost entirely localised on the oxygen atom and lies at nearly the same energy as the oxygen atomic orbital it came from. How is it classified, and why?
Nonbonding: it has negligible bonding or antibonding character, so occupying it neither strengthens nor weakens the O–H bonds; it holds one of the oxygen lone pairs. Bonding: every occupied molecular orbital in a stable molecule is a bonding orbital. Antibonding: an orbital localised on one atom cannot hold electron density between the nuclei. It is still an atomic orbital, because it did not combine with anything. Ethene, CH₂=CH₂, is planar. Where does the nodal plane of the occupied π molecular orbital lie?
In the plane of the molecule: the π orbital is formed from the two 2p orbitals perpendicular to that plane, so it has lobes above and below the plane and contains the C–C axis in its nodal plane. Perpendicular to the C–C axis, halfway between the two carbon atoms. Perpendicular to the molecular plane and containing the C–C axis. The π orbital has no nodal plane, because it is a bonding orbital. N₂ has the valence configuration (σ2s)²(σ*2s)²(π2p)⁴(σ2p)². One electron is removed to form N₂⁺. What happens to the bond?
The electron is lost from σ2p, the HOMO, so the bond order falls from 3 to 2.5 and the N–N bond becomes longer and weaker. The electron is lost from σ*2s, so the bond order rises from 3 to 3.5 and the bond becomes shorter. The electron is lost from σ2p, but the bond order stays at 3 because a triple bond is already drawn between the nitrogen atoms. The electron is lost from π2p, so the bond order falls to 2.5 and N₂⁺ becomes diamagnetic. Buta-1,3-diene has four π molecular orbitals, ψ₁ to ψ₄ in order of increasing energy, and four π electrons. Which statement is correct?
ψ₁ has no nodal plane perpendicular to the carbon chain and ψ₂ has one; the four π electrons fill ψ₁ and ψ₂, so ψ₂ is the HOMO (one node) and ψ₃ is the LUMO (two nodes). The four π electrons fill ψ₁ and ψ₂, so ψ₂ is the HOMO with two nodes and ψ₃ is the LUMO with three. The four π electrons fill ψ₁ to ψ₄ singly, so ψ₄ is the HOMO. ψ₂ and ψ₃ are degenerate, so the HOMO–LUMO gap of butadiene is zero. Rank the typical spacing between adjacent energy levels in a molecule for electronic, vibrational and rotational motion.
Electronic > vibrational > rotational, so the same molecule absorbs ultraviolet or visible radiation for electronic changes, infrared for vibrational changes and microwaves for rotational changes. Rotational > vibrational > electronic, because a whole molecule rotating carries more energy than one electron moving. Vibrational > electronic > rotational, because bond stretching involves the whole bond rather than a single electron. All three spacings are about equal, so any photon can excite any of the three modes. A molecule absorbs a photon of energy 3.98 × 10⁻¹⁹ J. Using h = 6.63 × 10⁻³⁴ J s, what is the frequency of the radiation?
6.00 × 10¹⁴ Hz 1.67 × 10⁻¹⁵ Hz 1.33 × 10⁻²⁷ Hz 2.64 × 10⁻⁵² Hz
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In an MO description of water, one occupied molecular orbital is almost entirely localised on the oxygen atom and lies at nearly the same energy as the oxygen atomic orbital it came from. How is it classified, and why?
Nonbonding: it has negligible bonding or antibonding character, so occupying it neither strengthens nor weakens the O–H bonds; it holds one of the oxygen lone pairs. Bonding: every occupied molecular orbital in a stable molecule is a bonding orbital. Antibonding: an orbital localised on one atom cannot hold electron density between the nuclei. It is still an atomic orbital, because it did not combine with anything. Ethene, CH₂=CH₂, is planar. Where does the nodal plane of the occupied π molecular orbital lie?
In the plane of the molecule: the π orbital is formed from the two 2p orbitals perpendicular to that plane, so it has lobes above and below the plane and contains the C–C axis in its nodal plane. Perpendicular to the C–C axis, halfway between the two carbon atoms. Perpendicular to the molecular plane and containing the C–C axis. The π orbital has no nodal plane, because it is a bonding orbital. N₂ has the valence configuration (σ2s)²(σ*2s)²(π2p)⁴(σ2p)². One electron is removed to form N₂⁺. What happens to the bond?
The electron is lost from σ2p, the HOMO, so the bond order falls from 3 to 2.5 and the N–N bond becomes longer and weaker. The electron is lost from σ*2s, so the bond order rises from 3 to 3.5 and the bond becomes shorter. The electron is lost from σ2p, but the bond order stays at 3 because a triple bond is already drawn between the nitrogen atoms. The electron is lost from π2p, so the bond order falls to 2.5 and N₂⁺ becomes diamagnetic. Buta-1,3-diene has four π molecular orbitals, ψ₁ to ψ₄ in order of increasing energy, and four π electrons. Which statement is correct?
ψ₁ has no nodal plane perpendicular to the carbon chain and ψ₂ has one; the four π electrons fill ψ₁ and ψ₂, so ψ₂ is the HOMO (one node) and ψ₃ is the LUMO (two nodes). The four π electrons fill ψ₁ and ψ₂, so ψ₂ is the HOMO with two nodes and ψ₃ is the LUMO with three. The four π electrons fill ψ₁ to ψ₄ singly, so ψ₄ is the HOMO. ψ₂ and ψ₃ are degenerate, so the HOMO–LUMO gap of butadiene is zero. Rank the typical spacing between adjacent energy levels in a molecule for electronic, vibrational and rotational motion.
Electronic > vibrational > rotational, so the same molecule absorbs ultraviolet or visible radiation for electronic changes, infrared for vibrational changes and microwaves for rotational changes. Rotational > vibrational > electronic, because a whole molecule rotating carries more energy than one electron moving. Vibrational > electronic > rotational, because bond stretching involves the whole bond rather than a single electron. All three spacings are about equal, so any photon can excite any of the three modes. A molecule absorbs a photon of energy 3.98 × 10⁻¹⁹ J. Using h = 6.63 × 10⁻³⁴ J s, what is the frequency of the radiation?
6.00 × 10¹⁴ Hz 1.67 × 10⁻¹⁵ Hz 1.33 × 10⁻²⁷ Hz 2.64 × 10⁻⁵² Hz
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What happens inside a molecule when its chromophore absorbs a photon of ultraviolet or visible radiation?
An electron is promoted from a filled molecular orbital to an empty one of higher energy, and the photon energy matches that HOMO–LUMO type gap exactly. A bond in the chromophore vibrates more strongly, because the photon sets the bonded atoms oscillating. An electron is removed from the molecule altogether, leaving a positive ion. The whole molecule rotates faster, which is detected as absorption. Hexane, propan-1-ol and propanone are each scanned from 200 nm to 800 nm in a hexane solution. Which compound shows an absorption band, and why?
Propanone, because its C=O group is a chromophore whose n → π* transition needs relatively little energy; hexane and propan-1-ol have only σ bonds and lone pairs whose transitions lie below 200 nm. Propan-1-ol, because its O–H group is polar and polar bonds absorb ultraviolet radiation. All three, because every organic compound absorbs ultraviolet radiation somewhere. Hexane, because it has the most C–H bonds and therefore the most electrons to excite. Lycopene, which contains eleven conjugated C=C bonds, is deep red; hex-1-ene, with one C=C bond, is colourless. Which explanation is correct?
Extended conjugation lowers the energy gap for the π → π* transition until it matches visible photons, so lycopene absorbs blue-green light and transmits red; hex-1-ene's much larger gap places its absorption in the far ultraviolet. Lycopene contains more electrons in total, so it absorbs more strongly at every wavelength, including the visible. Extended conjugation raises the energy gap, so lycopene absorbs the most energetic visible photons and transmits red. Lycopene is red because it emits red light after absorbing ultraviolet radiation. A solution of concentration 2.00 × 10⁻⁵ mol dm⁻³ is measured at its λmax in a 1.00 cm cell. The molar absorptivity there is 1.50 × 10⁴ dm³ mol⁻¹ cm⁻¹. What is the absorbance?
Why is a quantitative ultraviolet/visible determination normally made at the wavelength of maximum absorbance, λmax?
The absorbance is largest there, giving the greatest sensitivity, and the curve is flattest there, so a small drift in the selected wavelength changes the absorbance least. Only at λmax does the Beer-Lambert law hold; at other wavelengths absorbance is not proportional to concentration. λmax is the wavelength at which the solvent absorbs least, so no blank is needed. At λmax the molar absorptivity is zero, so the reading depends only on concentration.
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In an alkene, both σ → σ* and π → π* transitions are possible. Which comparison is correct?
The π → π* transition needs less energy, because the π–π* gap is smaller than the σ–σ* gap, so it occurs at a longer wavelength and is the one seen in an ordinary ultraviolet spectrum. The σ → σ* transition needs less energy, because σ bonds are longer-range than π bonds. Both transitions need the same energy, because both promote one electron in the same molecule. The π → π* transition needs less energy but is never observed, because π electrons are held too tightly. Two compounds are dissolved in ethanol: cyclohexene, and an azo dye containing two benzene rings joined by –N=N–. Which is coloured, and why?
The azo dye, because its extended chromophore absorbs visible light; cyclohexene's isolated C=C absorbs in the far ultraviolet and so transmits all visible wavelengths. Cyclohexene, because a C=C bond is the strongest chromophore in organic chemistry. Both, because both contain π bonds and all π bonds absorb visible light. Neither, because colour in organic compounds comes from transition metal ions. Dye P has five conjugated C=C bonds and dye Q has eight. Which absorbs at the longer wavelength, and what follows?
Q, because its longer conjugated system has the smaller HOMO–LUMO gap, and E = hc/λ then places its absorption at a longer wavelength than P's. P, because a shorter conjugated chain confines the π electrons less and so needs less energy to excite them. Q, because more conjugation means a larger HOMO–LUMO gap and therefore a longer wavelength. Neither: both absorb at the same wavelength, because both are conjugated polyenes. A solution measured in a 1.00 cm cell gives an absorbance of 0.620 at a wavelength where ε = 2.48 × 10⁴ dm³ mol⁻¹ cm⁻¹. What is its concentration?
2.50 × 10⁻⁵ mol dm⁻³ 1.54 × 10⁴ mol dm⁻³ 4.00 × 10⁴ mol dm⁻³ 2.50 × 10⁻⁵ mol A student must find the concentration of a coloured solute in an unknown solution using a spectrophotometer. Which procedure is correct?
Prepare several standards of accurately known concentration, measure each one's absorbance at λmax against a solvent blank, plot absorbance against concentration, then read the unknown's concentration from the straight-line region of that graph. Measure the unknown's absorbance and divide it by the absorbance of pure solvent at the same wavelength. Measure the unknown's absorbance at several wavelengths and take the mean, then look that mean up in a table of absorbances. Measure the absorbance of the unknown before and after dilution and take the difference as the concentration.
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In an alkene, both σ → σ* and π → π* transitions are possible. Which comparison is correct?
The π → π* transition needs less energy, because the π–π* gap is smaller than the σ–σ* gap, so it occurs at a longer wavelength and is the one seen in an ordinary ultraviolet spectrum. The σ → σ* transition needs less energy, because σ bonds are longer-range than π bonds. Both transitions need the same energy, because both promote one electron in the same molecule. The π → π* transition needs less energy but is never observed, because π electrons are held too tightly. Two compounds are dissolved in ethanol: cyclohexene, and an azo dye containing two benzene rings joined by –N=N–. Which is coloured, and why?
The azo dye, because its extended chromophore absorbs visible light; cyclohexene's isolated C=C absorbs in the far ultraviolet and so transmits all visible wavelengths. Cyclohexene, because a C=C bond is the strongest chromophore in organic chemistry. Both, because both contain π bonds and all π bonds absorb visible light. Neither, because colour in organic compounds comes from transition metal ions. Dye P has five conjugated C=C bonds and dye Q has eight. Which absorbs at the longer wavelength, and what follows?
Q, because its longer conjugated system has the smaller HOMO–LUMO gap, and E = hc/λ then places its absorption at a longer wavelength than P's. P, because a shorter conjugated chain confines the π electrons less and so needs less energy to excite them. Q, because more conjugation means a larger HOMO–LUMO gap and therefore a longer wavelength. Neither: both absorb at the same wavelength, because both are conjugated polyenes. A solution measured in a 1.00 cm cell gives an absorbance of 0.620 at a wavelength where ε = 2.48 × 10⁴ dm³ mol⁻¹ cm⁻¹. What is its concentration?
2.50 × 10⁻⁵ mol dm⁻³ 1.54 × 10⁴ mol dm⁻³ 4.00 × 10⁴ mol dm⁻³ 2.50 × 10⁻⁵ mol A student must find the concentration of a coloured solute in an unknown solution using a spectrophotometer. Which procedure is correct?
Prepare several standards of accurately known concentration, measure each one's absorbance at λmax against a solvent blank, plot absorbance against concentration, then read the unknown's concentration from the straight-line region of that graph. Measure the unknown's absorbance and divide it by the absorbance of pure solvent at the same wavelength. Measure the unknown's absorbance at several wavelengths and take the mean, then look that mean up in a table of absorbances. Measure the absorbance of the unknown before and after dilution and take the difference as the concentration.
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How does a stretching vibration differ from a bending vibration, and which usually absorbs at the higher wavenumber?
A stretch changes the length of a bond while the bond angles are kept; a bend changes a bond angle while the bond lengths are kept. Stretching a bond costs more energy than bending it, so stretching absorptions appear at higher wavenumbers. A stretch changes a bond angle and a bend changes a bond length; bending absorbs at the higher wavenumber. A stretch breaks the bond momentarily while a bend does not; bending therefore absorbs at the higher wavenumber. A stretch involves all the atoms of the molecule and a bend involves only two; stretching absorbs at the lower wavenumber. A water molecule, H₂O, is non-linear and contains three atoms. How many vibrational modes does it have?
3, from 3N − 6 with N = 3: two stretches and one bend. 4, from 3N − 5 with N = 3. 9, from 3N with N = 3. 2, one stretch for each O–H bond. An infrared spectrum of an organic compound shows a sharp band of medium intensity at 2250 cm⁻¹ and nothing else between 1800 cm⁻¹ and 3000 cm⁻¹. Which group is present?
A nitrile group, C≡N, whose stretching absorption lies at about 2200–2260 cm⁻¹. A carbonyl group, C=O, whose stretching absorption lies at about 2250 cm⁻¹. An amine group, N–H, whose stretching absorption lies at about 2250 cm⁻¹. An alkene group, C=C, whose stretching absorption lies at about 2250 cm⁻¹. A compound of formula C₃H₆O gives an infrared spectrum with a strong band at 1715 cm⁻¹ and no absorption at all above 3100 cm⁻¹. Which structure fits?
Propanone, CH₃COCH₃: the strong 1715 cm⁻¹ band is its C=O stretch, and the absence of any band above 3100 cm⁻¹ rules out an O–H group. Propan-2-ol, CH₃CH(OH)CH₃: the 1715 cm⁻¹ band is its C–O stretch. Prop-2-en-1-ol, CH₂=CHCH₂OH: the 1715 cm⁻¹ band is its C=C stretch. Propane-1,2-diol: the 1715 cm⁻¹ band comes from its two C–O bonds together. Carbon dioxide absorbs infrared radiation strongly, but nitrogen and oxygen, which make up most of the atmosphere, do not. Why?
A vibration absorbs infrared radiation only if it changes the molecule's dipole moment. N₂ and O₂ are homonuclear diatomics whose only vibration keeps the dipole moment at zero, while the bending and asymmetric stretching vibrations of CO₂ create a changing dipole moment. N₂ and O₂ have no vibrations at all, because their atoms are identical. CO₂ absorbs because it is heavier, and heavier molecules absorb lower-energy radiation. CO₂ absorbs because it is a polar molecule, while N₂ and O₂ are non-polar.
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In one vibration of a water molecule the H–O–H angle opens and closes while both O–H bond lengths stay essentially unchanged. How is this vibration classified, and where does it absorb relative to the O–H stretches?
It is a bending vibration, and it absorbs at a lower wavenumber than the O–H stretches, near 1600 cm⁻¹ rather than above 3000 cm⁻¹. It is a symmetric stretching vibration, and it absorbs at a higher wavenumber than the bends. It is a bending vibration, and it absorbs at a higher wavenumber than the O–H stretches. It is a rotation, and rotations are seen in the infrared region. Sulfur dioxide, SO₂, is a bent triatomic molecule. How many vibrational modes does it have, and how does this compare with a linear triatomic molecule?
3, from 3N − 6, which is one fewer than the 4 modes of a linear triatomic, because a linear molecule has only two rotational degrees of freedom instead of three. 4, from 3N − 5, the same as a linear triatomic molecule, because both have three atoms. 3, from 3N − 6, which is one more than the 2 modes of a linear triatomic. 6, one stretch and one bend for each bond. Which infrared absorptions would you predict in the 3000–3500 cm⁻¹ region for ethylamine, CH₃CH₂NH₂?
Two medium, fairly sharp N–H stretching bands between about 3300 and 3500 cm⁻¹, from the symmetric and asymmetric stretches of the –NH₂ group, together with C–H stretches just below 3000 cm⁻¹. One very broad band between 3200 and 3600 cm⁻¹, like that of an alcohol. No bands at all in that region, because nitrogen is not electronegative enough to give an infrared-active stretch. A single strong band near 3300 cm⁻¹ from the C–N stretch. A compound of formula C₃H₆O shows a strong broad band centred near 3350 cm⁻¹, a band near 1645 cm⁻¹ and no absorption between 1680 and 1800 cm⁻¹. Which structure fits?
Prop-2-en-1-ol, CH₂=CHCH₂OH: the broad 3350 cm⁻¹ band is a hydrogen-bonded O–H stretch and the 1645 cm⁻¹ band is a C=C stretch, while the empty carbonyl region rules out C=O. Propanone, CH₃COCH₃: the 1645 cm⁻¹ band is its C=O stretch. Propanal, CH₃CH₂CHO: the broad 3350 cm⁻¹ band is its aldehyde C–H stretch. Methoxyethene, CH₂=CHOCH₃: the broad 3350 cm⁻¹ band is its C–O stretch. Methane contributes to the greenhouse effect; argon, which is also present in the atmosphere, does not. Why?
Methane is a polyatomic molecule with bending and stretching vibrations that change its dipole moment, so it absorbs the infrared radiation the Earth emits; argon is monatomic and has no vibrations at all. Argon is chemically inert, and only reactive gases absorb infrared radiation. Methane absorbs because it is less dense than argon, so its molecules move faster. Argon does not absorb because it is present in too small a proportion.
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In one vibration of a water molecule the H–O–H angle opens and closes while both O–H bond lengths stay essentially unchanged. How is this vibration classified, and where does it absorb relative to the O–H stretches?
It is a bending vibration, and it absorbs at a lower wavenumber than the O–H stretches, near 1600 cm⁻¹ rather than above 3000 cm⁻¹. It is a symmetric stretching vibration, and it absorbs at a higher wavenumber than the bends. It is a bending vibration, and it absorbs at a higher wavenumber than the O–H stretches. It is a rotation, and rotations are seen in the infrared region. Sulfur dioxide, SO₂, is a bent triatomic molecule. How many vibrational modes does it have, and how does this compare with a linear triatomic molecule?
3, from 3N − 6, which is one fewer than the 4 modes of a linear triatomic, because a linear molecule has only two rotational degrees of freedom instead of three. 4, from 3N − 5, the same as a linear triatomic molecule, because both have three atoms. 3, from 3N − 6, which is one more than the 2 modes of a linear triatomic. 6, one stretch and one bend for each bond. Which infrared absorptions would you predict in the 3000–3500 cm⁻¹ region for ethylamine, CH₃CH₂NH₂?
Two medium, fairly sharp N–H stretching bands between about 3300 and 3500 cm⁻¹, from the symmetric and asymmetric stretches of the –NH₂ group, together with C–H stretches just below 3000 cm⁻¹. One very broad band between 3200 and 3600 cm⁻¹, like that of an alcohol. No bands at all in that region, because nitrogen is not electronegative enough to give an infrared-active stretch. A single strong band near 3300 cm⁻¹ from the C–N stretch. A compound of formula C₃H₆O shows a strong broad band centred near 3350 cm⁻¹, a band near 1645 cm⁻¹ and no absorption between 1680 and 1800 cm⁻¹. Which structure fits?
Prop-2-en-1-ol, CH₂=CHCH₂OH: the broad 3350 cm⁻¹ band is a hydrogen-bonded O–H stretch and the 1645 cm⁻¹ band is a C=C stretch, while the empty carbonyl region rules out C=O. Propanone, CH₃COCH₃: the 1645 cm⁻¹ band is its C=O stretch. Propanal, CH₃CH₂CHO: the broad 3350 cm⁻¹ band is its aldehyde C–H stretch. Methoxyethene, CH₂=CHOCH₃: the broad 3350 cm⁻¹ band is its C–O stretch. Methane contributes to the greenhouse effect; argon, which is also present in the atmosphere, does not. Why?
Methane is a polyatomic molecule with bending and stretching vibrations that change its dipole moment, so it absorbs the infrared radiation the Earth emits; argon is monatomic and has no vibrations at all. Argon is chemically inert, and only reactive gases absorb infrared radiation. Methane absorbs because it is less dense than argon, so its molecules move faster. Argon does not absorb because it is present in too small a proportion.
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Why can a ¹H nucleus absorb radio-frequency radiation when the sample is placed in a strong magnetic field?
A ¹H nucleus has spin ½, so in the field it takes one of two orientations — aligned with the field or opposed to it — that differ slightly in energy; a radio-frequency photon whose energy matches that difference flips the nucleus to the higher state. The magnetic field pulls electrons away from the nucleus, and the radio-frequency radiation excites those electrons. The magnetic field makes the molecule rotate, and radio-frequency photons match the rotational spacings. The nucleus is ionised by the radio-frequency radiation once the field has weakened its attraction for its electrons. Why is tetramethylsilane, Si(CH₃)₄, used as the reference at δ = 0 in ¹H NMR?
All twelve of its hydrogen atoms are equivalent, so it gives one sharp signal; silicon is less electronegative than carbon, so those protons are unusually shielded and the signal lies to one side of almost every other proton signal; and it is inert, volatile and easily removed. It contains silicon, and silicon nuclei have no spin, so the compound produces no signal of its own. Its protons are the most strongly deshielded of any common compound, so its signal lies beyond all others at the high-δ end. It reacts with the sample to fix the concentration, so the signal heights can be compared. How many signals appear in the ¹H NMR spectrum of ethanol, CH₃CH₂OH?
3, because the three methyl protons are equivalent to each other, the two methylene protons are equivalent to each other, and the hydroxyl proton is in a third environment. 6, one for each hydrogen atom in the molecule. 2, because the hydroxyl proton exchanges too quickly to be seen. 1, because all the hydrogen atoms are attached to the same carbon chain. A methyl group has exactly two hydrogen atoms on the adjacent carbon atom. Into how many lines is its signal split, and why?
3, a triplet, because the n + 1 rule gives 2 + 1 = 3 for two equivalent neighbouring protons. 4, a quartet, because the methyl group has three protons and 3 + 1 = 4. 2, a doublet, because there are two neighbouring protons and each gives one line. 1, a singlet, because all three methyl protons are equivalent to one another. In 1-chloropropane the protons of the CH₂ group bonded to chlorine appear near δ 3.5, while those of the terminal CH₃ group appear near δ 1.0. Why?
Chlorine is strongly electronegative and withdraws electron density along the σ bonds, so the electrons that would shield the nearby protons from the applied field are pulled away; the deshielded protons resonate at a higher δ. Chlorine donates electron density to the nearby protons, shielding them so strongly that they move to a higher δ. The CH₂ protons are heavier than the CH₃ protons because they sit next to a chlorine atom. The CH₂ group has fewer protons, and signals from fewer protons always appear at higher δ. Why is a ¹H NMR spectrum normally run in a deuteriated solvent such as CDCl₃ rather than in ordinary CHCl₃?
Deuterium does not resonate at the frequencies used for ¹H, so the solvent contributes essentially no signal; ordinary CHCl₃ is present in vast excess and its own proton signal would swamp the sample's. Deuterium has no nucleus, so it cannot interact with the magnetic field at all. CDCl₃ dissolves organic compounds better than CHCl₃ does. CDCl₃ reacts with any labile protons in the sample and so simplifies the spectrum.
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A spectrometer is replaced by one with a stronger magnetic field. What happens to the energy gap between the two ¹H spin states and to the frequency at which resonance occurs?
Both increase: a stronger field separates the two spin states further, and since ΔE = hf the matching photon frequency rises in proportion. The gap increases but the resonance frequency is unchanged, because the frequency is a property of the nucleus alone. Both decrease, because a stronger field holds the nuclei more firmly in the lower state. The gap is unchanged, because it is fixed by the nucleus; only the signal becomes stronger. A ¹H NMR spectrum contains a one-proton signal at δ 9.7. Which environment does this indicate?
The hydrogen atom of an aldehyde group, –CHO, which is strongly deshielded and characteristically appears between about δ 9.3 and δ 10.5. A hydrogen atom on a carbon atom bonded to oxygen in an alcohol, –CH₂–O–, which appears near δ 9.7. An aromatic ring hydrogen atom, which appears near δ 9.7. A methyl hydrogen atom next to a carbonyl group, which appears near δ 9.7. The three signals in the ¹H NMR spectrum of ethanol, CH₃CH₂OH, have integration trace heights in the ratio 3 : 2 : 1. What does this tell you?
The signals arise from 3, 2 and 1 protons respectively, so they can be assigned to the CH₃, the CH₂ and the OH group; integration gives the relative number of protons in each environment, not their absolute number. The three environments differ in chemical shift by a factor of 3 : 2 : 1. The molecule contains exactly six hydrogen atoms, since 3 + 2 + 1 = 6, and this can be read directly from the trace. The CH₃ signal is three times as tall as the OH signal, so peak height is what the integration measures. In the ¹H NMR spectrum of an ethyl ester, CH₃CH₂O–, the ethyl group gives a quartet and a triplet. Which signal is the quartet, and where does each lie?
The quartet is the –CH₂– signal, split by the three protons of the adjacent CH₃ group (3 + 1 = 4), and it lies further downfield because it is attached to oxygen; the CH₃ signal is the triplet, split by the two CH₂ protons (2 + 1 = 3). The quartet is the CH₃ signal, because it contains three protons and 3 + 1 = 4. The quartet is the –CH₂– signal, but it lies upfield of the CH₃ signal because it has fewer protons. Both signals are quartets, because each group has three protons on its far side. The methyl protons of CH₃F, CH₃Cl, CH₃Br and CH₃I resonate at δ 4.3, 3.1, 2.7 and 2.2 respectively. Which explanation fits?
Electronegativity falls from fluorine to iodine, so less σ electron density is withdrawn from the methyl protons along the series, the shielding increases and δ falls. The halogen atoms get heavier down the group, and heavier neighbouring atoms shield protons more effectively by their mass. The C–halogen bond gets longer down the group, so the halogen deshields more strongly and δ falls. The number of methyl protons falls along the series, and fewer protons resonate at lower δ. A spectrum is rerun after the sample has been shaken with a few drops of D₂O. One signal disappears. What does this show?
That signal came from a labile proton — one bonded to oxygen or nitrogen — which has exchanged with deuterium from the D₂O; deuterium does not resonate in the ¹H window, so the signal vanishes. That signal came from a proton on a carbon atom, which has been washed out of the sample by the D₂O. The D₂O has destroyed the molecule, so one of its fragments is no longer present. The D₂O has diluted the sample so much that the smallest signal has fallen below the detection limit.
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A spectrometer is replaced by one with a stronger magnetic field. What happens to the energy gap between the two ¹H spin states and to the frequency at which resonance occurs?
Both increase: a stronger field separates the two spin states further, and since ΔE = hf the matching photon frequency rises in proportion. The gap increases but the resonance frequency is unchanged, because the frequency is a property of the nucleus alone. Both decrease, because a stronger field holds the nuclei more firmly in the lower state. The gap is unchanged, because it is fixed by the nucleus; only the signal becomes stronger. A ¹H NMR spectrum contains a one-proton signal at δ 9.7. Which environment does this indicate?
The hydrogen atom of an aldehyde group, –CHO, which is strongly deshielded and characteristically appears between about δ 9.3 and δ 10.5. A hydrogen atom on a carbon atom bonded to oxygen in an alcohol, –CH₂–O–, which appears near δ 9.7. An aromatic ring hydrogen atom, which appears near δ 9.7. A methyl hydrogen atom next to a carbonyl group, which appears near δ 9.7. The three signals in the ¹H NMR spectrum of ethanol, CH₃CH₂OH, have integration trace heights in the ratio 3 : 2 : 1. What does this tell you?
The signals arise from 3, 2 and 1 protons respectively, so they can be assigned to the CH₃, the CH₂ and the OH group; integration gives the relative number of protons in each environment, not their absolute number. The three environments differ in chemical shift by a factor of 3 : 2 : 1. The molecule contains exactly six hydrogen atoms, since 3 + 2 + 1 = 6, and this can be read directly from the trace. The CH₃ signal is three times as tall as the OH signal, so peak height is what the integration measures. In the ¹H NMR spectrum of an ethyl ester, CH₃CH₂O–, the ethyl group gives a quartet and a triplet. Which signal is the quartet, and where does each lie?
The quartet is the –CH₂– signal, split by the three protons of the adjacent CH₃ group (3 + 1 = 4), and it lies further downfield because it is attached to oxygen; the CH₃ signal is the triplet, split by the two CH₂ protons (2 + 1 = 3). The quartet is the CH₃ signal, because it contains three protons and 3 + 1 = 4. The quartet is the –CH₂– signal, but it lies upfield of the CH₃ signal because it has fewer protons. Both signals are quartets, because each group has three protons on its far side. The methyl protons of CH₃F, CH₃Cl, CH₃Br and CH₃I resonate at δ 4.3, 3.1, 2.7 and 2.2 respectively. Which explanation fits?
Electronegativity falls from fluorine to iodine, so less σ electron density is withdrawn from the methyl protons along the series, the shielding increases and δ falls. The halogen atoms get heavier down the group, and heavier neighbouring atoms shield protons more effectively by their mass. The C–halogen bond gets longer down the group, so the halogen deshields more strongly and δ falls. The number of methyl protons falls along the series, and fewer protons resonate at lower δ. A spectrum is rerun after the sample has been shaken with a few drops of D₂O. One signal disappears. What does this show?
That signal came from a labile proton — one bonded to oxygen or nitrogen — which has exchanged with deuterium from the D₂O; deuterium does not resonate in the ¹H window, so the signal vanishes. That signal came from a proton on a carbon atom, which has been washed out of the sample by the D₂O. The D₂O has destroyed the molecule, so one of its fragments is no longer present. The D₂O has diluted the sample so much that the smallest signal has fallen below the detection limit.
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How is the molecular ion produced in an electron-impact mass spectrometer, and why do other peaks appear as well?
A high-energy electron knocks one electron out of the molecule, giving a positively charged species with an odd number of electrons; the energy left over can break bonds in that ion, so smaller charged fragments appear at lower m/z. A high-energy electron is captured by the molecule, giving a negative ion that then breaks up. A proton is added to the molecule, and the extra proton then migrates and breaks bonds. The molecule is heated until it decomposes, and the neutral products are separated by mass. A mass spectrum plots relative abundance against m/z. What does m/z mean, and what does it equal for the ions usually recorded?
The mass of the ion divided by its charge, both in units of the proton's charge and the unified atomic mass unit; almost all recorded ions carry a single positive charge, so m/z is numerically equal to the ion's mass. The mass of the molecule divided by the number of atoms it contains. The mass of the ion multiplied by its charge, so a doubly charged ion appears at twice its mass. The relative abundance of the ion divided by that of the most abundant ion. From which peak in a mass spectrum is the relative molecular mass of a compound read?
The molecular-ion peak, which is the peak of highest m/z apart from any isotope peaks lying above it; it corresponds to the intact molecule having lost one electron. The tallest peak in the spectrum, called the base peak, because the most abundant ion must be the intact molecule. The peak of lowest m/z, because that ion has not been broken up. The mean of the m/z values of all the peaks, weighted by their abundances. The molecular-ion region of a compound's mass spectrum contains these peaks, given as m/z (relative intensity): 78 (100), 79 (3.2), 80 (32.0) and 81 (1.0). There are no peaks above m/z 81. What do the peaks at m/z 78 and 80 indicate?
One chlorine atom in the molecule: chlorine occurs as ³⁵Cl and ³⁷Cl in an abundance ratio close to 3 : 1, so molecules differing by two mass units appear in the same ratio. One bromine atom: bromine's two isotopes differ by two mass units. Three chlorine atoms, one for each unit in the 3 : 1 ratio. Two carbon atoms, whose ¹³C isotope raises the mass by two units. The mass spectrum of a simple organic compound contains a peak at m/z 15. Which fragment ion is this?
CH₃⁺, whose formula mass is 12 + 3 = 15. NH⁺, whose formula mass is 14 + 1 = 15. CH₃CH₂⁺, whose formula mass is 15. OH⁺, whose formula mass is 15.
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When a molecular ion fragments it gives a smaller cation and a neutral radical. Why does only one peak appear from each such fragmentation?
Only charged species are accelerated by the electric field and deflected by the magnetic field, so the neutral radical is never recorded; the peak comes from the cation alone. The neutral radical recombines with the cation before reaching the detector, so neither is recorded. The neutral radical is too light to be deflected and passes straight through undetected. Both fragments are detected, but they always have the same m/z and so give one peak. An ion of mass 128 carries a charge of 2+. At what m/z value does it appear?
The mass spectrum of propanone shows significant peaks at m/z 15, 43 and 58, with nothing above m/z 59. What is the relative molecular mass of propanone?
58, read from the molecular-ion peak, which is the highest-m/z peak apart from the small isotope peak just above it. 43, because that is the tallest peak in the spectrum. 116, because the molecular ion has lost one electron and must be doubled to recover the molecule. 15 + 43 = 58, because the relative molecular mass is always the sum of the two largest fragments. The molecular-ion region of a compound's mass spectrum contains these peaks, given as m/z (relative intensity): 136 (100), 137 (4.3), 138 (97.3) and 139 (4.2). There are no peaks above m/z 139. Which element accounts for the peaks at m/z 136 and 138?
Bromine, because ⁷⁹Br and ⁸¹Br are present in almost equal abundance and differ by two mass units. Chlorine, because ³⁵Cl and ³⁷Cl also differ by two mass units. Carbon, because ¹²C and ¹³C are almost equally abundant. Oxygen, because ¹⁶O and ¹⁸O differ by two mass units. Ethanol, C₂H₅OH, has a molecular ion at m/z 46 and a strong fragment peak at m/z 31. Which fragment gives that peak?
CH₂OH⁺, of formula mass 12 + 2 + 16 + 1 = 31, formed by loss of a methyl radical from the molecular ion. CH₃CH₂⁺, of formula mass 29, formed by loss of the hydroxyl group. CH₃O⁺, of formula mass 31, formed by loss of a CH₃ group from the other end. C₂H₅OH⁺ having lost two hydrogen atoms, of formula mass 44.
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When a molecular ion fragments it gives a smaller cation and a neutral radical. Why does only one peak appear from each such fragmentation?
Only charged species are accelerated by the electric field and deflected by the magnetic field, so the neutral radical is never recorded; the peak comes from the cation alone. The neutral radical recombines with the cation before reaching the detector, so neither is recorded. The neutral radical is too light to be deflected and passes straight through undetected. Both fragments are detected, but they always have the same m/z and so give one peak. An ion of mass 128 carries a charge of 2+. At what m/z value does it appear?
The mass spectrum of propanone shows significant peaks at m/z 15, 43 and 58, with nothing above m/z 59. What is the relative molecular mass of propanone?
58, read from the molecular-ion peak, which is the highest-m/z peak apart from the small isotope peak just above it. 43, because that is the tallest peak in the spectrum. 116, because the molecular ion has lost one electron and must be doubled to recover the molecule. 15 + 43 = 58, because the relative molecular mass is always the sum of the two largest fragments. The molecular-ion region of a compound's mass spectrum contains these peaks, given as m/z (relative intensity): 136 (100), 137 (4.3), 138 (97.3) and 139 (4.2). There are no peaks above m/z 139. Which element accounts for the peaks at m/z 136 and 138?
Bromine, because ⁷⁹Br and ⁸¹Br are present in almost equal abundance and differ by two mass units. Chlorine, because ³⁵Cl and ³⁷Cl also differ by two mass units. Carbon, because ¹²C and ¹³C are almost equally abundant. Oxygen, because ¹⁶O and ¹⁸O differ by two mass units. Ethanol, C₂H₅OH, has a molecular ion at m/z 46 and a strong fragment peak at m/z 31. Which fragment gives that peak?
CH₂OH⁺, of formula mass 12 + 2 + 16 + 1 = 31, formed by loss of a methyl radical from the molecular ion. CH₃CH₂⁺, of formula mass 29, formed by loss of the hydroxyl group. CH₃O⁺, of formula mass 31, formed by loss of a CH₃ group from the other end. C₂H₅OH⁺ having lost two hydrogen atoms, of formula mass 44.
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Beyond the syllabus: optional enrichment that does not count towards your progress.
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This assessment needs JavaScript. Its questions are chosen when you start an attempt, so they cannot be listed here.
Finish and submit Time is up, but your answers have not been submitted yet. Check your connection and try again.
Try again Answer 6 questions. You'll see your score, the answers and explanations at the end. Your result can count towards your course progress.
Check my progress Continue my progress check
Recent attempts History is stored only in this browser.
No completed attempts are saved yet.
Beyond the syllabus: optional enrichment that does not count towards your progress.
Previous question Go to question Next question Progress check complete
Start again
This assessment needs JavaScript. Its questions are chosen when you start an attempt, so they cannot be listed here.
Finish and submit Time is up, but your answers have not been submitted yet. Check your connection and try again.
Try again Answer 6 questions. You'll see your score, the answers and explanations at the end. A scheduled review counts towards your course progress only when it is due.
Start review Continue my review
Recent attempts History is stored only in this browser.
No completed attempts are saved yet.
Beyond the syllabus: optional enrichment that does not count towards your progress.
Previous question Go to question Next question Review complete
Start again
This assessment needs JavaScript. Its questions are chosen when you start an attempt, so they cannot be listed here.
Finish and submit Time is up, but your answers have not been submitted yet. Check your connection and try again.
Try again Answer 5 questions. You'll see your score, the answers and explanations at the end. Your result can count towards your course progress.
Check my progress Continue my progress check
Recent attempts History is stored only in this browser.
No completed attempts are saved yet.
Beyond the syllabus: optional enrichment that does not count towards your progress.
Previous question Go to question Next question Progress check complete
Start again
This assessment needs JavaScript. Its questions are chosen when you start an attempt, so they cannot be listed here.
Finish and submit Time is up, but your answers have not been submitted yet. Check your connection and try again.
Try again Answer 5 questions. You'll see your score, the answers and explanations at the end. Your result can count towards your course progress.
Check my progress Continue my progress check
Recent attempts History is stored only in this browser.
No completed attempts are saved yet.
Beyond the syllabus: optional enrichment that does not count towards your progress.
Previous question Go to question Next question Progress check complete
Start again
This assessment needs JavaScript. Its questions are chosen when you start an attempt, so they cannot be listed here.
Finish and submit Time is up, but your answers have not been submitted yet. Check your connection and try again.
Try again Answer 5 questions. You'll see your score, the answers and explanations at the end. A scheduled review counts towards your course progress only when it is due.
Start review Continue my review
Recent attempts History is stored only in this browser.
No completed attempts are saved yet.
Beyond the syllabus: optional enrichment that does not count towards your progress.
Previous question Go to question Next question Review complete
Start again
This assessment needs JavaScript. Its questions are chosen when you start an attempt, so they cannot be listed here.