Owen's material

Question 1

Electromagnetic radiation (EMR)

Answer 1

EMR consists of two perpendicular components (oscillating electric (E) and magnetic (B) fields). Interaction with matter requires a molecular response in E or B. The molecular response is in phase with the oscillation. Creates diffraction patterns (constructive and destructive interference) when monochromatic planar waves are passed through slits.

Question 2

Wavenumber

Answer 2

The distance between two successive crests (m)

Question 3

Frequency

Answer 3

The number of waves passing a point over a unit time (Hz or /s)

Question 4

Wavenumber

Answer 4

Number of wavelengths in a unit distance (/m). Inverse of wavelength.

Question 5

Photoelectric effect

Answer 5

When light hits as object, it immediately releases a photocurrent. The max Ek of the electrons released depends on the frequency of the wave (not intensity).
The threshold frequency is the frequency for a material below which no electrons will be released.
E=hc/λ - shorter wavelength = higher frequency and energy (gamma rays - UV)

Question 6

Quantisation of Energy

Answer 6

Discrete values
Only electrons bound by nuclei have discrete energies. Unbound electrons are free to have any values of energy.
Hydrogen emission spectrum - all electrons in their lowest state gives the ground state. If in higher E orbitals they are excited.
E = -hcR/n^2, where n=prinicpal qn

Question 7

Single photon transitions

Answer 7

Transitions result from interactions of E or B dipoles of the molecule with the electric or magnetic field of radiation.
Electrons can jump from one orbital to another, which requires E(photon)=difference in E levels.
The electronic energy in an atom can increase/decrease resulting in absorption/emission of a photon of suitable energy.

Question 8

UV-Vis Spectrophotometer - single beam

Answer 8

Light source - emits white light (all wavelengths)
Monochromator - separates the light into separate wavelengths and passes each wavelength in turn through the sample
Sample - absorbs radiation of certain wavelengths
Light detector - measures intensity of light at each wavelength after passing through the sample
Signal processing - computer to generate spectrum

Question 9

Monochromator

Answer 9

Separates light into different wavelengths, and passes each wavelength in turn through the sample.
Slit - 1) white light passes through and 2) allow a narrow range of wavelength out
Concave focussing mirror - 1) directs light to dispersing elements and 2) directs one wavelength out the slit
Dispersing element - diffracting grating and prism separate white light into different wavelengths

Question 10

Photomultiplier tube (detector)

Answer 10

Photon enters through a transparent window into a sealed and evacuated tube. A photocathode (negative high voltage) causes the photoelectric effect. Dynodes (increasing voltage towards anode) cause electron cascade. Strong signal (lots of electrons) by the anode. Current-voltage amplifier.

Question 11

UV-Vis Spectrophotometer - double beam

Answer 11

Source of EM radiation -> monochromator -> beam splitter (2 identical beams) -> one beam sent through sample the other sent through a reference -> detector -> computer (compares the 2 signals, eliminating background absorption)

Question 12

Absorption spectroscopy

Answer 12

Incident radiation (I0)
Transmitted radiation (IT) - reduced in intensity due to absorption
Transmittance: T=IT/I0
Absorbance: A=log(I0/IT)=-log(T)
  - if 90% of light is absorbed, 
   10% is transmitted (0.1)

Question 13

Beer-Lambert Law

Answer 13

A=log(I0/IT)=εcl

-ε = molar absorptivity coefficient = measure of how strongly the sample absorbs (depends on wavelength)

Question 14

Transition intensities

Answer 14

Depend on c and l
The transition probability - the intrinsic probability that a molecule in the appropriate state can interact with the radiation to undergoing a transition to the final state.
P(fi)=0 - forbidden state
P(fi)>0 - allowed transition
-determined by selection rules
Boltzmann distribution - determines the number of molecules in the initial state

Question 15

Selection rules

Answer 15

Transitions result from interactions between B or E dipoles of the molecule with the E or B field of the radiation

Question 16

Gross selection rules

Answer 16

Physical properties that a system must have to be able to undergo any such transitions.

• rotational - permanent dipole moment
• vibrational - magnitude of electric dipole moment must change during vibration
• NMR - non-zero magnetic dipole moment

Question 17

Specific selection rules

Answer 17

Determine the particular states between which transitions are allowed.
-photons carry one unit of angular moment. So atomic spectroscopy angular qn = +/-1
(sp not sd)

Question 18

Boltzmann distribution

Answer 18

Determines the number of molecules in the initial state
For a small energy difference (high T), molecular state are evenly distributed
For a large energy difference (low T), molecules are in lower states
(ni/nj) = (gi/gj)e^(-E/kBT)
-ni = number of molecules in state i
-gi = degeneracy of state i
-Eij=Ei-Ej
For nmr, the upper and lower state populations are almost equal
For electronic, the upper-state can usually be ignored

Question 19

Atomic spectra

Answer 19

Atomic transitions give sharp lines in the UV, visible or near IR.
The hotter the flame, the stronger the emission.
ni/nj=0 - no transition
ni/nj=1 - strong transition

Question 20

Atomic Emission Spectroscopy (AES)

Answer 20

Quantify the amount of an element by measuring its emission intensity in a hot flame, atomising any compounds and thermally exciting atoms.
Requires careful control of flow rates, flame T and geometry.
Sample -> mixing chamber -> burner head and flame -photons-> monochromator -> detector -> recorder

Question 21

Inductively coupled plasma - optical emission spectroscopy (ICP-OES)

Answer 21

Atomisation and excitation by RF induction in argon to create an extremely hot plasma
Plasma much hotter and steadier than flame (stronger intensity)
Sample -> plasma torch with RF coils -photon-> monochromator -> detector -> recorder

Question 22

Atomic absorption spectroscopy (AAS)

Answer 22

In a flame, the vast majority of atoms are in the ground state, so absorption is much stronger than emission.
Sample -> mixing chamber -> hallow-cathode lamp emitting photon to a burner head with a flame -photon-> monochromator -> detector -> recorder
Hallow-cathode lamp is filled with low P inert buffer gas and made of an alloy containing the desired elemental analyte.

Question 23

Molecular dynamics

Answer 23

Quantised:
Electronic energy - larger E gap
Vibrational energy - each vibrational mode has a rotational mode. Produced when electrons are excited. Linearly spaced
Rotational energy - microwave region (low E), spacings get further apart
Not quantised:
Translational energy

Question 24

Rotational motion

Answer 24

I = moment of inertia (rotational inertia) 
P(rot) = rotational angular momentum
J=rotational quantum number (0,1,2...)
h-bar = h/2pi 
g = degeneracy = 2J+1

Question 25

Rotational selection rules - diatomic molecule

Answer 25

Single-photon rotational spectroscopy involves an electric dipole mechanism. The magnitude or orientation of the electric dipole must be changed by the transition. GS rule - the molecule must have a permanent electric dipole moment (heteronuclear diatoms_ SP rule - photons carry one unit of angular moment, so angular momentum qns must change by +/-1. ΔJ=+/-1 (+absorption-emission) Measurements done in gas phase where molecules can freely rotate (nJ/nO)=(2J+1)e^(-EJ/kBT)

Question 26

Absorbance vs wavenumber graph

Answer 26

Increases from 0-20 /cm - increasing degeneracy of initial state Decreasing from 20 /cm - higher E so lower Boltzmann population

Question 27

Rotational spectroscopy - polyatomic molecules

Answer 27

3 orthogonal rotational axes; 3 moments of inertia. GSR - have to have a permanent electric dipole moment. Applications - determine molecular geometries by microwave absorption, detection of atmospheric molecules by microwave emission.

Question 28

Vibrations of Hooke's Law - diatomic molecule

Answer 28

Vibrations between atoms similar to spring. F=-kx V=1/2kx^2 - parabolic potential Force constant (k) - proportional to bond strength (stiffer bond has larger k - steeper parabola). v(vib) is proportional to k and 1/reduced mass. -more confined motions (stiffer bond = larger k) lead to greater separation (larger hvvib) -heavier particles have less separation of quantum levels Gaps separated by wavenumber (one vibrational frequency) Zero point energy - E=(v+0.5)hvvib, so a molecule can never have zero vibrational energy.

Question 29

Vibrational selection rules - diatomic molecule

Answer 29

Electric-dipole mechanism. GSR - the electric dipole moment of the molecule must change during the vibration, so it must have a permanent electric dipole moment (heteronuclear diatoms) SPR - for a harmonic oscillator Δv=+/-1 Most molecules will be in the zero-point level

Question 30

Vibration states of anharmonic oscillators

Answer 30

Anharmonic bands are commonly described by Morse potential. - a bond gets weaker as it is stretched (decreasing k). Ep decreases so the E gap decreases at higher vibrational modes. Eventually bond will snap (E goes to 0 - bond dissociation). The effects of anharmonicity are more prevalent at higher E and at larger displacements from r0. Vibrational energy levels are closer together at higher E. Transitions with ΔV=>1 (overtones) are weakly allowed.

Question 31

Vibrations - polyatomic molecules

Answer 31

Normal modes - combinations of localised vibrations (stretches, bends and torsions) Linear - 2 degenerate bending modes < symmetric stretch < asymmetric stretch (E and ṽ). 3N-5 -same E but differ in direction they bend Non-linear - bending < symmetric stretch < asymmetric stretch. 3N-6 GSR - vibrations must involve a change in dipole to appear in IR spectrum (does not need a permanent electric dipole) -larger dipole = stronger intensity. Fine structures in IR bands due to rotational energy levels. H2O - permanent electric dipole, so direction of dipole changes -> all IR active CO2 - no permanent dipole, symmetric stretch is IR inactive, but all others are IR active. SSR - Δv=+/-1 (fundamental is stongest), +/-2... (overtunes due to anharmonicity - weak)

Question 32

Isotropic substitution

Answer 32

Bond strength similar, but reduced mass can change significantly, reducing the vibrational energy, leading to a smaller wavenumber.

Question 33

Fingerprint region

Answer 33

Region in IR spectra where skeletal vibrations can be observed (unique to each molecule).

Question 34

Characteristic group frequencies

Answer 34

A group of atoms have very different vibration frequencies, it will absorb a narrow frequency range that is characteristic of the group. Higher v require large k (multiple bonds) or smaller reduced mass (H).

Question 35

IR regions

Answer 35

Near IR - overtones, combination bands Mid IR - fundamental vibrations and fingerprint region Far IR - fundamental vibrations of bonds with heavy atoms or weak bonds.

Question 36

IR spectra collection

Answer 36

Nujol mull - ground up sample is suspended in oil and held between KBr plates KBr pellet - ground up sample with crystalline KBr then compressed into a disk Reflectance - ground up pure sample or diluted with ground up KBr Solution - held between KBr plates Gas cell - IR invisible windows

Question 37

Molecular electronic spectroscopy

Answer 37

Electronic transitions typically occur in UV and visible. Transitions in the visible result in pigments, in UV they are colourless - ligand-field transition of transition metals (weak) - charge-transfer transitions between ligands and metal ions (strong) - molecular orbitals π->π* transition in conjugated organic molecules like dyes (strong)

Question 38

π->π* transitions

Answer 38

Conjugations arises in organic molecules with alternating double-single bonds, allowing the π electrons to "spread-out" over larger distances, increasing the size of the bound potential. This causes the energy levels to become closer together. Thus, the energy required to excite an electron from the HOMO to LUMO is less, moving the necessary wavelength from far-UV to near-UV.

Question 39

Vibrational progressions

Answer 39

Transitions can occur from ground-state to a number of vibrational levels in the electronic excited state. Bond is longer in π* (k smaller), thus E levels have smaller gap. Franck-Cordon principle - electronic transitions occur on much shorter time scales than vibrational motions. Bands are usually broad due to vibrational modes.

Question 40

NMR

Answer 40

Involves a magnetic-dipole mechanism (weaker than electric), thus a nucleus must have a non-zero spin (uneven #protons and/or #neutrons). 1H, 13C, 19F etc are commonly used. GSR - must have a non-zero magnetic dipole moment SSR - Δm(I)=+/-1 Larmor frequency (v) in MHz (xE6) Very nearly equally populations and intrinsic weaknesses of magnetic dipole mechanism makes NMR absorption spectroscopy extremely insensitive.

Question 41

Nuclear spin of a proton

Answer 41

Described by quantum numbers I (nuclear spin qn) and M(I) (nuclear magnetic qn). I=1/2 m(I)=+/-1/2

Question 42

Zeeman effect for a proton

Answer 42

In the absence of an applied magnetic field, the spins have no orientation and equal E. In the presence of an applied magnetic field of strength B, they can adopt 2I+1 orientations, corresponding to different values of m(I), with E: E(MI)(B) = -m(I)γ(N)ℏB -γ(N)=gyromagnetic ratio of a proton (constant) -ℏ = h/2π 1H (I=1/2) can adopt 2 orientations corresponding to m(I)=+/-1/2 (+aligned with field), with E separated by E(p) =γ(p)ℏB. -m(I)=-1/2 so E(-1/2)=+γ(p)ℏB/2

Question 43

NMR spectrometer

Answer 43

A compound in between a strong magnet. A strong pulse of RF from the input oscillator (coil) excites nuclei to higher-energy level, creating a non-equilibrium population. Population relaxes back to eqm by emitting RF radiation at Larmor frequency. Emitted radiation detected by output receiver (coil). Signal frequencies sorted by Fourier transform. Stronger field -> greater population difference -> stronger signal

Question 44

1H NMR spectrum

Answer 44

Reference signal at 0ppm - TMS Number of resonance sets - number of H environments (if there is a chiral C, then each C will have separate H environments) Integral - number of H associated with C Chemical shift - nature of chemical environment multiplicity - number of H on adjacent C (within 3 bonds)

Question 45

Timescale

Answer 45

NMR is a slow technique, so for planar molecules electronic transitions and vibrations are much faster, leading to an average spin for 2 methyl groups.

Question 46

Chemical shift

Answer 46

A magnetic field (B) induces electrons to circulate around a nuclei (diamagnetic current). This induces a secondary magnetic field that is proportional, but opposed to B. The net field at the nucleus is weaker than the applied field due to electrons shielding the nucleus and the resonant frequency (chemical shift) is reduced. This caused an upfield shift. Nearby electron-withdrawing groups reduce electron density, deshielding the atom and increasing the resonance frequency. Aromatic rings - diamagnetic ring currents augment the field experienced by in-plane nuclei (effectively deshielding the ring) - same for multiple bonds. CH, OH and NH signals - CH give sharp lines and multiples at distinct chemical shifts. NH and OH subect to H-bonding and exchange with solvent. Highly purified in very pure solvents give broad singlets (no splitting).

Question 47

Multiplicity (spin-spin coupling)

Answer 47

``` Spin-spin coupling arises from interactions between magnetic dipoles associated with spins and cause spin-spin splitting. Intensities follow pascal's triangle. Singlet - no adj H, 1 Doublet - one adj H, 1:1 -spin up and spin down Triplet - 2 adj H, 1:2:1 Quartet - 3 adj H, 1:3:3:1 ```

Question 48

Heteronuclear NMR spectroscopy

Answer 48

A nucleus with a spin of 1/2 is the best: 19F, 31 P etc. Coupling to quadrupolar nuclei (spin >1/2) gives multiple lines. 2H nuclear spin =1, splitting the signal into 3 peaks at -1,0,1. 2H solvents lock the spectrometer frequency (C-D seen in CNMR). Sensitivity of nuclei in NMR depends on gyromagnetic ratio and natural abundance (1H most sensitive).

Question 49

Mass spectrometry

Answer 49

Mass spectrometry vaporises a sample, then ionises it (charge often +1). During this process, the sample may undergo chemical reactions. Sorts according to m/z ratio. Ionisation processes depend on analyte volatility and the desired degree of subsequent reactions Hard ionisation (extensive fragmentation with volatile analytes) - electron impact (EI) Soft ionisation (minimal fragementation) -chemical ionisation (CI) -fast-atom bombardment (FAB) -Matrix-assisted laser desorption ionisation (MALDI) -Electrospray ionisation (ESI)

Question 50

EI ionisation

Answer 50

Ionisation by collision with high-energy electrons. M -electon impact -> M+. + e- M+. is molecular/parent ion (must be charged to be seen in spectra) m/s=M electron usually lost from weakly bonded/non-bonding orbitals (lp). Excess energy of ionising electrons breaks bonds (fragmentation). Fragments can provide information about isomers. M+. -> R+ + R. R+ - fragmented ion R. - neutral radical fragment (not detected) Gas phase -> high vacuum -> e- source -> extractor plates and ion lenses -> m/z analyser

Question 51

CI

Answer 51

Soft ionisation by reaction with reagent ions (NH4+ and CH3+), no significant excess E (minimal fragmentation). M + NH4+ -> [MH]+ + NH3 m/z: (M+1) MH + CH3+ -> M+ + CH4 m/z: (M-1) gas phase -> low vacuum -> reagent ion source -> extractor plates -> m/z analyser

Question 52

FAB

Answer 52

Ionisation from a matrix by collision with fast atoms (Ar/Xe). Matrix absorbs the extra E so fragmentation is weak. M + Ar -> M+ + e- + Ar If the matrix contain acids/salts, adduct ions are produced: [MH]+ (M+1) or [MK]+ (M+39) - matrix influences peaks Fast atom source -> matrix -> extractor plates -> m/z analyser

Question 53

MALDI

Answer 53

Ionisation from a matrix by ablation with a pulsed laser. Matrix absorbs excess E. M + hv -> M+ + e- If the matrix contains salts/acids, adduct ions can be produced Pulsed laser -> matrix -> extractor plates -> m/z analyser

Question 54

ES-MS

Answer 54

Ionisation from a liquid by evaporation and Coulomb fission. Produces unfragmented adduct ions. [MHn]n+ so m/z: (M+n)/n Cone voltage of the spray nozzle can be increased to increase fragmentation (good for non-volatile ions). Spray nozzle (in solution) -> charged droplet (spray solution through shared nozzle into vacuum) -> solvent evaporation (droplet evaporates and charge concentrates) -> coulomb fission (coloumb explosion - smaller droplets until you have a single +) -> extractor plates -> m/z analyser.

Question 55

m/z analysis methods

Answer 55

Time-of-flight mass analyser (TOF-MA) | Magnetic-sector mass analyser

Question 56

TOF-MA

Answer 56

Pulsed ionisation unite -> accerlating voltage (15kV) -< ion pulse -> drifts through tube of length d -> detector (smaller are faster and heavier are slower). m/z ration = K(TOF)t^2 -K(TOF)=2eV/d^2 For given v and d, t depends on m/z Resolution can be improved by increasing d (longer drift tube), which requires a physically larger analyser.

Question 57

Magnetic-sector mass analyser

Answer 57

Ionisation unit -> accelerating voltage -> magnetic field (centrifugal force up and Lorentz force down) -> detector (smaller m/z closer to ionisation unit) Lorentz force = f(L)=Bzev Centrifugal force = f(c)mv^2/r The radius of curvature of the ion trajectory is constant when f(c)=f(L) ->mv^2/r = Bzev -> r=mv^2/Bzev ->m/z=K(mag)B^2 -K(mag)=er^2/2V For given V and r, B depends on m/z (vary B) Resolution can be improved by increasing V, requires physically larger analyser or stronger B.

Question 58

Mass spectra

Answer 58

M+1 peak due to isotopes (C13) Base peak - most intense, assigned 100% (most stable ion) (M-15) peak is methyl m/z=15 peak is methyl radical M->M.+ -> R. + R+ Alkanes - form stable carbocations -> 2/3 C=O groups - form R-CO+ Ethers, amines and alcohols - form oxonium (O) and iminium (N) cations

Question 59

Isotope patterns

Answer 59

``` Major isotope: 1-p(x) Minor isotope: p(x) If P(x)<<1 (C13 or less), then N(x)~(p1/p0)/px For small organic molecules, peaks due to 2H will be weak since px=0.02% ```

Question 60

Halogen isotope peaks

Answer 60

``` The minor isotopes of Cl and Br have very substantial abundances and are two mass units heavier than the major isotopes. 35Cl:37Cl is 3:1 79Br:81Br is ~1:1 p1/p0=(Nxpx)/(1-px) Cl - 31.9% for 1 atom Br - 97.2% for 1 atom Multiple halogen isotope peaks: 2 Cl: 35Cl/35Cl - 100 35Cl/37Cl or 37Cl/35Cl - 63.8 37Cl/37Cl - 10.2 (M+4) 2Br: Br follows pascal's triangle 1:2:1 Isotope patterns for heavier elements can be very complex ```

Question 61

Different charges

Answer 61

For cations that form chargers such as +2 or +3, the m/z will not equate to the atomic mass. Divide molecular mass by 1/2 or 1/3.

Question 62

High-resolution mass spectra

Answer 62

Some ion masses conincide at low resolution, but are different under high resolution. Not a proof of purity (elemental analysis) instead proof of composition.