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Chemistry 2 > Introduction to spectroscopy > Flashcards

Introduction to spectroscopy Flashcards

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1
Q

(Intro) Spectroscopic methods to determine struture & purity of unknown compound

A
  1. Melting point
  2. Elemental analysis
  3. Electronic Absorption Spectroscopy (UV/Vis)
  4. Vibrational spectroscopy (IR)
  5. Nuclear Magnetic Resonance (NMR)
  6. X-Ray crystallography
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2
Q

(Intro) Advantages of spectroscopy

A

Fast, small sample size, non-destructive and can analyse mixtures/follow reaction progress

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3
Q

(Intro) How does spectroscopy identify compounds

A
  1. Bonds may only stretch, bend or break w/ certain frequencies
  2. e- may only switch orbitals w/ certain energy difference
  3. Matter exists in a series of quantised energy levels
  4. Difference between energy levels is measured in spectroscopy.
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4
Q

(UV-Vis) The 2 molecular states produced by the combination of 2 atoms in a covalent bond?

A

Bonding and anti-bonding

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5
Q

(UV-Vis) Bonding and antibonding: Bonding state relative energy level and why?

A

Bonding state is lowest energy state as it is a favourable condition for the 2e- to be combined

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6
Q

(UV-Vis) Bonding and antibonding: Antibonding state relative energy level and why?

A

Antibonding state is higher in energy as the condition of the 2e- is not favourable, for this state, so energy needs to be applied for the system to exist

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7
Q

(UV-Vis) Energy applied to covalent bond (in terms of bonding/antibonding)

A

If a bond receives enough energy, an e- from the bonding level can become excited into the empty antibonding level

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8
Q

(UV-Vis) Why can a bond absorb light/radiation only at certain frequencies

A
  1. E=hv
  2. Therefore, specific frequency at which the bond can absorb the wave at
  3. This abs. can be detected through its absence in the recorded results, as non-absorbed frequencies will pass through and be detected.
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9
Q

(UV-Vis) Why would a UV-Vis spectrum have a broad band?

A
  1. e- distribution is changed, nuclei are subjected to different forces than each other, molecule vibrates
  2. Vibrational structure of ELECTRONIC TRANSITIONS merge in liquids/solids which gives a broad band
  3. VIBRATIONAL TRANSITIONS are further complicated by ROTATIONAL TRANSITIONS
  4. UV absorption is broad as VIB./ROT. levels overlay the electronic levels.
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10
Q

(UV-Vis) What does Epsilon(E) represent? What is it a measure of

A

Molar extinction coefficient, measures how strongly a species abs. light of a specific wavelength

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11
Q

(UV-Vis) Beer-Lambert law, equation and each explanation of components of equation

A

Equation: A = Elc

A = absorbance
E = molar extinction coefficient @ a specific wavelength (L mol-1 cm-1)
l = path length (cm)
c = conc. of solution (mol L-1)
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12
Q

(UV-Vis) On a absorbance against concentration graph, what is the gradient equivalent to?

A

Gradient = E*l

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13
Q

(UV-Vis) What is the trend in energy between bonding states, lowest to highest?

A 
Sigma Bonding
Pi Bonding
Nonbonding (e.g. lone pairs e-)
Pi Antibonding
Sigma Antibonding
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14
Q

(UV-Vis) What do the Feiser-Woodward Rules do?

A

Relates conjugation and substituents to electronic transitions and can therefore be used to estimate the λmax.

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15
Q

(UV-Vis) Affects of increased conjugation length

A

Increase in conj. length leads to an increase in absorption maximum

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16
Q

(UV-Vis) Affects of substituents on a conjugation chain

A

Substituents can shift bathochromically, e.g. adding 1 double bond adds 30nm to absorption maximum

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17
Q

(UV-Vis) Bathochromatic shift

A

Shift of λmax towards red end of the colour spectrum

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18
Q

(UV-Vis) Hypsochromic shift

A

Shift of λmax towards blue end of the colour spectrum

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19
Q

(UV-Vis) Difference in application of Feiser-Woodward rules on conj. carbonyl compounds, compared to parent dienes

A

The change in λmax of the same added structural feature, added to a cyclic compound, is dependent on whether it is added directly to the ring or not

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20
Q

(UV-Vis) Are spectra of conj. carbonyl compounds effected by solvent choice? How is this combated?

A

Yes, a correction factor must be applied.

selection of correction factors can be found in notes from lecture 1

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21
Q

(UV-Vis) What does HOMO stand for?

A

Highest Occupied Molecular Orbital

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22
Q

(UV-Vis) What does LUMO stand for?

A

Lowest Unoccupied Molecular Orbital

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23
Q

(IR) What are bending vibrations?

A

Lower energy vibrations where the bond moves side to side

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24
Q

(IR) What are Stretching vibrations?

A

Higher energy vibrations where the bond moves in and out (shrinks and expands)

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25
(IR) What affects vibrations?
Strength of bond and mass of atoms
26
(IR) Why does independent stretching occur?
1. Bond is stronger/weaker than those nearby | 2. Bond is between atoms that are much heavier/lighter than their neighbours
27
(IR) What does Hooke's law observe?
Movement of 2 masses attached to a spring
28
(IR) Hooke's law equation
v = [1/(2πc)]*√(k/μ) ``` v = frequency of vibration k = force constant for spring strength μ = the reduced mass ```
29
(IR) Calculating the reduced mass, μ
μ = (m1*m2)/(m1 + m2)
30
(IR) What is resonance?
Absorption of energy resulting in an increase in the amplitude of the vibrations
31
(IR) What is meant by IR active?
Bond has a dipole and increases on stretch
32
(IR) What is meant by IR inactive?
Bond has no dipole and no change on stretch
33
(IR) Symmetric stretch
Dipoles cancel out and is therefore IR inactive
34
(IR) Asymmetric stretch
Dipoles differ and is therefore IR active
35
(IR) Bending (stretch) mode
Dipoles have same magnitude, but orientation changes and therefore it is IR active
36
(IR) Single bond IR region
1500-1000 cm-1
37
(IR) Double bond IR region
2000-1500 cm-1
38
(IR) Triple bond IR region
2500-2000 cm-1
39
(IR) Bonds to H region
4000-2500 cm-1
40
(IR) Saturated and unsaturated C-H regions
Saturated: 3000-2500 cm-1 Unsaturated: 4000-3000 cm-1
41
(IR) Primary Vs. Secondary amine regions
1ry: Give independent vibrations from neighbouring bonds, therefore have sharp peak @ 3300 cm-1 2ry: Independent of rest of molecule, however, the 2N-H bonds are identical & vibrate as a single unit. Gives 2 strong bands, a symmetric vib. band (3300 cm-1) and an asymmetric vib. band (3400 cm-1)
42
(IR) Why do hydrogen bonds cause a peak to broaden?
HBs vary in length and therefore weaken covalent bonds by varying amounts (weaker bonds = weaker frequency). This gives a range of frequencies about a mean value.
43
(IR) Double bond region: Carbonyl C=O
1. Intense band 1900-1500 cm-1 2. V. polar, so stretch means a large dipole moment change 3. Usually strongest peak on IR spectrum
44
(IR) Double bond region: Nitro NO2
1. 2 intense bands @ mid 1500 and mid 1300 cm-1 2. Highly polarized 3. 2 peaks due to symmetric and asymmetric vibrations
45
(IR) Double bond region: Alkene C=C
1. Weak band 1650 cm-1 2. Least polar of double bonds 3. May be symmetric and therefore IR inactive
46
(IR) Relative C=O bond strength in carbonyl functional groups (strongest to weakest)
Acid Chloride Anhydride Ester Amide
47
(IR) Conjugated carbonyls: Why do they have a lower frequency?
1. C=O strength lowered due to effect of conjugate giving a weaker bond, requiring less energy to vibrate 2. C=C stretch intensely increased due to conj. giving an increased dipole, requiring less energy to vib. 3. C=C polarised by conj. and therefore stronger signal and requires less energy to vib.
48
(IR) Non-conjugated carbonyls: Why do they have a higher frequency?
1. Less stretch in C=O so more energy required to vibrate | 2. C=C double bond is non-polar and therefore has a weaker signal and requires more energy to vib.
49
(IR) Position of IR band depends on?
1. Reduced mass of atoms - Light atoms give higher frequencies 2. Bond strength - Strong bonds give high frequency
50
(IR) Strength of band depends on?
Change in dipole moment - Large dipole change gives strong absorption
51
(IR) Width of band depends on?
Hydrogen bonding - Strong HBs give a wide peak
52
What is Carbon, Hydrogen and Nitrogen converted to when it is burned in oxygen?
C > CO2 H > H2O N > Nitrous oxides >reduced> Nitrous oxide
53
(MS) Elemental analysis
1. Sample is weighed and completely burned in O2 2. Conversion of C, H and N occurs 3. Analyser separates components using gas chromatography 4. Components are detected and the respective conc./amounts can be established 5. Weight of samples has been accurately measured, so % composition can be calculated
54
Spectroscopy
A technique involving the production and subsequent recording of a spectrum of electromagnetic radiation, usually in terms of wavelength or energy
55
Spectrometry
A technique involving the measurement of a spectrum
56
What is mass spectrometry?
Technique that measures the weight of a compound and some of its fragments
57
Process of mass spectrometry
1. Sample placed in chamber, under vacuum, & volatilised 2. Sample leaks into spectrometer & is ionised 3. Electrical field accelerates ionised molecule towards magnetic field 4. Magnetic field induces charged species to take a circular path 5. Charged species, of different masses, pass through the slit, at different times, and are recorded to give intensity peaks on spectrum
58
How does magnetic field, in MS, affect the particles movement?
Radius depends on the mass of ion and strength of magnetic field Change in strength of magnetic field changes which charged species pass through the slit
59
(MS) Ion's kinetic energy equation (1)
eV = 1/2 * m(v)^2 ``` e = charge of ion V = voltage m = mass of ion v = velocity ```
60
(MS) Force on ion from magnetic field equation (2)
BeV = (m(v)^2)/r ``` B = Force from magnetic field e = ion charge V = voltage m = ion mass v = velocity r = radius ```
61
(MS) Relationship between force on ion, radius and voltage equation (3)
m/e = ((B)^2 * (r)^2)/2V ``` B = Force from magnetic field e = ion charge V = voltage m = ion mass v = velocity r = radius ```
62
What are the x and y axis on a mass spectrum
x-axis: mass to charge ratio (m/e) | y-axis: relative abundance
63
(MS) Ionisation techniques: | Electron impact ionisation, EI-MS
1. e- are produced by heating a metal filament 2. e- behaves as waves & are accelerated to specific kinetic energy 3. These kinetic energies equate to specific wavelengths
64
(MS) Ionisation techniques: | Chemical ionisation, CI-MS
1. Ions are produced through collision of molecules 2. e- impact affords progression of +ve radicals to split into 1 radical and 1 +ve ion 3. The +ve charged species then collide to produce adducts or initiate proton transfer to substrate
65
(MS) Ionisation techniques: | Fast atom bombardment, FAB-MS
1. Substrate is used in non-volatile matrix 2. Ar is ionised and accelerated towards collision chamber then the substrate 3. Collision chamber: interact w/ other atoms & are neutralised: Ar+* (rapid) + Ar (slow) > Ar+* (slow) + Ar (rapid) 4. Neutral atoms w high energy hit substrate & ejected ions/molecules accelerated towards the analyser by POTENTIAL DIFFERENCE
66
(MS) When is FAB-MS useful?
Technique induces little/no ionisation as it ejects any ions that are present into the gas phase Good technique for salts & polar compounds w/ high Mr
67
(MS) Ionisation techniques: | Matrix assisted laser desorption, MALDI-MS
1. Laser pulses directed at substrate, volatilising itinto ionic/neutral molecules 2. Wavelength of laser can be tuned to give selective ionisation 3. Subs. is embedded into a matrix of small molecules that have strong absorption @ laser wavelength 4. After photoexcitation, analyse is desorbed from the bulk & proton transfer occurs from matrix to subs.
68
(MS) Ionisation techniques: | Electrospray ionisation, EI-MS
1. Charge accumulates @ liquid surface, @ one end of the capillary, which breaks forming highly charged droplets 2. As solvent evaporates, charge become conc. 3. Build up of coulombic forces causes droplets to explode into smaller droplets, allowing for the desorption of ions
69
(MS) Ionisation techniques: | How is electrospray formed?
Produced by applying strong electric field, @ atmospheric pressure, to liquid passing through a capillary tube. Liquid consists of: analyse and solvent
70
(MS) Ionisation techniques: | Difference between EI-MS and API-MS
API-MS is an adaptation of EI-MS where source chamber of spectrometer is kept @ atm. pressure instead of low pressure
71
(MS) Ionisation techniques: | Atmospheric pressure ionisation, API-MS, problems and how they are fixed
Problem: Coupling the source chamber @ atmospheric pressure to the analyser compartment, that's @ v. low pressure of (10)^-5 Torr Problem solution: Using a small opening between 2 chambers and high capacity pumps
72
Interpreting mass spectra: What is the molecular ion
The mass of the intact analyte
73
Interpreting mass spectra
1. Assume the highest molecular mass is the molecular ion 2. In CI-MS cases, consider the highest peak contains a proton (+1) or ammonium (+18) as an ADDUCT 3. Look for tell tale isotopes 4. Can the fragments be easily worked out
74
Interpreting mass spectra: Tell tale isotopes and easy fragements
Isotopes like Br = 79/81 with 50:50 abundance Easy fragments: H2O = -18 Phenyl = -77 methyl = -15
75
Use of MS in elemental analysis example
1. MS gives 69 mass units as main peak on spectrum 2. Elemental analysis gives: C = 69.2% H = Y% N = Z% 3. Calc. how much of C, H and N are in the sample by doing, e.g. C: 69.2% * 69/100 = 47.9 mass units Mass of C = 12 Therefore, no. of C atoms in formula: 47.9/12 = 3.99 = 4

Chemistry 2 (3 decks)

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