block 2 - spectroscopy and structure determination Flashcards

1
Q

molecular spectroscopy

A

IR, UV-VIS NMR (NOT MS)
depend on interaction of molecules with radiation of specific energy
spectroscopy = study of interaction of electromagnetic radiation with matter

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2
Q

electromagnetic spectrum

A

high energy cosmic waves to low energy radio waves
energy DECREASES with WAVELENGTH - inversely proportional to wavelength (e = hc/wavelength), energy INCREASES with FREQUENCY - directly proportional to frequency (e = hv)

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3
Q

fundamental equations of electromagnetic radiation

A
E = energy
c = velocity of light (constant)
h = Planck's constant
v = frequency
λ = wavelength 
v (with line on top) = wavenumbers
1. E = hv
2. c = vλ
3. E = hc / λ
4. v (with line on top) = 1 / λ
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4
Q

basis of molecular spectroscopy (energy levels)

A

when energy of electromagnetic radiation put onto a sample EXACTLY corresponds to the energy DIFFERENCE between 2 molecular energy levels, then molecule can ABSORB the energy - ground state –> excited state (release energy eg. as heat) –> go back down
transition occur when energy of electromagnetic radiation = E2 - E1 (∆E) therefore energy absorption at ∆E = hv
only one frequency of radiation will be applicable to particular transition

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5
Q

what is the spectrum of a compound in molecular spectroscopy

A

graph of fraction of radiation ABSORBED or TRANSMITTED by sample vs the WAVELENGTH, FREQUENCY or WAVENUMEBER of the light impinging on the sample

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6
Q

mass spectrometry (graph, how it works, what it determines, limitations etc.)

A

allows determination of mass of individual ions derived from compounds in gas phase
molecule bombarded with high energy electron beam
M (molecule) –> M+•(–> fragmentation –> other positive ions + neutral fragments) (becomes ionised - 1 e- removed so becomes positive, and other e- is unpaired) + e-
detects charged molecules
plot of relative abundance vs m/z (m = mass of ion, z = charge of ion)
look at mass that is HEAVIEST (M+• = ion of highest m/z - z = 1)
limitations = not able to distinguish isomers, diff molecular formulas (with same molar mass)

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7
Q

IR spec (graph, transmittance, benefits, limitations etc)

A

units used = wavenumber cm-1 (wavenumber equation)
vibrational excitation
absorption of IR radiation occurs when radiation freq. matches freq of bond vibration
transmittance (%, y) and wavenumber (x)
transmittance = how much energy is RELEASED, what’s left over
low transmittance = mol has absorbed most, little left over
relative masses of atoms+ bond strength determines position of IR absorption (how much absorbed)
BENEFITS = rapid identification of functional groups; “fingerprint” - no two compounds have identical spectra
LIMITATIONS = compounds w same fg –> similar spectra (eg isomers); no indication of no. of fg’s

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8
Q

UV-VIS & energy of e-, spectrum

A

UV-VIS raises e- in some molecules (π e-) from LOWER energy bonding (or NON-BONDING) molecular orbitals –> HIGHER energy ANTI-BONDING molecular orbitals
EXPLANATION of non-bonding, MOs, ANTI-BONDING (not examinable)
- MOs = 2 e- in bond between 2 atoms –> BONDING MOs (2 atomic orbitals = MO)
- ANTIBONDING = every atom w/ BONDING MO has ANTIBONDING MOS (on the side for C-C bond for eg 0C-C0 0 = antibonding)
- UV-VIS radiation –> one e- jumps from BONDING mo to ANTIBONDING mo –> bond between C-C (for eg.) gets longer and weaker (e- not as close to bond) - if BOTH e- jump to antibonding mo the molecule falls apart (no e- in bonding mo)
- NON-BONDING = O , N have lone pairs - can jump to antibonding orbitals too
antibonding has * next to it eg π*
UV-VIS SPECTRUM = plot of ABSORBANCE (y axis) vs WAVELENGTH (x axis, nm). WAVELENGTH at which MAX ABSORPTION = λmax

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9
Q

conjugation

A

when there is π bond, sigma bond, π bond

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10
Q

NON conjugated π bonds in UV-VIS (absorption, why)

A

C=C - ≈ 170nm - 180nm due to a π to π* transition
C=O (or N) - 170nm - 180nm also, but EXTRA absorption at ≈280nm due to n (non-bonding) to π* transition
note - n - π* requires less energy (as n is in between π and π), therefore appears at a larger nm (shorter bump, after the larger π-π transition before it)

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11
Q

conjugated systems in UV-VIS

A

∆E for e- excitement is smaller - absorption is observed at a LONGER wavelength than for non-conjugated systems
any molecule w/ conjugated double bonds = absorption at >200nm
the greater the number of double bonds in conjugation, the higher nm they will appear at - ∆E (energy gap) is smaller and energy inversely proportional to wavelength
* conjugation has NO RELATION with ε

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12
Q

UV-VIS and Beer’s law

A

for a particular compound at a specified λ - ABSORBANCE is PROPORTIONAL to CONCENTRATION and PATH LENGTH
the more effective a molecule is at absorbing light at that wavelength, the greater the absorbance
A = ε b c
A = absorption
ε = molar absorptivity, L mol-1 cm-1
b = path length, cm
c = concentration, mol L-1

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13
Q

uses of Beers law

A
  1. determine conc. in solution of UV-VIS absorber of known structure - find out λmax, sub into law if absorptivity is known at λmax
  2. make calibration plot for solutions (dilutions) at a specific wavelength, then at SAME wavelength determine absorbance for solution of UNKNOWN conc. –> read off calibration plot
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14
Q

UV-VIS, MS and IR all have what limitation?

A

give NO INFO w/ respect to HYDROCARBON SKELETON

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15
Q

NMR technique, nuclei and shielding

A

info of C-H framework of molecules
nuclei have nuclear spin - random, when magnetic field applied –> nuclei align either PARALLEL (with - LOWER energy) or ANTIPARALLEL (against) mf
radio frequency –> energy absorbed and NUCLEAR SPIN FLIP from low energy (p) –> high energy (anti-p)
energy of spin flip depends on shielding of nucleus
shielding = when e- are shared equally –> lower ppm as nucleus doesn’t feel effect of radiation as much
deshielding = when one (more en) atom pulls e- close to itself –> higher ppm other nucleus is deshielded and feels more of the radiation

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16
Q

13C NMR

A

ALL 13C = PROTON DECOUPLED (no splitting signals)
signal given for each C in UNIQUE environment ∴ if mol. has NO symmetry, no. of signals = no. or carbons in mol. formula
ppm 0-220;
sp3 = 0-90, sp2 = 100-220
C bonded to more en –> higher ppm (nucleus deshielded)
INTENSITY ≠ NUMBER of C giving the signal

17
Q

1H NMR - 4 parts to it

A
  1. number of absorption signals (same as 13C NMR)
  2. portion of absorption signals (same as 13C NMR)
  3. relative areas under absorption signals
  4. splitting pattern for absorption signals
18
Q

no. of absorption signals, position of absorption signals and relative areas under absorption signals for 1H NMR

A

NUMBER:
H in x diff mol. enviro give x diff absorption signals
POSITION:
ppm usually 0-15
if H bonded to C that’s deshielded, H is also deshielded
H bonded to sp2 C higher ppm
AREAS:
signal areas are proportional to the number to H giving rise to the signal.
when total of these no. do NOT = no. of hydrogens in mol. formula, then both need to be x by same integer