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Unit One Part 10:infrared spectroscopy and mass spectrometry
gjr-–-
• Describe the process that gives rise to an infrared spectrum• Use characteristic vibrations to identify functional groups• Look at certain functional groups in a (little) bit of detail• Look at the process in which molecular ions are formed & detected• Understand the effect of isotopes on the mass spectrum
1
dr gareth rowlands; [email protected]; science tower a4.12http://www.massey.ac.nz/~gjrowlan
Determining structuregjr-–-
• Spectroscopy provides information about molecular structure• 1H NMR gives information about the C–H framework• Today look at infrared spectroscopy, which tells about functional groups• And mass spectrometry that looks at the size of a molecule
2
UV
uv-visexcitation of an
electron
infraredbond vibration
nmrnuclear spin
radio wavesNMR
spectro-scopy
Infraredregion
Electromagentic radiation and energygjr-–-
• If we consider energy as a wave (because it is...sort of)• Then we get the following relationship...
3
molecule in excited energy state E2
E1
E2absorption of energy
molecule in energy state E1
E1
E2energy in the
form of electromagnetic
radiation
λ E = hν = hcλ
E = energy (J) of 1 photonh = Plank's constant (6.63x10–34Js)c = speed of light (3x108ms–1)ν = frequency (Hz)λ = wavelength (m)
long wavelength = low energyshort wavelength = high energy
So what is going on?gjr-–-
• When molecule irradiated with IR radiation it can absorb energy if the frequency of radiation and the frequency of the bond vibration are the same
• Absorption of energy will increase the amplitude & frequency of vibration• We can measure absorption...
4
molecule - vibrating (stretching) with frequency υto absorb light, photon must be υ = λ/c, the same frequency
EtO CH3
O
υ1
υ2
υ3
υ1
υ3
υ1 υ2 υ3Expose molecule to range of frequencies. It absorbs some (amplitude of bonds vibration increases). We measure amount of light
reaching detector.
An infrared spectrumgjr-–-
• Spectra recorded as a frequency measurement in wavenumbers (1/λ) cm–1
• Higher wavenumber corresponds to a higher energy absorption
• Looks nasty?• Only about 8 stretches (peaks) at the most are ever of any use...• Spectrum divided into four easy sections
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More detailed notes on IR in organic chemistry can be found at my website: http://www.massey.ac.nz/~gjrowlan/teaching.html (Intro to organic and bioorganic...)
Interpreting IR spectragjr-–- 6
O H C C C C C O
N H C N C O C FC H C O C Cl
bonds to hydrogen
triple bonds
double bonds
single bonds
4000 3000 2000 1500 1000 cm-1
chan
ge in
sca
le
energy to cause vibration
1500–400 cm–1 is the
fingerprint region
O
O
Functional group absorptionsgjr-–-
• Other versions of this kind of table can be found in the course notes
7
Examples of IR spectragjr-–-
NH H
benzeneamineaniline
8
NH234803395
NH3C H
N-methylbenzenamineN-methylaniline N–H
3443
Examples of IR spectra IIgjr-–- 9
OHH
O
Ph
HO
Ph
HO
Phphenol - H-bonding
O
H
OH
2,6-di-tert-butyl-4-methylphenol
O–H3224
O–H3627
Examples of IR spectra IIIgjr-–- 10
O
hex-5-en-2-one
C=O1718
C=C1642
O
(E)-hex-4-en-3-one
C=O1674
C=C1634
Examples of IR spectra IVgjr-–- 11
N
OH
H
butanamideC=O16621634
N–H33563184
O
OH
butanoic acidC=O1712
O–H3010
Mass spectrometrygjr-–-
• First a substance is bombarded with a beam of high-energy electrons• This kicks an electron out of the molecule are creates the molecular ion
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"a mass spectroscopist is someone who figures out what something is by smashing it with a hammer & looking at the pieces"
JEOL (manufacturer) website
M + e– M+ + 2e–
We measure these positive molecular ions
Molecular mass from mass spectrumgjr-–-
• The electron is lost from a bond or a lone pair of electrons• As mass of electron is insignificant the mass of ion is same as original• Record mass-to-charge ratio (m/z)• Many ions produced & separated by magnetic field (high m/z travel faster)
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NH3 + e– +NH3 + 2e–
Cyclohexane
C6H12 + e– [C6H12]+ + 2e–
m/z 85due to 13C
isotope
m/z 84C = 6x12H = 12x1
Isotopesgjr-–-
• The two isotopes of chlorine, 35Cl &37Cl, occur in the ratio 3:1 • Thus two lines in spectrum at a ratio of 3:1
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m/z 114due to 37Cl
m/z 112due to 35Cl
C6H5Cl + e– [C6H5Cl]+ + 2e–
C6H535Cl Mr = M+
C6H537Cl Mr = M+2
= 112= 1146
Cl
Isotopes IIgjr-–-
• Multiple chlorines lead to a distinct isotope pattern due to probability of a molecule containing each isotope
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m/z 14835Cl, 37Cl
m/z 14635Cl, 35Cl
m/z 15037Cl, 37Cl
Cl
Cl
C6H4Cl2 + e– [C6H4Cl2]+ + 2e–
C6H435Cl35Cl Mr = M+
C6H435Cl37Cl Mr = M+2
C6H437Cl37Cl Mr = M+4
= 146= 148= 1506
Isotopes IIIgjr-–-
• The two isotopes of bromine, 79Br & 81Br, are the same except ratio is 1:1 • Therefore, the two lines in spectrum are in ratio of 1:1 (more or less)
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m/z 15881Br
m/z 15679Br
C6H5Br + e– [C6H5Br]+ + 2e–
C6H579Br Mr = M+
C6H581Br Mr = M+2
= 156= 1586
Br
Overviewgjr-–-
What have we learnt?• Over the entire of Unit 1, hopefully quite a lot• We have looked at functional groups & nomenclature• We then had a bit of fun with resonance, conformation & everyone's
favourite, stereochemistry• We finished with some of the analytical techniques that have allowed us to
'look' at molecules & thus build the models we have used elsewhere
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What's next?• Nothing from me you'll be glad to know• All the rest of the course builds on these fundamentals• Next up will be organic reactions
dr gareth rowlands; [email protected]; science tower a4.12http://www.massey.ac.nz/~gjrowlan