Top Banner
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 a4.12 http://www.massey.ac.nz/~gjrowlan
17

Unit One Part 10: infrared spectroscopy and mass spectrometry

Feb 18, 2022

Download

Documents

dariahiddleston
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Page 1: Unit One Part 10: infrared spectroscopy and mass spectrometry

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

Page 2: Unit One Part 10: infrared spectroscopy and mass spectrometry

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

Page 3: Unit One Part 10: infrared spectroscopy and mass spectrometry

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

Page 4: Unit One Part 10: infrared spectroscopy and mass spectrometry

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.

Page 5: Unit One Part 10: infrared spectroscopy and mass spectrometry

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

5

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...)

Page 6: Unit One Part 10: infrared spectroscopy and mass spectrometry

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

Page 7: Unit One Part 10: infrared spectroscopy and mass spectrometry

Functional group absorptionsgjr-–-

• Other versions of this kind of table can be found in the course notes

7

Page 8: Unit One Part 10: infrared spectroscopy and mass spectrometry

Examples of IR spectragjr-–-

NH H

benzeneamineaniline

8

NH234803395

NH3C H

N-methylbenzenamineN-methylaniline N–H

3443

Page 9: Unit One Part 10: infrared spectroscopy and mass spectrometry

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

Page 10: Unit One Part 10: infrared spectroscopy and mass spectrometry

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

Page 11: Unit One Part 10: infrared spectroscopy and mass spectrometry

Examples of IR spectra IVgjr-–- 11

N

OH

H

butanamideC=O16621634

N–H33563184

O

OH

butanoic acidC=O1712

O–H3010

Page 12: Unit One Part 10: infrared spectroscopy and mass spectrometry

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

12

"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

Page 13: Unit One Part 10: infrared spectroscopy and mass spectrometry

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)

13

NH3 + e– +NH3 + 2e–

Cyclohexane

C6H12 + e– [C6H12]+ + 2e–

m/z 85due to 13C

isotope

m/z 84C = 6x12H = 12x1

Page 14: Unit One Part 10: infrared spectroscopy and mass spectrometry

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

14

m/z 114due to 37Cl

m/z 112due to 35Cl

C6H5Cl + e– [C6H5Cl]+ + 2e–

C6H535Cl Mr = M+

C6H537Cl Mr = M+2

= 112= 1146

Cl

Page 15: Unit One Part 10: infrared spectroscopy and mass spectrometry

Isotopes IIgjr-–-

• Multiple chlorines lead to a distinct isotope pattern due to probability of a molecule containing each isotope

15

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

Page 16: Unit One Part 10: infrared spectroscopy and mass spectrometry

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)

16

m/z 15881Br

m/z 15679Br

C6H5Br + e– [C6H5Br]+ + 2e–

C6H579Br Mr = M+

C6H581Br Mr = M+2

= 156= 1586

Br

Page 17: Unit One Part 10: infrared spectroscopy and mass spectrometry

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

17

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