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Spectroscopy Infrared Spectra.

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Presentation on theme: "Spectroscopy Infrared Spectra."— Presentation transcript:

1 Spectroscopy Infrared Spectra

2 Infrared spectra in this presentation are taken by permission from the SDBS web site:

3 ?????????????????

4 Spectroscopy “seeing the unseeable”
Using electromagnetic radiation as a probe to obtain information about atoms and molecules that are too small to see. Electromagnetic radiation is propagated at the speed of light through a vacuum as an oscillating wave.

5 electromagnetic relationships: λυ = c λ µ 1/υ E = hυ E µ υ
  E = hc/λ E µ 1/λ λ = wave length υ = frequency c = speed of light E = kinetic energy h = Planck’s constant λ c

6 Two oscillators will strongly interact when their energies are equal. E1 = E2 λ1 = λ2 υ1 = υ2 If the energies are different, they will not strongly interact! We can use electromagnetic radiation to probe atoms and molecules to find what energies they contain.

7 some electromagnetic radiation ranges
Approx. freq. range Approx. wavelengths Hz (cycle/sec) meters Radio waves x x10-4 Infrared (heat) x x x10-7 Visible light x x x x10-7 Ultraviolet x x x X rays x x Gamma rays > 3x < 10-11

8 Infrared radiation λ = 2.5 to 17 μm υ = 4000 to 600 cm-1  These frequencies match the frequencies of covalent bond stretching and bending vibrations. Infrared spectroscopy can be used to find out about covalent bonds in molecules. IR is used to tell: 1. what type of bonds are present 2. some structural information

9 IR source è sample è prism è detector
graph of % transmission vs. frequency => IR spectrum 100 %T v (cm-1)

10 toluene

11 Some characteristic infrared absorption frequencies
BOND COMPOUND TYPE FREQUENCY RANGE, cm-1 C-H alkanes and alkenes (m) and RCH=CH and R2C=CH cis-RCH=CHR (v) trans-RCH=CHR aromatic rings (m) and monosubst and ortho-disubst meta-disubst and (m) para-disubst (m) alkynes O-H alcohols or phenols (b) C=C alkenes (v) aromatic rings and 1600 (v) C≡C alkynes (v) C-O primary alcohols (b) secondary alcohols (b) tertiary alcohols (b) phenols (b) alkyl ethers aryl ethers (b) and (m) all abs. strong unless marked: m, moderate; v, variable; b, broad

12 IR spectra of ALKANES C—H bond “saturated” (sp3) cm-1 cm-1 -CH -CH “ and 1375 -CH(CH3) “ and 1370, 1385 -C(CH3) “ and 1370(s), 1395 (m)

13 n-pentane 2850-2960 cm-1 CH3CH2CH2CH2CH3 sat’d C-H 3000 cm-1

14 n-hexane CH3CH2CH2CH2CH2CH3

15 2-methylbutane (isopentane)

16 2,3-dimethylbutane

17 cyclohexane no 1375 cm-1 no –CH3

18 IR of ALKENES =C—H bond, “unsaturated” vinyl (sp2) cm-1 RCH=CH & R2C=CH cis-RCH=CHR (v) trans-RCH=CHR C=C bond cm-1 (v)

19 1-decene 3020-3080 cm-1 C=C 1640-1680 unsat’d C-H
& RCH=CH2 C=C

20 4-methyl-1-pentene & RCH=CH2

21 2-methyl-1-butene R2C=CH2

22 2,3-dimethyl-1-butene R2C=CH2

23 IR spectra BENZENEs =C—H bond, “unsaturated” “aryl” (sp2) cm-1 mono-substituted , ortho-disubstituted meta-disubstituted , (m) para-disubstituted (m) C=C bond 1500, 1600 cm-1

24 ethylbenzene 3000-3100 cm-1 Unsat’d C-H 690-710, 730-770 mono-
1500 & 1600 Benzene ring , mono-

25 o-xylene ortho

26 p-xylene (m) para

27 m-xylene meta , (m)

28 styrene no sat’d C-H 1640 C=C & RCH=CH2 mono

29 2-phenylpropene Sat’d C-H R2C=CH2 mono

30 p-methylstyrene para

31 IR spectra ALCOHOLS & ETHERS
C—O bond (b) cm-1 1o ROH 1050 2o ROH 1100 3o ROH 1150 ethers O—H bond (b) 

32 1-butanol (b) O-H C-O 1o CH3CH2CH2CH2-OH

33 2-butanol O-H C-O 2o

34 tert-butyl alcohol O-H C-O 3o

35 methyl n-propyl ether no O--H C-O ether

36 2-butanone  C=O ~1700 (s)

37 C9H12 1500 & 1600 benzene C-H unsat’d & sat’d mono
C9H12 – C6H5 = -C3H7 isopropylbenzene n-propylbenzene?

38 n-propylbenzene

39 isopropylbenzene isopropyl split

40 C8H6 C-H unsat’d 1500, 1600 benzene 3300 C8H6 – C6H5 = C2H C-H mono
phenylacetylene

41 C4H8 Unst’d C=C R2C=CH2 isobutylene CH3 CH3C=CH2

42 Which compound is this? 2-pentanone 1-pentanol 1-bromopentane 2-methylpentane 1-pentanol

43 What is the compound? 1-bromopentane 1-pentanol 2-pentanone 2-methylpentane 2-pentanone

44

45 In a “matching” problem, do not try to fully analyze each spectrum
In a “matching” problem, do not try to fully analyze each spectrum. Look for differences in the possible compounds that will show up in an infrared spectrum.

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