1. Vibrational Spectroscopy - IR
‘Near IR’
Most useful region for
Functional Groups,
especially cm-1
p. 15 Course Manual

2. Most organic compounds have C-H bonds:
not as useful for structural assignment
C-H stretch ~ 3000 cm-1
C-H bend ~ cm-1
scale
‘salt’ cut-off

3. Functional groups involve other vibrations
e.g. ALCOHOLS all have –O-H stretches
Alcohols also have –C-O stretches
~ cm-1
-O-H stretches are always strong and almost
always broad, ~ cm-1

4. p. 16
~3400cm-1
~1100 cm-1

5. Functional group regions
Alcohols,
amines,
acids,
amides
alkynes, nitriles
alkynes
carbonyl compounds
aldehydes, ketones, acids,
esters, amides,....
Functional group regions

6. BAND STRENGTH Depends upon the bond dipole: large dipole = strong band
C---O large vs. C---C small
d+ d-

7. From a ball and spring model…we will see this again
BAND POSITION
1: k, force constant depends on strength of bond, C=C > C-C
2: masses, easier to stretch bonds of heavy elements, C-H > C-O
From a ball and spring model…we will see this again
later so don’t worry too much about the details now

8. p. 19
ALKANES
cm-1
saturated-C-H stretch, <3000cm-1

9. ALKENES UNSATURATED =C-H stretch, >3000 cm-1 not generally useful
p.19
ALKENES
UNSATURATED =C-H stretch, >3000 cm-1
not generally useful
bends
~3050cm-1
in alkenes and aromatics!

10. ALKYNES 2150cm-1, WEAK 3300cm-1, SHARP but diagnostic
Compare: alcohol is normally broad

11. STRETCHES: LEARN THESE VALUES! C---H C 3300cm-1 3050cm-1 ~2900cm-1
p. 20
STRETCHES:
C---H C
3300cm-1
3050cm-1
~2900cm-1
2150 cm-1
1650 cm-1
1250 cm-1
LEARN THESE VALUES!

12. p. 21
BENDS:
1,1

13. p. 21

14. p. 21
1,1

15. p. 21
TRANS
~970
CIS
~700

16. Similarly for benzene derivatives
p. 22 2 3 1

17. p. 22
MONO
TWO
META
THREE

18. p. 22
ORTHO
<800
PARA
>800

19. CARBONYL C=O REGION: 1650-1800 cm-1
p. 23/24
CARBONYL C=O REGION: cm-1
ANHYDRIDES R-CO-O-CO-R O’S
1810, 1760 cm-1

20. ACID CHLORIDES R-CO-Cl
p. 24
ACID CHLORIDES R-CO-Cl
1800 cm-1

21. p. 24
ESTERS R-CO-O-R’
1735 cm-1

22. ALDEHYDES R-CO-H (RCHO) & KETONES R-CO-R’
p. 25
ALDEHYDES R-CO-H (RCHO) & KETONES R-CO-R’
Both 1 “O”
H-C=O
2750
H-C=O
1725
C=O
1715

23. Note: esters also have 2 “O’s”, are at 1735cm-1, but no –O-H
p. 25
ACIDS R-CO-OH
But also a huge –O-H
1710 cm-1
Note: esters also have 2 “O’s”,
are at 1735cm-1, but no –O-H

24. p. 24/25
ESTER
ACID

25. AMIDES have both a C=O str and an N-H str (unless 3°; not as broad
as an acid)
C=O str <1700,
~1690cm-1
p. 26
1o amide RCONH2
two NH peaks
2o amide RCONHR’
one NH peak
3o amide RCONR’R”
no NH peak

26. CONJUGATION: alternating single and double bonds
p. 27
CONJUGATION: alternating single and double bonds
Conjugation of a carbonyl group
LOWERS frequency by ~25-30 cm-1
oxygen is electronegative
removes electrons thru bonds
this minor resonance structure has
a C—O single bond so it should be
easier to stretch
The resonance hybrid has a lower frequency

27. p. 27
>1700
<1700

28. p. 28
ACIDS
>1700
saturated
<1700
conjugated

29. p. 28
ESTERS

30. Strain Generally, as you strain bonds,
p. 29
Strain
Generally, as you strain bonds,
it becomes harder to stretch them.

31. p. 29
conj on C=O
LOWER n
conj on O
HIGHER n
strain increases

32. ALCOHOLS & ETHERS C-O Stretch If it has ONE “O” no C=O at 1700
p. 30
If it has ONE “O”
no C=O at 1700
no OH at 3400
must be an ether
C-O Stretch
Ar-OH
R3C-OH
R2CH-OH 1100
RCH2-OH 1050
in simple cases

33. AMINES 3400 N-H stretch 1o R-NH2 - two peaks Amines often weak,
amides strong
(and no C=O)
2o R2NH - one peaks
3o R3N NO NH peak

34. NITRILES 2250 cm-1 Alkynes 2150cm-1 [medium to strong]
p. 32
NITRILES cm-1
[medium to strong]
Alkynes 2150cm-1
[weak, sometimes absent]

35. SOLVING PROBLEMS: WHERE TO START?
Acyclic saturated hydrocarbons have the formula CnH2n+2
so if the formula of your compound does not fit, it must have
unsaturation (double bonds) or rings present
When solving problems WE ALWAYS work out
the number of double bond equivalents (DBE’s) FIRST

36. So, if your compound is a hydrocarbon:
# DBE’s = {CnH2n+2 - your formula} / 2
n = number of carbon atoms in your formula
e.g. if your unknown is C3H4
# DBE’s = {C3H8 – C3H4}/2 = 4/2 = 2
There are then formally two double bonds,
two rings, or one of each
all fit, but alkyne is most likely!
alkynes count as two double bonds

37. Benzene has 3 double bonds + one ring so has 4 DBE’s
p. 34
Benzene has 3 double bonds + one ring
so has 4 DBE’s
If you have >6C’s and 4 or more DBE’s,
HIGHLY likely to be aromatic

38. p. 34
OTHER ELEMENTS:
If valency = 1 (e.g. halogens) count each halogen as a hydrogen
If valency = 2 (e.g. oxygen, sulfur) ignore in the count
If valency = 3 (e.g. nitrogen) add one to the CnH2n for each atom present so if one nitrogen use CnH2n+3, if two nitrogens use CnH2n+4 etc

39. -OH str PROBLEM A, p. 35 C4H10O DBE’s = {C4H2n+2=10 – C4H10}/2
ignore oxygen rule 2 = zero
what is main peak?
so its an alcohol
can we tell what type?
PRIMARY
1042 cm-1 so

40. PROBLEM B, p. 36 C3H3Br what is main peak? 3300 cm-1 sharp!
additional hint
2150 cm-1 C
C--H
2 DBE’s - OK
DBE = {C3H2n+2=8 – C3H3+1}/2 = 4/2 = 2

41. a Br but cannot join Br to alkyne so must be:
PROBLEM B, p C3H3Br
DBE = {C3H2n+2=8 – C3H3+1}/2 = 4/2 = 2 -C
C--H
2 DBE’s - OK
Subtract what we have found from what we started with: C3H3Br – C2H = CH2Br so we have CH2 and
a Br but cannot join Br to alkyne so must be:
H-CC-CH2Br

42. PROBLEM C, p. 37 C5H8O MAIN PEAK = 1675 = conjugated carbonyl
DBE = [(2x5 + 2)-8]/2 = 2
MAIN PEAK
= 1675
= conjugated carbonyl
nothing at 2750, so
KETONE
KETONE or
ALDEHYDE?

43. PROBLEM C, p. 37 C5H8O DBE = 2 so conjugated ketone
so what is other DBE?
peak at 970 cm-1 suggests?
trans-alkene
C3H2O here
C2H6 left to find

44. p. 37
unique solution

45. PROBLEM D, p. 38 C10H14 DBE = {(2 x10 + 2) – 14}/2 = 4
4DBE, >6C suggests?
other functional groups?
BENZENE ring NO
what is substitution pattern?
829 cm-1 =
PARA
BUT 4 C’s still to find
C C3
C C4

46. PROBLEM E, p. 39 C7H9N DBE = {(2 x7 + 2 + 1) – 9}/2 = 4 add one for N
BENZENE DERIV (6C, 4DBE)
What is the N?
PRIMARY AMINE R-NH2
so we still need to find one more carbon
benzene carbon -NH2
but how joined?

47. PROBLEM E, p. 39 C7H9N benzene carbon -NH2 but how joined?
750 cm-1 = ?
ortho-benzene
so only one choice:

48. PROBLEM F, p. 40 C7H8O DBE = {(2 x7 + 2) – 8}/2 = 4 = BENZENE
substitution pattern?
3 peaks = META (6C)
What is the oxygen?
broad, 3300 = -OH
meta-BENZENE + carbon + OH

49. PROBLEM G, p. 41 C9H10O <1700 DBE = {(2 x 9 + 2) – 10}/2 = 5
Benzene
+ 1 more DBE
aldehyde
or ketone?
What is the oxygen?
~1700 = carbonyl (C=O)
check 2750
none, so
KETONE
<1700 or >1700
<1700, so conjugated

50. PROBLEM G, p. 41 C9H10O so benzene + conjugated ketone,
but only 5DBE, so where is the ketone?
2 peaks, so
mono-sub
on the benzene ring: but what substitution?
C9H10O – C7H5O = C2H5

51. PROBLEM H, p. 42 C9H10O Same formula, peaks as last problem!
BUT difference is
C=O stretch is >1700
so not conjugated to benzene

52. PROBLEM I, p. 43 C4H4O3 DBE: {(2 x 4 + 2)- 4}/2 = 3 1854, 1781, 3”O”s
ANHYDRIDE
what other atoms to find?
only H’s (4)
but what other info have we not used?

53. p. 43
+ 4H
DBE: {(2 x 4 + 2)- 4}/2 = 3
the anhydride is only 2DBE, so one more DBE to find X
RING!

54. PROBLEM J, p. 44 C8H11N DBE: {(2 x 8 + 2 + 1) – 11}/2 = 4
What is the N ?
2o amine
i.e. C-NH-C
4 DBE?
benzene
mono-substituted
all atoms, but how connected? OR

55. PROBLEMS K & L, pp. 45-46 C3H5O2Cl DBE: {(2 x 3 + 2) – (5 + 1)}/2 = 1
2 ‘O’ s,
huge OH AND C=O hence ACID -COOH
this is the DBE
so we have: -Cl -COOH and C2H4
this can be CH3CH< or -CH2CH2-
so:
Cl-CH2CH2-COOH or CH3CHClCOOH

56. p K
1718 L
1710

57. Halogen close to carbonyl INCREASES the frequencyp
Halogen close to carbonyl
INCREASES the frequency
Small but noticeable effect: caused by decreased electron density at C which results in more O to C donation

58. p K
CH3CHClCOOH
1718 L
ClCH2CH2COOH
1710

59. PROBLEM M, p. 47 C5H10N2 R3N DBE: {(2 x 5 + 2 + 2) – 10}/2 = 2
2250 cm-1
add 2 for 2 N’s
What are the N’s?
NITRILE
2 DBE
So what is other N ?
What atoms left?
1 C

60. Could be but not a common functional group and not met in Chem 231
What atoms left?
1 C
Could be but not a common functional group
and not met in Chem 231
Much more likely to be:

61. PROBLEM N, p. 48 C11H12O2 DBE : {(2 x11 + 2) – 12}/2 = 6 + 2 “O” s
BENZENE
MOST likely?
ESTER (1 DBE)
so still one more DBE
MOST likely?
-CH=CH- some sort of double bond
ester is likely conjugated
on carbonyl side

62. p. 48
NEED MORE INFORMATION! ?
980
trans-alkene?

63. p. 22
MONO
TWO
META
THREE

64. p. 48
mono?
980
meta?
Could be mono-substituted OR Meta-substituted: ambiguous?
NMR can sort this out easily as we will see later
ASSIGNMENT 2