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Ochem part 4 > IR/UV spectroscopy, Mass spectrometry, NMR > Flashcards

IR/UV spectroscopy, Mass spectrometry, NMR Flashcards

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1
Q

Mass spectrometry is used to determine 3 things

A

1- overall mass of the molecule
2- determine what functional groups are present
3- how molecule is structurally constructed

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2
Q

What part of a molecule is most susceptible to e- loss?

A

1- (most important) non-bonding e- pairs (probably from the fx group)
2- where the best carbocation would form
3- where the best radical would form

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3
Q

Bonds break either via heterolytic or homolytic cleavage

A

Heterolytic cleavage- e- move towards electronegative atom

Product with positive charge will be the thing detected via mass spec, probably the carbocation

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4
Q

Homolytic cleaveage

A

Occurs between atoms with comparable electronegativity
One form is called alpha cleavage because it occurs on the alpha carbon
Product with positive change will be detected via mass spec, probably on the molecule that lost a single e- in radical formation

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5
Q

M+ peak

A

mass of the intact molecule

** alcohols don’t have M+ peak

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6
Q

M/Z ratio

A

mass/charge ratio, almost always 1

M/Z values of 31, 45, 59, etc. indicate presence of alcohol

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7
Q

M+1 and M+2

A

In some molecules, like bromine or chlorine have two different isotopes that differ by 1-3 g/mole

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8
Q

base peak

A

part present is greatest abundance
tells you about the structure or where the molecule is most likely to fragment
tells you how many Carbons are carved off of a fragment

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9
Q

1C

A

15

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10
Q

2C

A

29

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11
Q

3C

A

43

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12
Q

4C

A

57

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13
Q

5C

A

71

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14
Q

6C

A

85

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15
Q

7C

A

99

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16
Q

Bromine isotopes

A

79, 81

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17
Q

Chlorine isotopes

A

35, 37

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18
Q

In Alkyl Halides

A
  • e- most likely moved from halogen

* alpha cleavage would occur via radical mechanism and C next to the halogen will always stay intact

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19
Q

In Ethers

A
  • e- most likely removed from the oxygen
  • bond most likely broken= one that will form the best carbocation
  • base peak where bond to O breaks
  • alpha cleavage would occur at any other potential cutting away from the carbon groups
  • heavier group more likely to come off
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20
Q

In Alcohols (doesn’t work well with mass spec but will show up well with IR)

A
  • e- most likely to be removed from -OH
  • bond to -OH most likely to be broken
  • most likely base peak where the carbons are left off
  • Alpha cleavage will occur with additional cuts
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21
Q

Alcohol special notes

A
  • OH falls off so often that never reaches the detector

- OH will steal another H from the molecule to become water when it leaves, so the carbon count will be off by one

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22
Q

In Amines

A

Characteristic N peak
Can tell when N is present because M+ peak will be odd#
Alpha cleavage dominates in aliphatic amines (C-containing)

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23
Q

In Ketones and Aldehydes

A
  • e- most likely to be removed from O
  • O not lost in heterolytic cleavage because it is still hanging on by one bond
  • base peak most likely to occur with carbocation or heavier portion
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24
Q

Ketone and Aldehyde special cleavage

A

Called McLafferty cleavage, will occur if there is a H attached to the gamma carbon
This cleavage actually occurs between the alpha and beta carbons

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25
Q

Infrared Spectrometry is used to-

A

Determine what functional groups are present

- Does not tell you anything about structure

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26
Q

IR Theory

A
  • bonds in molecules vibrate, stretch and bend

- when frequency of vibrations matches with frequency of IR radiation, radiation is absorbed and vibrations intensify

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27
Q

IR Theory cont.

A

Shorter bond= faster vibration= higher wavenumber

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28
Q

IR Theory cont.

A

> polar the bond= > intensity of absorption

*** If bond is not polar, it will will be IR silent

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29
Q

Short bonds vibrate faster

A

Bond length is determined by the # bonds and size of the molecule

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30
Q

General breakdown of IR chart

A

4000-1500, Important

1500-800 = fingerprint region/iffy at best

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31
Q

4000-3000

A

bonds to H

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32
Q

3000-2000

A

triple bonds

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33
Q

2000-1500

A

double bonds

34
Q

1500-800

A

single bonds

35
Q

Wavenumbers for specific types of bonds

A

All approximate per Dr.v

36
Q

3500 (broad and deep)

A

-OH

37
Q

3000 (broad and deep)

A

COOH group, characteristic of -OH group

38
Q

3500 (narrow and shallow)

A

N

39
Q

3300

A

C—C-H (— represents triple bond)

40
Q

3100

A

C=C-H

41
Q

2900

A

C-C-H

42
Q

2700

A

Aldehyde O=C-H

43
Q

2500

A

skip this

44
Q

2300 (actually 2250)

A

C—N

45
Q

2100 (2150)

A

C—C

46
Q

1950

A

skip this

47
Q

1750

A

C=O

48
Q

1650

A

C=C

49
Q

1550

A

skip this

50
Q

1450 (in the fingerprint region)

A

benzene ring, but verify all other features

- should also have a 1650 C=C and 3100 C=C-H

51
Q

1350 (1375)

A

CH3 specifically

52
Q

1250 (specific to an ester)

A

C-O

53
Q

1050 (specific to an ester)

A

O-C

54
Q

If a molecule is symmetrical around a functional group-

A

the IR band for the group will be silent

55
Q

More polar =

A

deeper valley

56
Q

When using several types of spectra, use each for their unique strengths in this order:

A
  • 1st- look at IR for type of functional group
  • 2nd- then figure that fx group into the molar mass from the mass spec and use the base peak to determine where that functional group is located
57
Q

NMR=

A

(proton) nuclear magnetic resonance

58
Q

NMR applications

A
  • determine structure of molecule

- what fx groups are present

59
Q

NMR disadvantages

A
  • things drift

- most powerful, but still bad for determining presence of -OH and amines

60
Q

NMR theory

A
  • when something with a charge spins, it generates a magnetic field
  • nuclei of atoms spin and are charged, so they have a magnetic field if they have an odd # protons or neutrons
61
Q

NMR theory cont.

A
  • if an external magnetic field is applied, magnetic nuclei will line with or against the field
  • can be done with any magnetic nucleus, if not specific, it refers to 1^H NMR
62
Q

Value of the NMR signal and its placement will be determined by-

A
  1. how close something e- withdrawing is

2. how close you have pi bonds (double/triple bonds) or something that resonates

63
Q

ppm on an NMR plot

A

parts per million

- higher the ppm value = less shielded the atoms is (or closer to a magnetic field)

64
Q

Interpreting an NMR chart

A

closer to zero= easier to flip because it is protected from the magnetic field by electron cloud (ex. hydrocarbon)
closer to 10= not shielded

65
Q

Interpreting an NMR chart (most to least important)

A
  1. look at # signals
  2. splitting pattern
  3. # H’s expected based on 2N+2 rule
  4. if fewer signals, consider symmetry
  5. integration curves to determine #H’s that contributed to the signal
  6. position of signal (?fx group nearby)
  7. coupling constants to determine which H go together
66
Q

To determine how many signals a molecule gives:

A

look at how many unique kinds of H are present

- chemically different atoms will have their own unique signal

67
Q

When protons or carbons are chemically equivalent,

A

they give the same signal

68
Q

When designating a,b,c, etc.,

A

the a signal is the most shielded/furthest away from e- withdrawing molecule

69
Q

Most important NMR values

A

on a scale from 12- 0.9
12= electronegativity AND resonance
0.9= minding its own business with nothing e- withdrawing nearby

70
Q

0.9

A

CH3

71
Q

7-8

A

Aryl H on a benzene ring

72
Q

10

A

aldehyde

73
Q

12

A

COOH groups (carboxylic acid)

74
Q

Integration lines on NMR plots

A

used to determine relative number of H’s that contributed to the signal

75
Q

NMR signals will split when:

A

there is a neighboring H with different signal (a,b,c, etc)

76
Q

Singlet

A

Doesn’t split at all (no different signal neighboring H’s)

77
Q

Doublet

A

1H neighbor

78
Q

Triplet

A

2H neighbors split into 3 peaks

79
Q

Quartet

A

3H neighbors

80
Q

N+1 rule

A

1 neighbor H = 2 peaks

81
Q

Reduces # signals?

A

Branching or symmetry in molecular structure

Ochem part 4 (1 decks)

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