CHE 232

MtWTF 8-8:50pm

Chemical Identification

• Comparison of

Physical
Properties

– Boiling Point
– Melting Point
– Density
– Optical rotation
– Appearance
– Odor

• Chemical Test

–

Elemental Analysis

• Burn the compound

and measure the

amounts of CO

2

, H

2

O

and other components

that are produced to

determine the

empirical formula.

• Used today as a test

of purity of

compounds that have

already been

identified.

• Spectroscopy

– measures the interaction of a compound

with electromagnetic radiation of different wavelengths.

– Nuclear Magnetic Resonance Spectroscopy (NMR)

measures the absorption of radio waves by C and H in a

magnetic field. Different kinds of C and H absorb energy of

different wavelengths.

– Infrared (IR) Spectroscopy measures the absorption of

infrared (heat) radiation by organic compounds. Different

functional groups (C=O, -OH) absorb energy of different

wavelengths.

– Ultraviolet/Visible Spectroscopy (UV/Vis) measures the

absorption of ultraviolet and visible light by  bonds in an

organic compound. Bonds of different types and with

different extents of conjugation (C=C, C=O, C=C–C=C,

aromatic) absorb energy of different wavelengths.

• Mass Spectrometry

– Doesn’t involve the absorption of any type of light.
– Used in determining the molecular weight and formula of

a compound.

– A compound is vaporized and ionized by bombardment

with a beam of high-energy electrons.

– The electron beam ionizes the molecule by causing it to

eject an electron.

– When the electron bean ionizes the molecule, the species

formed is called a radical cation, and symbolized as

M

+•

.

– The radical cation

M

+•

is called the molecular ion or

parent ion.

– The mass of

M

+

•

represents the molecular weight of

M

.

• Because

M

is unstable it decomposes to form

fragments of radicals and cations that have a
lower molecular weight than

M

+

•

.

• The mass spectrometer measures the mass of

these cations.

• The mass spectrum is a plot of the amount of

each cation (relative abundance) versus its

mass to charge ratio (m/z, where m is mass

and z is charge)

• Since z is almost always +1, m/z actually

measures the mass (m) of the individual ions

Methane CH

4

Though most C atoms have an atomic mass of 12,
1.1% have a mass of 13. Thus,

13

CH

4

is responsible for

the peak at m/z = 17. This is called the M + 1 peak.

• Alkyl Halides and the M+2 peak

– Most elements have one major isotope.
– However some halogens have more than one. Iodine and

Fluorine are isotopically pure but….

– Chlorine has two common isotopes,

35

Cl and

37

Cl, which occur

naturally in a 3:1 ratio.

– The larger peak, M, which corresponds to the compound

containing

35

Cl. The smaller peak, the M+2 peak, corresponds

to the compound containing

37

Cl.

– Thus, when the molecular ion peak consists of two peaks (M,

M+2) in a 3:1 ratio, a Cl atom is present.

– Bromine also has 2 isotopes,

79

Br and

81

Br in a 1:1 ratio. Thus

when the molecular ion consists of two peaks (M, M+2) in a

1:1 ratio, the compound contains a Br atom.

Cl

Cl

-CH

3

-Cl

-CH

3

Br

-Br

-CH

3

Sample introduced into GC
inlet vaporized at 250 °C ,
swept onto the column by He
carrier gas & separated on
column. Sample components
emerge from column, flowing
into the capillary column
interface connecting the GC
col-umn and the MS (He
removed).

Gas Chromatography/Mass Spectrometry (GC/MS)

Mass
Spectrometry

Gas Chromatography-Mass Spectrometry (GC-
MS)

•To analyze a urine sample for tetrahydrocannabinol,
(THC) the principle psychoactive component of
marijuana, the organic compounds are extracted
from urine, purified, concentrated and injected into
the GC-MS.
•THC appears as a GC peak, and gives a molecular
ion at 314, its molecular weight.

Forensic Mass Spectrometry

J. Yinon, Ed., Forensic Applications of Mass Spectrometry, CRC Press, 1995

•

Analysis of Body Fluids for Drugs of Abuse

•

Analysis of Hair in Drug Testing

•

Sports Testing

•

Analysis of Accelerants in Fire Debris

•

Analysis of Explosives

•

Use of Isotope Ratios

pentane

1-pentene

1-pentyne

What is the mass of the molecular ion of
C

3

H

6

O?

3 C’s 3 x 12 = 36

6 H’s 6 x 1 = 6

1 O 1 x 16 =
16

36 + 6 + 16 =
58

M = M

+•

(m/z) =

58

What molecular ions would you expect for
C

4

H

9

F?

4 C’s 4 x 12 = 48

9 H’s 9 x 1 = 9

1 F 1 x 19 = 19

48 + 9 + 19 =
76

A molecular ion peak at m/z
= 76

What molecular ion peak would you expect for
C

5

H

11

Cl?

5 C’s 5 x 12
= 60

11 H’s 11 x 1
= 11

1 Cl 1 x 35
= 35

60 + 11 + 35 =
106

C

5

H

11

35

Cl (m/z) =

106 ?

5 C’s 5 x 12 = 60

11 H’s 11 x 1 =
11

1 Cl 1 x 37 =
37

60 + 11 + 37 =
108

C

5

H

11

37

Cl (m/z) =

108

So the molecular ion peak of C

5

H

11

Cl consist of

two peaks at 106 and 108 in a 3:1 ratio.

C

N

H

2N+2

C

3

H

4

C

N

H

2N+2

C

N

H

2N

C

N

H

2N-2

(2(3)+2-5-1)/2=1

(2(4)+2-
7+1)/2=2

O

Br

O

NH

2

Suggest possible formulas for a molecular ion
(m/z) of 72.
Step 1 – Determine the maximum
number of C’s.

72/12 = 6 carbons maximum
C

6

is not a reasonable formula

Subtract 1 carbon and add
12 H’s

C

5

H

12

Step 2 – Calculate Degrees of
Unsaturation

(2n+2-#H’s)/2

(2(5) + 2 – 12)/2 =
0

Step 3 – Incorporate O into the formula (-CH

4

when adding O)

C

4

H

8

O

O

(2(4) + 2 – 8)/2
= 1

O

Add another O
atom

C

3

H

4

O

2

O

O

O

O

(2(3) + 2 – 4)/2
= 2

Adding O adds one degree of
unsaturation.

Suggest possible formulas for a molecular ion
(m/z) of 105.

Step 1 – If the mass of the molecular ion is
odd it contains at least one N.

N = 14
amu

105 – 14 =
91

Step 2 – Determine max # C’s

91/12 = 7.5

C

7

NH?

Step 3 – Add enough H’s to make up the rest of the
mass.

C

7

NH?

7 x 12 =
84

1 x 14 =
14

105 – (84 + 14) =
7

7 H’s gives C

7

NH

7

.

(2(7.5) + 2 – 7)/2
= 5

HN

Step 4 – Add an O atom.

C

7

NH

7

 C

6

NOH

3

(2(6.5) + 2 – 3)/2
= 6

N

O

Suggest a structure for a molecular ion
peak that has 2 peaks 144 and 146 in a
1:1 ratio.

Step 1 – Since we have an M and M + 2
peak as the molecular ion, we know
that there is a halogen.

Also since they occur in a 1: 1 ration
we know it’s Br.

144 – 79 =
65

146 – 81 =
65

Step 2 – Determine max #
C’s

65/12 = 5
Carbons

Step 3 – Add enough H’s to make up the rest of
the mass.

5 x 12 =
60

144 – (60 + 79) = 5
H’s

C

5

BrH

5

(2(5) + 2 – 6)/2 = 3

Br

C

6

H

6

m/z = 78

b.p. = 80.1C

C

7

H

8

m/z = 92

b.p. =
110.6C

C

8

H

10

m/z = 106

b.p. =
138.3C

Since the sample consists of three
components, the GC spectrum will have 3
peaks. Their order will be benzene, toluene
and p-xylene in order of increasing boiling
point.

And the mass spectra of these
compounds will have molecular ion
peaks corresponding to their
molecular weights

Benzene

Toluene

p-xylene

Electromagnetic spectrum

Infrared region is 2.5 – 25 m

Absorption of infrared light (heat) by a compound.
Different functional groups absorb at different
wavelengths.

.

Absorptions are recorded in
wavenumber = 1/

Infrared Spectroscopy

Using this scale, the IR region is 4000-
400 cm

-1

.

Chemical bonds are not static, they have
different vibrational modes, such as bending
and stretching.

Different kinds of bonds vibrate at different
frequencies, therefore they absorb different
wavelengths of radiation.

IR spectroscopy distinguishes between the
different types of bonds, thus allowing the
identification of the functional groups present.

IR spectra are a plot of wavelength or
wavenumber (x axis) versus
transmittance (y axis). Transmittance is
a measure of the light that isn’t absorbed
by the sample.

There are two sections of the IR spectra, the
functional group region (greater than 1500) and
the fingerprint region (less than 1500). We will be
concerned with the functional group region..

Bond strength and the wavelength of
absorption are proportional. Thus, the
stronger the bond the higher the wavelength
of absorption.

4000-2500

2500-2000

2000-1500

1500-400

Bonds to H

C-H, N-H, O-H

Triple Bonds

Double
Bonds

Single Bonds

CC, CN

C=C, C=O,
C=N

CC, CO, CN, CX

O-H stretch appears at 3200-3600, C-H at 3000.

-OH

 -CH

N-H stretch appears at 3200-3500, C-H at
3000.

-NH

 -CH

CC stretch appears at 2250, C-H at 3000.

 -CH

 - CC

CN stretch appears at 2250, C-H at 3000.

- CN 

 -CH

C=O stretch appears at 1650-1800, C-H
at 3000.

- C=O 

 -CH

C=C stretch appears at 1650, C-H at 3000.

- C=C 

 -CH

Which of these two isomers, cyclopentane or 1-
pentene, is this the IR spectra of?

1650 cm

-1

3000
cm

-1

What IR peaks would you expect for the two
molecules with the formula C

2

H

6

O?

OH

O

Ethan
ol

Dimethyl
ether

-OH

3200 - 3600

cm

-1

-CH

3000 cm

-1

-CH

3000 cm

-1

IR spectra of ethanol. The spectra of dimethyl
ether would look similar minus the large –OH
stretch.

3200 - 3600
cm

-1

3000
cm

-1

O

Which of the three compounds below
matches the IR spectra?

O

OH

1650 -
1800
cm

-1

3000
cm

-1

Using the mass and IR spectra, determine the
identity of the compound containing only C,
H and O?

100

58

Determine the # of C’s.

100/12 = 8 C’s 8 x 12 = 96
100 – 96 = 4

So C

8

H

4

We know there is at least one oxygen so
subtract a –CH

4

group.

C

7

O DOUS=8 also not a likely

formula

So get rid of 1 C and add 12 H’s

C

6

H

12

O DOUS=1 looks much

better

Now add one more O for more
possibilities.

C

5

H

8

O

2

DOUS=2

DOUS
=7

O

C

6

H

12

O

O

O

C

5

H

8

O

2

O

O

O

O

1650-1800
cm

-1

1650 cm

-1

O

O

OH

OH

7

7

4

4

7

2

3

4

3

4

2

3

10

5