Transcript Document

CHE 232
MtWTF 8-8:50pm
Chemical Identification
• Comparison of
Physical Properties
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Boiling Point
Melting Point
Density
Optical rotation
Appearance
Odor
• Chemical Test
– Elemental Analysis
• Burn the compound and
measure the amounts of
CO2, H2O 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 CH4
Though most C atoms have an atomic mass of 12, 1.1% have a mass
of 13. Thus, 13CH4 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, 35Cl and 37Cl,
which occur naturally in a 3:1 ratio.
– The larger peak, M, which corresponds to the
compound containing 35Cl. The smaller peak, the M+2
peak, corresponds to the compound containing 37Cl.
– 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, 79Br and 81Br 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
-CH3
-Cl
-CH3
Br
-Br
-CH3
Gas Chromatography/Mass Spectrometry (GC/MS)
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).
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
•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
J. Yinon, Ed., Forensic Applications of Mass Spectrometry, CRC Press, 1995
pentane
1-pentene
1-pentyne
What is the mass of the molecular ion of C3H6O?
3 C’s
3 x 12 = 36
6 H’s
6x1=6
1O
1 x 16 = 16
36 + 6 + 16 = 58
M = M+•(m/z) = 58
What molecular ions would you expect for C4H9F?
4 C’s
4 x 12 = 48
9 H’s
9x1=9
1F
1 x 19 = 19
48 + 9 + 19 = 76
A molecular ion peak at m/z = 76
What molecular ion peak would you expect for C5H11Cl?
5 C’s
5 x 12 = 60
5 C’s
11 H’s
11 x 1 = 11
11 H’s
11 x 1 = 11
1 Cl
1 x 35 = 35
1 Cl
1 x 37 = 37
60 + 11 + 35 = 106
C5H1135Cl (m/z) = 106 ?
5 x 12 = 60
60 + 11 + 37 = 108
C5H1137Cl (m/z) = 108
So the molecular ion peak of C5H11Cl consist of two peaks at
106 and 108 in a 3:1 ratio.
CNH2N+2
CNH2N+2
CNH2N
CNH2N-2
C3H4
(2(3)+2-5-1)/2=1
O
Br
(2(4)+2-7+1)/2=2
O
NH2
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
C6 is not a reasonable formula
Subtract 1 carbon and add 12 H’s
C5H12
Step 2 – Calculate Degrees of Unsaturation
(2n+2-#H’s)/2
(2(5) + 2 – 12)/2 = 0
Step 3 – Incorporate O into the formula (-CH4 when adding O)
O
C4H8O
O
(2(4) + 2 – 8)/2 = 1
Add another O atom
O
C3H4O2
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
C7NH?
Step 3 – Add enough H’s to make up the rest of the mass.
C7NH?
7 x 12 = 84
7 H’s gives C7NH7.
(2(7.5) + 2 – 7)/2 = 5
HN
1 x 14 = 14
105 – (84 + 14) = 7
Step 4 – Add an O atom.
C7NH7  C6NOH3
O
(2(6.5) + 2 – 3)/2 = 6
N
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
C5BrH5
(2(5) + 2 – 6)/2 = 3
Br
C6H6
C7H8
C8H10
m/z = 78
m/z = 92
m/z = 106
b.p. = 80.1C
b.p. = 110.6C
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
Infrared Spectroscopy
Absorption of infrared light (heat) by a compound. Different
functional groups absorb at different wavelengths.
Absorptions are recorded in wavenumber = 1/
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
Bonds to H
C-H, N-H, O-H
2500-2000
Triple Bonds
2000-1500
Double Bonds
CC, CN
C=C, C=O, C=N
1500-400
Single Bonds
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.
 - CC
 -CH
CN stretch appears at 2250, C-H at 3000.
 -CH
- CN 
C=O stretch appears at 1650-1800, C-H at 3000.
 -CH
- C=O 
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?
3000
cm-1
1650 cm-1
What IR peaks would you expect for the two molecules with the
formula C2H6O?
O
OH
Ethanol
-OH
3200 - 3600 cm-1
-CH
3000 cm-1
Dimethyl ether
-CH
3000 cm-1
IR spectra of ethanol. The spectra of dimethyl ether would
look similar minus the large –OH stretch.
3000 cm-1
3200 - 3600 cm-1
Which of the three compounds below matches the IR
spectra?
O
O
3000 cm-1
OH
1650 1800 cm-1
Using the mass and IR spectra, determine the identity of
the compound containing only C, H and O?
58
100
Determine the # of C’s.
100/12 = 8 C’s
So C8H4
8 x 12 = 96
100 – 96 = 4
DOUS=7
We know there is at least one oxygen so subtract a –
CH4 group.
O
C7O
DOUS=8 also not a likely formula
So get rid of 1 C and add 12 H’s
C6H12O DOUS=1 looks much better
Now add one more O for more possibilities.
C5H8O2 DOUS=2
O
C6H12O
O
C5H8O2
O
O
O
O
O
O
1650-1800 cm-1
1650 cm-1
OH
OH
7
4
7
4
2
7
3
3
4
3
2
4
10
5