Download CHEM 3780 Organic Chemistry II Infrared Spectroscopy and Mass

CHEM 3780 Organic Chemistry II
Infrared Spectroscopy and Mass Spectrometry Review
More information can be found in Chapter 12 in your textbook for CHEM 3750/ 3770 and on pages 13-28 in your
laboratory manual.
Identify the functional group(s) in the following molecules:
O
O
O
O
OH
OH
Functional
Group(s)
O
OH
C
O
N
H3C
C
C
CH3
O
OH
Functional
Group(s)

Primary Use of IR spectroscopy is functional group identification. Organic compounds absorb energy in the infrared
region (IR) of the electromagnetic spectrum.

Bonds are not rigid links—rather, one can compare them to tiny “springs” with varying degrees of stiffness that
connect the atoms together.

These vibrations include: stretching, bending, twisting, etc.

Infrared energy can cause the atoms of organic compounds to vibrate faster and with increased amplitude about the
covalent bonds that connect them.

The most important factors determining where a chemical bond will absorb in the IR region are:
o The bond order (single, double, or triple bond)—i.e., the stiffness of the spring
o The types of atoms joined by the bond (i.e., the mass (m1, m2)
Step 1: Unsaturation / Elements of Unsaturation / Degrees of Unsaturation:
Calculate the number of unsaturations (rings or multiple bonds) to limit the number of possible structures. The degree of
unsaturation indicates whether or not the compound has one or more double bonds or rings, or even a triple bond. There
are many different ways to calculate it:
*Note the following relationship between carbon and hydrogen: alkanes - CnH2n+2; alkenes and cyclohexane - CnH2n;
alkynes - CnH2n-2. Try these formulae by analyzing ethane, ethene, and ethyne.
*Note the following rules when heteroatoms (atoms other than carbon and hydrogen) are present:
organohalogens, containing C, H, X where X = F, Cl, Br, I; oxygen containing compounds, containing C, H, O; and
nitrogen containing compounds, containing C, H, N. Of course there are molecules that have all of the above, i.e. C, H, O,
X, N.
Take the formula that you are given and

Add the number of halogens to the number of hydrogens

Ignore the number of oxygens

Subtract the number of nitrogens from the number of hydrogens
For example: C5H9N is equivalent to “C5H8”. Compare the latter with a fully saturated molecule. According to CnH2n+2 an
alkane with 5 carbons should have 12 hydrogens. So, half of the hydrogen difference gives you the element(s)/degree(s)
of unsaturation: (12-8) / 2 = 2. Thus, C5H9N has two degrees of unsaturation and could contain one of the following: a) a
triple bond; b) two double bonds or c) one double bond and a ring.
Or take the formula and use the equation below

Number of Unsaturations = {2(#C) + 2 - #H - #Hal + #N}/2
For example: C5H9N would yield: {2x5 + 2 – 9 – 0 + 1}/2 = 2
Step 2: In even rather simple, mono-functional molecules there are so many IR bands that it is not feasible to assign every
band in an IR spectrum. Don’t try to assign or interpret every peak in the spectrum. Concentrate on learning the major
bands and recognizing their presence and absence in any given spectrum. It will be helpful to you to have the following
band regions committed to memory if you have not already done so.
3500-3300 cm-1
3500-3200 cm-1
3100-3000 cm-1
3000-2850 cm-1
1760-1665 cm-1
1680-1640 cm-1
N–H stretch
O–H stretch
C–H stretch
C–H stretch
C=O stretch
C=C stretch
primary and secondary amines
alcohols, a broad, strong band
alkenes
alkanes
ketones, aldehydes, esters
alkenes
Some things to look for:
Indicates:
alkyl groups (present in most organic molecules)
a C=C bond or aromatic group in the molecule
Are any or all to the right of 3000?
Are any or all to the left of 3000?
Look for a carbonyl in the region 1760-1690. If there is such a band:
Is an O–H band also present? (3000-2500 cm-1)
Is a C–O band also present? (1290-1180 cm-1)
Is an aldehydic C–H band also present? (2830-2700 cm-1)
Is an N–H band also present? (3500-3300 cm-1)
Are none of the above present?
Indicates:
a carboxylic acid group
an ester
an aldehyde
an amide
a ketone
Look for a broad O–H band in the region 3500-3200 cm-1. If there is such a band:
Indicates:
Is an O–H band present?
an alcohol or phenol
Look for a single or double sharp N–H band in the region 3400-3250 cm-1. If there is such a band:
Indicates:
Are there two bands?
a primary amine
Is there only one band?
a secondary amine
Other structural features to check for:
-1
Are there C–O stretches? (1290-1180 cm )
Is there a C=C stretching band? (1635-1675 cm-1)
Are there aromatic stretches? (860-680, 1500-1600 cm-1)
Is there a C≡C band? (2260-2150 cm-1)
Are there -NO2 bands? (1570-1550, 1380-1360 cm-1)
Indicates:
an ether (or an ester if there is a
carbonyl band too)
an alkene
an aromatic
an alkyne
a nitro compound
MOST IMPORTANT THING TO REMEMBER.....
When analyzing an IR spectrum avoid the temptation to try to assign every peak…
IR Spectroscopy
C-C and C-H Bonds
sp3 C—C
weak, not useful
sp2 C=C
1600-1700 cm-1
2
sp C=C (aryl) 1450-1600 cm-1
sp C≡C
2100-2250 cm-1
3
sp C—H
2800-3000 cm-1
2
sp C—H
3000-3300 cm-1
sp C—H
3300 cm-1
C(CH3)2
1360-1385 cm-1 (two peaks)
Alcohols and Amines
O—H or N—H 3000-3700 cm-1
C—O or C—N 900-1300 cm-1
Ethers
C—O
1050-1260 cm-1 strong
Ketones (saturated)
C=O
1640-1820 cm-1
Aldehydes
C=O
C(O) —H
1640-1820 cm-1
2820-2900 cm-1 (weak)
2700-2780 cm-1 (weak)
Carboxylic acids
C=O
1700-1725 cm-1
C(O)—OH
3330-2900 cm-1
Acid Chlorides
C=O
1800 cm-1
Acid anhydrides
C=O
1740-1800 cm-1
C—O
900-1300 cm-1
Esters
C=O
C(O)—OR
1735-1800 cm-1
1100-1300 cm-1
Amides
C=O
C(O)N—H
1630-1680 cm-1
3200-3400 cm-1
Nitrile
C≡N
2200-2250 cm-1
Interpreting Mass Spectra
A simple spectrum that of methanol is shown below. CH3OH+. (the molecular ion) and fragment ions appear in this
spectrum. Major peaks are shown in the table next to the spectrum. The x-axis of this bar graph is the increasing m/z
ratio. The y-axis is the relative abundance of each ion, which is related to the number of times an ion of that m/z ratio
strikes the detector. Assignment of relative abundance begins by assigning the most abundant ion a relative abundance of
100% (CH2OH+ in this spectrum). All other ions are shown as a percentage of that most abundant ion. For example, there
is approximately 64% of the ion CHO+ compared with the ion CH2OH+ in this spectrum. The y-axis may also be shown as
abundance (not relative). Relative abundance is a way to directly compare spectra produced at different times or using
different instruments.
Molecular ion (M .+): If the molecular ion appears, it will be the highest mass in the spectrum (except for isotope peaks
discussed below). This peak will represent the molecular weight of the compound. Its appearance depends on the stability
of the compound (think carbocation stability). Double bonds, cyclic structures and aromatic rings stabilize the molecular
ion and increase the probability of its appearance.
Reference Spectra: Mass spectral patterns are reproducible. The mass spectra of many compounds have been published
and may be used to identify unknowns. Instrument computers generally contain spectral libraries which can be searched
for matches.
Fragmentation: General rules of fragmentation exist and are helpful to predict or interpret the fragmentation pattern
produced by a compound. Functional groups and overall structure determine how some portions of molecules will resist
fragmenting, while other portions will fragment easily. A detailed discussion of those rules is beyond the scope of this
introduction, and further information may be found in your organic textbook or in mass spectrometry reference books.
Isotopes: Isotopes occur in compounds analyzed by mass spectrometry in the same abundances that they occur in
nature. A few of the isotopes commonly encountered in the analyses of organic compounds are below along with an
example of how they can aid in peak identification.
Methyl Bromide: An example of how isotopes can aid in peak identification.
The ratio of peaks containing 79Br and its isotope 81Br (100/98) confirms the presence of bromine in the compound.
Steps to interpret a mass spectrum
1.
Look for the molecular ion peak.
 This peak (if it appears) will be the highest mass peak in the spectrum, except for isotope peaks.
 Nominal MW (meaning=rounded off) will be an even number for compounds containing only C, H, O, S, Si.
 Nominal MW will be an odd number if the compound also contains an odd number of N (1,3,...)
2.
Try to calculate the molecular formula:
 The isotope peaks can be very useful, and are best explained with an example.
 Carbon 12 has an isotope, carbon 13. Their abundances are 12C=100%, 13C=1.1%. This means that for every
100 (12)C atoms there are 1.1 (13)C atoms.
 If a compound contains 6 carbons, then each atom has a 1.1% abundance of (13)C.
 Therefore, if the molecular ion peak is 100%, then the isotope peak (1 mass unit higher) would be
6x1.1%=6.6%. If the molecular ion peak is not 100% then you can calculate the relative abundance of the
isotope peak to the ion peak.
 For example, if the molecular ion peak were 34% and the isotope peak 2.3%: (2.3/34)x100 = 6.8%. 6.8% is
the relative abundance of the isotope peak to the ion peak. Next, divide the relative abundance by the isotope
abundance: 6.8/1.1=6 carbons.
 Follow this order when looking for information provided by isotopes: (A simplified table of isotopes is
provided in the introduction, more detailed tables can be found in chemistry texts.)



Look for A+2 elements: O, Si, S, Cl, Br
Look for A+1 elements: C, N
"A" elements: H, F, P, I
Common Fragments and Ions for Mass Spectrometry
Fragment lost
(or ion present)
CH3
CN
CH3CH2, CHO
CH2OH
CH3CH2CH2
HO(C=O)
CH3CH2C(O)
C5H5
Ph(C=O)
m/z
loss
M-15
M-26
M-29
M-31
M-43
M-45
M-57
M-65
M-105
Fragment lost
(or ion present)
NH2
CH2=CH2
NO
CH2-CH=CH2
C(=O)NH2
CH3CH(OH)
CH2=C(OH)NH2
Ph
PhCH2CH2
m/z
loss
M-16
M-28
M-30
M-41
M-44
M-45
M-59
M-77
M-105
Fragment lost
(or ion present)
OH
CH=NH, C≡O
OCH3
CH3C=O
OCH2CH3
NO2
CH2=C(OH)2
PhCH2, tropilium ion
I
m/z
loss
M-17
M-28
M-31
M-43
M-45
M-46
M-60
M-91
M-127