Physical Properties Melting Point Boiling Point Density Solubility Refractive Index Chemical Tests...

$Page 1: Physical Properties Melting Point Boiling Point Density Solubility Refractive Index Chemical Tests Hydrocarbons Alkanes Alkenes Alkynes Halides Alcohols.$
Physical Properties

Melting Point

Boiling Point

Density

Solubility

Refractive Index

Chemical Tests

Hydrocarbons

Alkanes

Alkenes

Alkynes

Halides

Alcohols

Aldehydes

Ketones

Spectroscopy

Mass

(Molecular Weight)

Ultraviolet/Visual

(Conjugation, Carbonyl)

Infrared

Functional Groups

NMR

(Number, Type, Location of protons)

Gas Chromatography

(Identity, Mole %)

SpectroscopyBuilding A Toolset

ForThe Identification of Organic Compounds

04/21/23 1

Spectroscopy

04/21/23 2

Spectroscopy Tools

Spectroscopy Spectroscopy

The Absorption of Electromagnetic Radiation and the use of the Resulting Absorption Spectra to Study the Structure of Organic Molecules.

When continuous radiation passes through a transparent material, some of the radiation can be absorbed.

If the portion that is not absorbed is passed through a prism, a spectrum with gaps is produced.

This is called an:

ABSORPTION SPECTRUM04/21/23 3

Spectroscopy Energy States

Energy absorption by transparent materials in any portion of the electromagnetic spectrum causes atoms or molecules to pass from a state of low energy (ground state) to a state of higher energy (excited state).

There are 3 types of Energy States:

Electronic

Vibrational

Spin04/21/23 4

Spectroscopy Electromagnetic Spectrum

Cosmic (Gamma) X-Ray

Ultraviolet Visible Infrared

Microwave Radio

Energy States and the Electromagnetic Spectrum

Electronic – Ultraviolet

Vibrational – Infrared

Spin – Radio

04/21/23 5

MicrowaveInfraredX-RayVacuum

UV

VisibleNear Ultraviolet

VibrationalInfrared

NuclearMagnetic

Resonance

Radio Frequency

400 nm200 nm 800 nm 2.5 15

1 m 5 m

Blue Red

Cosmic&

Ray

0.01 nm

3 x 1019 Hz 3 x 1016 Hz 2 x 1013 Hz

10 nm 30 cm

1 x109cm-1

0.002 cm-1

10 cm-1 3 cm-1 0.01 cm-1

1 mm

Frequency ()

Energy (E)

High

High Low

Low

Wavelength ()Short Long

1 x107cm-1 5 x104cm-1

2.5 x104cm-1

1.25 x104cm-1

667cm-1

4 x103cm-1

6 x 107 Hz

3 x 108 Hz

1.5 x 1015 Hz 1 x 109 Hz3 x 1011 Hz

1.2 x 1014 HzFrequency

Wave Number

Wavelength

Spectroscopy

04/21/23 6

E = hc /

E = [E (excited) – E (ground)] = h

= Frequency (Hz) c = Velocity of Light (cm/sec) = Wavelength (cm) h = Planck’s Constant

= c /

Where:

SpectroscopyQuantization

The excitation process is quantized, in which only selected frequencies of energy are absorbed representing the energy difference (E) between the excited and ground states.

04/21/23 7

SpectroscopySpectroscopy Types:

Mass Spectrometry (MS) – Hi-Energy Electron Bombardment

Use – Molecular Weight, Presence of Nitrogen, Halogens

Ultraviolet Spectroscopy (UV) – Electronic Energy States

Use –Conjugated Molecules; Carbonyl Group, Nitro Group

Infrared Spectroscopy (IR) – Vibrational Energy States

Use – Functional Groups; Compound Structure

Nuclear Magnetic Resonance (NMR) – Nuclear Spin States

Use – The number, type, and relative position of protons (Hydrogen nuclei) and Carbon-13 nuclei

04/21/23 8

Mass Spectroscopy High energy electrons bombard organic molecules breaking

some or all of the original molecules into fragments.

The process usually removes a single electron to produce a positive ion (cation radical) that can be separated in a magnetic field on the basis of the mass / charge ratio.

Removal of the single electron produces a charge of +1 for the cation.

Thus, the cation represents the Molecular Weight of the original compound or any of the fragments that are produced.

The mass spectrum produced is a plot of relative abundance of the various fragments (positively charged cation radicals) versus the Mass / Charge (M/Z) ratio.

The most intense peak is called the “Base Peak”, which is arbitrarily set to 100% abundance; all other peaks are reported as percentages of abundance of “Base Peak.”

04/21/23 9

M + e- M+ + 2e-

Molecule High EnergyElectron

Molecular Ion(Radical Cation)

TypicalMass

Spectrum

Molecular Ion Peak (M+ 88)

M - H2O

M - (H2O and CH3)

M - (H2O and CH2 – CH2)

Base Peak

CH2OH

1-Pentanol - MW 88

CH3(CH2)3 – CH2OH

Mass Spectroscopy

04/21/23 10

Mass Spectroscopy Molecular Ion Peak (M+)

Largest mass/charge ratio Always the last peak on the right side of

spectrum May or may not be the base peak (usually

not)! Abundance can be quite small, i.e., very

small peaks The Molecular Ion Peak represents the

Molecular Weight of the Compound

04/21/23 11

Mass Spectroscopy

04/21/23 12

Methyl Propyl Ketone (C5H10O) (CAS 107-87-9)

M+

– 15

(CH3) lost

M+

86

M+

– 28

(CH2CH2) lost

M+

– 43

(C2C2CH3) lostPropyl Group

Molecular Ion Peak

Mass Spectroscopy The Presence of Nitrogen in the Compound

If the Mass / Charge (m/z) ratio for the Molecular Ion peak is “Odd”, then the molecule contains an Odd number of Nitrogen atoms, i.e., 1, 3, 5, etc.

Note: An “Even” value for the Mass / Charge ratio could represent a compound with an even number of Nitrogen atoms, i.e., 0, 2, 4 etc.

The actual presence of Nitrogen in the compound is not explicitly indicated as it is with an “Odd” value for the ratio.

04/21/23 13

Mass Spectroscopy Halogens in Organic Compounds

Most elements exist in several isotopic forms:Ex. 1H1, 2H1, 12C6, 13C6, 35Cl17, 37Cl17, 79Br35, 81Br35

“Average Molecular Weight”

The average molecular weight of “All” isotopes of a given element relative to the abundance of the each isotope in nature

“Integral Molecular Weight”

The Number of Protons and Neutrons in a specific isotope

Each fragment represented in a Mass Spectrum produces several peaks each representing a particular isotopic mixture of the elements in the compound, i.e., an “integral molecular weight.

04/21/23 14

Mass Spectroscopy The Presence of Chlorine in a Compound

The two (2) principal Chlorine Isotopes in nature areCl-35 and Cl-37 (2 additional Neutrons in Cl-37)

The relative abundance ratio of Cl-35 to Cl-37 is:

100 : 32.6 or 75.8 : 24.2 or 3 : 1

Therefore, a Molecule containing a single Chlorine atom will show two Mass Spectrum Molecular Ion peaks, one for Cl-35 (M+) and one for Cl-37 (M+2)

Note: M+2 denotes 2 more neutrons than M+

Based on the natural abundance ratio of 100 / 32.6 (about 3:1), the relative intensity (peak height) of theCl-35 peak will be 3 times the intensity of the Cl-37 peak

04/21/23 15

1-Chloropropane

Molecule contains 1 Chlorine atom resulting in two Molecular Ion Peaks of about 3:1 relative intensity,

based on the 3:1 natural abundance ratio ofCl-35 / Cl-37

Molecular Ion PeaksM+ 78: M+2 80

very small

Mass Spectroscopy The Presence of Chlorine in a Compound (Con’t)

04/21/23 16

Mass Spectroscopy The Presence of Bromine in a Compound

The two (2) principal Bromine Isotopes in nature areBr-79 and Br-81 (2 additional Neutrons in Br-81)

The relative abundance ratio of Br-79 to Br-81 is

100 : 97.1 or 50.5 : 49.5 or 1 : 1

Molecules containing a single Bromine atom will also show two molecular ion peaks one for Br-79 (M+) and one for Br-81 M+2

Based on the natural abundance ratio of 100 / 97.1 (about 1:1), the relative intensity of the Br-79 peak will be about the same as the Br-81 peak

04/21/23 17

3-Bromo-1-Propene

Molecular Ion PeaksM+ 120; M+2 122

Molecule contains 1 Bromine atom resulting in two Molecular Ion Peaks of about 1:1 relative intensity, based on the 50.5:49.5 (1:1) natural

abundance ratio ofBr-79 / Br-81

Mass Spectroscopy The Presence of Bromine in a Compound (Con’t)

04/21/23 18

Mass SpectroscopyThe Presence of Fluorine in a Compound

Fluorine exists in nature principally as a single isotope

19F9

A compound containing any number of Fluorine atoms will have a single Molecular Ion peak (assuming no other Halogens present)

04/21/23 19

Mass Spectroscopy Multiple Halogens in a Compound

Compounds containing two (2) Chlorine atoms will produce three (3) Molecular Ion peaks representing the 3 possible isotope combinations of Chlorine:

35Cl17 35Cl17 (Rel Peak Intensity - 100.0)



04/21/23 20

Mass Spectroscopy Multiple Halogens in a Compound

Compounds containing three (3) Chlorine atoms will produce four (4) Molecular Ion peaks representing the 4 possible isotope combinations for Chlorine:

35Cl17 35Cl17 35Cl17 (Rel Peak Intensity - 100.0)




04/21/23 21

Mass Spectroscopy & Molecular Formula

Information from the Mass Spectrum can used to determine the Molecular Formula of a compound

Ex. Molecular Ion Peaks – M+ 94; M+2 96 (95)

2 Molecular Ion Peaks (3:1) suggest: 1 Chlorine atom

Partial Analysis: C – 25.4%; H – 3.2 %

Use 95 as average molecular weight

Carbon: 95 x 0.254 = 24.1 / 12 = 2 C atoms

Hydrogen: 95 x 0.032 = 3.0 / 1 = 3 H atoms

95 – (24 + 3) = 68 unresolved mass

(Use oxygen, nitrogen, halides (Cl or Br) to resolve mass)

2 Oxygen (16 + 16) + 1 Chlorine (35.5) 68

Molecular Formula - C2H3O2Cl

04/21/23 22

Mass Spectroscopy Summary

Fragmentation of Organic Molecules by high energy electrons

Base Peak – 100 % abundance

Molecular Ion Peak – Highest Mass/Charge ratio

Molecular Ion Peak – Last peak(s) on right side of chart

Molecular Ion Peak – Represents Molecular Weight of compound

Molecular Ion Peak – If value is “Odd” the compound contains an odd number of “Nitrogen” atoms

Molecular Ion Peak – If two peaks occur and the relative abundance ratio is 3:1, then the compound contains a single Chlorine atom.

Molecular Ion Peak – If two peaks occur and the relative abundance ration is 1:1, then the compound contains a single Bromine Atom

04/21/23 23

Ultraviolet/Visible (UV) Spectroscopy UV-Visible Spectrum : 190 nm – 800 nm

In Ultraviolet and Visible Spectroscopy, the energy absorption transitions that occur are between electronic energy levels of valence electrons, that is, orbitals of lower energy are excited to orbitals of higher energy

Thus, UV / Visible spectra often called Electronic Spectra All organic compounds absorb Ultraviolet light to some

degree, but in many cases at such short wavelengths to make its utility of very limited value in organic chemistry

For the purpose of this course, the primary use of UV/Vis will be to confirm: The presence of conjugated molecules, either aliphatic

alkene structures or aromatic ring structures To a lesser degree, the presence of the Carbonyl group

and the Nitro group

04/21/23 24

Ultraviolet/Visible (UV) Spectroscopy When a molecule absorbs radiation a valence electron

is generally excited from its highest occupied molecular orbital (HOMO) to the lowest unoccupied molecular orbital (LUMO)

For most molecules, the lowest energy orbitals are thesigma () orbitals ( - bonds)

The electrons of sigma bonds are too tightly bound to be promoted by radiation in the 200-700 nm region.

Therefore alkanes, saturated alcohols, simple alkenes show no or very little UV absorption.

The orbitals occupy somewhat higher energy levels. Orbitals that hold unshared pairs of electrons, the

nonbonding (n) orbitals, lie at even higher energies. Unoccupied or antibonding orbitals (* and *) have

the highest energies.

04/21/23 25

Ultraviolet/Visible (UV) Spectroscopy Chromophores

The absorption of Ultraviolet radiation results from the excitation of electrons from ground to excited state

The Nuclei in molecules, however, determine the strength with which electrons are bound to the molecule, thus, influencing the spacing between ground and excited states

The characteristic energy of a transition and the wavelength of radiation absorbed are properties of a group of atoms rather than the electrons themselves.

The group of atoms producing such an absorption is called a Chromophore.

As the structure of the molecule (alkane, alkene, alkyne, alcohol, amine, nitrile, carbonyl, etc.) changes, the energy and intensity of the Ultraviolet absorption will change accordingly

04/21/23 26

(sigma)

(pi)

n (non-bonding)

C::C

C : H Sigma & pi bonds have antibonding (unocuupied)orbitals associatedwith them * & *

Ultraviolet/Visible (UV) Spectroscopy Radiation in the range 190nm – 800nm causes

valence electrons (those that participate in bonding) to be excited to a higher energy level.

The ground state of an organic molecule can contain valence electrons in three principal types of molecular orbitals:

04/21/23 27

Energy Transition Examplesn * in carbonyl compoundsn * in oxygen, nitrogen, sulfur, and halogen compounds * in alkenes, alkynes, carbonyl and azo compounds * in carbonyl compounds * in alkanesIn

crea

sin

g

En

erg

y

**

n

Antibonding (single bonds)Antibonding (double bonds)

Nonbonding (as in lone electron pairs or the propenyl (allyl) radical

Bonding (double bonds)Bonding (single bonds)

Incr

easi

ng

En

erg

y

Note:Electronic energy levels in aromatic molecules are more complicated than are presented here.

Ultraviolet/Visible (UV) Spectroscopy

04/21/23 28

Ultraviolet/Visible (UV) Spectroscopy Atoms produce sharp lines of absorption

Molecules have many excited modes of vibration and rotation at room temperature. The rotational and vibrational levels are superimposed on the electronic levels

Electron transitions may occur from any of several vibrational and rotational states of one electronic level to any of several vibrational and rotational states of a higher electronic level.

Thus, the UV spectrum of a molecule consists of a broad band of absorption centered near the wavelength of the major transition, i.e. where the radiation has its maximum absorption (max).

04/21/23 29

Ultraviolet/Visible (UV) Spectroscopy The Ultraviolet-Visible

spectrum is generally recorded as a plot of absorbance versus wavelength; but the plot is more often presented with the Absorptivity () or log plotted on the ordinate (y-axis) and the wavelength plotted on the abscissa (x-axis)

Ex: Cyclohexane

(A Conjugated Aromatic Molecule)

Wavelength of Maximum Absorbance

max – 230 nm

************************

Molar Absorptivity

– 15,000 cm-1

Log – 4.204/21/23 30

Ultraviolet/Visible (UV) Spectroscopy The Beer-Lambert Law

The Ultraviolet/Visible Spectrum is a plot of the Wavelength () in nanometers (nm) over the entire Ultraviolet / Visible region versus the Absorbance (A) of the radiation at each wavelength.

A = log (Ir / Is) = C L

Is = Intensity of light through sample solution

Ir = Intensity of incident light passing through

Reference cell

= Molar Absorptivity (Molar Extinction Coefficient) A measure of the strength or intensity of the absorption.

Units – l/(mol • cm) (m2 x 10-2 /mol) (mmol/dm3)

C = Concentration of solute (mol/L) or (g/L) if molecular mass is unknown

L = Length of cell (cm)

04/21/23 31


A = • C • l

= A / (C • l )

Values of are usually expressed as Log

Aliphatic (single

band) = 10,000 –

20,000 (Log = 4.0 – 4.3)

Aromatic (two bands

= 1,000 –

10,000 (Log = 3.0 – 4.0)

Carbonyl compounds

= 10 – 100

(Log = ~ 2)

Nitro compounds= 10 (Log = ~

1)04/21/23 32

Example Transitionmax

(nm)max

Log

n-Butyl Iodide n * 257 486 2.7

Acetone n * 279 15 1.2

Acrolein * 210 11,500 4.1

(C=C & C=O) n * 315 14 1.1

1,3-Butadiene * 217 21,000 4.3

1,3,5-Hexadiene * 258 35,000 4.5

Benzene(2 transitions)

Ar * Ar *

ca 200

255

8,000

215

3.9

2.3

Ultraviolet/Visible (UV) Spectroscopy Typical Transitions & Associated Wavelengths of Maximum Absorption and Molar Absorptivities

04/21/23 33


Typical Transitions and Absorptions

of Simple IsolatedChromophores

04/21/23 34

Class Transition max (nm) Log

R-OH n* 180 2.5

R-O-R n* 180 3.5

R-NH2 n* 190 3.5

R-SH n* 210 3.0

R2C=CR2 * 175 3.0

R-C=C-R * 170 3.0

R-CN n* 160 <1.0

R-N=N-R n* 340 1.0

R-NO2 n* 271 <1.0

R-CHO * 190 2.0

R-CHO n* 290 1.0

R2CO * 180 3.0

R2CO n* 280 1.5

RCOOH n* 205 1.5

RCOOR’ n* 205 1.5

RCONH2 n* 210 1.5


Computation Example:

An -unsaturated ketone of relative molecular weight 110 has an absorption band with max at 215 nm and = 10,000 (l / mol • cm)

A solution of this ketone showed absorbance A = 2.0 with a 1 cm cell. Calculate the concentration of the ketone in this solution expressed in grams per liter.

Ans: A = c L

c = A / L

c = 2.0 / ((10,000 l/mol • cm) * 1.0 cm)

c = 2 x 10-4 mol/l

c = 2 x 10-4 mol/l x 110 g/mol

c = 2.20 x 10-2 g/l04/21/23 35


Computation Example:

Calculate the Molar Absorptivity, , for a solution containing 1.0 mmol dm-3 (1.0 x 10-3 moles per liter) of solute, when the absorbance of a 1 cm cell was 1.5.

Ans: A = c L

= A / c L

= 1.5 / (1 x 10-3 mol / L) (1 cm)

= 1.5 x 103 L / mol • cm

What would be the Absorbance for a solution of double this concentration?

Ans: A = 1.5 x 103 L / mol • cm • 2.0 x 10-3 moles / L • 1 cm

A = 3.004/21/23 36


Alkanes

Contain single sigma bonds resulting in only * transitions which absorb ultraviolet radiation at wavelengths generally too short for use in UV spectroscopy.

Utility: None

Alcohols, Ethers, Amines, Sulfur Compounds

The n * transitions absorb UV radiation within the experimentally accessible range (>180 nm).

Utility: Very little

04/21/23 37


Alkenes and Alkynes

Absorb UV radiation in the range < 180 nm.

“Cumulated” alkenes ( * transitions), which have one or more “” sigma bonds between the double bounds usually have absorption maxima below 200 nm.

Utility: Very little

Compounds with Oxygen double bonds

Unsaturated molecules containing oxygen or nitrogen structures such as Carbonyl (C=O) and Nitro (NO2) have both n * (280 - 290 nm) and * transitions (188 nm).

Utility: Moderate

04/21/23 38

Ultraviolet/Visible (UV) Spectroscopy Conjugated unsaturated systems are molecules with

two or more double or triple () bonds each alternating with a single or sigma bond ().

Conjugated unsaturated systems have delocalized bonds, i.e., a p-orbital on an atom adjacent to a double bond producing * transitions. Single electron as in the allyl radical (CH2=CHCH2•) Vacant p orbital as in allyl cation (CH2=CHCH2

+) P orbital of another double bond

(CH2=CHCH=CH2

Conjugated systems include the Aliphatic Alkenes as well as the Aromatic ring structures.

Compounds whose molecules contain conjugated multiple bonds absorb strongly in the UV / Visible portion of the electromagnetic spectrum (> 200 nm).

Utility: Good04/21/23 39

The Wavelength of Maximum Absorption ( max ) is obtained from the Absorption Spectrum

Wavelength of Maximum Absorbance (max) – 242.5 nm

Molar Absorptivity ( ) – 13,100 M-1 cm-1 (Log = 4.1)

2,5-Dimethyl-2,4-Hexadiene (in Methanol)


Conjugated Unsaturated Systems

Conjugated systems consist of alternating sigma () bonds and pi () bonds) and the Ultraviolet absorptions show large values of

04/21/23 40


Conjugated Unsaturated Systems (Con’t)

, - Unsaturated ketones, Dienes, Polyenes

Transitions - *

High Intensity Bands

= 10,000 to 20,000 (log = 4.0 - 4.3)

max > 210 nm

Aromatic Conjugated Systems

Transitions - *

2 Medium Intensity Bands

= 1000 - 60,000 (log = 3.0 - 4.8)

max both bands > 200 nm

Note: Substitution on ring increases Molar Absorptivity above 10,000

04/21/23 41

Ultraviolet/Visible (UV) Spectroscopy Carbonyl (C=O), Nitro Group (NO2) (Resonance

effects on substituted benzene)

Transitions n - * & *

Single Low Intensity Band = 10 (log = 1) to = 300 (log =

2.5)

max (250 - 360 nm)

Nitro (NO2) log (~1.0)

Carbonyl (C=O) log (~2.0)

The presence of these functional groups should be used only as confirmations of species identified in the IR Spectra.

04/21/23 42


Practical Approach to Interpreting UV/Vis Information

If the problem you are working on provides an UV/Vis spectrum and it indicates “No” absorption in the 200 – 700 nm range, the following conclusions are applicable:

The compound is not conjugated, i.e., it does not contain alternating double/single bonds (including Benzene ring.)

The compound probably does not contain “Carbonyl” or “Nitro” groups (confirm with IR).

If the problem provides Log Absorptivity values (Log ) the following possibilities exist:

Log (> 4.0) - Conjugated , - Unsaturated ketones, Dienes, Polyenes

Log (3.0 – 4.0) - Aromatic ring (Check IR, NMR)

Log (1.5 – 2.5) - C=O (Check IR)

Log (1.0 – 1.5) - NO2 (Check IR)04/21/23 43

Infrared Spectroscopy Infrared Spectroscopy References

Pavia, et al - pp. 851 - 886 Solomon’s et al - pp. 79 - 84; 821 – 822

Infrared Radiation That part of the electromagnetic spectrum

between the visible and microwave regions

0.8 m (12,500 cm-1) to 50 m (200 cm-

1). Area of Interest in Infrared Spectroscopy

The Vibrational portion of infrared spectrum

2.5 m (4,000 cm-1) to 25 m (400 cm-1) Radiation in the vibrational infrared region is

expressed in units called wavenumbers ( )

04/21/23 44

Infrared Spectroscopy Wavenumbers are expressed in units of reciprocal

centimeters (cm-1) i.e. the reciprocal of the wavelength () expressed in centimeters.

(cm-1) = 1 / (cm)

Wave Numbers can be converted to a frequency () by multiplying them by the speed of light (c) in cm/sec

(Hz) = c = c / (cm /sec /cm = 1/sec)

Recall: E = h c /

Thus, wavenumbers are directly proportional to energy

04/21/23 45

Infrared Spectroscopy Polar Covalent Bonds & Dipole Moments

Organic compounds are organized into families of compounds on the basis of certain groupings of atoms, i.e., Functional Groups.

The Electrons between atoms in an organic compound are shared forming “Covalent bonds.”

Covalent bonds between atoms with different electronegativities have an unequal sharing of the bond electrons setting up an electrostatic charge difference between the atoms.

The atom with the greater Electronegativity pulls the electrons closer to it forming a “Polar Covalent Bond.”

04/21/23 46

Infrared Spectroscopy Polar Covalent Bonds & Dipole Moments (Con’t)

The relative strength of the Polar Covalent Bond impacts the ability of the molecule, i.e., a Functional Group, to attract or repel other polar entities (functional groups).

The separation of the positive and negative charges in a Polar Covalent Bond is referred to as a Dipole.

A dipole has a Dipole Moment defined as the product of the magnitude of the partial charges (in electrostatic units, esu) times the distance (in cm) of separation.

Only those Covalent bonds with Dipole Moments are capable of absorbing Infrared Radiation.

04/21/23 47

Infrared Spectroscopy The Radiation (Energy) Absorption Process

The absorption of Infrared Radiation by a Polar Covalent Bond raises the molecule to a higher energy state.

This is a Quantized process in which only selected frequencies are absorbed dependent on the relative masses of the atoms, the force constants of the bond (electronegativity), and the geometry of the atoms.

Covalent Bonds possess Rotational and Vibrational frequencies.

Every type of bond has a natural frequency of vibration.

The same bond in different compounds has a slightly different frequency of vibration.

04/21/23 48

Infrared Spectroscopy When the frequencies of Infrared Radiation match the

natural vibrational frequencies of a bond with a Dipole Moment, the radiation is absorbed increasing the amplitude of the vibrational motions of the covalent bonds.

Infrared radiation is absorbed and converted by organic molecules with polar covalent bonds and dipole moments into energy of molecular rotation and molecular vibration.

Rotation - Less than 100 cm-1 (Spectrum is lines)

Vibration - 10,000 cm-1 to 100 cm-1 (Spectrum is bands)

The vibrational bands appears because each vibrational energy change is accompanied by a number of rotational changes

Infrared Spectroscopy is concerned only with the vibrational spectrum (4,000 cm-1 to 400 cm-1)

04/21/23 49

Infrared Spectroscopy Molecular Vibrations

Absorption of infrared radiation corresponds to energy changes on the order of 8-40 KJ/mole (2-10 Kcal/mole

The frequencies in this energy range correspond to the stretching and bending frequencies of the covalent bonds with dipole moments.

Stretching (requires more energy than bending) Symmetrical Asymmetrical

Bending Scissoring (in-plane bending) Rocking (in-plane bending) Wagging (out-of-plane bending) Twisting (out of plane bending)

04/21/23 50

C—H HC H

HC H

SymmetricStretch

(2853 cm-1)

AsymmetricStretch

(2926 cm-1)

Infrared Spectroscopy Stretching – A rhythmical movement along the

bond axis such that the interatomic distance is increasing or decreasing.

In any group of three or more atoms – at least two of which are identical - there are two modes of stretching or bending: Symmetric and Asymmetric

For the Methylene Group (CH2):

04/21/23 51

H H C

H H C

HC H

HC H

Scissoring

~1450 cm-1

(In Plane)

Rocking

~750 cm-1

(In Plane)

Wagging

~1250 cm-1

(Out of Plane)

Twisting

~1250 cm-1

(Out of Plane)

Infrared Spectroscopy Bending – A change in bond angle between bonds

with a common atom or

A movement of a group of atoms with respect to the remainder of the molecule

04/21/23 52

Infrared Spectroscopy Thus, no two molecules of different structure will have

exactly the same natural frequency of vibration, each will have a unique infrared absorption pattern or spectrum.

Two Uses: IR can be used to distinguish one compound from

another. Absorption of IR energy by organic compounds will

occur in a manner characteristic of the relative strengths of the Polar Covalent Bonds in the Functional Groups present in the compound; thus, an Infrared Spectrum gives structural information about the functional groups present in a molecule.

The absorptions of each type of bond (N–H, C–H, OH, C–X, C=O, C–O, C–C, C=C, C≡C, C≡N, etc.) are regularly found only in certain small portions of the vibrational infrared region, greatly enhancing analysis possibilities.

04/21/23 53

The split beams pass into a Monochromator, which consists of a rapidly rotating sector that passes each beam to a diffraction grating or prism.

The slowly rotating diffraction grating varies the wavelength of radiation reaching the detector.

The detector senses the ratio in intensity between the reference (air) and sample beams and records the differences on a chart.

Detector

Slit

Monochromator

IR Source

Recorder

SplitBeams Air

LenzSample

Infrared Spectroscopy Instrumentation

Dispersive (Double Beam) IR Spectrophotometer

04/21/23 54

Infrared Spectroscopy Sample Preparation

Liquid Samples 1 to 2 drops of liquid sample are placed between

two single crystals of sodium chloride (Plates) Note: NaCL plates are water soluble – keep dry

Solid Samples soluble in Acetone

Dissolve sample in acetone Evaporate on Salt Plate

Solid Samples not soluble in acetone Make Potassium Bromide (KBR) pellet

Put plates in plate holder Place holder in IR Spectrometer Obtain IR Spectrum Clean Plates with Acetone

04/21/23 55

Infrared Spectroscopy Fourier Transform (FT) Single Beam IR

Set background (air) Press “Scan” button Press “Background” button Verify No. of Scans is “4”; if not, push soft key

to set “4” Press “Execute”

Obtain Sample Spectra Insert Cell Holder into beam slot Press “SCAN” button Select Memory location ( X, Y, or Z) Press “Execute”

04/21/23 56

Infrared Spectroscopy Fourier Transform (FT) Single Beam IR (Con’t)

If spectrum bottoms out (might have to check with instructor), then remove Cell Holder; remove top of Salt Plate; wipe lightly with tissue; reassemble; and insert cell holder into beam slot.

Rerun Scan again

Push “Plot” to produce chart

Remove Cell Holder and disassemble

Clean Salt Plate; dry; return to instructor; place in desiccator

04/21/23 57

Infrared Spectroscopy The Infrared Spectrum

A plot of absorption intensity (% Transmittance) on the y-axis vs. frequency on the x-axis.

Transmittance (T) - the ratio of the radiant power transmitted by a sample to the radiant power incident on the sample.

Absorbance (A) - the logarithm to base 10 of the reciprocal of the Transmittance.

A = log10 (1 / T)

Frequency - The x-axis is represented by two scales:

Wavelength(2.5 to 25 ) (Bottom) Wavenumber (4000 cm-1 to 400 cm-1)(Upper)

04/21/23 58

Methyl IsopropylKetone

C5H10O CAS – 563-80-4

C=OCarbonyl

AliphaticC-H Stretch

CH3

CH2

C=OCarbonylOvertone

Infrared SpectroscopyIR SpectrumKetone

04/21/23 59

Infrared Spectroscopy IR Spectrum Peak Characteristics

Primary Examination Regions of the Spectrum

High Frequency Region - 4000 to 1300 cm-1

Intermediate (Fingerprint Region) - 1300 to 900 cm-1

High Frequency Region (Functional Group Region)

Characteristic Stretching frequencies of such groups as:

=CH, OH, NH, C=O, CO, C≡N, C≡C, C=C

The Fingerprint Region - 1300 to 900 cm-1

Absorption patterns frequently complex

Bands originate from interacting vibrational modes

Valuable when used in reference to other regions

Absorption unique for every molecular species

Effective use comes from experience04/21/23 60

Infrared Spectroscopy IR Spectrum Peak Characteristics (con’t)

Shape

Sharp (narrow)

Broad

Intensity

Weak (w)

Medium(m)

Strong (s)

Note: Peak intensity is dependent on amount of sample and sensitivity of instrument; therefore, the actual intensity can vary from spectrum to spectrum

04/21/23 61

Infrared Spectroscopy Principal Frequency Bands

O-H 3600 cm-1 (Acids, Alcohols)

N-H 3300 - 3500 cm-1 (Amino)

(1o - 2 peaks, 2o - 1 peak, 3o – 0 peaks) NO2 1450 – 1650 cm-1 (2 absorptions)

C≡N 2250 cm-1 (Nitrile)

C≡C 2150 cm-1 (Acetylene)

-C≡C-H 3300 cm-1 (Terminal Acetylene)

C=O 1685 - 1725 cm-1 (Carbonyl)

C=C 1650 cm-1 (Alkene) 2 absorptions

C=C 1450 – 1600 cm-1 (Aromatic) 4 absorptions

04/21/23 62

Infrared Spectroscopy Principal Frequency Bands (Con’t)

CH2 1450 cm-1 (Methylene)

CH3 1375 & 1450 cm-1 (Methyl)

C-O 900 - 1100 cm-1 (Alcohol, Acid, Ester, Ether, Anhydride)

−C-H Right side of 3000 cm-1 (Saturated Alkane)

=C-H Left side of 3000 cm-1 (Unsaturated Alkene)

=C-H 1667 – 2000 cm-1 (Aromatic Overtones)

≡C-H 2150 cm-1 (Stretch)

04/21/23 63

Functional Type of FrequencyGroup Vibration cm-1Intensity

Alkanes (C-H) (stretch) 3000-2850 s -CH3 (bend) 1450 & 1375 m

-CH2 (bend) 1465 m

Alkenes (C=C) (stretch) 3100-3000 m

(bend) 1000-650 s

Aromatics (stretch) 3150-3050 s

(OOP bend) 1000-650 s

Alkyne (C) (stretch) 3300 s

Aldehyde (CHO) (stretch) 2900-2800 w

(stretch) 2800-2700 w

Infrared Spectroscopy

04/21/23 64

Infrared Spectroscopy Correlation Table

Functional Group Frequency (cm-1) Intensity

CC Alkane Not UsefulC=C Alkene 1680-1600 m-w

Aromatic 1600-1400 m-wC≡C Alkyne 2250-2100 m-wC≡C-H Alkyne (terminal) 3300 sC=O Anhydride ~1810 s

~1760 sEster 1750-1730 sAldehyde 1740-1720 sKetone (acyclic) 1725-1705 sCarboxylic Acid 1725-1700 sAmide 1700-1640 s

04/21/23 65

Infrared SpectroscopyCorrelation Table Functional Group Frequency(cm-1) Intensity

C-O Alcohols, Ethers 1300-1000 s Esters, Acids

O-H Alcohols, Phenols Free 3650-3600 m H-Bonded 3400-3200 m

Carboxylic Acids 3300-2500 mN-H Primary & Sec Amines ~3500

mC≡N Nitriles 2260-2240 mN=O Nitro (R-NO2) 1600-1500 s

1400-1300 sC-X Fluoride 1400-1000 s

Chloride 800-600 sBromide, Iodide <600 s

04/21/23 66

Infrared Spectroscopy Analyzing the Spectrum – A Suggested Approach

Step 1. Check for the presence of Carbonyl group (C=O) in the range 1660 – 1820 cm-1 (~1700 cm-1)

If the Carbonyl Group is present, one of the following types of compounds is present:

Carboxylic Acid Ester Amide Anhydride Aldehyde Ketone Acid Halide

If the molecule is conjugated (alternating double & single bonds), the strong (C=O) absorption will be shifted to the right by ~30 cm-1

04/21/23 67


Step 2. Check for the presence of Saturated Alkane structures Compounds containing just Methyl (CH3) & Methylene (CH2)

groups produce generally simple IR spectra

C–H sp3 absorption is a stretch in the range 3000 – 2840 cm-1

Note: It is important to remember that the Alkane sp3 stretch occurs on the right side of the 3000 cm-1 mark in the

IR spectrum and that Alkene and Aromatic sp2 stretches occur on the left side of the 3000 cm-1 mark (see next slide).

CH3 Methyl groups (CH3) have a characteristic bending at 1375 cm-1 and a smaller absorption at 1450 cm-1.

CH2 Methylene groups (CH2) have characteristic bending at approximately 1465 cm-1

04/21/23 68


Step 3. Check for the presence of unsaturated (=C–H) sp2 structures.

=C–H sp2 absorption is a stretch in the range 3000 – 3100 cm-1, i.e., on the left side of the 3000 cm-1

mark on the x-axis scale.

Step 4. Determine whether the =C–H bond is Aliphatic Alkene, Aromatic, or both.

For Alkene =C–H bonds, look for the C=C stretch at 1600 – 1650 cm-1, usually an unequal pair of absorptions.

Out-of-Plan (OOP) bending at 650 – 1000 cm-1

Note: See next slide or the table on page 895 of Pavia text for guide to substitution patterns on substituted alkenes.

04/21/23 69


Out of Plane (OOP) substitution patterns (substituted alkenes)

04/21/23 70

1-Hexene

C6H12 CAS 592-41-6

Unsat=C-H Stretch

Sat’d-C-H Stretch

AliphaticC=C

Stretch

CH3

CH2

OOP BendingMonosubstitution

Infrared SpectroscopyIR SpectrumAliphatic Alkene

04/21/23 71


04/21/23 72

IR SpectrumCyclic Alkene

Cyclohexene

Unsat=C-H Stretch Sat’d

-C-H Stretch

AliphaticC=C

Stretch

CH2

OOP BendingCIS

Disubstitution

C6H10 CAS 110-83-8


Step 4 (Con’t)

Aromatic =C-H bonds. Look for C=C stretch - (pair of absorptions at 1450 cm-1

and a pair of absorptions at 1650 cm-1

Overtone/Combination bands appear between1667 & 2000 cm-1

Out-of-Plain (OOP) bending between 650 – 1000 cm-1

Note: See next slide or the table on page 897 of Pavia text for guide to substitution patterns on Benzene ring.

Note: The substitution pattern information in the “Overtone” area and the OOP area is duplicative. Use both tables to confirm substitution pattern

04/21/23 73


04/21/23 74

OOP – Substitution Patterns (Aromatic)

Overtone Area Substitution Patterns (Aromatic)

Toluene (Methyl Benzene)

C7H8CAS 108-88-3

CH3

AromaticC=C

StretchOOP Bending

Mono-Substitution

AromaticOvertones

Mono-Substitution

Sat’nUnsat’d

Infrared SpectroscopyIR Spectrum(Aromatic)

04/21/23 75


Step 5. Carbonyl Compounds (Carboxylic Acids)

Strong band of C=O group appears in range 1700-1725 cm-1.

Very broad absorption band of the OH group in the range2400-3400 cm-1.

This broad band will usually obscure the Alkane C-H stretch bands from 2849-3000 cm-1.

Medium intensity C-O stretch (as in C-OH) occurs in the range 1210-1320 cm-1

04/21/23 76

Isobutyric Acid

C4H8O2CAS 79-31-2

C=OCarbonyl

OH Stretch

C-O

CH3

sp3 C-HStretch

Infrared SpectroscopyIR SpectrumCarboxylic Acids

04/21/23 77


Step 6. Carbonyl Compounds (Esters)

C=O stretch appears in the range 1730-1750 cm-1

Check for 2 or more C-O stretch bands, one stronger and broader than the other, in the range 1100-1300 cm-1

04/21/23 78

Methyl Benzoate

C8H8O2CAS 93-58-3

Infrared SpectroscopyIR SpectrumEsters

04/21/23 79

C=OCarbonyl


Unsat’d=C-H Stretch

AromaticOvertones

AromaticOOP

C-O

Aromatic RingC=C Absorptions

C-O


Step 7. Carbonyl Compounds (Anhydrides)

2 C=O stretch bands (1740-1775 cm-1 & 1800-1830 cm-1)

Conjugation will move these bands to lower frequency

Multiple C-O stretch bands in the range 900 – 1300 cm-1

04/21/23 80

Propionic Anhydride

C6H10O3 CAS 123-62-6

Pair ofC=O

Stretch bands

C-HAliphatic Stretch

C-O Stretch

CH3CH2

C=OOvertone

Infrared SpectroscopyIR SpectrumAnhydrides

04/21/23 81


Step 8. Carbonyl Compounds (Amides)

C=O stretch at approximately 1640-1700 cm-1

N-H stretch (medium absorptions) near 3500 cm-1

Primary Amino (-NH2) - 2 Peaks (3180 & 3350 cm-1)

Secondary Amino (-NH) - 1 Peak (3300 cm-1)

N-H Scissoring - 1550 - 1640 cm-1

N-H Bend - 800 cm-1

04/21/23 82

IR SpectrumAmides Benzamide

C7H7NO CAS 55-21-0

NH2 Stretch2 peaks

Primary Amino

C=OCarbonyl

N-HScissoring

AromaticOvertones

C=CAromatic

Unsat’d=C-H Stretch

{


04/21/23 83

-C-N str

Acetanilide(N-Phenylacetamide)

C8H9NO CAS 103-84-4

OOP BendAromatic

MonosubstitutionNH Stretch

1 PeakSec Amino

C=CAromatic

N-HBend

C=OCarbonyl

{

AromaticOvertonesUnsat’d

=C-H Stretch

CH3

Infrared SpectroscopyIR SpectrumAmides

04/21/23 84


Step 9. Carbonyl Compounds (Aldehydes)

C=O stretch appears in the range 1720 - 1740 cm-1

2 weak Aldehyde C-H stretch absorptions near 2850 and 2750 cm-1)

04/21/23 85

Nonanal

C9H18O CAS 124-19-6

Infrared SpectroscopyIR SpectrumAldehydes

04/21/23 86


AldehydeHydrogen

Stretch2 Peaks

C=OCarbonyl

CH2

CH3

C=OOvertone

The Ketone structure produces a medium to strong absorption in the 1100 – 1300 cm-1 range due to coupled Stretching and Bending vibrations


Step 10. Carbonyl Compounds (Ketones)

C=O stretch occurs at approximately 1705 – 1725 cm-1

Ketones are confirmed when the other five compound types containing a Carbonyl group have been eliminated.

Ketone IR Spectra can sometimes be confused with Ester spectra because of an absorption in the 1100 -1300 cm-1 range similar to the location of the C-O stretch in esters. Usually, however, the ester will have 2 or more of the C-O stretch absorptions.

04/21/23 87

Ethyl Isopropyl Ketone(2-Methyl-3-Pentanone)

C6H12O CAS – 565-69-5

C=OCarbonyl

CH2

CH3


C=OOvertone

Infrared SpectroscopyIR SpectrumKetones

04/21/23 88


Step 11. Triple Bonds

Alkynes

R – C ≡ C – R weak, sharp stretch near 2150 cm-1

R – C ≡ C – H (Terminal Acetylene)

Weak, sharp stretch near 2150 cm-1

and a second stretch at 3300 cm-1

Nitriles

C ≡ N Medium, sharp stretch near 2250 cm-1

04/21/23 89

IR SpectrumAlkynes (CC)

Propargyl Alcohol (2-Propyn-1-ol)

C3H4O CAS 107-19-7

OHH - Bonded

≡C-H Terminal AlkyneStretch

C-O

C≡CStretch

CH2



04/21/23 90

Benzonitrile

IR SpectrumNitriles

C7H5N CAS 100-47-0

-C≡NStretch

Unsat=C-H Stretch

Aromatic ringC=C Absorptions

AromaticOOP Bending

Monosubstitution

AromaticOvertones


04/21/23 91


Step 12. - Alcohols & Phenols

Broad absorption near 3600 - 3300 cm-1

Confirm presence of C–O (C–OH) near 1000 - 1300 cm-1

04/21/23 92

2-Naphthol (Nujol Mull)

IR SpectrumAlcohols & Phenols

C10H9O CAS 135-19-3


04/21/23 93

OHH - Bonded


Unsaturation=C-H Stretch

Saturation-C-H Stretch

2-Naphthol (CCl4 Soln)IR SpectrumAlcohols & Phenols

C10H9O CAS 135-19-3

OHH - Bonded


Unsat=C-H Stretch

C-O


04/21/23 94

2-Naphthol (KBr Disc)IR SpectrumAlcohols & Phenols

C10H9O CAS 135-19-3


04/21/23 95

OHH - Bonded Aromatic ring

C=C Absorptions

Unsat=C-H Stretch

C-O

IR SpectrumAlcohols & Phenols

C4H10O CAS 78-92-2

2-Butanol

OH

C-OCH2 CH3



04/21/23 96


Step 13. Ethers

C–O absorptions near 1000 - 1300 cm-1

Absence of OH

Absence of C=O group

Aliphatic Ethers give a single strong C-O band at1120 cm-1

Unbalanced Ethers will show 2 C–O groups

Phenyl Alkyl Ethers give two (2) strong bands at about 1040 & 1250 cm-1

04/21/23 97

IR SpectrumEthers

Butyl Ether(Balanced Ether)

C8H18O CH3(CH2)3 – O – (CH2)3CH3 CAS 142-96-1

C-O

CH2

CH3



04/21/23 98

IR SpectrumEthers

C8H10O CAS 103-73-1

Phenetole(Unbalanced Phenyl Alkyl Ether)


04/21/23 99


Unsat=C-H Stretch

C-O

C-O

CH3


AromaticOvertones

OOP BendingAromatic

Monosubstitution

CH2


Step 14. Amines

N-H stretch (Medium absorptions) near 3500 cm-1

Primary Amino - 2 Peaks

Secondary Amino - 1 Peak

Tertiary Amino - No peaks

N-H Scissoring at 1560 - 1640 cm-1

N-H Bend at 800 cm-1

04/21/23 100

n-Butylamine(Primary Amine)

IR SpectrumAmines

C4H11N CAS 109-73-9

H-N-H Stretch2 Peaks

Primary Amine

N-HScissoring

CH2

CH3

-C-NStretch

-N-HOOP BendingAliphatic

(sat’n)C-H Stretch


04/21/23 101

N-Methylbenzylamine(Sec Amine)

IR SpectrumAmines

C6H11N CAS 103-67-3


04/21/23 102

N-HScissoring

CH2

CH3 -N-HOOP Bending


AromaticOvertones

OOP BendingAromatic

Monosubstitution

Unsat =C-H Stretch


Sec-Amino

N-HScissoring

CH2

CH3Sat

– C-H Stretch

C-N Str


Step 15. Nitro Compounds

Two strong absorptions

Aliphatic Nitro Compounds

Asymmetric strong stretch 1530 -1600 cm-1

Symmetric medium stretch 1300 -1390 cm-1

Aromatic Nitro Compounds

Asymmetric strong stretch 1490 -1550 cm-1

Symmetric strong stretch 1315 - 1355 cm-1

04/21/23 103

Nitro BenzeneIR Spectrum

Nitro Compounds

C6H5NO2CAS 98-95-3


04/21/23 104

Unsat=C-H Stretch

NO2 (-N=O) Stretch2 Absorptions

C=CAromatic ringAbsorptions

AromaticOvertones

Mono-Substitution

1-Nitro PropaneIR SpectrumNitro Compounds

C3H5NO2CAS 108-03-2

NO2 (-N=O) Stretch2 Absorptions



04/21/23 105

Infrared Spectroscopy Step 16. If none of the above apply then the compound

is most likely a:

Hydrocarbon

Alkyl Halide (see slides 105 - 109).

Hydrocarbons

Generally, very simple spectrum

–C-H Sat’d Alkanes – 2900 - 3000 cm-1

Methyl (CH3) – 1370 cm-1

Methylene (CH2) – 1450 cm-1

t-Butyl Group – 525 cm-1

Long Alkane (CH2) Chain – 720 cm-1

04/21/23 106

Decane

CH3(CH2)8CH3

IR SpectrumAlkane

C10H22CAS 124-18-5

CH2

CH3


Long AlkaneChain (CH2)

Bending


04/21/23 107

Infrared Spectroscopy Step 17. Halogens

The Halogens as CH2 - X absorptions show up in the region (1000 – 1300 cm-1).

Halogens (Cl, Br, I) show in the Fingerprint region (485 – 800 cm-1) as one or two absorptions – see next slide.

Using IR to identify Halogens in this region can be difficult, especially if OOP Bending absorptions (used for “Substitution Pattern information) from Alkene and Aromatic unsaturated Pi () bond structures are present.

Halogen identification should be restricted to Aliphatic Alkane structures containing mainly CH2 & CH3 groups.

Iodide and Bromide absorptions in the range 485 – 650 cm-1 are generally out range on NaCL Salt Plates, however, if other substrates, e.g.,KBr pellets, are used, the absorptions are extended to this range.

04/21/23 108

Infrared Spectroscopy Step 17. Halogens (Con’t)

Fluoride 1000 – 1400 cm-1

Monofluorides 1000 – 1200 cm-1

Polyfluorides 1100 – 1300 cm-1

Aryl Fluorides 1100 – 1250 cm-1

Chloride (2 or more bands) 540– 785 cm-1

CH2-CL (Bend Wagging) 1230 – 1300 cm-1

t-Butyl Group – 525 cm-1

Bromine (KBr Pellets) 510 –650 cm-1

CH2-Br (Bend Wagging) 1190 – 1250 cm-1

Aryl Bromides 1030 – 1075 cm-1

Iodide (KBr Pellets) 485 –600 cm-1

CH2-I (Bend Wagging) 1150 – 1200 cm-1

04/21/23 109

IR SpectrumHalogens 2-Bromobutane

Br

CH2-Br

CH3

CH2

-C-HSat’n

C4H9Br CAS 78-76-2


04/21/23 110

IR SpectrumHalogens 1-Chloropropane

C3H7Cl CAS 540-54-5

CH3

CH2

-C-HSat’n Cl

CH2-Cl


04/21/23 111

IR SpectrumHalogens

C7H7Cl

o-Chlorotoluene

CAS 95-49-8

-C-HSat’n

=C-HUnsat’n

AromaticOvertones

O-Disubstitution

-C=C-Aromatic

CH3

CH2-Cl

Cl

OOPo-disubstitution

(750 cm-1)(missing)

{


04/21/23 112


04/21/23 113

IR SpectrumHalogens

C5H14CL CAS 594-36-5

T-Pentyl525 cm-1CH2

CH3

CH2-ClSaturatedAliphatic

C-H Stretch

T-Pentyl Chloride(2-Chloro-2-MethylButane

Carbonyl (C=O) @ 1715-1685(Conjugation moves absorption to right ~30 cm-1

Acid

Ester

Amide

Anhydride

Aldehyde

Ketone

Alcohol

Amine

Ether

Alkanes -C-HMethylene -CH2

Methyl -CH3

Alkenes (Vinyl) -C=CAlkynes (Acetylenes) -C≡CAromatic -C=C

Nitriles Nitro

Saturation< 3000 cm-1

Yes No

Unsaturation> 3000 cm-1

Hydrocarbons

IR Analysis Scheme

04/21/23 114

Carbonyl (C=O) is Present

Acid - Broad OH Absorption @ 3300-2500 cm-1

Ester - C-O Absorption @ 1300-1000 cm-1

Amide - NH Absorption @ 3500 cm-1 (1 or 2 peaks)

Anhydride - 2 C=O Absorptions 1810 & 1760 cm-1

Aldehyde - Aldehyde C-H Absorptions @ 2850 & 2750 cm-1

Ketone - None of the above except C=O

Carbonyl is Absent

Alcohol - Broad OH absorption @ 3300 - 3000 cm-1

Also C-O absorption @ 1300 - 1000 cm-1

Amine - 1 to 2 equal NH absorptions @ 3500 cm-1

Ether - C-O absorption @ 1300 - 1000 cm-1

IR Analysis Scheme

04/21/23 115

Saturation

Unsaturation

Alkanes -C-H Stretch – several absorptions to “right” of 3000 cm-1

Methylene -CH2 1450 cm-1

Methyl -CH3 1375 cm-1

Double Bonds =C-H Stretch – several absorptions to “left” of 3000 cm-1

OOP bending at 1000 – 650 cm-1

Alkenes (Vinyl) -C=C- Stretch (weak) @ 1675 – 1600 cm-1

Conjugation moves absorption to the rightAlkynes -C≡C-H Terminal Acetylene Stretch at 3300 cm-1

Alkynes (Acetylenes) -C≡C Stretch @ 2150 cm-1 Conjugation moves absorption to the right

Aromatic (Benzene) =C-H Stretch absorptions also to left of 3000 cm-1

OOP bending at 900 – 690 cm-1

OOP absorption patterns allow determination of ring substitution (p. 897 Pavia text)

-C=C 4 Sharp absorptions (2 pairs) @ 1600 & 1450 cm-1

Overtone absorptions @ 2000 – 1667 cm-1

Relative shapes and numbers of peaks permit determination of ring substitution pattern (p. 897 Pavia text).

IR Analysis Scheme

04/21/23 116

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