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Structure Elucidation And Molecular
Spectroscopy
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Introduction
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 Structure is the key to everything in chemistry.
 The properties of a substance depend on the atoms it contains and the way the atoms
are connected.
 Until the second half of the twentieth century, the structure of a substance—a newly
discovered natural product was determined using information obtained from
chemical reactions. That included the identification of functional groups by
chemical tests, along with the results of experiments in which the substance was
broken down into smaller, more readily identifiable fragments.
 Typical of this approach is the demonstration of the presence of a double bond in an
alkene by catalytic hydrogenation and subsequent determination of its location by
ozonolysis.

Introduction
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 Molecular spectroscopy is the study of the interaction between matter and
electromagnetic radiation as a function of the wavelength or frequency of the
radiation.
 Molecular spectra result from either the absorption or the emission of
electromagnetic radiation as molecules undergo changes from one quantized energy
state to another.

Introduction
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 Qualitative tests and chemical degradation have been supplemented and to a large
degree replaced by instrumental methods of structure determination. The most
prominent methods and the structural clues they provide are:
 Nuclear magnetic resonance (NMR) spectroscopy tells us about the carbon
skeleton and the environments of the hydrogens attached to it.
 Infrared (IR) spectroscopy reveals the presence or absence of key functional
groups.

Introduction
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 Ultraviolet-visible (UV-VIS) spectroscopy probes the electron distribution, especially in
molecules that have conjugated p electron systems.
 Mass spectrometry (MS) gives the molecular weight and formula, both of the molecules
itself and various structural units within it.
 The particles are called photons, and each possesses an amount of energy referred to as a
quantum. the energy of a photon (E) is directly proportional to its frequency (u).
E = hv
 The SI units of frequency are reciprocal seconds (s-1), given the name hertz The constant of
proportionality h is called Planck’s constant and has the value h =6.63 x 10_34 J.s

Introduction
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 Electromagnetic radiation travels at the speed of light (c = 3.0 x 108 m/s), which is
equal to the product of its frequency v and its wavelength l:
c =vl
 The range of photon energies is called the electromagnetic spectrum
 Frequency is inversely proportional to wavelength; the greater the frequency, the
shorter the wavelength.
 Energy is directly proportional to frequency; electromagnetic radiation of higher
frequency possesses more energy than radiation of lower frequency.

Introduction
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 When a molecule is exposed to electromagnetic radiation, it may absorb a photon,
increasing its energy by an amount equal to the energy of the photon.
 The particular photon energies absorbed by a molecule depend on molecular
structure and can be measured with instruments called spectrometers.

ULTRAVIOLET-VISIBLE (UV-VIS) SPECTROSCOPY
 The main application of UV-VIS spectroscopy, which depends on transitions
between electronic energy levels, is in identifying conjugated p electron systems.
 The energy required to promote an electron from one electronic state to the next
lies in the visible and ultraviolet range of the electromagnetic spectrum.
 We usually identify radiation in the UV-VIS range by its wavelength in nanometers
(1 nm = 10-9 m).
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 Thus, the visible region corresponds to 400–800 nm.
 Red light is the low-energy (long wavelength) end of the visible spectrum, violet
light the high-energy (short wavelength) end.
 Ultraviolet light lies beyond the visible spectrum with wavelengths in the 200–400-
nm range.
 As is typical of most UV spectra, the absorption is rather broad and is often spoken
of as a “band” rather than a “peak.”
 The wavelength at an absorption maximum is referred to as the l max of the band.
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 The structural unit associated with the electronic transition in UV-VIS spectroscopy
is called a chromophore.
 A phenomenon of interaction of molecules with ultraviolet and visible lights.
Absorption of photon results in electronic transition of a molecule, and electrons
are promoted from ground state to higher electronic states.
 In structure determination : UV-VIS spectroscopy is used to detect the presence of
chromophores like dienes, aromatics, polyenes, and conjugated ketones, etc.
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Electronic transitions
 There are three types of electronic transition of organic compound
Transitions involving p, s, and n electrons
Transitions involving charge-transfer electrons
Transitions involving d and f electrons
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Electronic transitions Molecular Orbital Theory

ss Transitions
 An electron in a bonding s orbital is excited to the corresponding antibonding
orbital. The energy required is large.
 For example, methane (which has only C-H bonds, and can only undergo s  s
transitions) shows an absorbance maximum at 125 nm.
 Absorption maxima due to s  s transitions are not seen in typical UV-VIS
spectra (200 -800 nm)
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n s Transitions
 Saturated compounds containing atoms with lone pairs (non-bonding electrons) are
capable of n  s transitions.
 These transitions usually need less energy than s  s  transitions.
 They can be initiated by light whose wavelength is in the range 150 - 250 nm.
 The number of organic functional groups with n  s peaks in the UV region is
small.
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n p and pp Transitions
 Most absorption spectroscopy of organic compounds is based on transitions of n
or p electrons to the p excited state.
 These transitions fall in an experimentally convenient region of the spectrum
(200 - 800 nm).
 These transitions need an unsaturated group in the molecule to provide the p
electrons.
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Terms describing UV absorptions
1. Chromophores: functional groups that give electronic transitions.
2. Auxochromes: substituents with unshared pair e's like OH, NH, SH ..., when
attached to π chromophore they generally move the absorption max. to longer λ.
3. Bathochromic shift: shift to longer λ, also called red shift.
4. Hypsochromic shift: shift to shorter λ, also called blue shift.
5. Hyperchromism: increase in ε of a band.
6. Hypochromism: decrease in ε of a band.
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 UV/Vis spectrometers collate the data over the required range and generate the
spectrum of the compound under analysis as a graph representing the transmittance
(or the absorbance) as a function of wavelength, given in nanometers.
 In spectroscopy, the transmittance T is a measure of the attenuation of a beam of
monochromatic light based upon the comparison between the intensities of the
transmitted light (I) and the incident light (I0) according to whether the sample is
placed, or not, in the optical pathway between the source and the detector.
 T is expressed as a fraction or a percentage:

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Instrumentation in the UV/Visible
 For the visible region of the spectrum, an incandescent lamp fitted with a tungsten
filament housed in a silica glass
 For the UV region a deuterium arc lamp working under a slight pressure to
maintain an emission continuum (<350nm)

Quantitative Use of UV Spectra
 Absorbance for a particular compound in a specific solvent at a specified
wavelength is directly proportional to its concentration
Beers’ law: absorbance = ecl
– “e” is molar absorptivity (extinction coefficient)
– “c” is concentration in mol/L
– “l” is path of light through sample in cm
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Infrared radiation
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 IR identify the presence of certain functional groups within a molecule.
 The fraction of the infrared region of most use for structure determination lies between
2.5 x 10-6 m and 16 x 10-6 m in wavelength.
 Two units commonly employed in infrared spectroscopy are the micrometer and the
wave number. One micrometer (mm) is 10-6 m, and infrared spectra record the region
from 2.5mm to 16 mm.
 Wave numbers are reciprocal centimeters (cm-1), so that the region 2.5–16 mm
corresponds to 4000–625 cm-1.
 An advantage to using wave numbers is that they are directly proportional to energy.
 Thus, 4000 cm-1 is the high-energy end of the scale, and 625 cm-1 is the low-energy
end.

Infrared radiation
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 Electromagnetic radiation in the 4000–625 cm-1 region corresponds to the separation
between adjacent vibrational energy states in organic molecules.
 Absorption of a photon of infrared radiation excites a molecule from its lowest, or
ground, vibrational state to a higher one.
 These vibrations include stretching and bending
 A single molecule can have a large number of distinct vibrations available to it, and
infrared spectra of different molecules, like fingerprints, are different.

Infrared radiation
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 The atoms in molecules are not static, but vibrate about their equilibrium positions.
 When molecules absorb infrared radiation, the absorbed energy causes an increase
in the amplitude of the vibration of the bonded atoms.
 IR spectrum is obtained by measuring the absorption of IR radiation of a sample
at different frequencies.

Infrared radiation
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Different types of groups of atoms (C-H, O-H, N-H, etc…) absorb infrared radiation at
different characteristic wavenumbers.
In general, the IR spectrum can be split into four regions for interpretation:
4000  2500 cm-1: Absorption of single bonds formed by hydrogen and other elements
e.g. CH, OH, NH
2500  2000 cm-1: Absorption of triple bonds e.g. C≡C, C≡N
2000  1500 cm-1: Absorption of double bonds e.g. C=C, C=O
1500  625 cm-1: This region often consists of many different, complicated bands.
This part of the spectrum is unique to each compound and is often called the fingerprint
region.

Infrared radiation
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 Almost all organic compounds exhibit a peak or group of peaks near 3000 cm-1
due to carbon–hydrogen stretching. The peaks at 1460, 1380, and 725 cm-1 are
due to various bending vibrations.
 In using infrared spectroscopy for structure determination, peaks in the range
1500–4000 cm-1 are usually emphasized because this is the region in which the
vibrations characteristic of particular functional groups are found.

Infrared radiation
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 The region 1500–625 cm-1 is known as the fingerprint region; it is here that the
pattern of peaks varies most from compound to compound. This part of the
spectrum is unique to each compound and is often called the fingerprint region.
 If the infrared spectrum of an unknown compound shows an absorption peak in
the region of 3200 to 3700 cm-1, it is reasonable to assume that the compound
may contain either OH or NH group.

Infrared radiation
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Characteristic Range
Bond Wavenumber (cm-1)
C=C Alkenes 1610 to 1680
C=O Aldehydes, ketones, acids, esters 1680 to 1750
C≡C Alkynes 2070 to 2250
C≡N Nitriles 2200 to 2280
OH Acids (hydrogen-bonded) 2500 to 3300
CH Alkanes, alkenes, arenes 2840 to 3095
OH Alcohols, phenols(hydrogen-bonded) 3230 to 3670
NH Primary amine 3350 to 3500
Correlation Table of Functional Groups and IR Absorption Wavenumbers.

Infrared radiation
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• Wavenumber / cm-1
• RCOCl Acyl chloride 1815 - 1790
• RCOOR’ Aliphatic ester 1750 - 1730
• RCHO Aliphatic aldehyde 1740 - 1720
• RCOOH Aliphatic acid 1725 - 1700
• RCOR’ Aliphatic ketone 1725 - 1700
• ArCHO Aromatic aldehyde 1715 - 1695
• ArCOR Aromatic ketone 1700 - 1680
• ArCOAr Diaromatic ketone 1670 - 1650
• RCONH2 Aliphatic amide 1680 - 1640
Characteristic Absorptions of Carbonyl Groups

Infrared radiation
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 Bending is easier than stretching-- happens at lower energy (lower wavenumber)
 Bond Order is reflected in ordering--triple>double>single (energy)with single
bonds easier than double easier than triple.
 Heavier atoms move slower than lighter ones
 Intensity of IR Absorptions
 In order for a vibration mode to absorb in the infrared, the vibrational motion
must cause a change in the dipole moment of the bond.
 The intensity of the IR “peaks” is proportional to the change in dipole moment
that a bond undergoes during a vibration.
 C=O bonds absorb strongly.
 C=C bonds generally absorb much less.

Infrared radiation
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The IR spectrum of pentanoic acid. The C—O absorption lies at the higher end of
the range of C—O stretches (1250–1050 cm –1 ) because the C—O bond in a
carboxylic acid has some double-bond character.

Infrared radiation
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Used for identifying organic, inorganic, and biological species.
Identification of an organic compound is a two-step process.
The first step involves determining what functional groups are most likely present by
examining the group frequency region.
The second step then involves a detailed comparison of the spectrum of the unknown
with the spectra of pure compounds that contain all of the functional groups found in
the first step.
Qualitative Analysis

Infrared radiation
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Step 1: Look first for the carbonyl C=O band.
Look for a strong band at 1820-1660 cm-1.
This band is usually the most intense absorption band in a spectrum. It will have a
medium width. If you see the carbonyl band, look for other bands associated with
functional groups that contain the carbonyl by going to step 2.
If no C=O band is present, check for alcohols and go to step 3.
Step 2: If a C=O is present you want to determine if it is part of an acid, an ester, or an
aldehyde or ketone. At this time you may not be able to distinguish aldehyde from
ketone.
Infrared interpretation

Infrared radiation
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ACID
Look for indications that an O-H is also present.
It has a broad absorption near 3300-2500 cm-1.
This actually will overlap the C-H stretch. There will also be a C-O single bond
band near 1100-1300 cm-1.
Look for the carbonyl band near 1725-1700 cm-1.
ESTER
Look for C-O absorption of medium intensity near 1300-1000 cm-1.
There will be no O-H band.

Infrared radiation
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ALDEHYDE
Look for aldehyde type C-H absorption bands. These are two weak absorptions to the
right of the C-H stretch near 2850 cm-1 and 2750 cm-1 and are caused by the C-H bond
that is part of the CHO aldehyde functional group.
Look for the carbonyl band around 1740-1720 cm-1.
KETONE
The weak aldehyde CH absorption bands will be absent. Look for the carbonyl CO
band around 1725-1705 cm-1.

Infrared radiation
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Step 3
If no carbonyl band appears in the spectrum, look for an alcohol O-H band.
ALCOHOL
Look for the broad OH band near 3600-3300 cm-1 and a C-O absorption band near
1300-1000 cm-1.
Step 4
If no carbonyl bands and no O-H bands are in the spectrum, check for double
bonds, C=C, from an aromatic or an alkene.
ALKENE
Look for weak absorption near 1650 cm-1 for a double bond. There will be a CH
stretch band near 3000 cm-1.

Infrared radiation
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AROMATIC
Look for the benzene, C=C, double bonds which appear as medium to strong
absorptions in the region 1650-1450 cm-1. The CH stretch band is much weaker
than in alkenes.
C-H stretching region
•Alkanes C-H sp3 stretch < 3000 cm-1
•Alkenes C-H sp2 stretch > 3000 cm-1
•Alkynes C-H sp stretch ~ 3300 cm-1
•C-H Bending region
•CH2 bending ~ 1460 cm-1
•CH3 bending (asym) appears near the same value
•CH3 bending (sym) ~ 1380 cm-1

Infrared radiation
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Step 5
If none of the previous groups can be identified, you may have an alkane.
ALKANE
The main absorption will be the C-H stretch near 3000 cm-1. The spectrum will be
simple with another band near 1450 cm-1.
Step 6
If the spectrum still cannot be assigned you may have an alkyl halide.
ALKYL BROMIDE
Look for the C-H stretch and a relatively simple spectrum with an absorption to the
right of 667 cm-1.

Introduction to Nuclear Magnetic Resonance (NMR) Spectroscopy
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 Nuclear magnetic resonance spectroscopy is a powerful analytical technique used
to characterize organic molecules by identifying carbon-hydrogen frameworks
within molecules.
 NMR spectroscopy depends on the absorption of energy when the nucleus of an
atom is excited from its lowest energy spin state to the next higher one.
 NMR results from resonant absorption of electromagnetic energy by a nucleus
(mostly protons) changing its spin orientation.
 The resonance frequency depends on the chemical environment of the nucleus
giving a specific finger print of particular groups.

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Nuclei that have an odd number of protons or an odd number of neutrons (or both)
have a property called spin that allows them (1H, 13C, 19F and 31P)
Nuclear spin angular momentum is a quantized property of the nucleus in each atom,
it is assigned based on the properties of neutrons and protons.
The nuclear spin angular momentum of each atom is represented by a nuclear spin
quantum number (I).

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Two common types of NMR spectroscopy are used to characterize organic
structure:
1H NMR is used to determine the type(environments) and number of H atoms in a
molecule;
13C NMR is used to determine the type of carbon atoms in the molecule.
The source of energy in NMR is radio waves which have long wavelengths, and
thus low energy and frequency.
When low-energy radio waves interact with a molecule, they can change the
nuclear spins of some elements, including 1H and 13C.

NMR Spectra – The Internal Reference Standard
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 A small amount of an inert reference compound is added to the sample tube
containing the compound whose NMR spectrum is to be taken.
 The most commonly used reference compound is tetramethylsilane (TMS). Because
it is highly volatile (bp = 26.5 C) , it can easily be removed from the sample by
evaporation after the NMR spectrum is taken.
 The methyl protons of TMS are in a more electron-rich environment than are most
protons in organic molecules because silicon is less electronegative than carbon

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Different protons give signals at different field strengths.
The dependence of the resonance position of a nucleus that results from its molecular
environment is called its chemical shift (d). The position at which a signal occurs in an
NMR spectrum is called the chemical shift.
The chemical shifts of various protons in a molecule can be different and are
characteristic of particular structural features. The universally accepted standard
used in NMR is Tetramethylsilane (TMS)
The 12 protons on the four carbon atoms have the same chemical and magnetic
environment and they resonate at the same field strength, i.e., one signal (1 peak) is
produced.

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We measure them with respect to a standard tetramethylsilane (CH3)4Si, (TMS)
The protons of TMS are more shielded than those of most organic compounds, so all of
the signals in a sample ordinarily appear at lower field than TMS reference.
Protons in a molecule are connected to other atoms carbon, oxygen, nitrogen by covalent
bonds.
 All of the protons of a molecule are shielded from the applied field by the electrons, but
some are less shielded than others.
 Sometimes the term “deshielded,” is used to describe this decreased shielding of one
proton relative to another.
 A more shielded proton absorbs rf radiation at higher field strength (upfield) compared
with one at lower field strength (downfield).

Nuclear shielding and 1H chemical shifts
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 Upfield means farther to the right-hand side of the spectrum, and downfield
means farther to the left-hand side of the spectrum.
Protons that have different chemical shifts are said to be chemically nonequivalent

Effect of molecular structure on 1H chemical shift
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For most all other Proton environments, the electron density is less than TMS.
The TMS signal appears on the far right hand side of the X-axis. Thus, most all
other protons will have Chemical Shifts > 0 and will be downfield from the TMS
signal.
 Protons that have different chemical shifts are said to be chemical-shift-
nonequivalent(or chemically nonequivalent).

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 Nuclear magnetic resonance spectroscopy is such a powerful tool for structure
determination because protons in different environments experience different
degrees of shielding and have different chemical shifts.
 In compounds of the type CH3X, for example, the shielding of the methyl protons
increases as X becomes less electronegative.
 The chemical shift depends on the degree to which X draws electrons away from
the methyl group.

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 Methyl (CH3) protons are more shielded than methylene (CH2) protons, and
methylene protons are more shielded than methine (CH) protons.
 These differences are small only about 0.7 ppm separates a methyl proton from a
methine proton of the same type.
 Overall, proton chemical shifts among common organic compounds encompass a
range of about 12 ppm.
 The protons in alkanes are the most shielded, and O-H protons of carboxylic acids
are the least shielded.

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 The 1 H NMR spectrum of 1-bromo-2,2-dimethylpropane (see above figure) has two
signals because the compound has two different kinds of protons.
 The methylene protons are in a less electron-rich environment than are the methyl
protons because the methylene protons are closer to the electron-withdrawing bromine.
Therefore, the methylene protons are less shielded from the applied magnetic field. As a
result, the signal for these protons occurs at a higher frequency than the signal for the
more shielded methyl protons.
 The 1 H NMR spectrum of 1-nitropropane to have three signals because the compound
has three different kinds of protons. The closer the protons are to the electron
withdrawing nitro group, the less they are shielded from the applied magnetic field, so
the higher the frequency at which their signal will appear.

The splitting of signals is described by the N+1
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 Splitting is caused by protons bonded to adjacent carbons. The splitting of a signal is
described by the N + 1 rule, where N is the number of equivalent protons bonded to
adjacent carbons that are not equivalent to the proton producing the signal.
 The signal may be split into two peaks (a doublet), three peaks (a triplet), four
peaks (a quartet), or even more.
 The number of peaks into which the signal for a particular proton is split is called its
multiplicity.
 In 1H NMR spectrum of 1,1-dichloroethane the signal for the methyl protons is split
into a doublet (N = 1, so N + 1 = 2). The carbon adjacent to the carbon bonded to the
methine proton is bonded to three equivalent protons (CH3 ), so the signal for the
methine proton is split into a quartet (N = 3, so N + 1 = 4).

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 1H NMR spectrum of 1,1-dichloroethane (CH3CHCl2), which is characterized by a doublet
centered at s 2.1 ppm for the methyl protons and a quartet at s 5.9 ppm for the methine proton.
 The 1H NMR spectrum of CH3OCH2CN both signals are singlets; that is, each one consists of a
single peak.

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 The integration of NMR signals reveals the relative number of protons causing each
signal, not the absolute number.
 The two signals in the 1H NMR spectrum in figure below are not the same size
because the area under each signal is proportional to the number of protons producing
the signal .
 The area under the signal occurring at the lower frequency is larger because the signal
is produced by nine methyl protons, whereas the smaller, higher-frequency signal is
produced by two methylene protons.
THE INTEGRATION OF NMR SIGNALS

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 The ratio of the integrals is approximately 1.6 : 7.0.
 Dividing by the smallest number gives a new ratio (1:4.4). We then need to multiply this ratio
by a number that will make all the numbers in the ratio close to whole numbers—in this case,
we multiply by 2. This means that the ratio of protons in the compound is 2 : 8.8, which is
rounded to 2 : 9.

13C NMR spectroscopy
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 The number of signals in a 13 C NMR spectrum tells how many different kinds of carbons a
compound has.
 Carbons in electron-rich environments produce low frequency Signals, whereas carbons close to
electron-withdrawing groups produce high frequency Signals.

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Mass spectrometry
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 Mass spectrometry differs from the other instrumental methods discussed in this
chapter in a fundamental way.
 It does not depend on the absorption of electromagnetic radiation but rather examines
what happens when a molecule is bombarded with high-energy electrons.
 Mass spectrometry is a technique used for measuring the molecular weight and
determining the molecular formula of an organic compound.
 In a mass spectrometer, a molecule is vaporized and ionized by bombardment with a
beam of high-energy electrons.
 The energy of the electrons is ~1600 kcal (or 70 eV). The electron beam ionizes the
molecule by causing it to eject an electron.

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A :B + e A +.B + 2e
Molecule Electron Cation radical Two electrons
 Mass spectrometry uses high energy electrons to break a molecule into fragments.
 Separation and analysis of the fragments provides information about:
 Molecular weight and Structure
 The impact of a stream of high energy electrons causes the molecule to lose an
electron forming a radical cation.
 A species with a positive charge and one unpaired electron
+ e
-
C H
H
H
H H
H
H
H
C + 2 e
-
Molecular ion (M+)
m/z = 16

Mass spectrometry
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 The species that results called the molecular ion, is positively charged and has an
odd number of electrons. It is a cation radical.
 The molecular ion has the same mass (less the negligible mass of a single
electron) as the molecule from which it is formed.
 Electrons this energetic not only cause ionization of a molecule but impart a large
amount of energy to the molecular ion, enough energy to break chemical bonds.
 Ionization and fragmentation produce a mixture of particles, some neutral and
some positively charged.
 The molecular ion as well as fragment ions are directed into an analyzer tube
surrounded by a magnet.

Mass spectrometry
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 This magnet deflects the ions from their original trajectory, causing them to adopt a
circular path, the radius of which depends on their mass-to-charge ratio (m/z).
 Ions of small m/z are deflected more than those of larger m/z.
 By varying either the magnetic field strength or the degree to which the ions are
accelerated on entering the analyzer, ions of a particular m/z can be selectively focused
through a narrow slit onto a detector, where they are counted.
 Scanning all m/z values gives the distribution of positive ions, called a mass
spectrum, characteristic of a particular compound.
 The most intense peak in the mass spectrum is called the base peak and is assigned a
relative intensity of 100. Ion abundances are proportional to peak intensities and are
reported as intensities relative to the base peak.

Mass spectrometry
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 The tallest peak in the mass spectrum is called the base peak.
 The base peak is also the M peak, although this may not always be the case.
 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.

Mass spectrometry
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 The mass spectrum of CH4 consists of more peaks than just the M peak.
 Since the molecular ion is unstable, it fragments into other cations and radical cations
containing one, two, three, or four fewer hydrogen atoms than methane itself.
 Thus, the peaks at m/z 15, 14, 13 and 12 are due to these lower molecular weight
fragments.

Combustion analysis
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Combustion analysis is a method used in both organic chemistry and analytical
chemistry to determine the elemental composition (more precisely empirical
formula) of a pure organic compound by combusting the sample under
conditions where the resulting combustion products can be quantitatively
analyzed.

Combustion analysis
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Calculations and determining chemical formulas
 Combustion analysis is a standard method of determining a chemical formula of a
substance that contains hydrogen and carbon. First, a sample is weighed and then
burned in a furnace in the presence of excess oxygen.
 All of the carbon is converted to carbon dioxide, and the hydrogen is converted to
water in this way. Each of these are absorbed in separate compartments, which are
weighed before and after the reaction. From these measurements, the chemical
formula can be determined.

Combustion analysis
1/23/2023 67
Example: After burning 1.333 g of a hydrocarbon in a combustion analysis
apparatus, 1.410 g of H2O and 4.305 g of CO2 were produced. Separately, the
molar mass of this hydrocarbon was found to be 204.35 g/mol. Calculate the
empirical and molecular formulas of this hydrocarbon.

Combustion analysis
1/23/2023 68
Step 1: Using the molar masses of water and carbon dioxide, determine the moles of
hydrogen and carbon that were produced.
Step 2: Divide the larger molar amount by the smaller molar amount. In some cases,
the ratio is not made up of two integers. Convert the numerator of the ratio to an
improper fraction and rewrite the ratio in whole numbers as shown
Step 3: To get the molecular formula, divide the experimental molar mass of the
unknown hydrocarbon by the empirical formula weight.

Combustion analysis
1/23/2023 69
Therefore, the empirical formula is C5H8
Therefore, the molecular formula is (C5H8)3 or C15H24.

Combustion analysis
1/23/2023 70
Example-2
 11.55 mg of a compound produced 16.57 mg of carbon dioxide and 5.09 mg of water
from combustion. Another 5.12 mg of the same compound was found to contain 1.97
mg of chlorine. The molecular weight of the compound is 368.084 g/mole.
Solution: The mg of C is determined by multiplying the weight of CO2 by the ratio of
the atomic weight of carbon to the molecular weight of carbon dioxide (C/ CO2 ).

Combustion analysis
1/23/2023 71
 The mg of H is determined by multiplying the weight of H2O by the ratio of the
atomic weight of two hydrogens to the molecular weight of water (2H/H2O).
 Since 1.97 mg of Cl was found in 5.12 mg of sample, then the percentage of
chlorine can be determined by the ratio of mg of Cl to the weight of the sample
used for the analysis.

Combustion analysis
1/23/2023 72
The percentage of oxygen is determined by the difference.
Each percentage is then divided by the atomic weight of that element to obtain
the ratio of the elements.

Combustion analysis
1/23/2023 73
 The ratios are then divided by the lowest ratio (in this example, 1.085). If a
fraction is obtained that is not close to a whole number, then all of the ratios are
multiplied by whatever integer is necessary to obtain whole-number ratios.

Combustion analysis
1/23/2023 74
 Thus the empirical formula is C6H9Cl2O2, with an empirical weight of 184.042
g/mole. The formula weight, which had been previously determined to be 368.084
g/mole, is then divided by the empirical weight to obtain the number of empirical
units, n. Then multiply the subscripts of the empirical formula by n to obtain the
molecular formula.

Combustion analysis
1/23/2023 75
From the molecular formula, the number of rings and/or pi bonds can be
determined. The unsaturation number is calculated by the formula
Oxygen and sulfur do not change the number of unsaturation's and thus do
not appear in the formula.

Combustion analysis
1/23/2023 76

Natural product
1/23/2023 78
 There is great similarity between the organic reactions chemists carry out in the
laboratory and those performed by nature inside the living cell.
 In other words, bioorganic reactions can be thought of as organic reactions that take
place in tiny flasks called cells.
 A natural product is a chemical compound produced by a living organism.
 The term natural product is also used for commercial purposes to refer to cosmetics,
dietary supplements, and foods produced from natural sources without added artificial
ingredients.
 Natural products are usually restricted to mean purified organic compounds isolated
from natural sources that are produced by the pathways of primary or secondary
metabolism.

1/23/2023 79
 Primary metabolites are components of basic metabolic pathways that are required
for life. They are associated with essential cellular functions such as nutrient
assimilation, energy production, and growth/development.
 Primary metabolites include carbohydrates, lipids, amino acids, and nucleic acids
which are the basic building blocks of life.
 Secondary metabolites are not essential to the growth and development of the
producing organism. It’s primary function is to increase the likelihood of an
organism’s survival by repelling or attracting other organisms.

1/23/2023 80
 Terpenes contain carbon atoms in multiples of five.
 Terpenes are a diverse class of compounds that contain 10, 15, 20, 25, 30, or 40
carbons.
 Many are found in oils extracted from fragrant plants.
 Terpenes can be hydrocarbons, or they can contain oxygen and be alcohols, ketones,
or aldehydes. Oxygen-containing terpenes are sometimes called terpenoids .
 Terpenes and terpenoids have been used as spices, perfumes, and medicines
TERPENES

1/23/2023 81
 The structures of terpenes are consistent with how they are made: by joining
together five-carbon isoprene units, usually in a head-to-tail fashion.
 Terpenes are classified according to the number of carbons they contain.
 Monoterpenes are composed of two isoprene units, so they have 10 carbons.
Sesquiterpenes , with 15 carbons, have three isoprene units.
 Many fragrances and flavorings found in plants are monoterpenes or
sesquiterpenes. These compounds are known as essential oils.
 Triterpenes (30 carbons) and tetraterpenes (40 carbons).
 For example, squalene is a triterpene and lycopene and carotene, compounds
responsible for the red and orange colors of many fruits and vegetables, are
tetraterpenes.

1/23/2023 83
 The five-carbon compound used for the biosynthesis of terpenes is 3-methyl-3-
butenyl pyrophosphate (isopentenyl pyrophosphate).
HOW TERPENES ARE BIOSYNTHESIZED

Natural product
1/23/2023 84
 The first step a Claisen condensation
 The second step is an aldol addition with a third molecule of acetyl-CoA, which is
followed by hydrolysis of one of the thioester groups.
 The thioester is reduced by NADPH to form mevalonic acid.
 A pyrophosphate group is added by means of two successive phosphorylations of
the primary alcohol with ATP
 The tertiary alcohol is phosphorylated with ATP. Subsequent decarboxylation and
loss of the phosphate group forms isopentenyl pyrophosphate.

Natural product
1/23/2023 85
 A cysteine side chain is in the proper position at the enzyme’s active site to donate a
proton to the sp2 carbon of the alkene that is bonded to the most hydrogens
 A glutamate side chain removes a proton from the b-carbon of the carbocation
intermediate that is bonded to the fewest.
 The enzyme-catalyzed reaction of dimethylallyl pyrophosphate with isopentenyl
pyrophosphate forms geranyl pyrophosphate, a 10-carbon compound.

Natural product
1/23/2023 86
 Experimental evidence suggests that this is an SN1 reaction. Thus, the leaving
group departs, forming an allylic cation.
 Isopentenyl pyrophosphate is the nucleophile that adds to the allylic cation.
 A base removes a proton, forming geranyl pyrophosphate.

Natural product
1/23/2023 88
Brooks H. Pate, ... Melanie Schnell, in Chiral Analysis (Second
Edition), 2018

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