Organic Chemistry. SHE… · Klein, Organic Chemistry 2e • Molecular bonds can vibrate by stretching or by bending in a number of ways. 15.2 IR Spectroscopy • This chapter will

Chapter 15Infrared Spectroscopy and Mass Spectrometry

Organic ChemistrySecond Edition

David Klein

Copyright © 2015 John Wiley & Sons, Inc. All rights reserved. Klein, Organic Chemistry 2e

15.1 Introduction to Spectroscopy• Spectroscopy involves an interaction between matter

and light (electromagnetic radiation)• Light can be thought of as waves of energy or packets

(particles) of energy called photons• Properties of light waves include wavelength and

frequency• Is wavelength directly or inversely proportional to

energy? WHY?• Is frequency directly or inversely proportional to

energy? WHY?

Copyright © 2015 John Wiley & Sons, Inc. All rights reserved. 15-2 Klein, Organic Chemistry 2e

15.1 Introduction to Spectroscopy• There are many wavelengths of light that can not be

observed with your eyes


15.1 Introduction to Spectroscopy• When light interacts with molecules, the effect depends

on the wavelength of light used

• This chapter focuses on IR spectroscopyCopyright © 2015 John Wiley & Sons, Inc. All rights reserved. 15-4 Klein, Organic Chemistry 2e

15.1 Introduction to Spectroscopy• Matter exhibits particle-like properties• On the macroscopic scale, matter appears to exhibit

continuous behavior rather than quantum behavior– Consider the example of an engine powering the rotation of a

tire. The tire should be able to rotate at nearly any rate

• Matter also exhibits wave-like properties as we learned in section 1.6

• Matter on the molecular scale exhibits quantum behavior– A molecule will only rotate or vibrate at certain rates

(energies)


15.1 Introduction to Spectroscopy• For each of the types of molecular motion/energy

below, describe how it is quantized– Rotation

– Vibration

– Energy of electrons


15.1 Introduction to Spectroscopy• For each different bond, vibrational energy levels are

separated by gaps (quantized)• If a photon of light strikes the molecule with the exact

amount of energy needed, a molecular vibration will occur

• Energy is eventually released from the molecule generally in the form of heat

• Infrared (IR) Light generally causes molecular vibration• HOW might IR light absorbed give you information

about a molecule’s structure


• Molecular bonds can vibrate by stretching or by bending in a number of ways

15.2 IR Spectroscopy

• This chapter will focus mostly on stretching frequencies• WHY do objects emit IR light?• WHY do some objects emit more IR radiation than

others?• WHERE does that light come from?


• Some night vision goggles can detect IR light that is emitted

• IR or thermal imaging is also used to detect breast cancer



• The energy necessary to cause vibration depends on the type of bond



• An IR spectrophotometer irradiates a sample with all frequencies of IR light

• The frequencies that are absorbed by the sample tell us the types of bonds (functional groups) that are present

• How do we measure the frequencies that are absorbed?

• Most commonly, samples are deposited neat on a salt (NaCl) plate. WHY is salt used?

• Alternatively, the compound may be dissolved in a solvent or embedded in a KBr pellet



• In the IR spectrum below, WHAT is % transmittance and how does it relate to molecular structure?



• Analyze the units for the wavenumber,• ν = frequency and c = the speed of light



• HOW are wavelength and wavenumber different? • HOW are wavenumbers and energy related?



• A signal on the IR spectrum has three important characteristics: wavenumber, intensity, and shape



• The wavenumber for a stretching vibration depends on the bond strength and the mass of the atoms bonded together

• Should bonds between heavier atoms require higher or lower wavenumber IR light to stretch?

15.3 IR Signal Wavenumber


• Rationalize the trends below using the wavenumberformula

1.

2.



• The wavenumber formula and empirical observations allow us to designate regions as representing specific types of bonds

• Explain the regions above



• The region above 1500 cm-1 is called the diagnostic region. WHY?

• The region below 1500 cm-1 is called the fingerprint region. WHY?



DIAGNOSTIC REGION FINGERPRINT REGION

• Analyze the diagnostic and fingerprint regions below



• Analyze the diagnostic and fingerprint regions below



• Compare the IR spectra



• Given the formula below and the given IR data, predict whether a C-H or O-H bond is stronger

• C-H stretch ≈ 3000 cm-1

• O-H stretch ≈ 3400 cm-1

• Practice with conceptual checkpoint 15.1



• Compare the IR stretching wavenumbers below

• Are the differences due to mass or bond strength?• Which bond is strongest, and WHY?



• Note how the region ≈3000 cm-1 in the IR spectrum can give information about the functional groups present



• Is it possible that an alkene or alkyne could give an IR spectra without any signals above 3000 cm-1?

• Predict the wavenumbers that would result (if any) above 3000 cm-1 for the molecules below



HO


• Resonance can affect the wavenumber of a stretching signal

• Consider a carbonyl that has two resonance contributors

• If there were more contributors with C-O single bond character than C=O double bond character, how would that affect the wavenumber?



• Use the given examples to explain HOW and WHY the conjugation and the –OR group affect resonance and thus the IR signal?




• The strength of IR signals can vary

15.4 IR Signal Strength


• When a bond undergoes a stretching vibration, its dipole moment also oscillates

• Recall the formula for dipole moment includes the distance between the partial charges,

• The oscillating dipole moment creates an electrical field surrounding the bond



• The more polar the bond, the greater the opportunity for interaction between the waves of the electrical field and the IR radiation

• Greater bond polarity = stronger IR signals



• Note the general strength of the C=O stretching signal vs. the C=C stretching signal

• Imagine a symmetrical molecule with a completely nonpolar C=C bond: 2,3-dimethyl-2-butene

• 2,3-dimethyl-2-butene does not give an IR signal in the 1500-2000 cm-1 region



• Stronger signals are also observed when there are multiple bonds of the same type vibrating

• Although C-H bonds are not very polar, they often give very strong signals, WHY?

• Because sample concentration can affect signal strength, the Intoxilyzer 5000 can be used to determine blood alcohol levels be analyzing the strength of C-H bond stretching in blood samples

• Practice with conceptual checkpoints 15.5 – 15.7



• Some IR signals are broad, while others are very narrow

• O-H stretching signals are often quite broad

15.5 IR Signal Shape


• When possible, O-H bonds form H-bonds that weaken the O-H bond strength

• The H-bonds are transient, so the sample will contain molecules with varying O-H bond strengths

• Why does that cause the O-H stretch signal to be broad?• The O-H stretch signal will be narrow if a dilute solution

of an alcohol is prepared in a solvent incapable of H-bonding


• WHY does H-bonding affect the O-H bond strength?


• In a sample with an intermediate concentration, both narrow and broad signals are observed. WHY?


• Explain the cm-1 readings for the two O-H stretching peaks


• Consider how broad the O-H stretch is for a carboxylic acid and how its wavenumber is around 3000 cm-1

rather than 3400 cm-1 for a typical O-H stretch



• H-bonding is often more pronounced in carboxylic acids, because they can forms H-bonding dimers



• For the molecule below, predict all of the stretching signals in the diagnostic region



O

OH


• Primary and secondary amines exhibit N-H stretching signals. WHY not tertiary amines?

• Because N-H bonds are capable of H-bonding, their stretching signals are often broadened

• Which is generally more polar, an O-H or an N-H bond?

• Do you expect N-H stretches to be strong or weak signals?

• See example spectra on next slide





15.5 IR Signal Shape• The appearance of two N-H signals

for the primary amine is NOT simply the result of each N-H bond giving a different signal

• Instead, the two N-H bonds vibrate together in two different ways


15.5 IR Signal Shape• A single molecule can only vibrate symmetrically or

asymmetrically at any given moment, so why do we see both signals at the same time?

• Similarly, CH2 and CH3 groups can also vibrate as a group giving rise to multiple signals



15.6 Analyzing an IR Spectrum• Table 15.2 summarizes some of the key signals that help

us to identify functional groups present in molecules• Often, the molecular structure can be identified from an

IR spectra1. Focus on the diagnostic region (above 1500 cm-1)

a) 1600-1850 cm-1 – check for double bondsb) 2100-2300 cm-1 – check for triple bondsc) 2700-4000 cm-1 – check for X-H bondsd) Analyze wavenumber, intensity, and shape for each signal


15.6 Analyzing an IR Spectrum• Often, the molecular

structure can be identified from an IR spectra

2. Focus on the 2700-4000 cm-1 (X-H) region

• Practice with SkillBuilder15.1


15.7 Using IR to Distinguish Between Molecules

• As we have learned in previous chapters, organic chemists often carry out reactions to convert one functional group into another

• IR spectroscopy can often be used to determine the success of such reactions

• For the reaction below, how might IR spectroscopy be used to analyze the reaction?

• Practice with SkillBuilder 15.2Copyright © 2015 John Wiley & Sons, Inc. All rights reserved. 15-46 Klein, Organic Chemistry 2e

15.7 Using IR to Distinguish Between Molecules

• For the reactions below, identify the key functional groups, and describe how IR data could be used to verify the formation of product

• Is IR analysis qualitative or quantitative?

1) H-Br

2) Et-OK

O3

(CH3)2SO

O


15.8 Into to Mass Spectrometry• Mass spectrometry is primarily used to determine the

molar mass and formula for a compound1. A compound is vaporized and then ionized2. The masses of the ions are detected and graphed

• Can you think of ways to get an organic molecule to ionize?

• Will the molecule need to absorb energy or release energy?


15.8 Into to Mass Spectrometry• The most common method of ionizing molecules is by

electron impact (EI)• The sample is bombarded with a beam of high energy

electrons (1600 kcal or 70 eV)• EI usually causes an electron to be ejected from the

molecule. HOW? WHY?

• What is a radical cation?


15.8 Into to Mass Spectrometry• How does the mass of the radical cation compare to the

original molecule?

• If the radical cation remains intact, it is known as the molecular ion (M+•) or parent ion

• Often, the molecular ion undergoes some type of fragmentation. WHY?


15.8 Into to Mass Spectrometry• The resulting fragments may undergo even further

fragmentation

• The ions are deflected by a magnetic field • Smaller mass and higher charge fragments are affected

more by the magnetic field. WHY?• Neutral fragments are not detected. WHY?


15.8 Into to Mass Spectrometry• Explain the units on the x and

y axes for the mass spectrumfor methane

• The base peak is the tallest peak in the spectrum

• For methane the base peak represents the M+•

• Sometimes, the M+• peak is not even observed in the spectrum, WHY?


15.8 Into to Mass Spectrometry• Peaks with a mass of less than M+• represent fragments

• Subsequent H radicals can be fragmented to give the ions with a mass/charge = 12, 13 and 14

• The presence of a peak representing (M+1) +• will be explained in section 15.10


15.8 Into to Mass Spectrometry• Mass spec is a relatively sensitive analytical method• Many organic compounds can be identified

– Pharmaceutical: drug discovery and drug metabolism, reaction monitoring

– Biotech: amino acid sequencing, analysis of macromolecules– Clinical: neonatal screening, hemoglobin analysis– Environmental: drug testing, water quality, food

contamination testing– Geological: evaluating oil composition– Forensic: Explosive detection– Many More


15.9 Analyzing the M+• Peak

• In the mass spec for benzene, the M+•

peak is the base peak• The M+• peak does not easily fragment


15.9 Analyzing the M+• Peak

• Like most compounds, the M+• peak for pentane is NOT the base peak

• The M+• peak fragments easily


15.9 Analyzing the M+• Peak• The first step in analyzing a mass spec is to identify the

M+• peak– It will tell you the molar mass of the compound– An odd massed M+• peak MAY indicate an odd number of N

atoms in the molecule– An even massed M+• peak MAY indicate an even number of N

atoms or zero N atoms in the molecule

• Give an alternative explanation for a M+• peak with an odd mass



15.10 Analyzing the (M+1)+• Peak• Recall that the (M+1)+• peak in

methane was about 1% as abundant as the M+• peak

• The (M+1)+• peak results from the presence of 13C in the sample. HOW?


15.10 Analyzing the (M+1)+• Peak• For every 100 molecules of decane,

what percentage of them are made of exclusively 12C atoms?

• Comparing the heights of the (M+1)+•

peak and the M+• peak can allow you to estimate how many carbons are in the molecule. HOW?

• The natural abundance of deuterium is 0.015%. Will that affect the mass spec analysis?

• Practice with SkillBuilder 15.3


15.11 Analyzing the (M+2)+• Peak• Chlorine has two abundant isotopes• 35Cl=76% and 37Cl=24%

• Molecules with chlorine often have strong (M+2)+•

peaks• WHY is it sometimes difficult to be absolutely sure which peak is the (M)+• peak?


15.11 Analyzing the (M+2)+• Peak• 79Br=51% and 81Br=49%, so molecules with bromine

often have equally strong (M)+• and (M+2)+• peaks

• Practice with conceptual checkpoints 15.23 and 15.24


15.12 Analyzing the Fragments• A thorough analysis of the molecular fragments can

often yield structural information• Consider pentane• Remember, MS only

detects charged fragments


15.12 Analyzing the Fragments

• WHAT type of fragmenting is responsible for the “groupings” of peaks observed?


15.12 Analyzing the Fragments• In general, fragmentation will be more prevalent when

more stable fragments are produced• Correlate the relative

stability of the fragments here with their abundances on the previous slide


15.12 Analyzing the Fragments• Consider the fragmentation below

• All possible fragmentations are generally observed under the high energy conditions employed in EI-MS

• If you can predict the most abundant fragments and match them to the spectra, it can help you in your identification


15.12 Analyzing the Fragments• Alcohols generally undergo two main types of

fragmentation: alpha cleavage and dehydration


15.12 Analyzing the Fragments• Amines generally undergo alpha cleavage

• Carbonyls generally undergo McLafferty rearrangement

• Practice with conceptual checkpoints 15.25 – 15.28


15.13 High Resolution Mass Spec• High Resolution Mass Spectrometry allows m/z to be

measured with up to 4 decimal places• Masses are generally not whole number integers

– 1 proton = 1.0073 amu and 1 neutron = 1.0086 amu

• One 12C atom = exactly 12.0000 amu, because the amuscale is based on the mass of 12C

• All atoms other than 12C will have a mass in amu that can be measured to 4 decimal places by a high-resolution mass spec instrument


15.13 High Resolution Mass Spec• Note the exact masses and natural abundances below


15.13 High Resolution Mass Spec• Why are the values in table 15.5 different from those on

the periodic table?• Imagine you want to use

high-res MS to distinguish between the molecules below

• Why can’t you use low-res?


15.13 High Resolution Mass Spec• Using the exact masses and natural abundances for each

element, we can see the difference high-res makes

• The molecular ion results from the molecule composed of the isotopes with the greatest natural abundance

• What if the molecular ion is not observed?• Practice with conceptual checkpoints 15.19 and 15.30


15.14 High Resolution Mass Spec• MS is suited for the identification of pure substances• However, MS instruments are often connected to a gas

chromatograph so mixtures can be analyzed


15.14 High Resolution Mass Spec• GC-MS gives two main forms of information

• GC-MS is a great technique for detecting compounds such as drugs in solutions such as blood or urine

1. The chromatogram gives the retention time

2. The Mass Spectrogram (low-res or high-res)


15.15 MS of Large Biomolecules• To be analyzed by EI mass spec, substances generally

must be vaporized prior to ionization• Until recently (last 30 years), compounds that

decompose before they vaporize could not be analyzed• In Electrospray ionization (ESI), a high-voltage needle

sprays a liquid solution of an analyte into a vacuum causing ionization

• HOW is ESI relevant for analyzing large biomolecules?• ESI is a “softer” ionizing technique. WHAT does that

mean?


15.16 Degrees of Unsaturation• Mass spec can often be used to determine the formula

for an organic compound• IR can often determine the functional groups present• Careful analysis of a molecule’s formula can yield a list

of possible structures• Alkanes follow the formula below, because they are

saturated

• Verify the formula by drawing some isomers of pentane


CnH2n+2

15.16 Degrees of Unsaturation• Notice that the general formula for the compound,CnH2n+2, changes when a double or triple bond is present

• Adding a degree of unsaturation decreases the number of H atoms by two

• How many degrees of unsaturation are there in cyclopentane?


15.16 Degrees of Unsaturation• Consider the isomers of C4H6

• How many degrees of unsaturation are there?• 1 degree of unsaturation = 1 unit on the hydrogen

deficiency index (HDI)


15.16 Degrees of Unsaturation• For the HDI scale, a halogen is treated as if it were a

hydrogen atom

• How many degrees of unsaturation are there in C5H9Br?• An oxygen does not affect the HDI. WHY?


15.16 Degrees of Unsaturation• For the HDI scale, a nitrogen increases the number of

expected hydrogen atoms by ONE

• How many degrees of unsaturation are there in C5H8BrN?

• You can also use the formula below


15.16 Degrees of Unsaturation• Calculating the HDI can be very useful. For example, if

HDI=0, the molecule can NOT have any rings, double bonds, or triple bonds

• Propose a structure for a molecule with the formula C7H12O. The molecule has the following IR peaks – A strong peak at 1687 cm-1

– NO IR peaks above 3000 cm-1



Chapter 16Nuclear Magnetic Resonance Spectroscopy

Organic ChemistrySecond Edition

David Klein


16.1 Intro to NMR Spectroscopy• What is spectroscopy?• Nuclear Magnetic Resonance (NMR) spectroscopy may

be the most powerful method of gaining structural information about organic compounds

• NMR involves an interaction between electromagnetic radiation (light) and the nucleus of an atom– We will focus on C and H nuclei. WHY?– The structure (connectivity) of a molecule affects how the

radiation interacts with each nucleus in the molecule

Copyright © 2015 John Wiley & Sons, Inc. All rights reserved. Klein, Organic Chemistry 2e 16-82

16.1 Intro to NMR Spectroscopy• Protons and neutrons in a nucleus behave as if they are

spinning• If the total number of neutrons and protons is an ODD

number, the atoms will have net nuclear spin• Examples: • The spinning charge in the nucleus creates a magnetic

moment• We saw in Chapter 15 how a dipole moment creates an

electric field• What does a magnetic moment create?


16.1 Intro to NMR Spectroscopy• Like a bar magnet, a magnetic moment exists

perpendicular to the axis of nuclear spin


16.1 Intro to NMR Spectroscopy• If the normally disordered magnetic moments of atoms

are exposed to an external magnetic field, their magnetic moments will align

• WHAT if the total number of neutrons and protons is an EVEN number?


16.1 Intro to NMR Spectroscopy• The aligned magnetic moments can be either with or

against the external magnetic field• The α and β spin states are not

equal in energy. WHY?


16.1 Intro to NMR Spectroscopy• When an atom with an α spin state is exposed to radio

waves of just the right energy, it can be promoted to a βspin state

• The stronger the magnetic field, the greater the energy gap


16.1 Intro to NMR Spectroscopy• The magnetic

moment of the electrons generally reduces the affect of the external field

• The more shielded a nucleus is with electron density, the smaller the α β energy gap. WHY?


16.1 Intro to NMR Spectroscopy• The amount of radio wave energy necessary for the α

β energy transition depends on the electronic environment for the atom

• When the α spins are flipped to β spins, the atoms are said to be in resonance

• The use of the term, “resonance” here is totally different from when we are talking about electrons in molecular orbitals

• How does NMR spectroscopy tell us about molecular structure?


16.2 Acquiring a 1H NMR Spectrum• NMR requires a strong magnetic field and radio wave

energy• The strength of the magnetic field affects the energy gap


16.2 Acquiring a 1H NMR Spectrum• The strong magnetic field is created when a high current

is passed through a superconducting material at extremely low temperature (≈4 Kelvin)

• The greater the current, the greater the magnetic field• In most current NMR instruments, a brief pulse of radio

energy (all relevant wavelengths) is used to excite the sample

• Each of the atoms is excited and then relaxes, emitting energy

• The emitted energy is recorded as a free induction decay (FID)


16.2 Acquiring a 1H NMR Spectrum• The FID contains all of the information for each atom• A mathematical treatment called a Fourier-transform

separates the signals so an individual signal can be observed for each atom that was excited

• Such an instrument is called an FT-NMR• Often multiple FIDs are taken and averaged together• Before analysis, NMR samples must be prepared neat or

in a liquid solution and placed in a small NMR tube• The sample is placed into the magnetic field and the

tube is spun at a high rate to average magnetic field variations or tube imperfections


16.2 Acquiring a 1H NMR Spectrum• Solvents are used such as chloroform-d. WHY?

• The magnet is super-cooled, but the sample is generally at room temp


• NMR spectra contain a lot of structural information– Number of signals– Signal location – shift– Signal area – integration– Signal shape – splitting pattern

16.3 Characteristics of a 1H NMR Spectrum


16.4 Number of Signals• Protons with different electronic environments will give

different signals• Protons that are homotopic will have perfectly

overlapping signals• Protons are homotopic if the molecule has an axis of

rotational symmetry that allows one proton to be rotated onto the other without changing the molecule

• Find the rotational axis of symmetry in each molecule below


16.4 Number of Signals• Another test for homotopic protons is to replace the

protons one at a time with another atom

• If the resulting compounds are identical, then the protons that you replaced are homotopic


16.4 Number of Signals• Protons that are enantiotopic will also have perfectly

overlapping signals• Protons are enantiotopic if the molecule has a plane of

reflection that makes one proton the mirror image of the other

• Find the mirror plane that splits each molecule below


16.4 Number of Signals• The replacement test is universal• It will work to identify any equivalents protons whether

they are homotopic or enantiotopic

• If the resulting compounds are enantiomers, then the protons that you replaced are enantiotopic


16.4 Number of Signals• If the protons are neither homotopic nor enantiotopic,

then the are NOT chemically equivalent• Perform the replacement test on the protons shown in

the molecule below

• How would you describe the relationship between the protons shown?

• Practice with SkillBuilder 16.1Copyright © 2015 John Wiley & Sons, Inc. All rights reserved. Klein, Organic Chemistry 2e 16-99

16.4 Number of Signals• There are some shortcuts you can take to identify how

many signals you should see in the 1H NMR1. The 2 protons on a CH2 group will be equivalent if there are

NO chirality centers in the molecule2. The 2 protons on a CH2 group will NOT be equivalent if there

is a chirality center in the molecule


• There are some shortcuts you can take to identify how many signals you should see in the 1H NMR3. The 3 protons on any methyl group will always be equivalent

to each other4. Multiple protons are equivalent if they can be interchanged

through either a rotation or mirror plane

16.4 Number of Signals



16.4 Number of Signals• Identify all the groups of equivalent protons in the

molecules below and describe their relationshipsO

O


• Recall that cyclohexane chairs have 6 equitorial and 6 axial protons

• Do the axial and equitorial protons have different electronic environments?

• Try the replacement test for an axial/equitorial pair• How many signals should you see for the molecule in

the 1H NMR?



• At room temperature, the chair interconversion occurs rapidly

• The NMR is not fast enough to see the individual structures, so the average is observed (1 signal)

• What might you expect to see if the temperature of the NMR sample were brought down to -100 C?



16.5 Chemical Shifts• Tetramethylsilane (TMS) is used as the

standard for NMR chemical shift• In many NMR solvents, 1% TMS is added

as an internal standard• The shift for a proton signal is calculated as a

comparison to TMS

• For benzene on a 300 MHz instrument


16.5 Chemical Shifts• The shift for a proton signal is calculated as a

comparison to TMS

• For benzene on a 60 MHz instrument

• The Hz of the signal is different in different instruments, but the shift relative to TMS (δ) is constant


16.5 Chemical Shifts• The shift for a proton signal is calculated as a

comparison to TMS

• The shift relative to TMS (δ) is a dimensionless number, because the Hz units cancel out

• Units for δ are often given as ppm (parts per million), which simply indicates that signals are reported as a fraction of the operating frequency of the spectrometer

• Most 1H signals appear between 0-10 ppm


16.5 Chemical Shifts• Early NMRs analyzed samples at a constant energy over

a range of magnetic field strengths from low field strength = downfield to high field strength = upfield

• Shielded protons required a stronger external magnetic field to be excited at the same energy as deshieldedprotons. WHY?


16.5 Chemical Shifts


• Current NMRs analyze samples at a constant field strength over a range of energies

• Shielded protons have a smaller magnetic force acting on them, so they have smaller energy gaps and absorb lower energy radio waves

Higher Energy Lower Energy

16-109

16.5 Chemical Shifts• Alkane protons generally give signals around 1-2 ppm• Protons can be shifted downfield when nearby

electronegative atoms cause deshielding. HOW?


16.5 Chemical Shifts• To predict chemical shifts, start with the standard ppm

for the type of proton (methyl, methylene, or methine)

• Use table 16.1 to adjust the ppm depending on proximity to certain function groups





• Handbooks can be used for functional groups beyond table 16.1



16.5 Chemical Shifts• Predict chemical shifts for all of the protons in the

molecule belowO

OH

Cl

Cl

Cl


16.5 Chemical Shifts• When the electrons in a pi system are subjected to an

external magnetic field, they circulate a great deal causing diamagnetic anisotropy


16.5 Chemical Shifts• Diamagnetic anisotropy means that different regions in

space will have different magnetic strengths• Why are some regions more shielded than others?


• The result of the diamagnetic anisotropy effect is similar to deshielding for aromatic protons. What about the other protons?

• Why does it appear that only one signal is given for all of the aromatic protons?



16.5 Chemical Shifts• The result of the diamagnetic anisotropy effect is

similar to shielding for protons that extend into the pi system

• Some of the protons in [14] Annulene appear at 8 ppmwhile others appear at -1 ppm. Which are which?


16.5 Chemical Shifts• Explain all

of the shifts in table 16.2

• Practice conceptual checkpoints 16.10


16.6 Integration• The integration or area under the peak quantifies the

relative number of protons giving rise to a signal• A computer will calculate the area of each peak

representing that area with a step-curve

• The curve height represents the integration Copyright © 2015 John Wiley & Sons, Inc. All rights reserved. Klein, Organic Chemistry 2e 16-120

16.6 Integration• The computer operator sets one of the peaks to a whole

number to let it represent a number of protons• The computer uses the integration ratios to set the

values for the other peaks

1.00 1.051.48 1.56


16.6 Integration• Integrations represent numbers of protons, so you must

adjust the values to whole numbers• If the integration of the first peak is doubled, the

computer will adjust the others according to the ratio

2.00 2.10

2.96 3.12


16.6 Integration• The integrations are relative quantities rather than an

absolute count of the number of protons• Predict the 1H shifts and integrations for tert-butyl

methyl ether

• Symmetry can also affect integrations• Predict the 1H shifts and integrations for 3-pentanone

• Practice with SkillBuilder 16.4Copyright © 2015 John Wiley & Sons, Inc. All rights reserved. Klein, Organic Chemistry 2e 16-123

16.7 Multiplicity• When a signal is observed in the 1H NMR, often it is split

into multiple peaks• Multiplicity or a splitting patterns results


16.7 Multiplicity• Multiplicity results from magnetic affects that protons

have on each other• Consider protons Ha and Hb

• We already saw that protons align with or against the external magnetic field

• Hb will be aligned with the magnetic field in some molecules. Other molecules in the sample will have Hbaligned against the magnetic field

• Some Hb atoms have a slight shielding affect on Ha and others have a slight deshielding affect


16.7 Multiplicity• The resulting multiplicity or splitting pattern for Ha is a

doublet

• A doublet generally results when a proton is split by only one other proton on an adjacent carbon


16.7 Multiplicity• Consider an example where there are two

protons on the adjacent carbon• There are three possible affects the Hb

protons have on Ha


16.7 Multiplicity• Half of the Ha atoms will not experience a

signal shift. WHY?• ¼ of the Ha atoms will be shielded and ¼

deshielded


16.7 Multiplicity• Ha appears as a triplet• WHY?• The three peaks in the triplet have

an integration ratio of 1:2:1• WHY?


• Consider a scenario where Ha has three equivalent Hb atoms splitting it

• Explain how the magnetic fields cause shielding or deshielding

16.7 Multiplicity


16.7 Multiplicity• Ha appears as a quartet• What should the integration ratios be

for the 4 peaks of the quartet?


16.7 Multiplicity• Table 16.3 shows how the multiplicity trend continues• By analyzing the splitting pattern of a signal in the 1H

NMR, you can determine the number of equivalent protons on adjacent carbons


16.7 Multiplicity• The trend in table 16.3 also allows us to predict splitting

patterns• Explain how the n+1 rule is used


16.7 Multiplicity• Remember three key rules

1. Equivalent protons can not split one another– Predict the splitting patterns observed for

1,2-dichloroethane2. To split each other, protons must be within a 2 or 3 bond

distance


16.7 Multiplicity• Remember three key rules

3. The n+1 rule only applies to protons that are all equivalent– The splitting pattern observed for the proton shown below

will be more complex than a simple triplet

– Complex splitting will be discussed later in this section


OH

H


16.7 Multiplicity• Predict splitting patterns for all of the protons in the

molecule below

O

O


16.7 Multiplicity• The degree to which a neighboring proton will shield or

deshield its neighbor is called a coupling constant• The coupling constant or J value is the distance between

peaks of a splitting pattern measured in units of Hz• When protons split

each other, their coupling constants will be equal

• Jab = Jba


• The coupling constant will be constant even if an NMR instrument with a stronger or weaker magnetic field is used

• Higher field strength instruments will give better resolution between peaks,

16.7 Multiplicity

because the coupling constant is a smaller percentage of the overall Hz available


16.7 Multiplicity• Sometimes recognizable

splitting patterns will stand out in a spectrum

• An isolated ethyl group gives a triplet and a quartet

• Note the integrations

• The triplet and quartet must have the same coupling constant if they are splitting each other


16.7 Multiplicity• A peak with an integration

equal to 9 suggests the presence of a tert-butyl group

• An isolated isopropyl group gives a doublet and a septet

• Note the integrations



16.7 Multiplicity• Complex splitting results when a proton is split by

NONequivalent neighboring protons

• If Jab is much greater than Jbc, the signal will appear as a quartet of triplets

• In the molecule shown, Hb is split into a quartet by Ha and into a triplet by Hc


16.7 Multiplicity• Complex splitting results when a

proton is split by NONequivalent neighboring protons

• If Jbc is much greater than Jab, the signal will appear as a triplet of quartets

• If Jbc is similar to Jab, the signal will appear as a multiplet


• If Jbc is equal to Jab, what type of patterns will be observed?


16.7 Multiplicity• Complex splitting results when a

proton is split by NONequivalent neighboring protons

• Predict the splitting patterns for (S)-pent-2-en-4-ol

OH

H


16.7 Multiplicity• Splitting is not observed for some protons. Consider

ethanol

• The protons bonded to carbon split each other, but the hydroxyl proton is not split


16.7 Multiplicity• The hydroxyl proton and other labile or exchangeable

protons undergo rapid exchange with trace amounts of acid. Show a reasonable mechanism

• Such exchange blurs the shielding/deshielding affect of the neighboring protons giving a singlet that is often broadened

• If ethanol is rigorously purified to remove traces of acid, then hydroxyl proton splitting is generally observed

• Aldehyde protons also often appear as singlet because their coupling constants are sometimes too small to cause observable splitting


16.7 Multiplicity• Signals for exchangeable protons such as those shown

below disappear completely when the 1H NMR sample is prepared for analysis in a deuterated solvent such as chloroform-d. WHY?

• Protic compounds have exchangeable protons

RO

H

O

H

O

RR

N

H

R


16.8 Predicting Expected 1H Spectra for a Compound

• Predict the chemical shift, integration, and splitting patterns for all of the protons in the following molecule

• Draw a spectrum for the molecule


O O


16.9 Using 1H Spectra to Distinguish Between Compounds

• The three molecules below might be difficult to distinguish by IR of MS. WHY?

• Explain how 1H NMR could distinguish between them



16.9 Using 1H Spectra to Distinguish Between Compounds

• Explain how 1H NMR could be used to distinguish between the two molecules below


16.10 Analyzing a 1H NMR Spectrum• With a given formula and 1H NMR spectrum, you can

determine a molecule’s structure by a 4-step process1. Calculate the degree or unsaturation or hydrogen deficiency

index (HDI). What does the HDI tell you?2. Consider the number of NMR signals and integration to look

for symmetry in the molecule3. Analyze each signal, and draw molecular fragments that

match the shift, integration, and multiplicity4. Assemble the fragments into a complete structure like puzzle

pieces



16.10 Analyzing a 1H NMR Spectrum• Consider the data below, and propose a structure for

the molecule• The formula is C7H13Cl• 1H NMR data: δ 5.3 (dq 1H); 5.1 (d 1H); 3.4 (s 2H); 2.0 (d

3H); 1.0 (s 6H) 1. Calculate

2. Consider the number of NMR signals and integration to look for symmetry in the molecule


16.10 Analyzing a 1H NMR Spectrum• Consider the data below, and propose a structure for

the molecule• The formula is C7H13Cl• 1H NMR data: δ 5.3 (dq 1H); 5.1 (d 1H); 3.4 (s 2H); 2.0 (d

3H); 1.0 (s 6H) 3. Analyze each signal, and draw molecular fragments that

match the shift, integration, and multiplicity

4. Assemble the fragments into a complete structure like puzzle pieces


16.11 Acquiring a 13C NMR Spectrum• Because 1H is by far the most abundant isotope of

hydrogen, 1H NMR signals are generally strong• 13C only accounts for about 1% of carbon atoms in

nature, so a sensitive receiver coil and/or concentrated NMR sample is needed

• In 1H NMR, shift, splitting, and integration are important• In 13C NMR, only the number of signals and the shift will

be considered


16.11 Acquiring a 13C NMR Spectrum• In 13C NMR, the 1H-13C splitting is often so complex that

the spectrum is unreadable• To elucidate the 13C spectrum and make it easier to

determine the total number of 13C signals, 13C NMR are generally decoupled

• In the vast majority of 13C spectra, all of the signals are singlets


16.12 Chemical Shifts in 13C NMR Spectra

• Compared to 1H, 13C atoms require a different frequency of energy to excite (resonate)

• Compared to the standard TMS, 13C NMR signals generally appear between 220 and 0 ppm

• Each signal on the 13C spectra represents a carbon with a unique electronic environment

• Planes and axes of symmetry can cause carbon signals to overlap if their electronic surroundings are equivalent



• Note how symmetry affect the number of signals for the molecules above

• How many 13C signals should be observed for the molecule below

OO



• Like 1H signals, chemical shifts for 13C signals are affected by shielding or deshielding



Documents

Organic Chemistry. SHE… · Klein, Organic Chemistry 2e • Molecular bonds can vibrate by stretching or by bending in a number of ways. 15.2 IR Spectroscopy • This chapter will