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• 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.

15.1 Introduction to Spectroscopy

• 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 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e15-1

• There are many wavelengths of light that cannot be observed with your eyes.



• When light interacts with molecules, the effect depends on the wavelength of light used.


• This chapter focuses on IR spectroscopy.Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e15-3

• 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

i Th i h ld b bl l


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).Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e15-4

• For each of the types of molecular motion/energy below, describe how it is quantized.– Rotation

Vib i


– Vibration

– Energy of electrons


• 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.


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

• This chapter will focus mostly

15.2 IR Spectroscopy

This chapter will focus mostly on stretching frequencies.

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


• WHY do objects emit IR light?

• WHY do some objects emit more IR radiation than others?

• WHERE does that light come from?


• 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



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• 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

15.3 IR Signal Wavenumber

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


• Rationalize the trends below using the wavenumber formula:


1.

2.Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e15-16

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


• Explain the regions above.Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e15-17

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


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


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• 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.Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e15-22

• 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.



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• 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.


• Practice with CONCEPTUAL CHECKPOINT 15.2.


• 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


j gthe –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


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.Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e15-30

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

I i i l


• 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 to 15.7.


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

15.5 IR Signal Shape

• O–H stretching signals are often quite broad.Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e15-33

• When possible, O–H bonds form hydrogen bonds (H‐bonds) that weaken the O–H bond strength.

• WHY does H‐bonding affect


gthe 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.


• 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.



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• 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.




• 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?




• 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.

I d h N H b d


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


• 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.



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• 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 spectra:

15.6 Analyzing an IR Spectrum

1. Focus on the diagnostic region (above 1500 cm‐1):a) 1600‐1850 cm‐1: check for double bonds.

b) 2100‐2300 cm‐1: check for triple bonds.

c) 2700‐4000 cm‐1: check for X–H bonds.

d) Analyze wavenumber, intensity, and shape for each signal.


• Often, the molecular structure can be identified from an IR spectra:2 F th

15.6 Analyzing an IR Spectrum

2. Focus on the 2700–4000 cm‐1

(X–H) region.

• Practice with SKILLBUILDER 15.1.


• 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 f h i

15.7 Using IR to Distinguish Between Molecules

success of such reactions.

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

• Practice with SKILLBUILDER 15.2.Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e15-45

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

15.7 Using IR to Distinguish Between Molecules

• Is IR analysis qualitative or quantitative?


1) H‐Br

2) Et‐OK

O3

(CH3)2SO

O

• Mass spectrometry (MS) is primarily used to determine the molar mass and formula for a compound:1. A compound is vaporized and then ionized.

2. The masses of the ions are detected and graphed.

f

15.8 Introduction to Mass Spectrometry

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

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

Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e15-47 Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e7 -48

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• 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?Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e15-49

• 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?


• The resulting fragments may undergo even further fragmentation.


• The ions are deflected by a magnetic field.

• Smaller mass and higher charge fragments as affected more by the magnetic field. WHY?

• Neutral fragments are not detected. WHY?


• Explain the units on the x and y axes for the MASS SPECTRUM for methane.

• The base peak is the tallest peak i h


in the spectrum.

• For methane the base peak represents the M+•.

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


• 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.


• MS 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


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: explosives detection

– Many More


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• In the mass spectrum 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.



• 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


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 an M+• peak with an odd mass.



• 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

15.10 Analyzing the (M+1)+• Peak

the presence of 13C in the sample. HOW?


• 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


ymolecule. HOW?

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

• Practice with SKILLBUILDER 15.3.


• 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?



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• 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.


• A thorough analysis of the molecular fragments can often yield structural information.

• Consider pentane.

15.12 Analyzing the Fragments

– Remember, MS only detects charged fragments.Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e15-62

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



• 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.


• 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.


• Alcohols generally undergo two main types of fragmentation: alpha cleavage and dehydration.



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• Amines generally undergo alpha cleavage:

• Carbonyls generally undergo McLafferty rearrangement:


• Carbonyls generally undergo McLafferty rearrangement:

• Practice with CONCEPTUAL CHECKPOINTs 15.25 through 15.28.


• High resolution (high‐res) MS 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

15.13 High Resolution Mass Spectrometry

– Using E=mc2, the more stable an atom’s nucleus, the less mass its nuclides exhibit. WHY?

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

• All atoms other than 12C will have a mass in amu that can be measured to four decimal places by a high‐res MS instrument.


• Note the exact masses and natural abundances below.



• 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 resolution (low‐res) MS?


• 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 with the highest natural abundance.

• What if the molecular ion is not observed?

• Practice with CONCEPTUAL CHECKPOINTs 15.19 and 15.30.


• MS is suited for the identification of pure substances.

• However, MS instruments are often connected to a gas chromatograph (GC) so mixtures can be analyzed.



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• GC‐MS gives two main forms of information:1. The chromatogram gives

the retention time.


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

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


• To be analyzed by EI‐MS, substances generally must be vaporized prior to ionization.

• Until recently (last 30 years), compounds that decompose before they vaporize could not be analyzed.

15.15 MS of Large Biomolecules

• 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?


• MS 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

15.16 Degrees of Unsaturation

of possible structures.

• Alkanes follow the formula below, because they are SATURATED.

• Verify the formula by drawing some isomers of pentane.


• Notice that the general formula for the compound,, 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?


• Consider the isomers of C4H6.

• How many degrees of unsaturation are there?

• 1 degree of unsaturation = 1 unit on the hydrogen deficiency index (HDI)



• 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?


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• 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:


• Calculating the HDI can be very useful. For example, if HDI=0, the molecule CANNOT have any rings, double bonds, or triple bonds.

• Propose a structure for a molecule with the formula C H O Th l l h h f ll i IR k


C7H12O. The molecule has the following IR peaks: – A strong peak at 1687 cm‐1

– NO IR peaks above 3000 cm‐1

• Practice with SKILLBUILDER 15.4.Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e15-80

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