Chapter 15
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?
15.1 Introduction to Spectroscopy
15.1 Introduction to Spectroscopy
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
• Molecular bonds can vibrate by stretching or by bending in a number of ways
15.2 IR Spectroscopy
• For each different bond, vibrational energy levels are separated by gaps (quantized)
• 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
• Rationalize the trends below using the wavenumber formula
1.
2.
• The wavenumber formula and empirical observations allow us to designate regions as representing specific types of bonds
• The region above 1500 cm-1 is called the diagnostic
region. WHY?
• The region below 1500 cm-1 is called the fingerprint
region. WHY?
15.3 IR Signal Wavenumber
• Analyze the diagnostic and
fingerprint regions in each spectrum.
• How are the two spectra similar/ different?
• 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
• 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
• 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
• 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
• 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
• Some IR signals are broad, while others are very narrow
• 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
15.5 IR Signal Shape
• In a sample with an intermediate concentration, both narrow and broad signals are observed. WHY?
15.5 IR Signal Shape
• Explain the cm-1 readings for the
• 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
• 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?
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
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?
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 spectra
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
15.6 Analyzing an IR Spectrum
• Often, the molecular
structure can be identified from an IR spectra
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
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
15.8 Into to Mass Spectrometry
• Mass spectrometry 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
• Can you think of ways to get an organic molecule to ionize?
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?
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
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?
15.8 Into to Mass Spectrometry
• Explain the units on the x and y axes for the mass spectrum for methane
• The base peak is the tallest peak in the spectrum
• For methane the base peak represents the M+•
• Sometimes, the M+• peak is
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
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
15.9 Analyzing the M
+•
Peak
• In the mass spec for benzene, the M+•
peak is the base peak
15.9 Analyzing the M
+•
Peak
• Like most compounds, the M+• peak for
pentane is NOT the base peak
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
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
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?
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
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
15.12 Analyzing the Fragments
• A thorough analysis of the molecular fragments can often yield structural information
• Consider pentane
• Remember, MS only detects charged
15.12 Analyzing the Fragments
• WHAT type of
fragmenting is
responsible for the “groupings” of
15.12 Analyzing the Fragments
• In general, fragmentation will be more prevalent when more stable fragments are produced
• Correlate the relative
15.12 Analyzing the Fragments
• Consider the fragmentation below
• All possible fragmentations are generally observed under the high energy conditions employed in EI-MS
15.12 Analyzing the Fragments
15.12 Analyzing the Fragments
• Amines generally undergo alpha cleavage
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 amu
scale is based on the mass of 12C
• All atoms other than 12C will have a mass in amu that
15.13 High Resolution Mass Spec
15.13 High Resolution Mass Spec
• Why are the values in table 15.5 different from those on the periodic table?
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
15.14 High Resolution Mass Spec
• MS is suited for the identification of pure substances
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
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?
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
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
15.16 Degrees of Unsaturation
• Consider the isomers of C4H6
• How many degrees of unsaturation are there?
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?
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?
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
• Explain why a completely nonpolar bond will not give a stretching signal in the IR spectra. Would you expect to see a signal for C-H stretching for a nonpolar molecule? Why or why not?
• Explain how IR might be used to qualitatively determine the degree of substitution when ammonia is treated
with excess bromoethane.
• How might you use EI GCMS to distinguish between constitutional isomers?
• Explain how an experiment involving isotopic labeling might be used to explore the type of fragmentation that occurs in the MS analysis of organic compounds.
