ch15 - Laney College4/11/2013 6 • Note the general strength of the C=O stretching signal vs. the...
Transcript of ch15 - Laney College4/11/2013 6 • Note the general strength of the C=O stretching signal vs. the...
4/11/2013
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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?
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• There are many wavelengths of light that cannot be observed with your eyes.
15.1 Introduction to Spectroscopy
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• When light interacts with molecules, the effect depends on the wavelength of light used.
15.1 Introduction to Spectroscopy
• 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
15.1 Introduction to Spectroscopy
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
15.1 Introduction to Spectroscopy
– Vibration
– Energy of electrons
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• 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
15.1 Introduction to Spectroscopy
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.
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• WHY do objects emit IR light?
• WHY do some objects emit more IR radiation than others?
• WHERE does that light come from?
15.2 IR Spectroscopy
• IR or thermal imaging is also used to detect breast cancer.
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• The energy necessary to cause vibration depends on the type of bond.
15.2 IR Spectroscopy
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• 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.
15.2 IR Spectroscopy
• 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.
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• In the IR spectrum below, WHAT is % transmittance and how does it relate to molecular structure?
15.2 IR Spectroscopy
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• Analyze the units for the wavenumber, .– ν = frequency and c = the speed of light
15.2 IR Spectroscopy
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• HOW are wavelength and wavenumber different?
• HOW are wavenumbers and energy related?
15.2 IR Spectroscopy
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• A signal on the IR spectrum has three important characteristics: WAVENUMBER, INTENSITY, and SHAPE.
15.2 IR Spectroscopy
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• 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?
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• Rationalize the trends below using the wavenumber formula:
15.3 IR Signal Wavenumber
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.
15.3 IR Signal Wavenumber
• 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?
15.3 IR Signal Wavenumber
• The region below 1500 cm‐1 is called the fingerprint region. WHY?
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• Analyze the diagnostic and fingerprint regions below.
15.3 IR Signal Wavenumber
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• Analyze the diagnostic and fingerprint regions below.
15.3 IR Signal Wavenumber
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• Compare the IR spectra.
15.3 IR Signal Wavenumber
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• Given the formula below and the given IR data, predict whether a C–H or O–H bond is stronger.
15.3 IR Signal Wavenumber
• 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.
15.3 IR Signal Wavenumber
• Are the differences due to mass or bond strength?
• Which bond is strongest, and WHY?
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• Note how the region ≈3000 cm‐1 in the IR spectrum can give information about the functional groups present.
15.3 IR Signal Wavenumber
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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.
15.3 IR Signal Wavenumber
• Practice with CONCEPTUAL CHECKPOINT 15.2.
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• Resonance can affect the wavenumber of a stretching signal.
• Consider a carbonyl that has two resonance contributors.
15.3 IR Signal Wavenumber
• If there were more contributors with C–O single bond character than C=O double bond character, how would that affect the wavenumber?
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• Use the given examples to explain HOW and WHY the conjugation and
15.3 IR Signal Wavenumber
j gthe –OR group affect resonance and thus the IR signal?
• Practice with CONCEPTUAL CHECKPOINT 15.3.
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• The strength of IR signals can vary.
15.4 IR Signal Strength
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• 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
15.4 IR Signal Strength
The oscillating dipole moment creates an electrical field surrounding the bond.
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• The more polar the bond, the greater the opportunity for interaction between the waves of the electrical field and the IR radiation.
15.4 IR Signal Strength
• 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
15.4 IR Signal Strength
• 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.
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• 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?
15.4 IR Signal Strength
• 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.
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• 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
15.5 IR Signal Shape
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.
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• 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 two O‐H stretching peaks.
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• 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.
15.5 IR Signal Shape
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• H‐bonding is often more pronounced in carboxylic acids, because they can forms H‐bonding dimers.
15.5 IR Signal Shape
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• For the molecule below, predict all of the stretching signals in the diagnostic region.
15.5 IR Signal Shape
• Practice with CONCEPTUAL CHECKPOINT 15.9.
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• Primary and secondary amines exhibit N–H stretching signals. WHY not tertiary amines?
• Because N–H bonds are capable of H‐bonding, their
15.5 IR Signal Shape
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?
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15.5 IR Signal Shape
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• 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
15.5 IR Signal Shape
• Instead, the two N–H bonds vibrate together in two different ways.
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• A single molecule can only vibrate symmetrically or asymmetrically at any given moment, so why do we see both signals at the same time?
15.5 IR Signal Shape
• Similarly, CH2 and CH3 groups can also vibrate as a group, giving rise to multiple signals.
• Practice with CONCEPTUAL CHECKPOINT 15.10.
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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.
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• 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.
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• 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?
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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?
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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).
15.8 Introduction to Mass Spectrometry
• 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?
15.8 Introduction to Mass Spectrometry
• 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?
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• The resulting fragments may undergo even further fragmentation.
15.8 Introduction to Mass Spectrometry
• 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?
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• Explain the units on the x and y axes for the MASS SPECTRUM for methane.
• The base peak is the tallest peak i h
15.8 Introduction to Mass Spectrometry
in the spectrum.
• For methane the base peak represents the M+•.
• Sometimes, the M+• peak is not even observed in the spectrum, WHY?
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• Peaks with a mass of less than M+• represent fragments:
15.8 Introduction to Mass Spectrometry
• 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.
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• 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
15.8 Introduction to Mass Spectrometry
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
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• Like most compounds, the M+• peak for pentane is NOT the base peak.
• The M+• peak fragments easily.
15.9 Analyzing the M+• Peak
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• 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
15.9 Analyzing the M+• Peak
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.
• Practice with CONCEPTUAL CHECKPOINT 15.19.
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• 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?
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• 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
15.10 Analyzing the (M+1)+• Peak
ymolecule. HOW?
• The natural abundance of deuterium is 0.015%. Will that affect the mass spectrum analysis?
• Practice with SKILLBUILDER 15.3.
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• 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
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• 79Br=51% and 81Br=49%, so molecules with bromine often have equally strong (M)+• and (M+2)+• peaks.
15.11 Analyzing the (M+2)+• Peak
• Practice with CONCEPTUAL CHECKPOINTs 15.23 and 15.24.
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• 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?
15.12 Analyzing the Fragments
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• In general, fragmentation will be more prevalent when more stable fragments are produced.
15.12 Analyzing the Fragments
• Correlate the relative stability of the fragments here with their abundances on the previous slide.
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• Consider the fragmentation below.
15.12 Analyzing the Fragments
• 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.
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• Alcohols generally undergo two main types of fragmentation: alpha cleavage and dehydration.
15.12 Analyzing the Fragments
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• Amines generally undergo alpha cleavage:
• Carbonyls generally undergo McLafferty rearrangement:
15.12 Analyzing the Fragments
• Carbonyls generally undergo McLafferty rearrangement:
• Practice with CONCEPTUAL CHECKPOINTs 15.25 through 15.28.
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• 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.
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• Note the exact masses and natural abundances below.
15.13 High Resolution Mass Spectrometry
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• 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
15.13 High Resolution Mass Spectrometry
below.
• Why can’t you use low resolution (low‐res) MS?
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• Using the exact masses and natural abundances for each element, we can see the difference high‐res makes.
15.13 High Resolution Mass Spectrometry
• 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.
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• 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.
15.14 High Resolution Mass Spectrometry
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• GC‐MS gives two main forms of information:1. The chromatogram gives
the retention time.
15.14 High Resolution Mass Spectrometry
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.
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• 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?
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e15-74
• 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.
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e15-75
• Notice that the general formula for the compound,, changes when a double or triple bond is
present:
15.16 Degrees of Unsaturation
• Adding a degree of unsaturation decreases the number of H atoms by two.
• How many degrees of unsaturation are there in cyclopentane?
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• 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
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• For the HDI scale, a halogen is treated as if it were a hydrogen atom.
15.16 Degrees of Unsaturation
• 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.
15.16 Degrees of Unsaturation
• How many degrees of unsaturation are there in C5H8BrN?
• You can also use the formula below:
Copyright 2012 John Wiley & Sons, Inc. Klein, Organic Chemistry 1e15-79
• 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
15.16 Degrees of Unsaturation
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