Infrared Spectroscopy, Mass Spectrometry, and NMR Spectroscopy

Infrared Spectroscopy, Mass Spectrometry, and NMR Spectroscopy

Introduction to Spectroscopy• Spectroscopy involves an interaction between matter

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

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

frequency

Introduction to Spectroscopy• Only a small segment of EMR spectrum is visible to us.

Introduction to Spectroscopy• Frequency & wavelength are inversely related: As wavelength (λ) get longer,

frequency (ν) gets lower, and vice-versa (c = λν). Higher frequency = higher energy.

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

on the wavelength of light used

Introduction to Spectroscopy• Matter exhibits both particle-like and wave-like properties• Matter on the molecular scale exhibits quantum behavior

– A molecule will only rotate or vibrate at certain rates (i.e., discrete energies)

• 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

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

IR Spectroscopy

We can’t see IR, but we have technology that can• 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

IR Spectroscopy

• 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

IR Spectroscopy

• IR energy is absorbed in some regions by bonds, but not others. The % transmittance decreases where energy is absorbed.

IR Spectroscopy

• Units for the wavenumber = inverse cm• ν = frequency and c = the speed of light• Wavenumber directly proportional to frequency: Higher frequency = higher

energy=higher wavenumber

IR Spectroscopy

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

IR Spectroscopy

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

• Shorter, stronger bonds, and smaller atoms, absorb at higher frequencies (ergo, at longer wavenumbers)

IR Signal Wavenumber

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

• As previously noted, shorter, stronger bonds and smaller atoms have higher wavenumber


• The region above 1500 cm-1 is called the diagnostic region, which has fewer absorptions and is easier to interpret.

• The region below 1500 cm-1 is called the fingerprint region. This region contains signals from stretching and bending of most single bonds and is more difficult to interpret. However, each compound has a unique fingerprint even if signals overlap.


DIAGNOSTIC REGION FINGERPRINT REGION

• Compare the IR spectra


• Compare the IR stretching wavenumbers below

• As s character of bond increases, orbitals get more compact, bonds get shorter and stronger, wavenumber gets larger.


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


• Tetrasubstituted alkenes and internal alkynes do not produce C-H signals in diagnostic region.

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


HO

• In general, conjugation of pi bonds weaken the carbonyl by increasing resonance structures, giving it more single bond character, resulting in lower wavenumber.


• The strength of IR signals can vary

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

IR Signal Strength

• 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 (increased intensity)• NP alkenes may produce no signal

IR Signal Strength

• 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, because there are many of them.

• 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

IR Signal Strength

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

• O-H stretching signals are often quite broad

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

• This leads to signal broadening• The O-H stretch signal will be narrow if a dilute solution

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

IR Signal Shape

• In a sample with an intermediate concentration, both narrow and broad signals are observed because some –OH groups are H-bonding, and some are not.

IR Signal Shape

• 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

• C.A.s produce an even broader –OH signal due to dimeric H-bonding.

IR Signal Shape

• Primary and secondary (but not tertiary) amines exhibit N-H stretching signals.

• 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?

IR Signal Shape

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, producing two different signals

Analyzing an IR Spectrum• Table 15.2 (p. 701) 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 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

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

• What can you conclude from this spectrum?

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?

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

Intro to Mass Spectrometry• Mass spectrometry (MS) used primarily to determine molar mass

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

• Electron impact (EI) ionization most common method for ionizing molecules

• The sample is bombarded with a beam of high energy electrons (1600 kcal or 70 eV). This ejects an electron and produces a radical cation.

Into to Mass Spectrometry• If the radical cation remains intact, it is known as the

molecular ion (M+•) or parent ion• Often, the molecular ion is unstable and may undergo

further fragmentation where 1 fragment carries the unpaired electron and the other fragment carries the charge.

• This process can generate many carbocation fragments

Into to Mass Spectrometry• The resulting fragments are then accelerated, and

charged ions are deflected by a magnetic field into a detector

• Radicals not detected• Smaller mass and higher charge fragments are affected

more by the magnetic field. • Cations separated by mass-to-charge ratio (m/z), and

usually have a charge of +1.

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 (100% relative abundance)

• The base peak isn’t always the same as the parent ion (M+•)

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

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

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

Analyzing the M+• Peak

• In the mass spec for benzene, the M+• peak is the base peak

• The M+• peak does not easily fragment for this compound


• 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– 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

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 isotope in the sample.

Analyzing the (M+1)+• Peak • The natural abundance of 13C is ~1.1%.

For every 100 molecules of CH4, ~1.1% contain 13C. For every 100 molecules of decane (10 C), ~11% contain 13C.

• Comparing the heights of the (M+1)+• peak and the M+• peak can allow you to estimate how many carbons are in the molecule, based on isotopic abundance of C.

𝑅𝑒𝑙 . 𝐴𝑏𝑢𝑛𝑑𝑎𝑛𝑐𝑒𝑜𝑓 (𝑀+1 )+•

𝑅𝑒𝑙 . 𝐴𝑏𝑢𝑛𝑑𝑎𝑛𝑐𝑒𝑀+• ×100=𝑋%

𝑋 %1.1%

=𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐶

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

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

Analyzing the (M+2)+• Peak

• 79Br=51% and 81Br=49%, so molecules with bromine often have equally strong (M)+• and (M+2)+• peaks

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

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

detects charged fragments

Analyzing the Fragments

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

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

more stable fragments (ions and radicals) are produced• Correlate the relative

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

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

• Tertiary carbocations more stable than secondary, etc. Should be evident in spectrum.


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

M-18 produced due to loss of water


• Amines generally undergo alpha cleavage

• Carbonyls generally undergo McLafferty rearrangement when H is present on gamma C

High Resolution Mass Spec• High Resolution MS allows m/z to be measured up to 4 decimal places• Masses generally not whole number integers

– 1 proton = 1.0073 amu and 1 neutron = 1.0086 amu• One 12C atom = exactly 12.0000 amu, because amu scale based on 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

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

element, high-res is better for distinguishing between compounds with nearly identical molar masses, when molecular ion is observed.

GC/MS

• MS is suited for the identification of pure substances• However, MS instruments are often connected to a gas

chromatograph so mixtures can be separated and analyzed

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)

1MS 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

• ESI is a “softer” ionizing technique.

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• Analysis of a molecule’s formula can yield a list of possible

structures, and indicate degrees of unsaturation• Alkanes follow the formula below, because they are saturated

• Adding 1 degree of unsaturation decreases the number of H atoms by two = 1 unit on the hydrogen deficiency index (HDI)

CnH2n+2

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?• A non-carbonyl oxygen does not affect the HDI.

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

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

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Infrared Spectroscopy, Mass Spectrometry, and NMR Spectroscopy