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Organic Chemistry:CHEM2322

Dr. Christopher J. O’Brien203 CRB, [email protected]

Structure DeterminationMS & IR

CHEM2322 © C. J. O’BrienSlide 1

• Classroom etiquette• Arrive on time if you are late you must quietly take the first available

seat• Please turn off your cell phones• If you wish to leave early you must inform me in advance and position

yourself close to the door. Then quietly leave• If you wish to ask a question you must raise your hand. If time allows I

will answer your question

• Make-up exams• There will be no make-up exams however allowances will be made for

serious documented reasons (family death, sickness etc.).• During exams you may not leave the examination room and return to

continue an exam (no bathroom breaks). If you leave you must hand in your exam.

Etiquette & Course Overview

© C. J. O’BrienSlide 2CHEM2322

Course Overview

© C. J. O’BrienSlide 3

• Please attend all the lectures• In future notes will be placed on WebCT

• My office hours are 2-330pm Tuesdays and Thursdays• I will not re-teach during office hours• You must bring worked examples so that I can trouble shoot

your problems• There will be 3 of 4 midterms (70%) and one final exam (30%)

• Please read each chapter before the lecture or immediately after• Also answer the questions in the text. During revision answer the

questions at the end of the chapter• Do not look at the answers until you have made significant effort• It is easy to trick yourself into thinking you know the solution by

looking at the answers

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How Do We Know?


• In CHEM 2321 I showed you structures of products which results from specificreactions. Some examples are:

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• However how do we know? Did I just guess? • Well no, there are tools we can use to determine the structures of a molecule• These are mass spectrometry (MS), infrared (IR) spectroscopy, nuclear magnetic

resonance spectroscopy (NMR) and ultraviolet spectroscopy• When these techniques are used in combination we will see we can get a great deal

of structural information and use it to identify the molecule

The Bigger Picture


• So why is it important to know the structure, why should we care?• In chemical research you will normally have a target molecule to make. However

you mostly likely will have to invent the route to make this molecule• This molecule may be a drug, catalyst or polymer for example, but how do you

know if you succeeded? Guess?• No: You’ll use the knowledge of the techniques just mentioned to find the structure

of the molecule(s) you just made• This brings me to the experiment cycle:• 1) Design an experiment to yield desired molecule• 2) Conduct the experiment• 3) Find out what happen!• This is where the techniques we will cover in the first two weeks or so will help

you!• 4) Plan your next move• 5) Design an experiment…and so on

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The Bigger Picture


• So what would we like to know about the molecule to draw a structure• 1) We’d like to know the mass (how big and what elements are in this thing!)• 2) What functional groups are present (is it an ester, amine etc..)• 3) Is there a conjugated system present (linked π-bonds, systems)• 4) How are these connected• Well each technique is like a piece of the jigsaw puzzle, which is the molecule’s

structure• So what technique tells us what•• Mass Spectrometry: Mass Spectrometry: What size and formula What size and formula •• Infrared Spectroscopy:Infrared Spectroscopy: What functional groups are thereWhat functional groups are there•• Ultraviolet spectroscopy: Ultraviolet spectroscopy: Is there a conjugated system presentIs there a conjugated system present•• Nuclear magnetic Resonance:Nuclear magnetic Resonance: What is the carbonWhat is the carbon--hydrogen frameworkhydrogen framework• Let’s begin with Mass Spectrometry………

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Mass Spectrometry


• Mass Spectrometry is a way to measure the mass and therefore the MW of a molecule. From this we can get the molecular formula

• In addition as the molecule fragments (breaks apart) we can get additional information about the structure

• There are many types of mass spectrometers based on how they determine the mass, below depicts one type called magnetic sector, others are quadrapole, ion trap and time of flight (TOF). How ever all contain an ion source, mass analyzerand a detector

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• In order to determine a mass of a molecular fragment it must be an ion, positive or negative

• Most commonly it is positive

• This is all I will cover

So How Do We Form the Ion?


• So how do we form this ion?• There are several method: Electron Impact ionization (EI), Fast Atom

Bombardment (FAB), Chemical Ionization (CI) and Electro-Spray (ESI) are the most common ones used by organic chemists

• However for this course we will only consider EI• So how does Electron Impact ionization work?

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• We take our compound and we vaporize it

• The we hit it with a stream of high energy electrons

• The energy of the beam is typically 70 eV (1 eV = ≈ 23 Kcal/mol or 90 kJ/mol)

• What is a C-C bond? 376 kJ/mol

Radical-cation is formed!

Consequences of Energy


• Due to the large amounts of energy transferred to the molecule it normally fragments into smaller pieces, some will be positively charged and some will be neutral

• The charged fragments can be controlled by the magnetic fields and are guided and focused so they can be detected. Neutral fragments cannot so they hit walls in the machine etc and are therefore not detected

• The spectrometry sorts the fragment but their mass to charge ration m/z since the charge is normally +1 this value is simply the mass of the fragment

• The data we get from a mass spectrometer is normally shown as a bar graph of m/z (X axis) and number of ions (intensity) on the y axis. The tallest peak is normally assign 100% intensity and is called the base peak

• A peak that corresponds to the unfragmented molecule ion is called the parent peak or molecular ion (M+)

• So let’s have a look at some..

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Mass Spectrum of Methane


Interpreting Mass Spectra


• So what information can we get from these spectra• Well the molecular weight, we can use this to see if it matches the product we

expect from the reaction. Hexane (MW = 86) 1-hexene (84) 1-hexyne (MW = 82) can easily be distinguished

• With some high resolution mass spectrometers (0.0001 amu, atomic mass units) we can determine the chemical formula

• For example C5H12 and C4H8O both have the MW of 72. However C5H12 has an exact mass of 72.0939 amu and C4H8O has an exact mass of 72.0575 amu

• A high resolution machine can easily tell them apart• Note that exact mass refers to molecules with specific isotopic compositions.

Therefore normally you use the mass of the most abundant isotope 1.00783 for 1H and 12.00000 for 12C etc…

• If you cannot see your M+ with EI then you would normally use a softer (gives the molecule less energy therefore less fragmentation) ionization technique for example CI or ESI

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Interpreting Mass Spectra


• Knowing the molecular weight means we can narrow down the choices of molecular formula

• A molecule with a m/z of 110 could yield a formula of C8H14, C7H10O or C6H10N2

• A further point is atoms have isotopes therefore we can use the exact mass to narrow the molecular formula to one. We can see the effect of isotopes if we look at the mass spectrum of methane again

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Small peak at M+1 due to 13C or 2H incorporation as their natural abundance is 1.10% and 0.015% respectively

• Once we know the formula we can calculate the level of unsaturation and from this we know how many double bonds, rings etc… we need in the structure and we can begin to think of structures

Interpreting Fragmentation Patterns


• Worked Problem 12.2

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Isotopic Patterns


• When these compounds mass spectra were measured the following M+ peaks were found

• Why do the ratio differ?• This is due to the natural abundance of

the isotopes of each of the halides• This is very useful for Br and Cl they

have indicative isotopic patterns in MSwhich can be used to confirm there presences in a molecule

• So what do these isotopic patterns look like?

• Let’s see

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Isotopic Patterns for Halogens


Fluorine19F (100%)

Chlorine (3:1):35Cl (75.78%) 37Cl (24.22%)

Bromine (1:1):79Br (50.69%) 81Br (49.31%)

Iodine:127I (100%)



• If all that MS gave was the mass, therefore the molecular formula, it would still be a great analytical method. However, how the molecule fragments also provides a great deal of information

• When we use EI each molecule has a unique fragmentation pattern like a finger print. The reason why the fragmentation patterns are so different is due to chemical structure so it is unlikely that two compounds with have identical patterns

• In fact the EI pattern is often used for identification (we’ve seen CSI). This works by a computer program comparing the spectrum against an know database for example National Institute of Standards and Technology (NIST) database

• When fragmentation occurs the radical cation in the case of EI breaks apart forming one cation fragment and one radical fragment

• These fragments can give you structural information about a molecule, unsurprisingly the way a molecule fragments, which part of the molecule becomes the cationic fragment and which the radical fragment, is govern by stability

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• Let’s consider the fragmentation of 2,2-dimethylpropane when it fragments it does so, so that the positive charge is on the tert-butyl group

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• Due to the stability of this cation it is the base peak and has an m/z of 57 • So how can we use fragmentation to get structural info: well you just learned that

the presence of a strong peak at an m/z of 57 could be a tert-butyl group• Let’s look at some more; hexane shows signals at 86 (molecular ion), 71, 57, 43,

and 29. So what are these fragments?



• MS Spectrum

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• Worked Problem 12.3

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Fragmentation Mechanisms


• Common fragmentation mechanisms: So we know that when we use EI we get a radical cation, most common forms are: [Detail mechanism on board: α-cleavage, C-Y cleavage and H-Y elimination]

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• Common fragmentation mechanisms of functional groups:• Alcohols: Fragment by α-cleavage and H-Y elimination; in this case dehydration

Fragmentation Mechanisms


• Common fragmentation mechanisms of functional groups:• Amines: Fragment by α-cleavage

• Carbonyl compounds: Fragment by α-cleavage and the McLaffertyrearrangement

Spectroscopy


• Infrared, ultraviolet, and NMR differ from MS in that they involve interaction of molecules with electromagnetic energy

• The next slides offer a quick reminder:

Spectroscopy


• Electromagnetic (light) radiation as you may remember has wave-particle duality. This means it can be considered (described) as a particle or a wave

• When we talk about light as a particle with use the term photon• As you may expect if you describe light as a wave, we use the terms wavelength (λ),

Frequency (ν), measured in Hz (Hz = s-1) and amplitude, the height of the wave

A Bit of Mathematics


• So how do we relate these values to each other?• Well remember from Gen. Chem. the speed of light in a vacuum is constant at 2.997

108 m/s with the overall equation λ (m) × ν (s-1) = c (m/s)

• We can rewrite this as λ = c/ν or ν = c/λ• Electromagnetic energy is transmitted in discrete amounts call quanta, this is govern

by the equation below

λhchc =∈=

• = Energy of one photon• h = Plank’s constants (J•s)• ν = Frequency (Hz)• λ = Wavelength (m)• c = Speed of light

A Bit of Mathematics


• The equation before basically says that the energy of a photon is directly proportional with its frequency (higher frequency higher energy). However inversely proportional with wavelength.

• Take home message high frequencies, small wavelengths mean high energy!• If we multiple the energy value by Avogadro’s number NA we arrive at numbers we

are more use to!

λλmolkJ

mhcNE A /1020.1

)(

4−×==

• E = Energy of mole photons at a specific wavelength• h = Plank’s constants (J•s)• ν = Frequency (Hz)• λ = Wavelength (m)• c = Speed of light

Shining Light on Molecules


• Ok so now we have a rough idea of the energy of light at variouswavelength/frequencies but as organic chemists we are interested on how this light interacts with organic molecules

• Remember light is one way we can a energy to a molecule!• So when we expose an organic molecule to light (shine light on it), the molecule may

absorb light of a specific wavelength and transmit light of other wavelengths. When this ocurrs in the visable range we see different colours!

• We can measure this and we produce an absorption spectrum below shows and absorption spectrum when we shine infrared light on ethanol



• So what happens when a molecule absorbs the radiation? • Well this energy is distributed over the molecule in some way, this may lead to a

bond stretching and bending, or for an electron to jump into a higher energy orbital• As you will see the different frequencies of light do different things and all tell us

structural information.• To answer that all important question, what did I just make!• Let’s first consider what happens when we shine infrared on a molecule• The region of the Infrared (IR) spectrum most interesting for organic chemists ranges

from 3 ×10-6 to 3 ×10-5 m• With frequencies expressed as wave numbers

)(1)(: 1

cmcmWavenumber

λυ =−



• Thus the IR region useful for chemists runs from 4000 cm-1 to 400 cm-1, using the equation on slide 27

• We can calculate (E=(1.20 kJ/mol)/λ) that the IR radiation ranges from 48 – 4.80 kJ/mol in energy



• Molecules have energy and this cause bonds to stretch and bend as well as other vibrations to occur some allowed vibrations are shown below

• Remember energy in molecules is quantized this means that a molecule can only stretch and bend at specific frequencies

• One way to picture this is shown below:

• When a molecule is irradiated with electromagnetic radiation energy is absorbed if the frequency of the radiation matches the frequency of the vibrating bond

• The result of the energy absorption results in the bonds stretching and compressing a little more that before absorption

• Because each frequency absorbed by a molecule corresponds to a specific molecular motion, we can identify these by measuring the IR spectrum

• With this information we can get structure information

A Bit of Theory


• We can think of a bond being a spring between two atoms as shown below

• This means the stretching frequency of the bond can be approximated by Hooke’s law.

wherek is the force constantm is the massν is the frequency of the vibration

A Bit of Theory


• However as mentioned vibration energy is quantized, it must follow certain rules

• This means that the lowest energy level E0 = ½ hν the next E1 = 3/2 hν and so on. However this rule can sometimes be over come and leads to overtones

A Bit of Theory


• However a molecule is not a harmonic oscillator it is an anharmonic oscillator• This is because a bond can come apart and cannot be compress beyond a certain

point. Note how the allowed transitions become closer together in the anharmonicoscillator as the energy goes up therefore calculations can over predict.

• For a diatomic the following equation has been derived,

• f is the force constant of the bond

A Bit of Theory


• Worked example:

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Interpretation of IR Spectra


• By obtaining the IR spectra we can see the molecule motion and the correlate this to functional groups. Therefore IR is used to see what functional groups are in a molecule

• The Full interpretation of and IR spectrum is difficult due to its complexity however the region between 4000 cm-1 and 1500 cm-1 has indicative bands of function groups

• Between 1500-400cm-1 is called the fingerprint-region and can be used to identify compounds

• So let’s have a look at some • C=O absorption of a ketone range 1680 - 1750 cm-1

• O-H absorption of a alcohol range 3400 - 3650 cm-1

• C=C absorption of an alkene ranges 1640 - 1680 cm-1

• Intensity of absorption depends on the dipole moment the higher the dipole moment the higher the absorption no dipole, no absorption!

• Other important shifts are in your handout and overleaf…..

IR Absorptions


IR Absorptions


IR Spectra


IR of Hydrocarbons


• Alkanes: IR spectra of these are fairly uniform as only C-H and C-C bonds are present

• Alkenes: Interestingly it is possible to use IR to determine substitution patterns on C=C as for mono and di-substitute alkene there is a =C-H out of plane bend between 700-1000 cm-1


IR Spectra


IR Spectra


Functional Group Regions


What Bands Can You See?

Other Functional Groups


• Alcohols: IR spectra of these are fairly uniform as only C-H and C-C bonds are present. Are broad if hydrogen-bonding

• Amines:N-H sharp less intense that OH

• Aromatic:

Other Functional Groups


• Carbonyls:

Summary

CHEM 2322 © C. J. O’BrienSlide 45

• The structure of a molecule is determined by spectroscopic methods• Two of these are Mass Spectrometry (MS) and Infrared Spectroscopy (IR)• In MS ions are first formed by collision with a high energy electron beam• The ion may fragment during this process, all ion are sorted according to their

mass-to-charge ratio m/z• The ionized molecule is called the molecular ion, M+, and its weight is that of the

molecule being analyzed• Structure clues can be obtained from the fragmentation patterns• Infrared spectroscopy (IR) involves the interaction of the molecule with

electromagnetic radiation, during this process certain frequencies are absorbed• Since every function group has a characteristic combination of bonds every

functional group has unique IR absorptions• For example C=O, C=C and C≡C-H all have unique bands• By running an IR it is possible to see what functional groups are present

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