Unit 11:Data processing and analysis.

A. Infrared spectroscopyB. Mass spectrometryC. X-ray diffraction/crystallographyD. H NMR

A. Infrared spectroscopy

What is infrared?

An electromagnetic wave

Energy of electromagnetic radiation is carried in discrete packets of energy called photons or quanta.

E = hνE = energy of a single photon of radiationh = 6.63 x 10-34 Js (Plank’s constant)ν = frequency of the radiation

Example: Calculate the energy of a photon of visible light with a frequency of 3.0 x 1014 s-1. Express in kJ mol-1.

E = hνE = (6.63 x 10-34 Js)(3.0 x 1014 s-1)E = 1.989 x 10-19 J

1000J1kJ

molphotons106.02

photonJ101.989 2319

1molkJ 120

Note: . ν = c = 3.0 x 108 ms-1

Type of em radiation Typical f (s-1) Typical (m)

Radio waves (low energy) 3 x 106 102

Microwaves 3 x 1010 10-2

Infrared (IR) 3 x 1012 10-4

Visible (ROYGBIV) 3 x 1015 10-7

Ultraviolet (UV) 3 x 1016 10-8

X-rays 3 x 1018 10-10

Gamma rays (high energy) > 3 x 1022 < 10-14

Thus, ν= c/

Wavenumber

c=λν 1/λ = ν/c

In IR spectroscopy, the frequency of radiation is often measured as number of waves per centimeter (cm-1), also called wavenumber.

Example: Calculate the wavenumber in cm-1 for an IR wave with a frequency of 3 x 1013 s-1.

Infrared spectroscopy

normal vibration vibration having absorbed energy

By measuring the IR spectrum of a molecule we can determine what kind of motion the molecule has and therefore what kind of bonds are present in the molecule.

A bond will absorb radiation of a frequency similar to its vibration

Light atoms vibrate at higher freq. than heavier atoms;

they absorb IR radiation of shorter wavelength ( more energy)

Multiple bonds vibrate at higher freq. than single bonds;

they absorb IR radiation of shorter wavelength ( more energy)

Symmetricalstretching

Antisymmetricalstretching

Scissoring

Rocking Wagging Twisting

Infrared has the right energy to be absorbed by the polar bonds of a molecule . The IR radiations ( specific energy or frequency ) make the bonds stretch, bend, or vibrate.

Using IR to excite molecules

Not all vibrations absorb IR.

For absorption, there must be a change in bond polarity (dipole moment) as the vibration occurs.

Thus, diatomic gas molecules such as H2, Cl2 and O2 do not absorb IR.

Vibrations of H2O, SO2 & CO2

Molecule Asymmetrical stretching

Symmetrical stretching

Symmetrical bending

H2O

SO2

CO2

O OS

- -

+

IR active

O OS

- -

+

IR active

O OS

- -

+

IR active

H HO

+ +

-

IR active

H HO

+ +

-

IR active

H HO

+ +

-

IR active

O OC- -+

IR active

O OC- -+

IR inactive

O OC- -+

IR active

Matching wavenumbers with bonds

“fingerprint region”lots of overlap, so

not very useful

broad and strong

very strong

broad and strong

Usually sharper than OH

Data Booklet -table 26

https://www.youtube.com/watch?v=DDTIJgIh86E

IR spectrum of ethanol, CH3CH2OH

IR spectrum of ethyl ethanoate, CH3COOCH2CH3

“fingerprint region”C=O

C-H

FINGERPRINT REGION

The 1400 cm-1 to 800 cm-1 range is the “fingerprint” region : its the pattern is characteristic of a particular compound

IR SPECTRUM OF A CARBONYL COMPOUND

carbonyl compounds show a sharp, strong absorption between 1700 and 1760 cm-1

this is due to the presence of the C=O bond

IR SPECTRUM OF AN ALCOHOL

alcohols show a broad absorption between 3200 and 3600 cm-1

this is due to the presence of the O-H bond

IR SPECTRUM OF A CARBOXYLIC ACID

carboxylic acids show a broad absorption between 3200 and 3600 cm-1

this is due to the presence of the O-H bond they also show a strong absorption around 1700 cm-1

this is due to the presence of the C=O bond

http://undergrad-ed.chemistry.ohio-state.edu/anim_spectra/index.html

Practice some more spectra

IR spetra of butanal and butanone

B.Mass spectrometryB. Mass spectrometry

The Mass Spectrometer

http://www.youtube.com/watch?v=J-wao0O0_qM&feature=related

How it works:The sample is bombarded with a stream of high energy electrons. The collision is so energetic that it causes the molecule to break up into different fragments (ions).

The fragments ( + ions) of a particular mass are detected and a signal is sent to a recorder. The strength of the signal is a measure of the number of ions with that charge/mass ratio that are detected.

Fragmentation Patterns : evidence for the structure of the compound

The largest mass peak corresponds to a parent ion passing through the instrument unscathed, but other ions produced as a result of this break up are also detected.

For each fragmentation, one of the products keeps the + charge and will be detected.

Generally the most stable + ion is formed

CH3-CH2-OH CH3-CH2+ + OH

peak 29 no peak 17

Example: ethanol

mass/charge

rela

tive

abun

danc

e

0 6030

100

0

1529

31

45

46

Example: ethanol

Example: ethanol

Note: This fragmentation will yield either CH3

+ and CH2OH or CH3 and CH2OH+, yielding peaks at both 15 and 31

Data table 28-Mass spectral of fragment lost.

Mr loss of…

15 CH3

17 OH

29 C2H5 or CHO

31 CH3O

45 COOH

C . X-ray diffraction / X-ray crystallography

http://www.theguardian.com/science/video/2013/oct/09/100-years-x-ray-crystallography-video-animation  

When X-rays shine on a crystal (orderly structure), they are reflected and produce an ordered diffraction pattern.

The diffraction pattern produced by X-rays helps determine

the electron density of the crystal.

As the electron densities are related to the element’s electron configuration, we can also determine the identity of the atoms.

Note: H atoms have a very low electron density ( 1e) and are not visible on the X-ray diffraction pattern.

http://www.rsc.org/learn-chemistry/resource/res00000020/computational-chemistry#!cmpid=CMP00001685

Nuclear magnetic resonance (NMR) spectroscopy

D. Nuclear magnetic resonance. H NMR spectroscopy.

Nuclear magnetic resonance (NMR) spectroscopy

 

When an external magnetic field is applied to a molecule the energy of its spinning protons splits in 2 separate levels. The spin aligned with the magnetic field will be at a lower energy state. The spin aligned against the magnetic field will be at a higher energy state.  We can then get a proton change its spin from with the magnetic field to against the magnetic field by providing the right amount of energy.

The moment the proton returns to its lower energy level and reverses its spin, it gives out the energy that shows as a peak - resonance- in the NMR spectrum. 

The HNMR spectrum: its interpretation

https://www.youtube.com/watch?v=gaGUpACXijE

Chemical shifts vs TMS standard (tetramethylsilane)

Different H groups ( types)

Integration Number of H in the group = area under the peak

Multiplicity of peak(splitting)

Number of adjacent H

See Data Table 27: H NMR data

The TMS standard-All 12 H’s are in identical chemical environments, so one signal is recorded = zero point on the scale

-The H are very shielded by the electrons of C that is more electronegative that silicon. Therefore they experience the strength of the magnetic field the least.

-The signal doesn’t interfere with the signal given by H bonded to Carbon. The less the H will be shielded in a group, the

greater the shift will be.

5 4 3 2 1 0 d

CHEMICAL SHIFTS

3 environments = 3 signals

Triplet d = 3.4Sextet d = 1.9Triplet d = 1.0

Signal for H’s on carbon 3 is shifted furthest downfield from TMS due to proximity of the electronegative halogen

1 2 3

TMS

1H-NMR spectroscopy: 1-bromopropane

5 4 3 2 1 0 d

INTEGRATION

Area ratio from relative heights of integration lines = 2 : 2 : 3

Carbon 1 3Carbon 2 2Carbon 3 2

1 2 3

2

2

3

TMS

1H-NMR spectroscopy: 1-bromopropane

5 4 3 2 1 0 d

SPLITTING

SPLITTING PATTERNCarbon 1

Chemically different hydrogen atoms on adjacent atoms = 2

2 + 1 = 3

The signal will be aTRIPLET

1 2 3

1

TMS

1H-NMR spectroscopy: 1-bromopropane

5 4 3 2 1 0 d

SPLITTING

SPLITTING PATTERNCarbon 2

Chemically different hydrogen atoms on adjacent atoms = 5

5 + 1 = 6

The signal will be aSEXTET

1 2 3

2

TMS

1H-NMR spectroscopy: 1-bromopropane

Peaks Three different signals as there are three chemically different protons.Shift Signals are shifted away from TMS signal, are nearer to the halogen ( H is deshielded more by Br ).Integration The integration lines show that the ratio of protons is 2:2:3Splitting Signals include a triplet (d = 1.0) sextet (d = 1.8) triplet (d = 3.4)

The signals due to the protons attached to carbon ...

C1 triplet (d = 1.0) coupled to the two protons on carbon C2 ( 2+1 = 3 ) C2 sextet (d = 1.8) coupled to five protons on carbons C1 and C3 ( 5+1 = 6 ) C3 triplet (d = 3.4) coupled to the two protons on carbon C2 ( 2+1 = 3 )

4 3 2 1 0 d

SUMMARY

TMS

1 2 3

1H-NMR spectroscopy: 1-bromopropane