Spectroscopy Chemistry 3.2: Demonstrate understanding of spectroscopic data in chemistry (AS 91388)

Acknowledgements: The following notes were put together using information primarily supplied by: The Chemistry department of Victoria University of Wellington & The “Year 13 Chemistry Spectroscopy” workbook, written by Dr. Stephen McCracken.

Learning Objectives Understand general concepts of Spectroscopy Be familiar with some common forms of spectroscopy, used in molecular structure determination Be able to correctly interpret spectra Be able to use spectra to solve the structures of simple organic molecules

Origins of Spectroscopy The word spectrum originates with Newton’s dispersion of white light into a rainbow of colours with a prism. The spectrum of white light is the range of colours that make it up.

Each individual colour from the prism cannot be further separated into other colours. Each part of the spectrum has a characteristic frequency, wavelength and quantum energy. The spectrum is a continuum or range of frequency.

The visible spectrum is a small part of a much larger spectrum of electromagnetic radiation. This radiation can be thought of as transverse waves in perpendicular electric and magnetic fields.

3 types of spectra being considered: IR (infrared) spectroscopy 13C NMR (nuclear magnetic resonance) spectroscopy Mass spectrometry Alkanes, alkenes, alcohols, haloalkanes, amines, aldehydes, ketones, carboxylic acids, amides, acid chlorides and esters. Organic molecules we will consider:

Spectra Examples: IR spectra for propan-2-ol

13C NMR spectra for ethanol

Mass spectra of ethanol

How IR Spectroscopy works (RSC) [Click Here] Watch the video on Moodle 2 about how infrared spectroscopy works, supplied by the “Royal Society of Chemistry”. How IR Spectroscopy works (RSC) [Click Here] IR radiation is used to analyse vibrations in molecules. The frequency of vibration directly relates to the strength of bonding and the mass of the atoms, therefore different types of bonds (single, double, etc.) between various atoms have distinct absorption profiles.

[Frequency is measured in wave numbers; unit = cm-1] When IR radiation is passed through a sample, certain frequencies will be absorbed. The absorbed frequencies are those in which the molecule also vibrates. The vibrations are generally quite distinct for each molecule, therefore these help us to identify features of the sample molecule. [Frequency is measured in wave numbers; unit = cm-1]

Some important frequencies to remember: Broad absorptions in the range of 3000 – 3500 cm-1 indicate either N-H (NH2) or O-H bonds. (2 types of H-bonding situations) Sharp, intense absorptions in the range of 1600 – 1800 cm-1 indicate C=O (a carbonyl). A peak (or series of peaks) in the range of 3000 cm-1 indicate C-H bonds. (Present in virtually all samples we will look at, as we are dealing with Organic molecules)

Tasks: Read the information and study the example spectra on pages 6 – 9 of your workbooks. Complete “IR Problems”, pages 10 – 14 of your workbooks.

How NMR Spectroscopy works (RSC) [Click here] 13C NMR Spectroscopy Watch the video on Moodle 2 about how NMR spectroscopy works, supplied by the “Royal Society of Chemistry”. How NMR Spectroscopy works (RSC) [Click here] If a sample of compound is exposed to an external magnetic field, it can cause the nuclei of the compound’s atoms to line up. We can then pass photons (the particles in electromagnetic radiation) through this sample. This causes the charged particles (electrons and protons – and hence nuclei) to “spin” (or flip); moving them from a lower energy state to a higher one.

As the nuclei return to the lower energy state, the energy that is released, is released at slightly different frequencies. This helps up to obtain the NMR spectra. The type of NMR, e.g. 13C, 1H, 15N, etc. is named so due to the type of nuclei in the sample we are causing to spin. We will only consider 13C NMR.

Chemical Environments & Chemical Shifts To determine the chemical environment of EACH carbon in a compound, we look at the number of each type of group that is attached to it. If two carbons have the EXACT same number and type of groups attached to it, then we say that they are in the “same chemical environment”, in other words, they are chemically equivalent.

e.g. Propane: C #1 & #3 are chemically equivalent as they both have (3 x H) + (1 x CH2CH3) attached to them. C #2 is chemically different as it has (2 x CH2) + (2 x CH3) attached to it. 1-chloropropane: C #1, 2 & 3 are all chemically different:- C # 1 has (1 x Cl) + (2 x H) + (1 x CH2CH3) attached to it; C #2 has (1 x CH2Cl) + (2 x H) + (1 x CH3) attached to it & C #3 has (3 x H) + (1 x CH2CH2Cl) attached to it.

e.g.2 Butane: C #1 & #4 are chemically equivalent as they both have (3 x H) + (1 x CH2CH2CH3) attached to them. C #2 & #3 is chemically equivalent to each other, but different from #1 & #4; as they have (2 x H) + (1 x CH3) + (1 x CH2CH3) attached to them. Butan-1-ol: C #1, 2, 3 & 4 are all chemically different to one another. Can you explain why?

Chemical Shifts: Groups in a compound that are in the same chemical environment (are chemically equivalent), will have the same “shift” on an NMR spectra. This is shown by the peaks on the spectra. [Shifts are recorded in parts per million – ppm] Some examples of NMR spectra, with their shifts outlined, are on the following slides.

Ethyl ethanoate

Ethanol C1 C2

Heptan-1-ol

Bromomethane

Butane C2 & C3 C1 & C4

Tasks: Read the information and study the examples on pages 15 – 17 of your workbooks. Complete “Chemical Environments Problems”, pages 18 – 19 of your workbooks. Read the information on pages 20 – 21 of your workbooks. Complete “Chemical Shift Problems”, pages 22 – 23 of your workbooks.

Tasks: Read the information and study the examples on pages 24 – 27 of your workbooks. Complete “C-13 Spectra Problems”, pages 28 – 34 of your workbooks.

How MS Spectrometry Works (RSC), [Click Here] Mass Spectrometry Watch the video on Moodle 2 about how Mass Spectrometry works, supplied by the “Royal Society of Chemistry”. How MS Spectrometry Works (RSC), [Click Here] Mass Spectrometry measures the mass of the molecular ion, (i.e. the mass of the molecule, once an electron has been lost.) Remember, most of a molecule’s mass comes from the nuclei, therefore the loss of an electron does not affect its mass. Knowing the mass helps us to determine the molecular formulae.

[M+] Mass of Decane (and therefore the molecular ion [M+] = 142 [M+] can be identified at the far right of the spectrum The base peak always has a 100 % relative intensity On this spectrum, peaks (fragments) are seen every 14 mass units, as each CH2 unit has a mass of 14 [M+]

The Nitrogen Rule If a molecular ion (molecule) has an ODD number of nitrogens in it, then it will have an ODD numbered MASS. However, if there are an even number of nitrogens in the molecule, or if there are no nitrogens in the molecule, then the mass of the molecular ion will be even.

Base peak [M+] = 59 (odd number due to only 1 N in aminopropane)

Isotopes Molecular ions exist with atoms of different isotopes. e.g. Alkanes can be made from 12C or 13C. In nature, we usually find the molecular ions in the ratios of which these isotopes exist; e.g. about 1.1 % of carbon is 13C, so on the MS spectra, [M+] represents the molecular ion made from 12C and a very, very small peak at [M+]+1 representing the molecular ion made from 13C can be seen; (13 is one mass unit heavier than 12, therefore [M+]+1).

e.g.2 e.g.3 79Br and 81Br occur in an approximate 1:1 ratio. (81 – 79 = 2) Therefore the [M+] and [M+]+2 peaks representing molecular ions made from both of these isotopes can be seen in a 1:1 ratio (i.e. they will be the same height). Note: [M+]+1 will still be present for the 13C isotope. Note 2: 81Br is 2 mass units heavier than 79Br, so the isotopes of Br are at [M+] and [M+]+2. e.g.3 35Cl and 37Cl occur in an approximate 3:1 ratio. (37– 35 = 2) Therefore the [M+] and [M+]+2 peaks, representing molecular ions made from both of these isotopes can be seen in a 3:1 ratio (i.e. The [M+] peak for 35Cl will be 3 x as tall as the [M+]+2 peak for 37Cl).

Iodine in molecules Iodine does not have multiple isotopes and so we should not expect to see [M+]/ [M+]+1/ [M+]+2 peaks like in the last examples. Iodine has a mass of 127, so instead we can look for: A peak at 127 for the I+ fragment A peak that is 127 mass units less than [M+] (because I+ has fragmented) Sometimes a peak at 128 for HI (I+ has combined with a proton)

Tasks: Read the information and study the examples on pages 35 AND 38 – 39 of your workbooks. Complete “Mass Spectrum Problems”, pages 36 – 37 AND 40 – 41 of your workbooks. Read the information and study the examples on pages 42 – 44 of your workbooks about “Fragmentation Analysis” and in particular “Mass Spectrometry - What should you take away?”.