11.3 Spectroscopy

*CONCEPT: The degree of unsaturation (IHD) can be used to determine from a molecular formula the number of rings or multiple bonds in a molecule.

*unsaturated compound = a compound that contains C-C double/triple bonds

*The IHD can be determined from:

A. the structure; or,

B. the molecular formula of the compound

C

*In order to deduce the IHD for the generic molecular formula, C_cH_hN_nO_oX_x, where X is a halogen atom (F, Cl, Br, or I), we can use the following expression:

IHD = (0.5) (2c + 2 – h – x + n)

So, for C₄H₈O_2;

c = 4 ; h = 8 ; n = 0 ; o = 2 ; x = 0 à IHD = (0.5)(8 + 2 – 8 – 0 + 0) = 1….therefore, the molecule contains either one double bond or one ring.

Deduce the IHD of the following molecules using the formula.

(a) C₁₇H₂₁NO₄ (Cocaine) (b) C₂₇H₄₆O (Cholesterol)

(c) C₆H₇N (aniline) (d) C₁₅H₁₀ClN₃O₃ (clonazepam)

Mass Spectrometry gives information about the molecular mass of a compound and can also give important structural information if used with other spectroscopic techniques like IR and NMR.

When organic molecules are put into the mass spec, electrons cause the molecule to ionize forming a molecular ion. Unlike atoms, though, those whizzing electrons also cause the molecules to break down into smaller pieces.

These pieces can give us structural information about the molecule which can be very useful.

A mass spectra for a molecule like Pentane: CH₃CH₂CH₂CH₂CH₃ might look like:

· From the spectra you can figure out that the molecular mass is about 72 g/mol since this is the largest fragment formed. (This isn’t always true. If the molecular ion is unstable than this could be just a fragment.)

· The other peaks in the spectra are fragments.

The relative molecular weight of common fragments (Mr-X)⁺ are:

1. (M_r-15)⁺, which is methyl group (CH₃)

• The mass of one carbon and 3 hydrogen atoms add up to 15 amu

• If two peaks differ by 15 units, this is often an indication that the fragments only differ by a methyl group

2. (M_r-29)⁺, which is either C₂H₅ or CHO

• If you add up the mass of 2 carbons and 5 hydrogens you get 29 or

• If you add up the mass of 1 carbon, one Hydrogen and one oxygen you get 29

• If you have a peak at 29 you can’t know for sure if it’s C2H5 or CHO unless you’ve also done an Infrared spectra and identified the presence (or lack) of a carbonyl group (C=O)

3. (M_r-31)⁺, CH₃O

4. (M_r-45)⁺, COOH

5. Fragments that differ by only 1 or 2 mass unit differ by a hydrogen atom or 2.

· Since we know the molecule is pentane, which doesn’t have a carbonyl group, the peak at 29 represents a C₂H₅⁺ fragment.

· The peak at 57 differs from 72 by 15 mass units, so probably the fragment at 57 has lost a methyl group and is the fragment CH₃CH₂CH₂CH₂⁺

· The peak at 43, we kind of have to infer…it has to have less than 4 carbons since the mass of 4 carbons is 48 mass units. If it’s 3 carbons, it must have 7 Hydrogen atoms to add up to 43 mass units. Therefore, the fragment is probably CH₃CH₂CH₂⁺

· The smaller peaks that are only 1 or 2 different from the larger peaks are the same fragments that have lost one or 2 hydrogen atoms.

Let say you figured out that you have a pentanone molecule but you don’t know on which carbon the carbonyl group is located. You can use the fragmentation patter from a mass spec to figure it out.

Let’s say that the spectra gave peaks at 29, 57, and 86

If the carbonyl was on the 2^nd carbon: CH₃-CO-CH₂CH₂CH₃

Common fragments would be:

1. CH₃CO = M_r-43

2. CH₃CH₂CH₂CO = M_r-71

If the carbonyl was on the 3^rd carbon: CH₃CH₂-CO-CH₂CH₃

The most common fragment would be:

1. CH₃CH₂CHO = M_r-57

Since the peak at 57 matches the peak expected for pentan-3-one, this carbonyl must be on the third carbon.

What does the height of the peak tell you?

· The height of the peak tells you which fragment is the most stable.

· This is useful for determining the presence of branched structures. For example, a tertiary molecular ion would be more stable than a primary ion.

- the study of the way matter interacts with radiation

- Various regions of the EMS and techniques involving spectroscopy are used to identify the structure of substances

- Widely used in forensic science

A) *INFRARED SPECTROSCOPY (IR)

- IR radiation causes certain bonds in a molecule to vibrate and thereby provide information about the type of functional groups / bonds present in the molecule.

B) *MASS SPECTROSCOPY (MS)

- Used to determine relative atomic and molecular masses.

- The fragmentation pattern can be used as a fingerprint technique to identify unknown substances or for evidence for the arrangements of atoms in a molecule.

C) *PROTON NUCLEAR MAGNETIC RESONANCE (¹H NMR)

- Used to show the chemical environments of hydrogens / protons in a molecule and so gives vital structural information.

- Radiowaves can cause nuclear transitions in a strong magnetic field because radiowaves can be absorbed by certain nuclei, which causes their spin state to change. This forms the basis of NMR.

D) X-RAY CRYSTALLOGRAPHY

- High energy X-rays remove electrons from inner energy levels of atoms.

- Resulting diffraction patterns can lead to information about bond distances and bond angles in a structure.

E) ULTRAVIOLET-VISIBLE SPECTROSCOPY

- This type of radiation gives rise to electron transitions, and thereby provide information about the energy levels in an atom or molecule.

F) MICROWAVES

- This type of radiation causes molecular rotations and can give information on bond lengths.

*Infrared Spectroscopy, Mass Spectroscopy and Proton Nuclear Magnetic Resonance Spectroscopy = focus of unit

OVERVIEW:

- Infrared radiation = __________ energy radiation…not enough energy to result in _______________ transitions

- Enough energy, however, to cause molecular vibration of certain groups of molecules and their ______________.

- The absorption of particular _____________________ (frequencies) helps the chemist identify the bonds/functional groups in a molecule.

- Different molecules absorb at different (infrared) frequencies because the energy required to execute a vibration will depend on the bond ______________.

- IRS is often the starting point for the organic chemist into the detective work of analytical chemistry.

- The basis of IRS is the spring model.

MAIN IDEA:

Absorption of a particular wavenumber (frequency) of infrared radiation helps the chemist identify the bonds in a molecule.

BONDS AS SPRINGS

- Every covalent bond is considered as a ___________________.

- Springs can be stretched (both symmetrically and asymmetrically), bent or twisted, giving rise to a distortion.

- Hooke’s Law:

*lighter atoms vibrate at higher frequencies

-IR absorptions are typically cited as the ________________ of the wavelength (). This is the _____________ and has units of ________.

-Therefore: smaller the wavelength absorbed = larger frequency absorbed = higher energy absorbed = LARGER WAVENUMBER

Polyatomic Species – may have several different modes of vibration

However, for a covalent bond to absorb IR radiation (and have a wavenumber) there must be a change in the molecular dipole moment.

Binary, nonpolar molecules do not absorb IR radiation and are deemed IR inactive.

For example.

- An IR spectrum is a plot of the percentage______________________, %T, versus the ________________ (in cm^-1).

-The lower the %T, the ___________ the absorption.

- Downward spikes in the IR spectrum represent absorbed _________________. The C=C absorption in _____________ typically occurs in the range 1620-1680 cm^-1.

For example.

DATA BOOKLET INFRARED DATA

BOND ORGANIC MOLECULES WAVENUMBER (cm^-1) INTENSITY

C-I Iodoalkanes 490-620 Strong

C-Br Bromoalkanes 500-600 Strong

C-Cl Chloroalkanes 600-800 Strong

C-F Fluoroalkanes 1000-1400 Strong

C-O Alcohols, esters, ethers 1050-1410 Strong

C=C Alkenes 1620-1680 Medium-weak, multiple bands

C=O Aldehydes, ketones, carboxylic acids, esters 1700-1750 Strong

C≡ C Alkynes 2100-2260 Variable

O-H Hydrogen bonding in carboxylic acids 2500-3000 Strong, very broad

C-H Alkanes, alkenes, arenes 2850-3090 Strong

O-H Hydrogen bonding in alcohols/phenols 3200-3600 Strong, broad

N-H Primary amines 3300-3500 Medium, two bands

a) IR Spectrum of BUTANOIC ACID

1. The infrared spectrum was obtained from a compound and showed absorptions at 2100 cm^-1, 1700 cm^-1, and 1200 cm^-1. Identify the compound.

A. CH₃COOCH₃ B. C₆H₅COOH

C. CH₂=CHCH₂OH D. CH≡CCH₂CO₂CH₃

2. A molecule absorbs IR at a wavenumber of 1720 cm^-1. Which functional group could account for this absorption?

I. ALDEHYDES II. ESTERS III. ETHERS

A. I only B. I and II C. I, II, and III D. none of the above

3. An unknown compound has the following mass composition: C = 40%; H = 6.7%; O = 53.3%. The largest mass recorded on the mass spectrum of the compound corresponds to a relative molecular mass of 60.

(a) Determine the empirical and molecular formulas of the compound.

(b) Deduce the IHD of the compound.

(c) The IR spectrum shows an absorption band at 1700 cm^-1 and a very broad band between 2500 and 3300 cm^-1. Deduce the molecular structure of the compound.

4. Cyclohexane and hex-1-ene are isomers. Suggest how you could use infrared spectroscopy to distinguish between the two compounds.

5. The intoximeter, used by the police to test the alcohol levels in the breath of drivers, measures the absorbance at

2900 cm^-1. Identify the bond which causes ethanol to absorb at this wavenumber.

6. A molecule has the molecular formula C₂H₆O. The infrared spectrum shows an absorption band at 1000-1300 cm^-1, but no absorption bands above 3000 cm^-1. Deduce its structure.

OVERVIEW:

- Another analytical tool used to help determine the structure/identity of organic compounds.

- Perhaps the most important structural technique available to an organic chemist.

- Gives information regarding different chemical environments of ___________________ atoms in a molecule.

- Protons in water molecules within human cells can be detected by magnetic resonance imaging (MRI), giving a three-dimensional view of organs in the human body.

-A spinning proton behaves like a very tiny bar_________________– it produces a _________________ field with a corresponding ‘North’ and ‘South’ pole.

-When a proton (a tiny ‘bar magnet’) is placed in a stronger magnetic field it can exist in _______ possible spin states

(i.e., ________________ energy levels); it will either:

*Note that prior to exposure to an external magnetic field the spin states are random and all have equal energy.

-The energy difference, ∆E, between the alpha and beta spin states is very small – wavelengths in the ___________________region of the EMS. The stronger the applied magnetic field, the larger the ∆E. The ∆E also depends on the chemical environments of the hydrogen atoms.

- In practice, a sample is placed in an ___________________. The field strength is varied until the radio waves have the exact frequency needed to make the nuclei flip over and spin in the ________________direction. This is called _________________ and can be detected electronically and recorded in the form of a spectrum.

*See Data booklet for common shift values.

1. CHEMICAL SHIFTS / # OF PEAKS

Sample Shift Values

- Hydrogen nuclei in different chemical environments have different chemical shifts (due to varying ΔE b/w and spin states).

- Therefore the number of signals on a ¹H NMR spectrum shows the different chemical environments in which the hydrogen atoms are found.

- In a ¹H NMR spectrum, the position of the NMR signal relative to a standard (tetramethylsilane, TMS) is termed the chemical ________________, δ.

- The chemical shift of the proton is expressed in parts per million, ppm.

- δ for TMS is ______ ppm.

Example1: Ethanoic acid, H₃CCOOH Example 2: Ethanol, CH₃CH₂OH

- The position of the ¹H NMR signal depends on the strength of the magnetic field, therefore frequencies can be variable, as no two magnets will be identical. The TMS signal is used as the universal reference standard and its use effectively overcomes this issue.

- contained within the ¹H NMR spectrum

- shows the relative number of hydrogen atoms present in each chemical environment

- e.g., For methanoic acid, HCOOH, the ratio will be _______.

*Features of High-Resolution ¹H NMR:

-In practice, most ¹H NMR spectra do not consist of sets of single peaks/signals – though this may seem the case at low resolution. At high-resolution, ____________ patterns can be seen at certain absorptions.

-Splitting patterns result from spin-spin coupling.

*What is spin-spin coupling?

e.g., Consider the spectrum for 1,1,2-trichloroethane under high-resolution

Features:

- The compound has two different types of hydrogens (H_a and H_b) in two different chemical environments.

*RULE 1

If a proton, H_a, has n protons as its nearest neighbours, that is n x H_b, then the peak of H_a will be split into (n + 1) peaks.

*RULE 2

PROBLEM #2

*An unknown compound, X, of molecular formula, C₃H₆O, with a characteristic fruity odour, has the IR and ¹H NMR spectra.

What is the structure of the compound?