DUNCANRIG SECONDARY ADVANCED HIGHER CHEMISTRY

UNIT TWO

BOOKLET 4 Experimental determination

of structure

From a chemical point of view, the most fundamental data about a compound

are: which elements are present and how much of each are in the compound?

This information leads to the empirical formula for the material.

The empirical formula shows the simplest whole number ratio of the different

atoms in a compound.

The alkenes C2H4, C3H6, C4H8 all have the same empirical formula - CH2.

A hundred years ago, chemists didn’t have much in the way of instruments that

could analyze substances. They had a balance and a Bunsen burner. So the

natural thing to do was to weigh something first, then burn it and then weigh

the ashes.

The percentage of carbon, hydrogen, nitrogen and sulphur in a compound can

be found by burning a known mass of the material in an oxygen atmosphere and

measuring the mass of carbon dioxide, water, nitrogen and sulphur dioxide

produced.

Using this apparatus, a known

mass of the compound is

burned. Any hydrogen turns to

water which is trapped in the

first canister. The mass of

water is found and from this

the mass of hydrogen is

calculated.

Any carbon in the compound is converted to

carbon dioxide which is trapped by the

second canister. The mass of the carbon

dioxide is found and from this the mass of

carbon in the compound is calculated.

Other elements are analysed in a similar

manner.

modern combustion analyser.

1. Complete combustion of 1.75 g of an organic compound containing only carbon,

hydrogen and oxygen gave the following results.

Mass of carbon dioxide produced = 3.51 g

Mass of water produced = 1.43 g

The relative molecular mass of the compound was found to be 88 g.

Use the results to calculate the empirical formula and the molecular formula of the

organic compound.

Step 1 – calculate masses of C,H and O in the compound

The mass of any element in a compound is found using the formula

Mass of C = {C/CO2} x 3.51 = {12/44} x 3.51 = 0.957 g

Mass of H = {H/H2O} x 1.43 = {2/18} x 1.43 = 0.159 g

In general if oxygen is present in a compound its mass can be determined by

subtracting the total mass of the other elements from the mass of the

original sample.

Mass of O = mass of sample – {mass of C + mass of H}

= 1.75 – {0.957+0.159} = 0.634 g

Step 2 – determine the empirical formula

This is found by determining the number of moles of each element and converting the values

to a whole number mole ratio.

Element C H O

Mass(g) 0.957 0.159 0.634

Moles 0.957/12 = 0.0798 0.159/1 = 0.159 0.634/16 = 0.0396

Mole ratio 0.0798/0.0396 = 2 0.159/0.0396= 4 0.0396/0.0396 = 1

The empirical formula is C2H4O

Step 3 – determine the molecular formula

The molecular formula can be found using the formula........

Multiplying the empirical formula by the

constant will produce the molecular formula.

Constant = 88/{GFM C2H4O} = 88/44 = 2

Molecular formula = 2 x C2H4O = C4H8O2

2. A hydrocarbon has a composition by mass of

88.25% carbon and 11.75% hydrogen.

Calculate the empirical formula of the compound.

3. Salicylic acid is used to make aspirin. It contains only carbon, oxygen, and hydrogen.

Combustion of a 43.5 mg sample of this compound produced 97.1 mg of CO2 and 17.0 mg

of H2O. The molar mass of salicylic acid is 138. What is its molecular formula?

Mass of C = {C/CO2} x 97.1 = {12/44} x 97.1 = 26.48 mg

Mass of H = {H/H2O} x 17.0 = {2/18} x 17.0 = 1.889 mg

Mass of O = 43.5 – {26.48+1.889} = 15.131 mg

Element C H

Mass(%) 88.25 11.75

Moles 88.25/12 = 7.354 11.75/1 = 11.75

Mole ratio 7.354/7.354 = 1 11.75/7.354= 1.6

Element C H O

Mass(g) 26.46 1.889 15.131

Moles 26.46/12 = 2.205 1.889/1 = 1.889 15.131/16 = 0.946

Mole ratio 2.205/0.946 = 2.33 1.889/0.946= 2 0.946/0.946 = 1

In this example the % mass is given.

These numbers can be used directly to

find the empirical formula.

note.... the scaling is often the

tough part, here we have an

initial ratio of 1 : 1.6, so you

need to find the first multiple of

1.6 that is a whole number

(or very close to it).

Avoid the temptation to round

the first two decimal places.

The empirical formula is C5H8

Empirical formula mass, C7H6O3,is 138 138/138 = 1 molecular formula is C7H6O3

Multiply by 3 to give empirical formulaC7H6O3

1. Dianabol is one of the anabolic steroids that has been used by some athletes to

increase the size and strength of their muscles. The molecular formula of dianabol,

which consists of carbon, hydrogen, and oxygen, can be determined using the data

from two different experiments.

The apparatus shown below was used for the first experiment.

A 14.765 g sample of dianabol was burned, and 43.257 g CO2 and 12.395 g H2O

formed in the absorption chambers.

a. (i) What practical steps need to be undertaken to obtain the values for the masses

of CO2 and H2O formed?

(ii) Calculate the empirical formula of dianabol.

b. In the second experiment, the molecular mass of dianabol is found to be 300.

What is the molecular formula for dianabol?

2. Caffeine, a compound found in coffee, tea, and cola drinks that has a marked

stimulatory effect on mammals. The combustion analysis of caffeine shows that it

contains 49.5% carbon, 5.2 % hydrogen, 28.9 % nitrogen, and 16.5 % oxygen by mass,

and its experimentally determined molecular mass is 195.

Use this information to calculate the molecular formula of caffeine.

3. An antibiotic compound containing the elements C,H,N,S and O was completely burned

in oxygen. Combustion of 0.3442 g of the antibiotic gave 0.528 g of carbon dioxide,

0.144 g of water, 0.128 g of sulfur dioxide and 0.056 g of nitrogen.

a. Calculate the mass of carbon, hydrogen, sulfur and oxygen in the compound.

b. Calculate the empirical formula of the compound.

c. The molecular mass of the compound is 172.1.

Calculate the molecular formula of the compound.

Instrumental analysis is a field of analytical chemistry that investigates

compounds using scientific instruments.

Spectroscopy is a technique that uses the interaction of energy with a sample

to perform an analysis. Nowadays with modern computer technology

spectroscopy can identify almost any organic molecule.

In our study of instrumental analysis you will be expected to interpret the

graphs , called spectra, produced by three different types of instrument

Mass spectroscopy

This technique provides information about MOLECULAR MASSES and the

structure of organic molecules.

Infra - red (IR) spectroscopy

This technique provides information about the FUNCTIONAL GROUPS

in organic molecules.

Proton nuclear magnetic resonance (NMR) spectroscopy

This technique uses radio-waves to provide information about the number of

HYDROGEN ATOMS and their environments in an organic molecule.

For all these different techniques you should be able to interpret simple

spectra (usually with the help of the data booklet) and identify the species

responsible for the various peaks present.

Given a simple organic molecule you should be able to suggest what its various

spectra might look like.

Mass spectroscopy

Points 1. Used to determine molecular mass and structure.

2. Molecules are changed into POSITIVE IONS - mostly +1

3. Ions are deflected into separate ion paths by a magnetic field

according to their mass/charge ratio (m/z).

4. As molecules ionise

they tend to break

into smaller ion

fragments .

Questions on mass spectrometry are likely to involve; identifying the formula for

ion fragments (including the molecular ion); deducing the molecular mass;

deducing s structure for the compound.

Remember to include the positive charge in the formula for any ion fragment.

1. The mass spectrum of methyl butanoate,

C5H10O2, is shown opposite.

a. How can the mass spectrum be used to confirm

the molecular mass of methyl butanoate?

b. Write the formula for the molecular ion of

methyl butanoate.

c. Draw the full structural formula for methyl butanoate.

d. Draw the skeletal formula for methyl butanoaote.

e. Write a possible formula for the ions which cause the peaks at m/z 71 and m/z 59.

f. Butyl methanoate is an isomer of methyl butanoate. Suggest an m/z value for a peak

which would appear on the spectrum for butyl methanoate which does not appear on

the spectrum for methyl butanoate.

2. The mass spectra of propanal and propanone are shown below.

a. For compound A, which group of atoms could be lost when the ion of m/z 43 forms f

a. Write the formula for the molecular ion of propanal.

b. Explain which spectrum, A or B, is the spectrum of propanone and which is the

spectrum of propanal.

Your answer must include reference to the ion fragments formed.

m/z

102

10 20 30 40 50 60 70 80 90 100 110

59

43

29 87

71

3. Compound L is a sweet smelling liquid containing the elements carbon, hydrogen and

oxygen. When 0.4080g of the liquid is completely burned in oxygen, 1.0560g of

carbon dioxide and 0.2160g of water were the only products.

The mass spectrum of compound L is shown below.

a. Calculate the empirical formula of compound L.

b. Calculate the molecular formula of compound L.

c. Write the formula for the molecular ion of compound L.

4. Fragmentation of the molecular ion of methylbutanone, (CH3)2CHCOCH3, gives rise to

dominant peaks at m/z = 71 and m/z = 43.

a. Draw a skeletal formula for methyl butanone.

b. Write a balanced equation to show how fragmentation of the molecular ion gives rise to

the peak at m/z = 71

c. Two fragmented ions are responsible for the peak at m/z = 43.

Give the formula for both these ion fragments.

5. The mass spectrum of a straight chain alkane is shown below.

a. State the name of the alkane.

b. Draw the skeletal formula for the

alkane.

c. Write the formula for the ion

fragments at m/z = 57 and m/z = 29.

6. The mass spectrum of methylene chloride, CH2Cl2, is shown below.

a. Draw a diagram of the structure of methylene chloride which clearly shows

the shape of the molecule.

b. Write the formula for the ion fragment responsible for the peak at m/z = 49.

c. The molecular ion, M+, is found at m/z = 84 indicating the molecular mass of

methylene chloride is 84.

(i) Write the formula for the molecular ion, M+.

(ii) Suggest what might be responsible for the peaks at any of the values higher

than 84. These peaks have values of 85, 86, 87 and 88.

7. Consider the mass spectrum shown below.

Use your knowledge of chemistry to discuss information displayed in the diagram.

m/z

When you sit in front of a fire, which gives

out infra-red (IR) radiation, you feel warm. This is because the IR radiation

absorbed by molecules in your skin is of a suitable frequency to increase their

vibrational energy. The greater a molecule's vibrational energy, the hotter it is.

Atoms in a molecule are always vibrating, and

the frequency of this vibration depends on the

atoms mass and the length or strength of its

bonds. The vibrational frequency of atoms lies

in the IR region of the electromagnetic

spectrum and so they can absorb IR radiation.

IR radiation causes parts of a molecule to

vibrate, and the wavelengths absorbed will depend on the type of chemical

bond and the group of atoms at the end of these bonds.

Obviously it is difficult to show molecules

vibrating on this paper.

The diagrams show some of the vibrations

that are possible. Molecules can bend, twist,

rotate and stretch. If the molecule is

subjected to energy of the correct

frequency (i.e. IR frequencies) then certain

bonds will absorb this energy – this is what an

IR spectrometer detects and displays.

The approximate frequency range for

infra-red radiation is from 1x1013 – 4x10 14 Hz.

This equates to wavenumber values of around

400 cm-1 to 4000 cm-1. Wavenumbers are used

on the x-axis of IR spectra, while the Y-axis is

usually % transmission. (Wavenumber=1/wavelength)

The higher the wavenumber the higher the energy of the radiation.

Infra-red (IR) Spectroscopy

IR spectra are often used to

identify the functional groups

in an organic molecule.

Functional groups are seen at

characteristic frequencies

(wavenumbers) no matter

which compound they are in.

The spectrum above is for butanal and so it has

the C=O functional group at 1750 cm-1 and the

C-H stretch at 2900 cm-1.

For this spectrum, the fingerprint region will be unique to butanal.

This IR correlation table is found in the data booklet.

This is the IR spectrum of ethanol. Important features are the C-O stretches

and the hydrogen bonded O-H stretch due to the hydroxyl functional group.

Notice that the O-H stretch is a

broad peak unlike the C-O

stretches which are relatively

sharp. This broad OH peak is very

characteristic of alcohols as the

H bonds cause more interaction between molecules, which in turn, causes more

possible vibrations and these will have a variety of frequencies and so the peak

appears over a wider wavenumber range.

Detailed analysis of IR spectra is a high order skill as many peaks for

different groups can occur at very similar wavenumbers.

For this course you only need to distinguish between the major functional

groups and correctly identify their presence on a given IR spectrum.

The examples on the next page highlight this.

1. By categorising appropriate peaks on each spectra, identify which spectra could be an

alkane, which could be an ester, which could be an alcohol and which a carboxylic acid.

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2. The infrared spectra shown below are those of the four compounds, A, B, C and D.

a. Write skeletal formula for A, B, C and D.

b. Match A,B, C and D to their appropriate spectra.

c. Calculate the energy, in kJ mol-1, of the vibration associated with a wavenumber of

3000 cm-1.

3. Spectrum of an organic compound A are shown below

a. Compound A has the empirical formula C2H4O.

Deduce the molecular formula of compound A.

b. The absorption peak at 1745 cm−1 in the infra-red spectrum can be used to help

identify A.

(i) Which bond is responsible for this absorption?

(ii) Which type of compound is A?

c. Draw a possible skeletal formula for compound A.

This type of spectroscopy works due to the fact that a spinning hydrogen

nucleus behaves like a tiny magnet. Normally the direction of the spinning

nuclei will be random but if the nuclei placed in a magnetic field the nuclei will

line up in the direction of the applied magnetic field, rather like how a compass

needle lines up with the Earth’s magnetic field

If energy of the correct frequency is now applied, the tiny magnet can flip

over so that it aligns against the magnetic field and will therefore move to a

higher energy state.

The energy required for this transition is in the radio frequency region of the

electromagnetic spectrum, typically from 60 mega hertz, MHz to 1000

megahertz, MHz.

When the energy is removed the nuclei

“flip back” to their lower energy state

and release the small amount of energy

they absorbed. It is this emitted energy

that a nuclear magnetic resonance

spectrometer measures.

This is a picture of one of the world’s strongest NMR spetrometers.

Strength 21.1 Tesla

Type Superconducting

Cost $16 million

Weight 36,287 kg (40

tons)

Height 4.7 meters (15

feet, 6 inches)

Operating

temperature

-271.45 ° C

(-456.61 ° F)

Length of

superconducting

cable

153 km (95 miles)

Proton nuclear magnetic resonance (NMR) spectroscopy

The spectrum produced in NMR gives detailed information about the

hydrogen atoms present in organic molecules.

1. Hydrogen atom environment

2. Number of atoms in each environment.

What is meant by hydrogen environment?

Consider the molecules shown below...............

a. Propan-1-ol has FOUR different b. Propan-2-ol has THREE

hydrogen environments. hydrogen environments.

c. Benzene has ONE hydrogen d. Ethanal has TWO

environment. hydrogen environments.

e. But-2-ene has TWO hydrogen f. 1,4-dimethyl benzene has TWO

environments. hydrogen environments.

C C C O

H

H

H H

H

H

H

HC C C

H

H

H H H

HO

H

H

C C C C

H

H

H H H H

H

H

Consider the NMR low resolution NMR spectrum for ethanol.

Reading the spectrum

1. There are three peaks indicating there are three types of hydrogen

environment.

2. The relative areas under the peaks are proportional to the number of

hydrogen atoms giving rise to each signal – in this case 1:2:3

To make this easier to evaluate most modern NMR spectrometers

integrate and record the area of each signal. The lines A,B and C on this

spectrum are called INTEGRATION lines.

Measuring the height of these lines will also give the 1:2:3 ratio.

Tetramethylsilane

{TMS}

This NMR spectrum of an ester has an

integration signal ratio of 23:67

which cancels down to 1:3. This ratio

shows the 3 H atoms in one

environment and the 9 H atoms in the

second environment.

3. One molecule can contain many hydrogen environments. Each environment

will release a different frequency of energy when it drops down from its

excited state to line up with the magnetic field. Each different hydrogen

environment will appear in a different position in the NMR spectrum. This

is called Chemical Shift which has the symbol delta, δ, and is measured in

parts per million. (parts per million of what I hear you ask – we do not need

to go into this for AH chem.)

4. The peak at zero ppm is for the compound tetramethylsilane(TMS). This

substance is used as a reference chemical against which all other chemical

shifts are compared. This substance is used because it is inert, can be

easily removed from the test sample and it has a signal frequency lower

than almost all other compounds.

5. There is a table of chemical shift

values on page 15 of the data

booklet and this should always be

referred to when attempting NMR

questions. The table shown here is a

simplified version of the data

booklet table.

You will be expected to be able to

assign peaks to individual spectra,

identify compounds form there

spectra and to sketch simple

spectra for given compounds

Functional Group Chemical

Shift

Alkane 0.8-1.2

1.6

Benzyl 2.3

Carbonyl 2.2

Amine 2.3

Alcohol 3.3

Alkyl Halide 3.6

Alkene 4.5-6.0

Benzene 6.0-9.0

Alcohol 0.5-4.5

Very Broad

Carbox. acid 9.0-15.0

R CH3

CHC CH3

CH3

R C CH3

O

R N CH3

HO CH3

H3C Cl

H2C CH2

H

R OH

R

O

OH

1. How many different hydrogen environments do each of the following molecules have?

a. b. c.

d. ` e. f.

2. Use the data booklet to find the approximate chemical shift values for the hydrogen

atoms in the molecules in Q1 and then draw the NMR spectrum for each of them. You

should draw a vertical line for each signal and remember to have them at the

appropriate height ratios

Here is the spectrum for but-2-ene as an example.

a.

b.

c.

d.

e.

f.

3. An NMR spectrum was recorded for each of these straight-chain hydrocarbons:

CH4, C2H6, C3H8 and C4H10.

a. (i) Which compound(s) would only show one peak ?

(ii) How many peaks would you expect for both the structural isomers of C4H10 ?

b. A compound of formula C5H12 has only one peak on its NMR spectrum.

Draw the structural formula of this compound.

4. The approximate values and integrals for the NMR spectrum for one of the isomers

of C4H10O is given in the table below.

a. Draw the skeletal formula of this isomer.

b. Using the information in the table draw the NMR spectrum of this isomer.

c. Draw the isomer of this molecule which is a tertiary alcohol.

5. Phenylethanal is one of the aldehyde molecules responsible for the smell of chocolate.

The skeletal formula for phenylethanal is shown opposite.

3-methylbutanal and 3-methylbutanal are another two

aldehydes which contribute to the smell of chocolate.

a. Draw a skeletal formula for both 3-methylbutanal and 2-methylbutanal

b. The low resolution proton NMR spectrum shown is for one of these three aldehydes.

Explain which of the three aldehydes

would give this proton NMR spectrum.

Isomer Group Approximate

value Integral

CH3CH

2CH

2OCH

3 CH

3 0.8–1.3 3

CH

2 1.2–1.8 2

CH

2O 3.5–4.0 2

OCH

3 3.0–4.0 3

All the NMR spectra we have looked at so far have been low resolution. With

high resolution NMR the spectra are more complicated but provide a clearer

picture of molecular structure. In this sense NMR instruments are like digital

cameras and HDTVs: more megapixels means better resolution which means

more information and clearer pictures (and much higher price tags!)

Consider the molecule 1-bromopropane

This low resolution spectrum has three peaks

indicating three different hydrogen environments.

The ratio of the peaks is 2:2:3 which reflects the

number of each type of hydrogen.

The major difference seen in a high resolution spectrum is the phenomenon

known as multiplicity or coupling

This occurs because the magnetism of a hydrogen atom in one environment can

affect the magnetism of a hydrogen atom in a different environment on an

adjacent carbon atom.

When this happens the NMR signals splits into several signals forming new

signals which are termed doublets, triplets, quartets, etc

These split signals indicate the number of hydrogen atoms on carbon atoms

adjacent to the hydrogen atoms responsible for the signal.

This is the high resolution spectrum of 1-bromopropane

Analysis

Peaks There are three different signals so there are three

chemically different protons.

Splitting The signals include a triplet ( = 1.0)

sextet (= 1.8)

triplet ( = 3.4)

In general the individual signal seen on low resolution will split according to the

(n+1) rule where n = the number of hydrogen atoms on adjacent carbon atoms.

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

C1 triplet ( = 1.0) coupled to the two hydrogens on carbon C2 ( 2+1 = 3 )

C2 sextet ( = 1.8) coupled to five hydrogens on carbons C1 and C3( 5+1 = 6 )

C3 triplet ( = 3.4) coupled to the two hydrogens on carbon C2 ( 2+1 = 3 )

Integration The integration lines show that the ratio of protons is 2:2:3

The high resolution spectrum of pentan-2-one

Which peaks are due to hydrogens a,b,c and d?

Summary An NMR spectrum provides several types of information :-

• number of signal groups tells you ... the number of different

hydrogen environments

• chemical shift the general environment of the hydrogens

• multiplicity how many hydrogens are on adjacent atoms

• peak area the number of hydrogens in each environment

1. A compound with empirical formula C2H4O gives a molecular ion peak at m/z = 88.

The compound is neutral and has the following high resolution NMR spectrum.

Using all the information deduce the full structural formula for the compound.

2. The two spectra on the right were

obtained from a sample of propane.

a. Draw the full structural formula for

propane.

b. How many different types of hydrogen

atom do these spectra show.

c. Explain why the spectra appear different.

3. The high resolution proton NMR spectrum of butanone is shown below.

a. Name the compound responsible for the

peak at zero PPM

b. Draw a structure for butanone and

allocate the remaining peaks to the

correct hydrogen atoms in the molecule.

4. The high resolution spectrum of a straight chain ether with molecular formula C4H10O2

is shown below.

a. Deduce the name of the ether and draw its full structural formula.

b. Indicate on the spectrum which hydrogens in the ether give rise to each cluster of

signals.