Percentage yield:

                Is the measure of conversion of reactants to products.  It is a measure of the waste of the reactants

• When we think about reactions, we always think of them as going 100% to products.
• This is usually not the case due to:

• Percentage yield is like a score in a test. It is an indication of what you achieved out of what you could have got:

The rules:

Examples:

A)  Preparation of ethanoic acid: 

A student reacted 9.20g of ethanol with and excess of sulphuric acid and sodium dichromate (the oxidising agent).  The student obtained 4.35g of ethanoic acid.

Calculate the % yield:

1)  Write a balanced chemical equation:

2)  Identify the limiting reactant

• You are told in the question that you have an excess of sulphuric acid and sodium dichromate.
• This means that the limiting reactant is ethanol.

3)  Calculate the theoretical amount of moles of product starting from the limiting reactant:

• Calculate the amount of moles of ethanoic acid you could have made:

4)  Calculate the actual amount of moles of product obtained:

• Calculate the number of moles you actually made:

5)  Calculate % yield using the formula:

B)  Preparation of propyl methanoate

A student prepared propyl methanoate from propan - 1 - ol and methanoic acid. 

The students reacted 3.00g of propan - 1 - ol witth 2.50g of methanoic acid in the presence of a sulphuric acid catalyst.  He was disappointed to obtain only 1.75g of propyl methanoate.

Calculate the % yield of propyl methanoate:

1)  Write a balanced chemical equation:

2)  Identify the limiting reactant

• You are given 2 starting amounts which means you have to work out which one is the limiting reactant:

• Propan - 1 - ol is the limiting reactant so the theoretical calculation must be made using this

3)  Calculate the theoretical amount of moles of product starting from the limiting reactant:

• Calculate the amount of moles of ethanoic acid you could have made:

4)  Calculate the actual amount of moles of product obtained:

• Calculate the number of moles you actually made:

5)  Calculate % yield using the formula:

Questions  P163 1-2

Atom economy:

               Is the measure of the waste associated with the products

• % yield tells us how much of our product is made from our starting materials but it doesn't take into account any undesirable or side products.
• Atom economy takes into account any wasteful by products too
• By products are considered wasteful as they are usually disposed of.  This is costly and can cause environmental problems.
• A more efficient way of dealing with by products would be to sell them on to companies that would make use of them.

Calculating atom economy:

A)  Bromination of propene:

• Any reaction that gives only one product is very atom economic, addition reactions for example.

B)  Preparation of butan - 1 - ol:

• This means that most of the starting materials ended up as waste.

Atom economy and type of reaction:

• Reactions having only one product have a high atom economy.  The type of reactions giving only one product are addition reactions.
• Reactions giving more than one product have a low atom economy.  The type of reactions giving more than one product are substitution / elimination reactions.
• To improve the atom economy for substitution / elimination reactions, a use for the undesired product should be found.
• If the undesired product is toxic, we have even bigger problems -disposal.

Questions  P165  1-2

Infrared spectroscopy

Infrared radiation and molecules:

• All molecules absorb IR light.
• The IR light makes the bonds in a molecule vibrate (like the engine of a bus making the windows vibrate).
• Vibrations occur in one of 2 ways, a stretching vibration or a bending vibration:

• Every bond vibrates at its own unique frequency depending on:

        1)  Bond strength

        2)  Bond length

        3)  Mass of atom at either end of the bond

How it works:

• The full IR spectrum is passed through a sample.

• The frequencies are called wavenumbers - 300 - 4000cm^-1

• Some frequencies make some of the bonds in a molecule vibrate

• When these bonds vibrate they absorb energy from the IR light source.

• This means less IR light gets through the sample to the detector.

• Each absorbance peak is characteristic of a particular bond / atoms vibrating.

• A trace which we call a spectrum is produced.

What does the spectrum look like:

• The spectrum gives us 'peaks' which are actually absorbance troughs.

• These troughs are caused by a frequency of IR light being absorbed from a bond vibrating bond.

• Each 'peak' is characteristic to a specific bond / atoms

Applications of IR spectroscopy:

• It is used widely in forensic science analysing:

- Paint fragments in hit and run offences

- Monitor unsaturation in polymerisation

- Drug analysis  P167

- Perfume quality control

Questions  P167  1-3

Infrared spectroscopy:  Functional groups

Identification of functional groups:

• We have just seen that the peak on an IR spectra are due to specific bonds (and atoms) vibrating or stretching.
• The frequency at which you find an absorbance peak is therefore unique to bonds and atoms at each end of the bond.
• This means that functional groups will give specific peaks.
• The groups you need to know are:

• Do not get the peaks for C - H bonds confused with O - H bonds.

Alcohols:

• The IR spectrum for methanol, CH₃OH is shown below:

• The peak at 3230 - 3500 represents an O - H group in alcohols.

Aldehydes and ketones:

• The IR spectrum for propanal, CH₃CHO is shown below:

• The peak at 1680 - 1750 represents a C=O group in aldehydes and ketones.

Aldehydes and ketones:

• The IR spectrum for propanoic acid, CH₃CH₂COOH is shown below:

• The peak at 2500 - 3300 represents an O - H group in a carboxylic acid.
• The peak at 1680 - 1750 represents a C=O group in a carboxylic acid.

Questions  P169 1-2

Mass Spectrometry

Uses of mass spectroscopy:

• First developed by JJ Thompson at the start of the 20th Century.
• It is used:
• To identify unknown compounds.
• To determine the abundance of isotopes
• To gain further information about the structure and chemical properties of molecules.

Examples:

• To examine patients breath while under anaesthetic.
• Detecting banned substances - steroids in athletes.
• Detecting traces of toxic chemicals in contaminated marine life

How a mass spectrometer works:

Mass spectra of elements:

• One of the most important uses is to determine the isotopes present in a natural sample of an element.
• A mass spectrum shows the mass charge (the Ar) and the abundance as a %.
• This information can be used to determine the relative atomic mass:

From mass spectra	From table of data
	RAM of silicon isotopes % Abundance 28 92.2 29 4.7 30 3.1
Use the formula:-   RAM = (%  x  Ar)  +  (%  x  Ar)  +  ....etc                               100	Use the formula:-            RAM = (%  x  Ar)  +  (%  x  Ar)  +  ....etc                                         100
RAM = (90.9  x  20) + (0.2  x  21) + (8.9  x  22)                                       100	RAM = (92.2  x  28)  +  (4.7  x  29)  +   (3.1  x  30)                                                    100
RAM = 20.8	RAM = 28.1

Questions 1-2  P171

Mass spectrometry in organic chemistry

Mass spectrometry and molecules:

• Ionisation in a mass spectroscope is usually done by electron bombardment.
• Electron bombardment knocks another electron out of the molecule producing a positive molecular ion:

• This is called the molecular ion, M⁺.
• The mass of the electron lost electron is negligible.
• The molecular ion has the same mass as the Mr of the molecule.
• As we have a mass and a charge we can use a mass spectrometer to determine the Mr (m/z).

Fragmentation:

• Excess energy from the ionisation process causes bonds in the organic molecule to vibrate and weaken.
• This causes the molecule to split or fragment into smaller pieces.
• Fragmentation gives a positively charged molecular fragment ion and a neutral molecule:

• The fragment ion, CH₂OH⁺ has a mass and charge so we can use a mass spectrometer to determine the Mr (m/z) of that fragment.
• Fragment ions can be broken up further to give a range of m/z values.
• The m/z values correspond to the Mr's of the molecule and its fragments.
• The Mr of the molecule is always the highest m/z value - ie this molecule has not been fragmented so it must have the highest Mr.
• The one below is for ethanol.  It has a m/z of 46 which is also its Mr.

Fragmentation patterns:

• Mass spectroscopy is used to identify and determine the structures of unknown compounds.
• Although 2 isomers will have exactly the same M⁺ peak, the fragmentation patterns will be unique to that molecule, like a fingerprint.
• In practice mass spectrometers are linked to a database and the spectra is compared until an exact match is found:

• These are the mass spectra for pentane and a structural isomer of pentane, 2 methyl butane.
• The M⁺ peak is the same for each but the fragmentation patterns are different.

Questions  1-2  P173

Mass spectrometry:  Fragmentation patterns

Identifying fragment ions:

• When you look at a mass spectrum, other peaks seem to look more important than the M⁺ peak.
• These fragment peaks give clues to the structure of the compound.
• Even simple structures give common peaks that can be identified:

• Functional groups are a good place to start, OH = m/z of 17
• Some fragments are more difficult to identify as these will have undergone molecular rearrangement.

Identification of organic structures:

• A mass spectrum will not only tell you the Mr (from the M⁺ peak), but it can also tell you some of the structural detail.
• These peaks have been labelled with a letter:

• The mass spectrum above has been produced from hexane.
• The following reactions show how the molecule could fragment to form the fragment ions 57 and 43:

Questions  1-2  P175  /  14  P179  /  3  P181