AS Summary Notes

The following documents are summary files covering a wide range of topics.

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Provided by David K, this article covers OCR Chem A – F321
The topics covered in this superb revision document are:
Atomic Structure, Ionisation Energy, Bonding, Intermolecular Forces, Periodicity, Group 2 and Group 7.
Chemistry – Atoms, Bonds Groups – F321 Summary Notes by David K

Mass Spectrometry

Mass spectrometry
by Tom. L.

In order for us to create a mass spectrum, a mass spectrometer is used; the whole purpose of mass spectrometry is to try to establish, firstly the molecular mass of a compound, but also the possible structure of this compound using the fragmentation peaks on the mass spectrums. Mass spectrums are normally used in co-ordination with infrared spectrums, so that you can establish not just the structure of the compound but the functional groups present in the compound therefore proving further evidence for the possible structure of the compound present.

KEY:

  • Mass spectrometry – identifying possible structure of the compound
  • Infrared spectrometry – identifying possible functional groups in the compound

How is a mass spectrum made? :

  • When a compound is placed in a mass spectrometer it is ionised, meaning that the compound looses one electron to form a positive ion.
  • This positive ion formed can be detected; this positive ion will have a molecular mass equal to the molecular mass of the compound and is called the molecular ion.
  • There are also other peaks detected on the mass spectrum; these are caused by the compound fragmenting, and these are what are referred to as the fragmentation peaks.
  • The mass spectrum is recorded as a line diagram (see below) showing m/e ratio against relative abundance.
  • m/e = mass to charge ratio

There are 2 important peaks to identify in the mass spectra, these are always found at the highest mass to charge ratio of the spectra (i.e. furthest to the right):

  • The molecular ion peak (M), which provides us with the molecular mass of the compound present
  • M+1 peak – which shows the existence of 13C – note this may not be present in all spectra, it is sometimes left out of exams, but also may be present so must be aware of this and the reasons why it is present, if it is present it is obvious to see.

The M+1 peak exists in many spectra’s and at very small amounts, hence is very small in height compared to the peak representing the molecular mass of the compound (M peak). The reason we see this M+1 peak is due to presence of carbon-13, is makes up 1.11% of all carbon atoms hence why it’s present in very small amounts.

Mass Spectrum:  - example question worked through

1.) Above is the mass spectrum of compound A, with an empirical formula of C2H4O

a.) what is the relative molecular mass of compound A?

  • The answer is 88 – can see that the M peak is at 88 on the M/z ratio axis, which is the same of the molecular mass of the compound.
  • Note that the M+1 peak is present at 89 on the axis, but is obviously much smaller than the molecular ion peak

b.) molecular formula of compound A?

  • The molecular formula for compound A is C4H8O2
  • As we know the empirical formula is C2H4O, we work of the mr of the empircal formula which is 44, we then divide 88 the mr of compound a by 44, which gives us answer of 2.
  • Use this answer to times the empirical formula by this factor, so in this case 2, so molecular formula is C4H8O2

c.) write an equation to show the formation of the molecular ion?

  • C4H8O2 + e- à C4H8O2+ + 2e-

d.) suggest possible structures for the ions that give rise to the peaks at the following m/z values:

29 – CH3CH2+

43 – CH3CH2 CH2+

45 – CHO2+

Now that you have given possible structures of the fragmentation peaks, you can begin to piece together these fragments of the molecule to suggest a possible structure of this molecule, could be:

Simple process:

  1. Identify molecular ion peak and M+1 peak if present
  2. Identify fragmentation peaks and structures
  3. Piece together to form possible structure of the compound present

However you would normally then go on to use an infrared spectrum of the compound to help identify the functional groups present which would confirm whether it has the functional groups of a carboxylic acid present

Once you know what you’re looking for within a mass spectra you just need to keep practicing, and the best way if to do past papers, as most of these questions are used within the long answer questions, linked with empirical formulas and infrared spectroscopy

Uses of mass spectrometry:

  • Differentiate between 2 different compounds – 2 similar compounds with the same atoms but different structural arrangement, will have the same mr hence Molecular ion peak, however there fragmentation peaks will be different as they will break down into different ions, hence the two mass spectrums will be different
  • Probes to mars – study composition of surface of mars and look for particular molecules
  • Levels of pollutants in the environment – can look for presence of certain pesticides of lead within a food chain.

Stereoisomerism

Stereoisomerism
by Joe. G

Stereoisomerism is where molecules have the same structural formula, but different spacial arrangements in space.

Many alkanes have stereoisomers, due to the lack of rotation around the C=C bond.

When the double-bonded Carbons each have 2 different groups or atoms attached, you get a ‘Cis’ (Same side) isomer or a ‘Trans’ (opposite sides) isomer. You can remember this by thinking of the word ‘Transfer’ – a football player moves across from one team to another when he’s Transferred!

For example:

Both of these molecules are C2H2Br2. However, the Bromines are on opposite sides of the double bond. We therefore have to call the molecule on the left Trans-1,2-Dibromoethene, and the one on the right Cis-1,2-Dibromoethene.

It also works when the atoms or groups aren’t the same:

On the left is Cis-1-Fluoro-2-Iodoethene, whilst on the right is Trans-1-Fluoro-2-Iodoethene. The Cis/Trans system, however, falls down when the groups attached to the Carbons are different, for example:

The Cis Trans system doesn’t work here  – it’s impossible to say whether it refers to the Hydrogen or the Iodine on the Left, or the Fluorine or Chlorine on the Right. 

Instead, we use the E/Z system. Simply:

1)      Look at the atomic number of the 2 atoms directly attached to the Carbon on the Left. Whichever is higher is the atom to be considered.
2)      Do the same on the right.
3)      If your 2 atoms are on the same side (vertically), it’s a Z isomer. If they’re on opposite sides of the double bond, it’s an E isomer.

The Atomic Mass of I > that of H, and the Atomic Mass of Cl > that of F. This means that we look at the I and Cl when talking about E & Z isomers.

The molecule on the left is therefore Z-1-Chloro-1-Fluoro-2-Iodoethene, whilst on the right we have E-1-Chloro-1-Fluoro-2-Iodoethene.

Alkenes – Polymerisation

Polymerisation
by Daniela F.

 Addition polymerisation is simply turning an alkene, which is a monomer into a polymer. A polymer is long chained molecule which has a large molecular mass. For example: 

Addition polymerisation produces saturated chains which contain no double bonds.  Polymerisation is carried out under high temperature and pressure.

You may get asked to draw 3 repeat units of the polymer. A repeat unit is the product shown above, where the specific arrangement of atoms is shown in square brackets which an n on the outside of the square brackets. So to draw 3 repeat units you simply remove the brackets and draw out the sequence 3 times:

Another thing you need to bear in mind is the naming of both the monomer and the polymer. You may be given the polymer and have to name the monomer which formed it. In this case, it would be fluoro-ethene for the monomer and (poly)fluoro-ethene for the polymer.

Structure of Alkenes

Structure of Alkenes

by Ryan B

Alkenes are UNSATURATED HYDROCARBONS (They contain a c=c double covalent bond)

They have a general formula of Cn H₂n.

 

When two p orbitals overlap at the C=C, a pi bond is formed with two parts. Above and below the molecular axis.

This is very electron dense and it makes alkenes very reactive.
The pi bond sticks above and below the molecule and so is easily attacked by electrophiles.

The C=C in an alkene cannot rotate, and so the C=C is flat and planar.
(This causes cis-trans and E-Z isomerism.)

Alcohols – Oxidation

By Kirsty H.

Primary Alcohols

Primary alcohols can be oxidised, under distillation, to produce an aldehyde.

General Equation:
°1 alcohol + [O]   —>   aldehyde + water

Example:
Ethanol + [O]     —> ethanal + water

Conditions: acidified potassium dichromate & distillation

The aldehyde produced can then be further oxidised, under reflux, to form a carboxylic acid.

General Equation:
Aldehyde + [O] —> carboxylic acid

Example:
Ethanal + [O] —> ethanoic acid

Conditions: acidified potassium dichromate & reflux

It is also possible to go straight from the °1 alcohol to the carboxylic acid. For this reaction you have to use reflux and two lots of the oxidising agent.

General Equation:
°1 alcohol + 2[O] —> carboxylic acid + water

Example:
Ethanol + 2[O] —> ethanoic acid + water

Conditions: acidified potassium dichromate & reflux

Secondary Alcohols
Secondary alcohols are oxidised to ketones under reflux.

General Equation:
°2 alcohol + [O] —> ketone + water

Example:
Propanol + [O] —> propanone + water

Conditions: acidified potassium dichromate & reflux

Tertiary Alcohols
Tertiary alcohols cannot be oxidised because there are no available hydrogens to be removed.

General Equation
°3 alcohol + [O] —> no reaction

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