Sunday, 27 November 2011


1)      Addition polymers

  • The monomer is an alkene which has a p bond
  • The p bond breaks and is used to join many monomers into a long chain
  • The polymer has single bonds along its backbone
  • Addition polymers are not biodegradable
  • The monomers come from crude oil which is a non-renewable resource

Example – poly(phenylethene), commonly called polystyrene, whose monomer is phenylethene                                       

Condensation polymers

  • The monomer has a functional group at each end
  • The functional groups react to form a link; the reaction also produces a small molecule, such as water
  • The links in condensation polymers (ester, amide and peptide) can be hydrolysed by heating with dilute acid or alkali

a)   Polyesters

·         The monomers are a di-carboxylic acid and a diol
·         An ester link is formed between an acid group and an alcohol group with a water molecule eliminated
·         Example 1: Terylene whose monomers are benzene-1,4-dicarboxylic acid and ethane-1,2-diol

·         Polyesters are used as fibres in clothing e.g. poly-cotton shirts

·         Example 2: Poly(lactic acid) whose monomer is lactic acid (2-hydroxypropanoic acid)

b)   Polyamides

·         The monomers are a di-carboxylic acid and a diamine
·         An amide link is formed between an acid group and an amine group
·         Example 1: Nylon-6,6 whose monomers are hexane-1,6-dicarboxylic acid and 1,6-diaminohexane

·         Example 2: Kevlar whose monomers are benzene-1,4-dicarboxylic acid and benzene-1,4-diamine

·         Polyamides are used as fibres in clothing e.g. Kevlar is used in bullet-proof vests and in extreme sports equipment
·         Hydrogen bonds form between polyamide chains from the H in –NH to the O in C=O
·         Polyamide chains are regular because they are made from one or two monomers
·         Polyamides contain the same link as polypeptides/proteins but polyamides are synthetic (man-made) and polypeptides/proteins occur naturally

c)   Polypeptides and proteins

·         The monomers are amino acids
·         A peptide link is formed between the acid group of one amino acid and the amine group of another amino acid e.g. glycine and alanine forming a dipeptide

·         Hydrogen bonds form between polypeptide and protein chains from the H in –NH to the O in C=O
·         Polypeptide and protein chains are irregular because they are made from many different monomers
·         Polypeptides and proteins will have chiral centres (unless they are formed solely from glycine)

3)      Going from polymer to monomer

  • Identify the type of link. If there is no ester or amide link it’s an addition polymer
  • For a polyester, put in the –OH’s on the acid groups and the H’s on the alcohol groups
  • For a polyamide, put in the –OH’s on the acid groups and the H’s on the amine groups
  • For an addition polymer, identify the repeating unit and put in the double bond

4)      Hydrolysis and degradable polymers

  • Acid hydrolysis of a polyester produces an alcohol and a carboxylic acid
  • Base hydrolysis of a polyester produces an alcohol and a carboxylate salt
  • Acid hydrolysis of a polyamide produces an amine and a carboxylic acid
  • Base hydrolysis of a polyester produces an amine and a carboxylate salt
  • Chemists have helped to minimise environmental waste by developing degradable polymers similar to poly(lactic acid). This can be used to make packaging, waste sacks, disposable eating utensils and internal stitches. All of these products will degrade over time
  • Condensation polymers may be photodegradable as the C=O bond absorbs radiation
  • Condensation polymers may be hydrolysed at the ester or amide link
  • Another advantage of degradable polymers is that the monomer is made from renewable resources

Chromatography Questions 1

Chromatography is a versatile technique that may be used to separate and identify compounds.
(i)      Name a type of chromatography that is used to separate and identify dissolved substances.
(ii)      State what quantitative value may be determined from the chromatogram to identify the substances present in the solution.
(iii)     Sketch a chromatogram to show how the value in (ii) is determined.


          Gas-liquid chromatography is used to separate and identify gases and liquids.
(i)      State what quantitative value is normally used to identify the components in this type of chromatography.
(ii)     Sketch the chromatogram to show how the value in (i) is determined.

(b)     State the physical process on which the separation used in gas-liquid chromatography depends.
[Total 3 marks]

In this question, one mark is available for the quality of the use and organisation of scientific terms.
          Describe and explain the different ways that a high resolution n.m.r. spectrum can give information about a molecule.

Synthesis and Stereoisomerism

1)      Stereoisomerism

ê                                  ê
                        Structural                      Stereoisomers
Isomers                                    ______ï_______
ê                       ê
                                                E/Z isomers       Optical isomers

  • Stereoisomers have the same structural formulae but different 3-dimensional arrangements
  • There are two types of stereoisomerism: E/Z isomerism, which occurs in alkenes, and optical isomerism, which occurs when a carbon atom is attached to four different groups

2)      E/Z isomers

  • E/Z isomers occur in alkenes because of restricted rotation about a double bond
  • For E/Z isomers to occur, the two groups at each of the double bond must be different e.g. but-2-ene has E and Z isomers because there is a CH3- and an H- at each end of the double bond.


Z-but-2-ene                               E-but-2-ene

3)      Optical isomers

  • An optical isomer is one that cannot be superimposed on its mirror image
  • This occurs when a carbon is attached to four different groups. Such a carbon is known as a chiral centre
  • Optical refers to the ability of these molecules to rotate plane-polarised light. D isomers rotate light to the right and L isomers rotate light to the left.
  • Most amino acids display optical isomerism e.g. glutamic acid

The central carbon is a chiral centre because it is attached to four different groups:
 -NH2, -H, -COOH and –CH2CH2COOH

1)      General

  • A synthesis is a series of reactions giving a desired chemical product
  • Answering questions on synthesis requires a good knowledge of the reactions in the current module and the relevant parts of Module F322 (first year)

2)      Synthesis of pharmaceuticals

  • Compounds produced naturally in living systems will often be present as one optical isomer only e.g. our bodies only contain L-amino acids
  • The synthesis of pharmaceuticals often requires the production of chiral drugs containing a single optical isomer
  • Molecules synthesised in a laboratory often contain a mixture of optical isomers whereas molecules of the same compound produced naturally by enzymes in living systems will often be present as only one optical isomer
  • Synthesis of a single optical isomer of a pharmaceutical increases costs due to the difficulty in separating the optical isomers but has the benefits of reducing side effects and improving pharmacological activity
  • Modern synthesis of a single optical isomer of a pharmaceutical is often carried out:
    • Using enzymes or bacteria which promote stereoselectivity
    • Using chemical chiral synthesis or chiral catalysts
    • Using natural chiral molecules, such as L-amino acids or sugars, as starting materials

Carbonyl compounds


  • Aldehydes and ketones are carbonyl compounds
  • They contain the carbonyl group C=O
  • The functional group in aldehydes is –CHO on the end of a chain e.g. ethanal CH3CHO
  • The functional group in ketones is C=O not at the end of a chain e.g. propanone CH3COCH3

2.       AS Recap

  • Primary alcohols are oxidised by acidified potassium dichromate. An aldehyde is produced first and this can be further oxidised to a carboxylic acid. To get the aldehyde, it must be distilled off as it is formed. To get the acid, heat under reflux.
  • Secondary alcohols are oxidised to ketones by acidified K2Cr2O7. Colour change is orange to green.
  • The C=O bond in aldehydes, ketones, carboxylic acids and esters can be identified by infrared spectroscopy. It produces a large peak around 1700 cm-1.

3.       Reduction of Aldehydes and Ketones

Reduction, here, means addition of hydrogen.

a)         Reduction using NaBH4

A specific reducing agent for aldehydes and ketones is sodium borohydride, NaBH4. In equations the reducing agent is represented by [H].

Aldehydes are reduced to primary alcohols by NaBH4 e.g.

CH3CHO + 2[H] ® CH3CH2OH

Ketones are reduced to secondary alcohols by NaBH4 e.g.

CH3COCH3 + 2[H] ® CH3CH(OH)CH3

Other points to note about this reaction are:
·         It is an addition reaction (there is only one product)
·         The mechanism is called nucleophilic addition
The nucleophile is H- which is provided by NaBH4

)           Comparison with hydrogen gas

            NaBH4 will reduce C=O double bonds but it will not reduce C=C double bonds

e.g.      CH2=CH-CHO + 2[H] ® CH2=CH-CH2OH

            To reduce both C=O and C=C use H2 with Ni catalyst

e.g.      CH2=CH-CHO +2H2 ® CH3CH2CH2OH

4.       Reaction with 2,4-dinitrophenylhydrazine (2,4-DNPH)

  • Aldehydes and ketones react with 2,4-DNPH to give an orange precipitate (solid)
  • This is used a test for the presence of an aldehyde or ketone
  • The particular aldehyde or ketone can be identified by purifying (recrystallisation) the precipitate, measuring its melting point and comparing this with the melting points of known compounds (from a data book).

5.       Reaction with Tollen’s Reagent

  • Tollen’s reagent is ammoniacal silver nitrate.
  • It is a mild oxidising agent that is used to distinguish between aldehydes and ketones. The compound to be tested is warmed with Tollen’s reagent.
  • Aldehydes are oxidised by Tollen’s reagent which is reduced to silver metal

CH3CHO + [O] ®        CH3COOH                    Oxidation
Ag+(aq) + e-    ®         Ag(s)                           Reduction
silver mirror on inside of test tube

  • Ketones do not react with Tollen’s reagent because they are not easily oxidised.

Topic 8b – Carboxylic acids and esters
Revision Notes

1.       Carboxylic acids

  • Carboxylic acids contain the functional group –COOH on the end of a chain.
  • They are weak acids (H+ donors). The acidic H is in the –COOH group e.g.

CH3COOH ¾ CH3COO- + H+    (note – reversible reaction so ¾ not ®)

  • They are soluble in water because they can hydrogen bond to water molecules

            a)         Reactions of carboxylic acids

As they are acids they will react with metals, carbonates and bases e.g.  

            CH3COOH + Na ®       CH3COONa + ½H2                  Fizzing seen
            Ethanoic acid                 sodium ethanoate                       Sodium dissolves

            2CH3COOH + Na2CO3 ® 2CH3COONa + H2O + CO2   Fizzing seen                                                                                                       Carbonate dissolves

            CH3COOH + NaOH ® CH3COONa + H2O

2.       Esters

  • Esters contain the functional group –COOR on the end of a chain
  • Making esters is called esterification
  • Esters can be made in two ways: carboxylic acid + alcohol or acid anhydride + alcohol
  • Esters are sweet smelling and are used as flavourings and perfumes in food.

            a)         Esterification of carboxylic acid with alcohol

Carboxylic acids react with alcohols to make and ester and water e.g.

            CH3COOH + C­2H5OH ¾ CH3COOC2H5 + H2O
            Ethanoic acid                   ethyl ethanoate

Conditions:        Reflux with concentrated H2SO4 (acts as a catalyst)

            b)         Esterification of acid anhydride with alcohol

  • Acid anhydrides can be thought of as 2 molecules of acid that have lost a molecule of water e.g. propanoic anhydride, (CH3CH2CO)2O

Propanoic anhydride                   methyl propanoate   propanoic acid

            c)         Acid hydrolysis of esters

  • This is the reverse of esterification

            CH3COOC2H5 + H2O ¾ CH3COOH + C­2H5OH
            Ethyl ethanoate 

  • For acid hydrolysis, heat the ester with a dilute acid such as HCl   

d)         Alkaline hydrolysis

  • This is similar to acid hydrolysis but produces the carboxylate salt of acid rather than acid itself. This is not reversible.

CH3COOC2H5 + NaOH ®         CH3COONa + C­2H5OH
            Ethyl ethanoate                          sodium ethanoate

  • For alkaline hydrolysis, heat the ester with dilute NaOH.   

Topic 8c – Triglycerides, unsaturated and saturated fats

Revision Notes

1.       Triglycerides

  • Triglycerides are more commonly known as fats and oils
  • A triglyceride is a tri-ester of glycerol and 3 fatty acids e.g.

  • A fatty acid is an unbranched, long chain carboxylic acid e.g. octadecanoic acid. The shorthand formula for a fatty acid shows the number of carbons and the number and position of any double bonds. Octadecanoic acid has 18 carbons and no double bonds so its shorthand formula is 18, 0

  • Glycerol is propane-1,2,3-triol

  • In a triglyceride the 3 fatty acids do not have to be the same

2.       Saturated and unsaturated fats

  • In a saturated fat there are no double bonds in the fatty acids from which the triglyceride was formed
  • Unsaturated fats are formed from one or more fatty acids that contain a double bond
  • The systematic name for oleic acid, shown below, is octadec-9-enoic acid which indicates that the double bond starts on carbon 9. The shorthand formula for this acid is 18, 1(9)
  • Linoleic acid has 2 double bonds starting on carbons 9 and 12 so its systematic name is octadec-9,12-dienoic acid and the shorthand formula is 18, 2(9,12)

  • The presence of double bonds means that unsaturated fatty acids can have cis and trans isomers
  • Note that trans fatty acids are less kinked than cis fatty acids
  • The presence of trans fatty acids in the diet raises the level of LDL (‘bad’) cholesterol and reduces the level of HDL (‘good’) cholesterol. This increases the risk of coronary heart disease and strokes

3.       Biodiesel

  • Biodiesel consists of the esters of fatty acids
  • Biodiesel can be made from cooking oil. The oil is mixed with methanol and potassium hydroxide is added as a catalyst
  • The use of biodiesel is increasing because of reduced greenhouse gas emissions, deforestation and pollution compared with petro-diesel (made from crude oil)

Titration calculations

Titration calculations

  • Work out the moles of acid and base at the start
  • Work out the excess moles of acid or base (the rest will be neutralised)
  • Work out the new [H+] or new [OH-] and then the pH

Calculate the pH of the solution formed when 20 cm3 of 0.10 mol dm-3 HCl is added to 30 cm-3 of 0.04 mol dm-3 NaOH

Moles HCl                = 0.10 x 20/1000
                              = 0.0020 mol
Moles NaOH            = 0.04 x 30/1000
                              = 0.0012 mol
Excess HCl               = 0.0020 – 0.0012
                              = 0.0008 mol
Total volume           = 50 cm3
New [H+]                = moles/volume
                              = 0.0008/(50/1000)
                              = 0.016 mol dm-3
pH                          = -log(0.016)
                              = 1.80

4)      Buffer solutions

  • A buffer solution minimises pH changes on addition of an acid or base
  • Buffer solutions are important for controlling pH in blood (so that enzymes are not denatured) and shampoos (so that eyes do not sting and skin is not damaged)

a)   Acidic buffers

  • An acidic buffer consists of a weak acid and the salt of a weak acid (e.g. ethanoic acid & sodium ethanoate)
  • For ethanoic acid/sodium ethanoate, the following equilibrium exists:

CH3COOH(aq) ¾ CH3COO-(aq) + H+(aq)

  • If a small amount of acid is added, equilibrium will shift to the left to remove the added H+. The following reaction occurs: CH3COO- + H+ ® CH3COOH
  • If a small amount of base is added, the OH- will react with H+ to form water. The equilibrium will shift to the right to replace the H+ that has been removed. The following reaction occurs: CH3COOH ® CH3COO- + H+

b)   Calculating the pH of a buffer solution

  • The pH of an acidic buffer can be calculated using the Ka expression for the weak acid e.g. for ethanoic acid/sodium ethanoate:

Ka = [CH3COO-][H+]

      Rearranging gives:
                                          [H+] = Ka x      [CH3COOH]

Calculate the pH of a buffer solution containing equal volumes of 2.5 mol dm-3 HCOONa and 1.0 mol dm-3 HCOOH (Ka = 1.6 x 10-4 mol dm-3)

      [H+]      = Ka x [HCOOH]/[HCOONa]
                  = 1.6 x 10-4 x 1.0/2.5
                  = 6.4 x 10-5 mol dm-3
      pH        = 4.19

Source:           OCR June 2003 paper
  • The pH of a particular buffer depends on the value of Ka and the ratio of [CH3COOH] to [CH3COO-]

d)   pH of Blood

  • The pH of blood is kept in the range 7.35-7.45
  • This is achieved by dissolved carbon dioxide acting as a buffer solution
  • The relevant equations are:

CO2 + H2O ¾ H2CO3
H2CO3 ¾ HCO3- + H+

  • H2CO3 is carbonic acid. HCO3- is the hydrogencarbonate ion
  • If the blood pH is less than 7.35 (too acidic), more CO2 is breathed out. Both equilibria shift to the left to produce more CO2 and [H+] is reduced
  • If the blood pH is more than 7.45 (too alkaline), less CO2 is breathed out. Both equilibria shift to the right to remove CO2 and [H+] is increased

5)      Enthalpy of neutralisation

            H+(aq) + OH-(aq) ® H2O(l)

  • This is a calorimetry calculation e.g.


50 cm3 of 1.0 mol dm-3 hydrochloric acid was added to 50 cm3 of 1.0 mol dm-3 sodium hydroxide solution.   The temperature rose by 6.8°C.  Calculate the enthalpy of neutralisation for this reaction.  Assume that the density of the solution is 1.00 g cm-3, the specific heat capacity of the solution is 4.18 J g-1 K-1.

                  HCl + NaOH ® NaCl + H2O

m   = mass of solution = total of acid and alkali = 100g
q    = -mcDT/1000
      = -100 x 4.18 x 6.8/1000
      = -2.8424 kJ

n    = moles HCl = moles NaOH = 1.0 x 50/1000 = 0.05 mol

DH  = q/n
      = -2.8424/0.05
      = -56.8 kJ mol-1