Chemistry: Colour by Design

a, b(iii) How dyes attach to fibres:

Test yourself
• Most fibres (e.g. nylon, wool) have functional groups which dyes bind to, such as:
  - COOH
  - NH
  - CONH (peptide)
• Hydrogen bonds are formed between OH groups on fibre molecules and some dyes (like those with a NH₂ group)
• Ionic bonds may bind a dye to a fibre. Usually, this is with an -NH₃⁺ group on the fibre and -SO₃^- on the dye
• Covalent bonds are strongest, and dyes which bond covalently have a functional group which will react with the fibre
• Instantaneous dipole-induced dipole bonds may also bind a dye to a fibre. This usually only happens with small dye molecules since the bonds are very weak

Colourfast dyes:

• Colourfast dyes don't fade or wash out very easily
• This is due to strong bonds between the dye and fabric - typically only ionic bonds and covalent bonds are strong enough to be considered highly colourfast

Fibre reactive dyes:

• Dyes which form covalent bonds with the fabric are called fibre reactive

b(i) Dye chromophores:

Test yourself
• The chromophore is the part of a dye which gives it its colour
• It usually contains double/triple bonds, lone pairs and benzene rings
• The colour can be changed by removing/adding functional groups such as phenol, nitro (NO₂) and amine

b(ii) Dye solubility:

Test yourself
• Dyeing is done by soaking the fibre in a solution of the dye. Therefore, dyes need to be water soluble (otherwise they would be called a pigment)
• To make a dye more water soluble, solubilising functional groups can be added, usually ionic groups
• For example, sulfonate ions (SO₃^2-) can be added (often as a salt, SO₃^- Na⁺)

c Structure of fats/oils:

Test yourself
• propane-1,2,3-triol (glycerol) has the following structure:
  
• Fatty acids contain a COOH group attached to a long chain
  - saturated fatty acids have single C-C bonds only
  - unsaturated fatty acids have one or more C=C double bonds
• A fatty acid's COOH group will react with the OH group on propane-1,2,3-triol, joining them together. If three fatty acid molecules are added, a triglyceride is created, alongside water, in a condensation/esterification reaction
• In the condensation reaction, the OH is lost from the propane-1,2,3-triol and H is lost from the fatty acid, making one H₂O molecule (per fatty acid)
• An example of a triglyceride:
  
  This was made by reacting propane-1,2,3-triol with pentanoic acid. 6 hydrogen atoms and 3 oxygen atoms were lost, so 3H₂O would also be produced for every molecule of the triglyceride

Fats and oils:

• Fats and oils are names for triglycerides. The only difference between the two is melting point:
  - Fats have a melting point above room temperature (so they're solids at RTP)
  - Oils have a melting point below room temperature (so they're liquids at RTP)
• So short-chain triglycerides (such as the one above) are usually oils since they have a low melting point
• Most fats/oils have three different fatty acid groups with varying degrees of unsaturation

d Arenes:

Test yourself
• Arenes are cyclic hydrocarbons with a planar shape, where the π electrons are stabilised by electron delocalisation
• The number of π electrons must also be exactly 4n + 2 (i.e. 6, 10, 14, 18, ...)
• A phenyl group is a benzene ring with one hydrogen replaced by another group

Substitution:

• Usually, Br₂ will react with an alkene to produce a dibromoalkane
• However, benzene does not react with bromine in an addition reaction like with straight chain alkenes. Instead, it reacts very slowly by electrophilic substitution
• This is explained by the increased stability with delocalised rings

e Benzene representations:

Test yourself
• Benzene can be represented in two ways:
  
• The one on the left is the current model, with a delocalised ring of electrons. This delocalised ring is above and below the plane of the molecule, made from p-orbitals
• The model on the right is an older suggestion. If this were true, the bond lengths would be different, as double bonds are shorter. However, we now have evidence that benzene is a regular hexagon

Hydrogenation evidence for the delocalised model:

• Hydrogenation is a reaction where H₂ is added to an alkene to remove its double bonds
• Cyclohexene is a six-carbon ring with one double bond
• Its hydrogenation enthalpy change is -120
• With the three double-bond structure of benzene, it would have a hydrogenation enthalpy change of 3× this
• However, experimental evidence suggests that it's much less exothermic than this
• This is explained by the delocalised ring's increased stability - more energy would be needed to hydrogenate it

f Polyfunctional molecules:

Test yourself
• Polyfunctional molecules have multiple functional groups
• The highest precedence functional group in the molecule according to the list below forms the end of the name. For example, any molecule with a COOH group will likely end in 'oic acid' because this is one of the highest precedence functional groups
• This list is created by IUPAC so that all chemists use just one name for a specific compound. I only listed groups relevant at A Level here
    [Highest precedence]
  - Cations (e.g. NH₄⁺, suffix -ammonium)
  - Carboxylic acids (-oic acid)
  - Carboxylic acid derivatives (e.g. esters (-oate), amides(-amide))
  - Aldehydes (-al)
  - Ketones (-one)
  - Alcohols (-ol)
  - Amines (-amine)
    [Lowest precedence]
• All other functional groups should be listed in alphabetical order as a prefix

Naming example:

• Example structure:
  
• Its highest precedence group is COOH, so it must end in 'oic acid'
• Its other groups are 'phenyl' (-C₆H₅, benzene with one less hydrogen) and 'amino' (-NH₂)
• Position number 1 is on the COOH group's carbon atom, so it has '3-phenyl' (because the benzene ring is bonded to the 3^rd carbon) and '2-amino' (because the amine group is bonded to the 2^nd carbon)
• Now put the prefixes in alphabetical order, separated by dashes: 2-amino-3-phenyl
• And add the suffix: 2-amino-3-phenylpropanoic acid

Tests:

• Tests for identifying functional groups on monofunctional molecules (e.g. the iron(III) chloride test for phenol) work on polyfunctional molecules too
• If you do two tests for two different functional groups and they are both positive, it would either suggest that you have an impure sample or a polyfunctional molecule. To find out which is correct, use melting point apparatus - if it all melts at one temperature and passes 2 or more tests, it must be polyfunctional

g Arene electrophilic substitution:

Test yourself
• Benzene mainly reacts with electrophilic substitution reactions
• The A Level mechanisms are listed below

Nitration:

• Nitrobenzene is formed if you gently heat benzene, concentrated nitric acid and concentrated sulfuric acid (catalyst). Keep the temperature under 55 °C if you want only one NO₂ group per benzene ring
  1) HNO₃ + H₂SO₄ → H₂NO₃⁺ + HSO₄^-
  2) H₂NO₃⁺ → NO₂⁺ + H₂O
       The nitronium ion (NO₂⁺) is the electrophile
  3) The nitronium ion attacks the benzene ring to bond with it. This is unstable
  4) The H⁺ ion is lost, and then reacts with the HSO₄^- to reform the catalyst
  

Sulfonation:

• Benzenesulfonic acid is formed if you:
  - Heat benzene under reflux with concentrated sulfuric acid for a few hours
  - Warm benzene under reflux at 40 °C with fuming sulfuric acid for 20-30 minutes
     - Fuming sulfuric acid is SO₃ (sulfur trioxide) dissolved in sulfuric acid, so it contains more SO₃ than concentrated sulfuric acid
  1) H₂SO₄ → H₂O + SO₃
       SO₃, sulfur trioxide, is the electrophile
  2) SO₃ attacks the benzene and takes two electrons (the second goes to an O)
  3) The new O^- takes a H atom from the benzene
  4) The electron pair from the old C-H bond join the delocalised ring
  

Friedel-Crafts alkylation:

• This method is used to add alkyl groups to benzene
• It uses a halogen carrier as a catalyst, e.g. AlCl₃
• Halogen carriers work by polarising halogen molecules, making them into electrophiles
• The alkyl group must be bonded to chlorine - e.g. CH₃Cl
  1) CH₃Cl + AlCl₃ → CH₃⁺ + AlCl₄^-
  2) Now the CH₃⁺ reacts with benzene with electrophilic substitution (under reflux)
  3) The halogen carrier now gets regenerated by reacting with the H⁺ from the benzene
  

Friedel-Crafts acylation:

• Acyl groups (with C=O) can also be added to benzene to create a phenylketone. Again, this reaction requires reflux
  1) CH₃COCl + AlCl₃ → CH₃CO⁺ + AlCl₄^-. The CH₃CO⁺ is called an acylium ion
  2) Now the CH₃CO⁺ reacts with benzene with electrophilic substitution (under reflux)
  3) The AlCl₃ is now regenerated with a reaction between the H⁺ and AlCl₄^-
  

Halogenation:

• Halogenation is the adding of a group 7 atom to a benzene ring

  Halogenation with bromine:

• Use reflux and an iron catalyst (e.g. iron filings)
• Add Br₂(l)
• You will get a mixture of bromobenzene and HBr
• The mechanism is:
  - 2Fe + 3Br₂ → 2FeBr₃
  - Br₂ + FeBr₃ → Br⁺ + FeBr₄^-
  - A H⁺ on the ring is replaced by the Br⁺
  - H⁺ + FeBr₄^- → HBr + FeBr₃ (the catalyst is regenerated)

  Halogenation with chlorine:

• Use an aluminium chloride catalyst at room temperature and pressure
• Make sure to use anhydrous conditions because aluminium chloride reacts violently with water
• The mechanism is similar to with bromine:
  - Cl₂ + AlCl₃ → AlCl₄^- + Cl⁺
  - A H⁺ on the ring is replaced by the Cl⁺
  - H⁺ + AlCl₄^- → HCl + AlCl₃

h Diazonium compounds:

Test yourself
• Azo dyes are dyes containing an azo group (-N=N-). Usually, there is an aromatic group bonded to each nitrogen. The electrons delocalise across the entire molecule

Making a diazonium salt:

• A diazonium salt contains a -N≡N⁺-. This is often ionically bonded to a chloride ion, as shown below:
  
• To make one, add sodium nitrate and hydrochloric acid with a phenylamine
• This makes nitrous acid (HNO₂) with the reaction NaNO₂ + HCl → HNO₂ + NaCl
• The nitrous acid then reacts with the phenylamine (benzene ring and NH₂ group)
• The reason you add all three reactants at once is because nitrous acid is unstable, and if it's in a phenylamine solution, they can react almost instantly
• Water is also produced in the reaction
• The temperature must be kept under 10 °C to prevent a phenol from forming

Coupling to make an azo dye:

• The second step is to couple the diazonium salt with a sodium phenoxide solution (C₆H₅O^-Na⁺; made by dissolving phenol in sodium hydroxide). In ice, the salt and sodium phenoxide will react to create a precipitate - the azo dye. This step must be done in alkaline conditions
  
• The diazonium ion attacks the phenol electrophilically. It works with phenol because the oxygen's lone pairs lower the electron density of the area near to it with repulsion. This makes the side of the phenol opposite to the OH group a nucleophile
• Sodium chloride and water will be produced as byproducts

i Fehling's solution:

Test yourself
• Fehling's solution is made by dissolving a specific copper(II) complex in sodium hydroxide. It is blue, due to the Cu²⁺
• It is used to distinguish between an aldehyde and ketone
• Add a few drops of the unknown carbonyl compound to a test tube with about 2 of Fehling's solution. Put it in a hot water bath for a few minutes (we don't use a Bunsen flame for these tests because aldehydes and ketones are flammable)
• With the presence of an aldehyde, it will form a brick red precipitate, Cu₂O. H₂O and RCOO^- are also formed. We don't need to know an equation, only the products
• With a ketone, nothing will happen - the solution will remain blue

Tollen's reagent:

• Tollen's reagent is also used to distinguish between aldehydes and ketones. It contains diamminesilver(I) ions
• You can make Tollen's reagent by:
  - Putting 2 of silver nitrate solution in a test tube
  - Add some sodium hydroxide solution with a pipette to create a brown precipitate
  - Add dilute ammonia solution with a pipette until the brown precipitate dissolves
• Perform the test in a hot water bath as before. Add the aldehyde/ketone with a pipette (a few drops) and wait a few minutes
• Aldehydes reduce the diamminesilver(I) to elemental silver, giving the solution a shiny silver 'mirror' precipitate
• RCOO^-, 4NH₃ and H₂O are also formed

i(iii), k Formation of cyanohydrins:

• Hydrogen cyanide, HCN, is a weak acid, so it partially dissociates in water: HCN ⇌ H⁺ + CN^-
• HCN reacts with carbonyl compounds in a nucleophilic addition reaction, producing cyanohydrins (molecules with -CN and -OH)
• The CN^- attacks the partially positive carbon. The oxygen gains two electrons and these bond to the H⁺ from the HCN(aq). This mechanism is shown below:
  

j Organic synthesis:

• Below are the pathways you need to be familiar with. They are all covered elsewhere in the course, but it much easier to memorise them from this page
• There are five organic reactions given in the data sheet. Those are not listed here

Routes:

• Acyl chloride → ester: alcohol
• Aldehyde → carboxylic acid: K₂Cr₂O₇, H₂SO₄ catalyst, reflux
• Aldehyde → cyanohydrin: HCN
• Alkane → bromoalkane: Br₂, UV light
• Alkene → alkane: Ni catalyst, 150 °C, 5 atm
• Alkene → bromoalkane: HBr, 20 °C
• Alkene → dibromoalkane: Br₂, 20 °C
• Alkene → primary alcohol: H₂SO₄, cold H₂O. Or H₃PO₄, steam, 300 °C
• Bromoalkane → primary amine: concentrated NH₃ solution in ethanol
• Carboxylic acid → ester: alcohol, H⁺ catalyst, reflux
• Ester → carboxylic acid: H⁺(aq), reflux
• Ketone → cyanohydrin: HCN
• Primary amine → n-substituted amide: acyl chloride
• Primary alcohol → aldehyde: K₂Cr₂O₇, H₂SO₄ catalyst, heat then distil
• Primary alcohol → alkene: conc H₂SO₄ catalyst, 170 °C, reflux
• Primary alcohol → bromoalkane: conc H₂SO₄, NaBr, warm then distil
• Primary amide → carboxylic acid: H⁺(aq), heat
• Primary bromoalkane → primary alcohol: NaOH(aq), reflux
• Secondary alcohol → ketone: K₂Cr₂O₇, H₂SO₄ catalyst, reflux
• Secondary bromoalkane → secondary alcohol: NaOH(aq), reflux

Adding groups to benzene:

• Add a carbon chain (alkylation): RCl, AlCl₃ catalyst, reflux
• Add NO₂ (nitration): conc H₂SO₄, conc HNO₃, 55 °C
• Add SO₃H (sulfur bonded to -OH and double bonded to two oxygen atoms) (sulfonation): fuming H₂SO₄, 40 °C
• Add RCO (acylation): RCOCl, AlCl₃ catalyst, reflux
• Add Cl (chlorination): Cl₂, warm AlCl₃ catalyst

Other aromatic reactions:

• Phenol to phenyl ethanoate (shown below): CH₃COCl
  
• Phenol to sodium phenoxide (shown below): NaOH
  

l Organic reaction types:

Test yourself
• Every organic reaction is one or more of the following:
• Addition: two molecules join, breaking a double bond to make one product
• Condensation: two molecules join, with a small molecule being lost (e.g. H₂O, HCl) [opposite of hydrolysis]
• Elimination: a functional group is lost, released as part of a small molecule
• Substitution: a functional group on a molecule is replaced by another group
• Oxidation: the loss of electrons - usually by losing a hydrogen or gaining an oxygen
• Reduction: the gain of electrons - usually by gaining a hydrogen or losing an oxygen
• Hydrolysis: water splits a molecule into two [opposite of condensation]

m Colour in organic substances:

Test yourself
• Substances absorb photons (radiation) and this provides the energy for electrons to move from their ground state to an excited state
• If this photon is visible light, that colour will be absorbed. Because all other colours pass through the substance, it will appear the complementary colour of the absorbed colour
• The complementary colours are (you don't need to remember these):
  - red and green
  - orange and blue
  - yellow and violet

Delocalisation:

• Double bonds have closer together energy levels (when comparing the energy levels of all four of their orbitals) than single covalent bonds. Single bonds will absorb UV and not visible light, and double bonds on their own will absorb low-frequency UV
• Delocalisation, such as in benzene, will decrease this frequency further, usually into the visible part of the spectrum
• As well as in aromatic molecules, delocalisation also occurs in conjugated systems - a chain of alternating double and single bonds. In these chains, the π bonds will spread across the chain, leading to delocalised p-electrons
• Conjugation can also happen with C=C, C≡C, C=O, N=N and other similar double/triple bonds
• Usually, about 5 or more π bonds are needed in a conjugated system for the molecule to absorb visible light

n Gas-liquid chromatography (GLC):

Test yourself
• The sample is injected with a syringe into a stream of carrier gas, an inert gas such as helium
• It enters an oven and turns into a gas. It passes through a coiled column (tube). It contains a very porous material and a liquid with a high boiling point (e.g. oil)
• Compounds with a high boiling point will spend a lot of time condensed in liquid form at the start of the column, so they will spend a long time in the machine. This time is known as the retention time

GLC chromatograms:

• Each substance other than the carrier gas detected by the detector is plotted on the GLC chromatogram with its corresponding time. The height shows how much of it was detected at that particular moment
• The retention time is the distance from 0 to the centre of the peak
• Retention times can be compared to data tables to determine which substance caused that peak
• The area under each peak is proportional to the amount of that substance
  

Adding to the results:

• Sometimes a mass spectrometer is used in addition to GLC. The GLC separates the initial substance, so these can be individually fed into a mass spectrometer. This allows us to more accurately determine the composition of the input, as well as the composition of each