Colour By Design

Question 1

Aromatic compounds are?

Answer 1

Arenes/aromatic compounds contain benzene rings.

Question 2

Carbon structure?

Answer 2

Carbon has 4 valent electrons.

Each carbon is bonded to 2 other carbons and 1 hydrogen.

Lone electron is on p-orbital which sticks out above and below the planar ring.

Drawing on notes.

Question 3

Different benzene structure?

Answer 3

Can be shown using the:

• Kehule structure (single and double bonds alternating).
• delocalised structure (a circle in the middle of the 6 sided shape).

C6H6 formula.

It’s a cyclic, planar molecule.

Question 4

Why was the Kehule structure proven wrong using carbon length?

Answer 4

Benzene couldn’t have been a simple carbon chain because there aren’t enough hydrogens.

August Kehule solved this, proposing the carbons were arranged in a planar ring with single and double alternating bonds.

However, the X-ray diffraction studies showed that all carbon-carbon bonds were the same length (Kehule proposed that 3 would be shorter and 3 would be longer because of the double and single bonds).

Kekules structure is called cyclohaxa-1,3,5-triene.

The bond length for a single bond is 154pm.
For double bond is 134pm.
Benzene’s c-c- lengths are right in the middle.

Question 5

The delocalised model of benzene?

Answer 5

All bonds are the same length - between a double and single.

Each carbon donated an electron from its p-orbital. The p-orbitals therefore combine to form a ring of delocalised electrons.

Draw circle in middle.

Question 6

Why does benzene react via substitution?

Answer 6

You would expect from Kekules structure that benzene would react via electrophilic addiction because of the double c-c bonds (act as alkenes). It was expected that a quick reaction would occur.

This did not happen. Benzene doesn’t react via addition. And it reacts very slowly, needing a catalyst begin to react at all.

Benzene reacts via electrophilic substitution instead.

The electrons in the delocalised ring have more room than if they were squeezed into localised double bonds. They can get further away from eachother, spreading out the negative charge so that the molecule is more stable.

An addition reaction would need to take electrons from the stable delocalised ring to form new bonds. Substitution reactions don’t do this - a hydrogen atom just gets swapped for something else and the stability of the delocalised electrons is preserved.

Question 7

Enthalpy changes of benzene?

Answer 7

We work out the enthalpy change by undergoing hydrogenation of benzene.

Cyclohexane has one double bond and the enthalpy change of that bond is -120kj mol-1.

In kekules model, you would expect the enthalpy change to be -360 (3x that^) because it has 3 double bonds.

The enthalpy change was actually -208kj mol-1 - far less exothermic than expected. This is the experimental value.

Energy is put in to break bonds and that energy is released again when bonds are made. So more energy must have been put in to break the bonds in benzene than would be needed to break the bonds in the Kehule structure.

This difference indicated that benzene is more stable than the kekule structure. This is because of the delocalised ring of electrons.

The symbol of hydrogenation is triangle, H then a little circle at top.

Question 8

What is an electrophile?

Answer 8

The compound that accepts the electrons from the delocalised ring.

They need to be strongly positively charged to be able to attack the stable ring of benzene.

Only strongly polarised compounds or positive ions can do this.

Question 9

Nitration of benzene?

Answer 9

Arenes/aromatic compounds including benzene undergo electrophilic substitutions with nitronium ions as the electrophile.

If you warm benzene with concentrated nitric and sulfuric acids, you get nitrobenzene.

Sulfuric acid is the catalyst - it helps make the nitronium ion (NO2^+) which is the electrophile.

Mechanism for this is on paper.

1. Nitronoum ion attacks the benzene ring.
2. An unstable intermediate forms.
3. The H+ ion is lost.

The catalyst - sulfuric acid - is then reformed:
HSO4^- + H+ —> H2SO4

If you only want one substitution (mononitration) do it below 55 degrees. If you want lots of substitutions (lots of NO2 groups added), do it above 55 degrees.

Question 10

Nitration of benzene equation?

Answer 10

1. HNO3 + H2SO4 —> H2NO3^+ + HSP4^-
2. H2NO3^+ —> NO2^+ + H2O

Overall:
C6H6 + HNO3 (concentrated) —(conc H2SO4)—> C6H5NO2 + H2O

Conc H2SO4 is sulfuric acid as catalyst.

Question 11

Sulfonation of benzene?

Answer 11

Sulfur trioxide ions are SO3. They are the electrophile (the thing that attacks the benzene ring).

If you wanted to make benzenesulfonic acid:
1. Boil benzene with concentrated sulfuric acid for several hours.
Or,
2. Warm benzene to 40 degrees with fuming sulfuric acid for half and hour.

The electrophile in these reactions is SO3 because concentrated sulfuric acid breaks like this:
H2SO4 —> H2O of SO3
And fuming sulfuric acid is lots of SO3 molecules dissolved in sulfuric acid. It’s richer in SO3 than conc. This is why it needs less heat and it’s quicker.

The mechanism is on paper.

1. The SO3 attacks the benzene, drawing a pair of electrons from the delocalised ring.
2. The -ve O atom on the SO3^- takes an H atom from the benzene. The pair of electrons in the C-H bond move to the delocalised ring.
3. Benzenesulfornic acid is formed.

Question 11

How do halogen carriers help make good electrophiles?

Answer 11

An electrophile has to have a strong positive charge to be able to attack the stable benzene ring (only positive ions or strongly polar compounds will attack the benzene ring).

Most compounds are not polarised enough - but some can be made into stronger electrophiles using a catalyst called a ‘halogen carrier’.

A halogen carrier accepts a lone pair of electrons from the polar molecule containing a halogen - the electrophile. As the lone pair of electrons is pulled away, the polarisation in the electrophile increases and sometimes a carbocation forms.

A carbocation is an organic ion with a positively charged carbon atom.

This makes the electrophile a lot stronger.

Halogen carriers include alimony halides, iron halides and iron.

Mechanism is on paper.

Question 12

Formation of benzenesulfornic acid overall equation?

Answer 12

C6H6 + H2SO4 —> C6H5SO3H + H2O

Question 13

Why do we use acetylation and alkylation?

Answer 13

Aceyl group (R-CO) or alkyl group (R-) can be added to benzene via substitution reactions.

When it’s added, the benzene structure is weaker and so it’s easier to modify the benzene further and make useful products.

Question 15

Friedel-Crafts reactions?

Answer 15

There’s two types:
Alkylation and acylation.

They follow the electrophilic substitution mechanism.

Question 16

What does reflux mean?

Answer 16

Reflux means boiling reactants in a flak fitted with a condenser to stop them boiling away.

Question 17

Alkylation of benzene?

Answer 17

This is where electrophilic substitution takes place to substitute a alkyl group onto a benzene so it becomes a ‘methylbenzene’ or something similar.

It involves reaction of benzene under reflux with a halogenalkane (which provides the alkyl group) and a halogen carrier as a catalyst.

The hydrogen carrier can be AlCl3 or FeCl3.

Since the halogen carriers don’t get used up, they’re sometimes called Friedel-Crafts catalysts.

This general reaction can be represented by equation:
C6H6 + RCl —(reflux and AlCl3^)—> C6H5R + HCl.

R is an alkyl group.

This is basically how an alkyl group can be added to a benzene ring.

The electrophile is R+ (the alkyl group).
Formation of electrophile:
R-Cl + AlCl3 —> R+ + AlCl4^-

The mechanism is on paper.

1. The carbocation is the electrophile. It attracts t he electrons in the delocalised ring.
2. An unstable intermediate forms.
3. Methylbenzene is made an the H+ ion is lost.

AlCl3 is regenerated here.

Produces an alkylarene.

Question 18

Acylation of benzene?

Answer 18

Used to substitute an acyl group onto a benzene ring producing a phenylketone.

Involves benzene under reflux with an acyl chloride (provides the acyl group) and a halogen carrier catalyst.

The halogen carrier catalyst can be AlCl3 or FeCl3.

The mechanism is on paper.

The electrophile is the R-C+—O.
Paper.

Formation of electrophile:
R-C—O-Cl + AlCl3 —> R-C+—O + AlCl4^-
On paper.

Overall:
On paper.

Phenylethanone is produced - this is when the acyl group is on the benzene ring.

AlCl3 is also regenerated here.

Question 19

Regeneration of AlCl3?

Answer 19

Regeneration:

AlCl4- + H+ —> AlCl3 + HCl

Question 20

Formation of a carboncation?

Answer 20

The carbocation if formed from the chloroalkane and AlCl3:

CH3Cl + AlCl3 —> CH3 + AlCl4^-

Question 21

Chlorination/Bromination of benzene?

Answer 21

Mechanism is on paper.

Question 22

How do halogen carriers help halogens substitute into the benzene ring?

Answer 22

Halogen carriers polarise halogen molecules, such as Br2 or Cl2.

The positively charged end of the halogen molecule then acts as an electrophile and reacts with the benzene ring in the usually electrophilic substitution reaction.

Mechanism on paper.

Question 23

How does benzene react with bromine?

Answer 23

The electrophile is Br+.

C6H6 + Br —> C6H5Br + HBr.

This is an electrophilic substitution reaction.

It requires a halogen-carrier (catalyst), e.g. FeBr3.

It is less reactive than when alkenes react with bromine. Why?
1. Electrons in benzene are less reactive than alkenes. Therefore, the electron density is lower (because the electrons are more spaced out).

2. This means that the Br-Br covalent bond (which is non-polar) cannot attract the pie electrons from the benzene because there is not enough repulsion from the pie electron pair and the benzene to generate a dipole which would split the Br-Br bond.
3. A catalyst therefore needs to be used to generate an electrophile (Br+) which created a dipole between the Br-Br bond. This creates a sufficient attar film between the pie electrons and the Br-Br bond is split. This allows for the benzene to have a Br on it.

A catalyst is needed and therefore, it’s less reactive than alkenes which don’t need catalysts.

Question 24

How do alkenes react with bromine?

Answer 24

Must know the mechanism for this reaction. Saved to notes.

Alkenes react with bromine water and a colour change occurs - brown-orange to colourless. Bromine (brown-orange) is the electrophile and adds to the alkene forming a dibromoalkane (colourless).

In this reaction, Br2 is polarised (a delta positive is formed) as the electrons in the double bond repels the electrons in the Br2.

An electron pair in the double bond is attracted to the delta positive bromine and forms a bond. This breaks the Br-Br bond.

A carbocation intermediate is formed and Br is attracted to C+.

Couolourless 1,2-dibromoethane is formed.

This is an electrophilic addition reaction.

Alkeke is C2H4. Double bond between C’s. On paper.

This reaction is more reactive than benzene.

Why?
1. Electrons in alkene between the C double bond C are localised. This means the electron density is higher (less spread out).

2. The electrons between the Br-Br are repelled by pie electron in alkene, creating a dipole bond (slight positive and negative charge between atoms) between the Br-Br.
3. This means the pie electrons in the alkene are attracted to the bromine and the Br-Br bond breaks.
4. A carbocation intermediate is formed and Br- is attracted to C+.
5. Colourless 1,2-bromoethane is formed.

There’s no need for a catalyst so this is more reactive.

This turns the brown-orange bromine to colourless when reacting with alkene.

Question 25

What does it mean when something is delocalised?

Answer 25

The electrons are shared between 2 or more carbons. They’re spread out.

Question 26

What does it mean when a bond is non-polar?

Answer 26

The electronegativity of the two atoms either side of the bond is the same.

Question 27

What are azo dyes?

Answer 27

Azo dyes are man made.

They contain -N—N- (azo groups) which link two benzene rings together.
^ double bonded.

They’re stable because of the benzene delocalised electron rings. Their stability is useful for dying fabrics because the dye doesn’t compose.

They’re colourful because light is absorbed by delocalised electrons - the electrons vary depending on what amines/aromatic groups are used to make the dye and therefore this changes the colour.

Question 28

How to make an azo dye - process 1?

Answer 28

There are 2 separate processes to making a azo dye: make the diazonium salt and then couple it with phenol.

Process 1: making diazonium salt.
1. Make nitrous acid (HNO2) - react sodium nitrate and hydrochloric acid together.

2. Then make the diazonium salt - react nitrous acid, hydrochloric acid and phenylamine together.

NaNO2 + HCl —> HNO2 + NaCl

Nitrous acid is unstable and so steps 1 and 2 should be done in situ (same reaction at same time in same beaker).

Mechanism is on paper.

This reaction will only occur if the reaction is kept under 5 degrees Celsius. If over 5 degrees, phenol is made instead. Not what we want.

Question 29

Making an azo dye: process 2?

Answer 29

Process 2: making azo dye.
1. Add phenol to sodium hydroxide to make sodium phenoxide.

2. Then add the diazonium salt (e.g. benzenediazonium chloride).
3. The azo dye will precipitate out quickly.

Make sure the reaction vessel is kept under 5 degrees (ice bath) for this whole thing.

Phenol is a coupling agent because it has a high electron density. Some electrons on the O of the -OH actually integrate into the benzene ring on phenol. This makes carbon 2,4,6 on benzene ring susceptible to an attack from the weak electrophile (which is the diazonium ion from salt).

This means the -N—N- will always attach to the phenol on the 2nd, 4th or 6th carbon (the OH is always on 1st).

Mechanism on paper.

Question 30

What does it mean if a dye is colourfast?

Answer 30

For it to be practical, it must be colourfast. This means that it doesn’t fade by light or can’t be washed out. Some dyes are more colourfast than others.

The more colourfast it is, the better it is at bonding to fibres.

Dyes have functional groups that allow the dye to interact with groups on fibres (like -OH groups on cellulose fabrics, like cotton).

Question 31

How do dyes bond using amine groups?

Answer 31

Amine groups (NH2) are attached to dyes to form hydrogen bonds with cellulose fabrics.

Cellulose fabrics have the functional group (-OH) in them. E.g. cotton.

Problem with this is that hydrogen bonds (H+ bonds) are not very strong compared to other bonds used in dyes. This means they’re not very colourfast.

Colour can bleed out fabric.

Question 32

How do dyes bond using ionic salts?

Answer 32

Ionic salts are easy to recognise because they’ll have a small + on one part of the molecule and a small - on another part.

E.g. SO3-Na+ is a ionic salt.

Ionic salts are attached to azo dyes and dissolved in H2O/acid to form SO3- ions and Na+ ions.

They attach to -NH (found in nylons).

This forms Na2+ ions with the N’s on the NH.

The SO3- is then attached to the dye which is attracted to the Na2+ ion attached to fabric.

Other examples of ionic salts are -SO3H which form SO3- ions and H+ ions.
COOH which form COO- ions and H+ ions. They both donate protons to the -NH- on fabric to form -NH2+ attractions.

Question 33

Fibre reactive dyes?

Answer 33

Form covalent bonds between -NH- or -OH- on fibre.

They’re the most permanent dyes on the market.

Question 34

Why are azo dyes colourful?

Answer 34

The chromophores are the benzene ring-N—N-benzene ring. Nothing else.

Electrons are delocalised in benzene ring which extends into azo link - this is the chromophore and what makes the dyes colourful. The chromospore can get become larger because of more delocalisation of electrons.

Chromosomes absorbs certain wavelengths of light.

Colours that are NOT absorbed are the ones we observe.

A chromosphere can be effected by features which have high electron density, such as:

• benzene rings.
• lone pairs of electrons.
• double bonds,
• triple bonds.

These features delocalise electrons more. If the chromophore is adapted, the frequencies absorbed are changed and the observed colour changes too.

E.g. lone pairs of electrons on functional groups change the distribution of electrons and can incorporate the electrons into a whole chromophore. This means a different colour will be seen because a different frequency of light is being absorbed by the chromophore.

They can also change colour due to electron orbitals - later flashcard.

Question 35

Why and how are azo dyes soluble in water?

Answer 35

Solubility functional groups are added to azo dyes to help them be soluble.

Sodium sulfate salts are solubility functional groups. Look at paper to see how to draw. Ionic compounds are generally soluble too - so look out for them as solubility functional groups.

Question 36

How can colour of dyes be impacted by molecular orbitals?

Answer 36

Colours of dyes are also effected by molecular orbitals - they only exist on molecular compounds.

Molecular orbitals are created from linking atomic orbitals (p/s/s orbitals are atomic). The atomic orbitals collapse together and overlap to form molecular orbitals.

How a molecular orbital is filled:

• the molecular orbital fills from the lowest energy level first.
• electrons absorb light energy (either UV of visible) so they move from a lower energy level (ground state) to a higher energy level (excited state). This excited state is a molecular orbital.
• in order for this to happen, the AE (the energy gap between each energy level) must be equal to the frequency of light energy.

If AE corresponds to frequency of visible light, we will be able to see the colour. But if AE corresponds to frequency of UV light, we won’t be able to see the colour change.

Colour observed is complementary to AE. (A is a triangle).

Question 37

Determining how many molecular orbitals are formed?

Answer 37

A single bond - 1 orbital from each atom with 1 electron in each forms 2 molecular orbitals.

Double bond - 2 orbitals from each atom form 4 molecular orbitals.

High frequency UV is needed to excite electrons from single and double bonds because the AE (gap) is greater. A bit lower of a frequency is needed for double.

We can’t see these changes because the AE is complementary to UV.

Triple bonds - the AE is closer. So low frequency UV is needed OR visible light.

Delocalised system - visible light frequency.

We can see these colour changes.

Summary - the more delocalisation, the closer the molecular orbitals and the lower the frequency absorbed. If it’s absorbing visible light, we can see it.

Question 38

How do energy levels effect the colour of a molecule?

Answer 38

Electrons move to a higher energy level.

Absorbing em radiation Δ E = hv

Yellow dye molecule has more delocalisation / larger chromophore AND energy levels are closer together / Δ E is smaller

Energy absorbed is at lower visible frequencies and so we can observe colour.

Complementary colour is transmitted.