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F324: Rings, Polymers and Analysis 1. Rings, Acids and Amines Structure of Benzene - Kekulé suggested that benzene was a cyclic molecule with alternating C=C double bonds. BUT, - Benzene was more stable than expected - Bromine doesn’t decolourise bromine water - All bonds in Benzene are the same length - Different enthalpy change than expected reacting with hydrogen SO THE CURRENT MODEL: - Each carbon atom contributes 1 electron to π-delocalised ring of electrons above and below the carbons. - Each carbon has one p-orbital perpendicular to the carbon plane. Each p-orbital overlaps with adjacent ones so the delocalisation is spread over 6 carbon atoms. MEANING - The π-delocalised ring accounts for increased stability of benzene. Electrophilic Substitution reactions of Benzene NITRATION OF BENZENE with Conc. H 2 SO 4 and HNO 3 - Reagents are H 2 SO 4 and HNO 3 and conditions of 60 o c OVERALL: C 6 H 6 + HNO 3 C 6 H 5 NO 2 + H 2 O MECHANISM: The electrophiles is generated: The Electrophiles attack at the benzene ring: Arenes - Aromatic hydrocarbons containing one or more benzene rings

Rings, Polymers and Analysis (Unit 4) - OCR Chemistry Notes- Robbie Peck

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Page 1: Rings, Polymers and Analysis (Unit 4) - OCR Chemistry Notes- Robbie Peck

F324: Rings, Polymers and Analysis

1.Rings, Acids and Amines

Arenes- Aromatic hydrocarbons containing one or more benzene rings

Structure of Benzene- Kekulé suggested that benzene was a cyclic molecule with alternating C=C double bonds.

BUT,- Benzene was more stable than expected - Bromine doesn’t decolourise bromine water- All bonds in Benzene are the same length- Different enthalpy change than expected reacting with hydrogen

SO THE CURRENT MODEL:- Each carbon atom contributes 1 electron to π-delocalised ring of electrons above

and below the carbons. - Each carbon has one p-orbital perpendicular to the carbon plane. Each p-orbital

overlaps with adjacent ones so the delocalisation is spread over 6 carbon atoms.MEANING

- The π-delocalised ring accounts for increased stability of benzene.

Electrophilic Substitution reactions of Benzene

NITRATION OF BENZENE with Conc. H2SO4 and HNO3

- Reagents are H2SO4 and HNO3 and conditions of 60ocOVERALL: C6H6 + HNO3 C6H5NO2 + H2OMECHANISM:

The electrophiles is generated:

The Electrophiles attack at the benzene ring:

The catalyst is re-generatedH+ + HSO4 H2SO4

Remember this can be made into an aromatic amine. (See amines)

Arenes- Aromatic hydrocarbons containing one or more benzene rings

Page 2: Rings, Polymers and Analysis (Unit 4) - OCR Chemistry Notes- Robbie Peck

HALOGENATION OF BENZENE- Reagents are Cl2 and AlCl3 and anhydrous conditions

OVERALL: C6H6 + Cl2 C6H5Cl + HClMECHANISM:

The electrophiles is generatedCl2 + AlCl3 Cl+ + AlCl4

-

The electrophile attacks the benzene ring

Regeneration of the catalystCl2 + AlCl3

- AlCl3 + HCl.

The relative ease of reaction with a cycloalkenes rather than benzene is of the high electron density in the C=C double bonds which can induce a dipole in the Br-Br bond making an electrophile. The electrophile is then attracted to the high electron density in the C=C bond.

Benzene doesn’t undergo bromination easily.

Phenols- In phenols, the OH is directly attached to the benzene

Reactions with sodium hydroxide and sodium- When dissolved in water, phenol forms a weak acid solution by losing the H+ from the –OH group.- C6H5OH + aq C6H5O- + H+

- Phenol is neutralised by aqueous sodium hydroxide to form the salt sodium phenoxide.

C6H5OH + NaOH C6H5O-Na+ + H2O- With sodium, phenol gives sodium phenoxide and hydrogen gas

Na + C6H5OH C6H5O-Na+ + 12

H2

Electron- pair donation of the oxygen p orbital- The lone-pair electrons on the oxygen p-orbital of the –OH

group can be delocalised on the benzene ring.- This increases the electron density on the ring (especially at sites 2,4,6) ,

making phenols more reactive than Benzene!

- Therefore Substitution occurs at the sites indicated on the diagram because of the increased electron density at these sites compared with the others.

- Substitution usually occurs at ALL THREE of these sites because of the extra reactivitye.g. REACTS WITH BROMINE to form 2,4,6-tribromophenol

C6H5OH + 3Br2 C6H2Br3OH + 3HBr where the product is 2,4,6-tribromophenolPhenolic Compounds; the uses!

- Production of plastics- Antiseptics

Page 3: Rings, Polymers and Analysis (Unit 4) - OCR Chemistry Notes- Robbie Peck

RECAP: The presence of this C=O carbonyl group means the molecule is unsaturated.The position of the carbonyl group determines whether it is an aldehyde or ketone.

Aldehydes are formed by oxidation of primary alcohols (not too long though!)2CH3CH2OH + [O] 2CH3CHOH + H2O

Ketones are formed by oxidation of secondary alcoholsCH3CHOHCH3 + [O] CH3COCH3 + H2O

Both use acidified potassium dichromate K2Cr2O7/H2SO4 is the ‘in’ one to remember. Orange Green.

- Disinfectants - Resins/PaintsCarbonyl compounds (C=O)

ReductionAldehydes and ketones can be reduced to their respective alcohols. NaBH4 is a suitable reducing agent. [H] is used to represent this reducing agent.

Aldehyde: CH3CH2COH + 2[H] CH3CH2CH2OHKetones: CH3COCH3 + 2[H] CH3CHOHCH3

This is a nucleophillic addition reactionThe carbonyl group is unsaturated and polar and consequently undergoes nucleophillic addition reactions.This occurs in the reduction reactions.NaBH4 is the source of the hydride ion H- which is the nucleophile for the reaction. The intermediate formed reacts with the solvent H2O to form the alcohol.

If the compound was a ketone it would make no difference.

Reactions with 2,4-dinitrophenulhydrazine (we can call it 2,4-DNPH ) These reactions are important because:

1. Carbonyls react with 2,4-DNPH to produce distinctive orange precipitates. Therefore this reaction can be used to identify an carbonyl group.

2. The organic product is relatively easy to purify by recrystillisation. Therefore the melting point of the brightly coloured precipitate can be measured. Each derivative has a different melting point and the value of the melting point can be used to identify the specific carbonyl.e.g. Ethanal at 142-43 oc, 2-Methylpropanal at 180-81oc etc.

Dude; is this chemical here an aldehyde?1. Aldehydes are readily oxidised to carboxylic acids. We take advantage of this with our Tollen’s test.2. Infra-red spectrum.

Page 4: Rings, Polymers and Analysis (Unit 4) - OCR Chemistry Notes- Robbie Peck

3. Tollen’s reagent identifies an aldehyde group, (It is an aqueous solution of Ag+ ions in an excess of ammonia, Ag(NH3)2

+). When Tollen’s reagent is reacted with aldehyde and warmed gently in a water bath at about 60oc, silver metal is precipitated which forms a silver mirror. This is a redox reaction whereby the Ag + is reduced to Ag metal and the aldehyde oxidised to a carboxylic acid.

Ag+ + e- ---reduction---> Ag (silver mirror)H3CCOH + [O] ---oxidation---> H3CCOOH

Tollen’s reagent does not react with a ketone or carboxylic acids, because ketones are not oxidised.

Carboxylic acids and estersCarboxylic acids are soluble in water. This is due to the hydrogen bonds formed between the OH in the carboxylic acid and the water molecule.

The solubility of carboxylic acids decreases with increasing molecular mass.Carboxylic acids are weak acids so they donate protons, but they only partially dissociate into their ions:CH3COOH(aq) ↔CH3COO-

(aq) + H+(aq)

The carboxylic acid group can be attached to a chain (aliphatic) or a ring (aromatic). Makes sense!

Carboxylic acids display typical reactions of an acid and form salts. Acid + base salt + water

CH3COOH(aq) + NaOH(aq) CH3COO-Na+(aq) + H2O(l)

Acid + metal Salt + Hydrogen

CH3COOH(aq) + Na(s) CH3COO-Na+ + 12

H2

Acid + Carbonate Salt + Water + HydroxideCH3COOH(aq) + Na2CO3(aq) 2CH3COO-Na+

(aq) + H2O(l) + CO2(g)

This reaction with a carbonate can be used as a test for a carboxylic acid. When an acid is added to a solution of carbonate, bubbles of carbon dioxide are seen.

Carboxylic acids can also react with alcohols to form esters. This type of reaction is esterification. Reversible and carried out with acid catalyst such as concentrated H2SO4. Esters are used in flavourings and perfumes.

Esters , trigylercerides , unsaturated and saturated fats

Esters react with water. The hydrolysis react is slow and is carried out in presence of either an acid H+ or a base OH-. (Hot aqueous acid) Acid-catalysed hydrolysis leads to the formation of the carboxylic acid and the alcohol.H3CCOOCH3 + H2O ---H+---> H3CCOOH + HOCH3

(Hot aqueous alkali) Base-catalysed hydrolysis forms the salt (of the carboxylic acid) and the alcohol.

Page 5: Rings, Polymers and Analysis (Unit 4) - OCR Chemistry Notes- Robbie Peck

H3CCOOCH3 + OH-

- Fats/Triglycerides are esters of long-chain carboxylic acids (fatty acids) and the alcohol glycerol.- Fats are hence also called triglycerides.- The fatty acids that form the esters with glycerol can be either saturated or unsaturated.

- The greater the number of carbon-carbon double bonds, the lower the melting point of the fatty acid and hence the triglycerides.

- Unsaturated fatty acids make up triglycerides found in vegetable and fish oils whilst saturated fatty acids make up triglycerides in animal fats.

- Esters are used in Biodiesel as they increase the quality of the fuel.Trans fatty acids (Transfats) are formed when manufacturers add hydrogen to vegetable oil.

Amines (derivatives of ammonia)The functional group is the amino (-NH2) group.

The naming goes like this:CH3CH2NH2 + H+ CH3CH2NH3

+

Ethylamine ethylammonium ionC6H5NH2 + H+ C6H5NH3

+

Pheylamine phenylammonium ion

Because of the nitrogen lone pair, they are proton acceptors, they are bases.With water, they NICK the H+ and so they form OH- ions so they are weak alkalis

CH3CH2NH2 + H2O CH3CH2NH3+ + OH-

Because they act as bases, primary amines react with acids to form salts. The names of these salts come from the ions formed from the amine and the acid. For example, ethylammonium chloride are formed by the reactions of the bases ethylamine and phenylamine with hydrochloric acid.e.g.NH3 + HCl NH4

+Cl-

Ammonium chloride CH3CH2NH2 + HCl CH3CH2NH3+Cl-

Ethylammonium chloride C6H5NH2 + HCl C6H5NH3+Cl-

Phenylammonium chloridePreparation of aminesThere are 2 methods; one for straight-chain amines and one for aromatic amines.

1. Preparation of straight-chain (aliphatic) amines:

Excess ammonia is refluxed with a halogenoalkanes (e.g. bromoalkane) with ethanol as the solvent.

RBr + NH3 RNH2 + HBr

2. Preparation of aromatic amines:

Page 6: Rings, Polymers and Analysis (Unit 4) - OCR Chemistry Notes- Robbie Peck

Nitrobenzene is reduced by refluxing with the reducing agent; tin and concentrated

hydrochloric acid.C6H6NO2 + 6[H] C6H6NH2 + 2H2O (Aromatic amine; AZO DYE!)

Azo DyesPhenylamine and other aromatic amines are the stating compounds for azo dyes.

Stage 1Nitrous acid (HNO2) (which is formed from NaNO2 and excess HCl) and temperature below 10oc. The reaction is kept below 10oc because otherwise the product, benzenediazonium chloride decomposes.

Stage 2The benzenediazonium chloride is added to phenol in the presence of an alkali; an azo dye is formed.

The azo group N=N is responsible for the colour of the dye.The formation of azo-dye compounds is the basis of the dye industry

Polymers and SynthesisAmino acids, polypeptides and proteinsThe general formula for an α -amino acid is RCH(NH2)COOH

If R is not H, the C is attached to 4 different atoms/groups so it is a chiral carbon and hence amino acids exhibit optical isomerism (Linkage!).

- All amino acids contain the base amine (-NH2) group which accepts protons and contains the carboxylic acid (COOH) group which donates protons.

Amino acids have higher melting points than expected. Zwitterions are formed because of the movement of the hydrogen from the carboxylic acid to the amine. At solid state, the amino acid exists as this.

If we add an basic solution (OH- ions), it forms a negative ion. The proton (H+) moves to the hydroxide ion to form water.

If we add an acidic solution (H+) ions, it forms a positive ion. The proton moves to the O- ion.

Page 7: Rings, Polymers and Analysis (Unit 4) - OCR Chemistry Notes- Robbie Peck

Suppose we start with the positive ion produced under acidic conditions and slowly add alkali to it until you get back to the zwitterion.

So when you have added just the right amount of alkali, the amino acid no longer has a net positive or negative charge. That means that it wouldn't move towards either the cathode or anode during electrophoresis.

The pH at which this lack of movement during electrophoresis happens is known as the isoelectric point of the amino acid. This pH varies for each amino acid and can depend on the R group.

Peptide bonds and polypeptides/proteins!

Amino acid react to form peptides with peptide bonds.

If two different amino acids such as glycine and alanine react, two different dipeptides could be formed depending on which gives the OH (from COOH) and H form (NH2) in the peptide bond.

All dipeptides react further with other amino acids, extending their chain length. Proteins are polypeptides; chains of amino acids linked by peptide bonds. Because of the loss of H2O when proteins are put together, proteins are considered condensation polymers (Like polyesters and polyamides as well)

Hydrolysis back to amino acids

To get the individual amino acids back from a polypeptide, water has to be added back to the polypeptide and the peptide bonds are broken.

This process requires HCl (to mimic stomach conditions) and H2O . The process can also be carried out with alkalis, but a COO- is formed rather than the COOH.

Stereoisomers; E/Z and optical

Stereoisomers have the same structural formula but a different arrangement of atoms in space. There are two types; E/Z isomerism and optical isomerism.

E/Z isomerism requires a C=C double bond.If the 2 highest priority groups are on the zame side, they are Z isomersIf the 2 highest priority groups are oppositE, they are E isomers.

If 2 of substituent groups one on each carbon are the same, it is cis-trans isomerism. If the groups are on the cis same side they are cis.If the groups are on the opposite side they are trans; TRANSEXUAL

Page 8: Rings, Polymers and Analysis (Unit 4) - OCR Chemistry Notes- Robbie Peck

Optical isomerism requires a chiral carbon (4 different groups). Optical isomers are non-superimposable mirror images of each other.The optical isomers react in exactly the same way but show different biological activity.The solutions or crystals of optical isomers can be distinguished from one another because they rotate in plane-polarised light. The isomer rotating the light clockwise is called the (+) isomer, whilst the anticlockwise one is the (-) isomer.

Polyesters and PolyamidesWe studied addition polymerisation at AS where alkenes became polymers.In a condensation reaction, two large molecules join to make a larger molecule + water. The two main types are polyesters and polyamides.

PolyestersFormed when a di-alcohol and dicarboxylic acid. Remember the

H2O made as a product. Used as fibres in clothing.

PolyamidesFormed when a diamine and dicarboxylic acid.e.g. Nylon-6,6 is made from 1,6-diaminohexane and hexane-1,6-dicarboxylic acid.

Nylon-6,6 forms a strong flexible fibre when strung. Kevlar is another polyamide. It is stronger than steel and used for making bulletproof vests. Used as fibres in clothing.

Polyesters and polyamides can be acid or base hydrolysed like the polypeptides before.Polyesters (-ol) are more easily hydrolysed by bases and Polyamides by .

- Polyamides acid hydrolyse by reacting with H2O with H2SO4 (as a catalyst) to form the constituent dicarboxylic acid and diamine.

- Polyesters base hydrolyse with NaOH. A metal salt of the carboxylic acid is formed and the diol.

Minimising environmental waste

- The C=O bond absorbs radiation. When they are in the polymer chain, the energy causes the breakdown of the chain. This makes them photodegradable which is desirable for chemists.

- The esters and amide bonds can be hydrolysed by acids and alkalis helping their breakdown (See above)

- Producing polymers formed from plant feedstock such as starch. E.g. Lactic acid which can be extracted from corn starch and sugar cane. Degradable polymers

Page 9: Rings, Polymers and Analysis (Unit 4) - OCR Chemistry Notes- Robbie Peck

Pharmaceuticals and organic synthesis- Enzymes in the body only recognise one optical isomer of a chiral compound. Therefore

pharmaceutical products acting on living systems require synthesis of these single optical isomers.- But the molecules prepared synthetically in the laboratory often contain a mixture of optical

isomers, whereas the molecules of the same compound produced naturally by enzymes in living systems will only be on one optical isomer.

- Separating the single optical isomer has costs due to separation, but increases the pharmacological activity and rules out possible side effects from the other optical isomer .

- Nowadays , modern synthesis of a pharmaceutical is carried out:

Using enzymes or bacteria to promote stereoselectivity.

Using natural chiral molecules as starting materials

Using chemical chiral synthesis (Using carefully chosen reagents and conditions so as only 1 isomer made)

Organic Synthesis

Page 10: Rings, Polymers and Analysis (Unit 4) - OCR Chemistry Notes- Robbie Peck

3. Analysis

Chromatography; Thin Layer (TLC) and Gas (GC)Small scale analytical technique that separates components in a mixture between a mobile phase (liquid in TLC and gas in GC) and a stationary phase (solid in TLC and liquid on solid support in GC). The mobile phase moves through the stationary phase and the components in the mixture separate out between the phases.

Different types of chromatography separate the components in a mixture by either adsorption or partition.

Thin layer (TLC); Solid stationary phase separating by adsorption.- The mobile phase is a solvent e.g. ethanol that passes over the stationary phase.- The stationary phase is a thin layer of solid e.g. Silica Gel or alumina on a

glass/plastic plate.As the solvent spreads up the plate, the different substances in the mixture move with it but at different rates so they separate out.

A value of R f is worked out where R f=distance travelledby SPOTdistance travelledby solvent

and compared to the

known value of R f for different substances.

How far each part of the mixture moves depends on how strongly it’s attracted to the stationary phase. The attraction between a substance and the surface of the stationary phase is called absorption. A substance that is strongly absorbed will move slowly and not travel very far as one that’s weakly absorbed.

Gas (GC); Liquid stationary phase separating by relative solubility.- The stationary phase is a viscous liquid, such as an oil, which coats the inside of the long tube.- The mobile phase is an unreactive carrier gas, such as nitrogen or helium.

Sample injected into stream of carrier gas, so passes over stationary phase.Components of the mixture dissolve in the stationary phase, evaporate into the mobile phase, dissolve again etc.

The solubility of each component in the mixture determines how long it spends. The time taken is called the retention time and can identify the substance.

Page 11: Rings, Polymers and Analysis (Unit 4) - OCR Chemistry Notes- Robbie Peck

Each peak corresponds to the substance with the particular retention time.

Retention times are measured from zero to the centre of each peak, and can be looked up in a reference table to identify the substances present.

The area under each peak is proportional to the amount of substance. Tallest peak doesn’t always have the greatest area.

Limitations- Similar compounds often have similar retention times so difficult to identify. A mixture of 2 similar

substances may only produce 1 peak so you can’t tell how much there is.- There must already be a recorded reliable reference retention time.

Combining Gas chromatography and mass spectroscopy (GC-MS)The combination of these two A2 and AS techniques makes a powerful analytical took that is used in forensics.Gas chromatography is good at separating a mixture into its individual components, but not so good at identifying those compounds. Mass spectroscopy is good at identifying the unknown compounds.

The mixture is separated using gas chromatography and the different components are sent to the mass spectroscopy to be identified.

Spectroscopy

NMR spectroscopy involves interaction of atomic nuclei with radio waves that are at the low-energy end of the electromagnetic spectrum.

- The sample is placed in a strong magnetic field and exposed to a range of different frequencies of low-energy radio waves.

- The nuclei of certain atoms within the molecule absorb energy from the radio waves.- The amount of energy that a nucleus absorbs at each frequency depends on the environment they

are in.- The 2 types of NMR spectroscopy are carbon-12 NMR (tells you about the number of carbon atoms

that there are in a molecule and the environments they are in) and high resolution proton NMR (Information about the number of hydrogen atoms that are in a molecule and the environments they are in)

An Atom’s environment depends on all the groups it’s connected to. The environment makes the nuclei absorb different amounts of energy at different frequencies.

The chemical shift, δ is the difference in radio frequency and runs along the x-axis. => The number of peaks determines the number of environments. The differences in adsorption are relative to tetramethylsilane, TMS which produces a single peak because it has 1 environment and is used as a reference point.

Page 12: Rings, Polymers and Analysis (Unit 4) - OCR Chemistry Notes- Robbie Peck

13 C NMR spectra tells us about the carbon environments.

- The number of peaks (excluding TMS peak) is the number of environments for the carbons.

- The chemical shift tells us what bond the C (or pair of C’s) belongs to

- The area under the peak tells us the number of carbon atoms in that environment.

1 H (Proton) NMR Spectra tells us about hydrogen environments.

- Works as carbon-13 NMR works but with respect to hydrogen’s.

- The number of peaks (excluding TMS peak) is the number of environments for the hydrogen’s.

- The chemical shift tells us what bond the H involved belongs to

- The area under the peak tells us how many hydrogen atoms are in the environment.

- The splits is caused by hydrogen atoms bonded to neighbouring carbons. This effect is called spin-spin coupling. Only hydrogen nuclei on adjacent carbon atoms effect. Follows n+1 rule.

The solvent the thing is dissolved in WOULD add peaks to NMR (both proton and carbon-13) spectra. Hence, deuterated solvents, CDCl3 is used which doesn’t absorb wave energy.

In Proton NMR OH and NH protons can be identified using D2O. The problem is OH and NH have a huge range so are hard to spot. Hence making another spectra with D2O will get rid of the OH or NH by replacing the O or N with D, which does not show up. Clever.

NMR spectroscopy is used is MRI (Magnetic resonance imaging) to obtain information about internal structures in body scanners in casualty.

Infrared Spectroscopy from F322A beam of infra-red radiation is passed through the chemical. The IR radiation is absorbed by the covalent bonds in the molecules increasing their vibrational energy. Bonds between different atoms absorb different frequencies of IR radiation. So the bonds in a molecule can be determined.

Using the combined techniques of infrared spectroscopy, mass spectrometry, thin layer chromatography, gas chromatography and both NMR spectrums, it is easy to deduce the molecule you have.

F324 ; Rings, Polymers and Analysis; Robbie Peck ‘09/’10