A2 Chemistry

1. A2 Chemistry Unit 1

2. The Structure of Benzene Hydrogenation of Benzene: Cyclohexene enthalpy change of hydrogenation = -120kJmol⁻₁ . Therefore benzene must have an enthalpy change of hydrogenation should be -360kJmol⁻₁ . (3x cyclohexene, 3x C=C) The enthalpy change is actually -208kJmol⁻₁ . The real structure of benzene is more stable than Kekulé’s structure. This energy is known as the resonance energy of benzene. Kekulé’s Equilibrium Model of Benzene • Kekulé’s structure failed to explain benzene’s low chemical reactivity. • If C=C bonds were present, benzene should react similarly to alkenes. • Each C=C bond would be expected to decolourise bromine water. • Benzene does NOT take part in electrophilic addition reactions as expected from the C=C bonds. C-C single bonds and C=C bonds have different bond lengths. Kathleen Lonsdale found that all of the carbon bonds were the same length -0.139nm. (between the lengths of C-C and C=C).

3. The Delocalised Model of Benzene • The Delocalised Model has the following features: • Cyclic hydrocarbon – 6 C molecules and 6 H molecules. • Arranged in a planar hexagonal ring where each C is bonded to 2 other C atoms and 1 H atoms. • The shape is a trigonal planar with a bond angle of 120°. • Each C atom has 4 outer shell electrons. 3 of these e¯ bond to 2 other C atoms and 1 H atom. The bonds in this plane are called sigma bonds. The 4th outer shell e⁻ in a 2p orbital above and below the plane of the carbon atoms. • The e⁻ in the p orbital overlap creating a ring of electron density above and below the plane of carbons. • The pi-bonds spread over all 6 carbons and the ring is said to be delocalised. Instead, Benzene takes part in substitution reactions – a hydrogen (H) is replaced with another group. The organic product retains the delocalised structure. • Under normal conditions, benzene does not: • Decolourise bromine water • React with strong acids such as HCl • React with halogens such as bromine, chlorine or iodine. Addition reactions will disrupt the delocalisation of the ring structure.

4. Reactions of Benzene Nitration of Benzene Formation of NO₂⁺: HNO₃ + H₂SO₄ NO₂⁺ + HSO₄⁻ + H₂O Benzene’s high electron density attracts electrophiles. The H⁺ reacts with the HSO₄⁻ to reform H₂SO₄. This is acting as a catalyst. To preserve the ring’s stability, benzene takes part in ELECTROPHILIC SUBSTITUION reactions. C₆H₆ + HNO₃ C₆H₅NO₂ + H₂O H⁺ + HSO₄⁻ H₂SO₄ Conditions: conc. HNO₃, conc. H₂SO₄, 50°C

5. Reactions of Benzene Halogenation of Benzene Benzene will react with halogens in the presence of a HALOGEN CARRIER. Halogen Carriers Include: FeCl₃ FeBr₃ AlCl₃ AlBr₃ Iron Metal Formation of Br⁺ (or Cl): Br₂ + FeBr₃ Br⁺ + FeBr₄⁻ Bromobenzene is used in the preparation of pharmaceuticals. Chlorobenzene is used as a solvent and in pesticides. Regeneration of Br⁺ (or Cl): H⁺ + FeBr₄⁻ FeBr₃ + HBr

6. Reactivity of Alkenes and Benzene Cyclohexene and Bromine Water: The pi-bond is localised – this gives cyclohexene HIGH ELECTRON DENSITY. The pi-bond repels the electrons in the Br-Br bond inducing a dipole. The Br₂ molecule becomes polar. The electrons in the double bond attract to the Br+ causing the double bond to break. This forms a positive carbocation. The Br-Br bond breaks via heterolytic fission forming Br⁻. The Br⁻ is attracted to the intermediate carbocation forming a covalent bond. Benzene and Bromine: Benzene has delocalised electrons spread over a ring structure. Alkenes have localised electrons. Benzene has LOWER ELECTRON DENSITY and CANNOT POLARISE Br₂. Benzene is therefore resistant to reactions with non-polar halogens. A halogen carrier is needed to generate a more powerful electrophile. The greater charge on Br₂ can attract the pi-electrons from benzene so the reaction can take place. Bromine +Benzene = orange Bromine + Benzene + Iron fillings = decolourised and white fumes of hydrogen bromide gas.

7. Phenols O⁻ Na⁺ Sodium Phenoxide + H₂O NaOH O⁻ Na⁺ OH Sodium Phenoxide Na + H₂ Phenol Solid at room temperature and pressure. Slightly soluble in water as OH group can make hydrogen bonds with water. Benzene ring makes it less soluble than alcohols. Only a phenol when an OH group is directly linked to the ring.

8. Bromination and Uses of Phenols OH Bromination of Phenol: Lone pair of electrons on the O group of phenol. This creates a HIGHER ELECTRON DENSITY. This POLARISES the Br₂ which are more strongly attracted towards the ring structure. Na Phenol 3Br₂ (aq) OH Br Br + 3HBr Br 2,4,6-tribromophenol

9. Carbonyl Compounds ᶞ⁻ O O = = ᶞ⁺ - C - - - C H Ketone In the middle of the chain Aldehyde At the end of the chain The functional group is the part of the molecule responsible for its chemical reactions. Oxygen is more ELECTRONEGATIVE so electrons are more attracted to it than the carbon.

10. Reduction of Carbonyl Compounds Reducing Agent [H] = NaBH₄ (Sodium Borohydride) Water is the solvent NaBH₄ readily generates hydride ions. Ketone + [H] = Secondary Alcohol Aldehyde+ [H] = Primary Alcohol H H O H H :O O = H C C H C C H H H H H :H⁻ H OH C C + OH⁻ H H H H

11. Oxidation of Carbonyl Compounds Reflux is the CONTINUAL BOILING AND CONDENSING of a mixture to ensure that it does not dry out. Oxidising Agent [O] = Acidified Potassium Dichromate Ions K₂Cr₂O₇/H⁺ Primary Alcohol Aldehyde Carboxylic Acid Heated under reflux before distillation. Secondary Alcohol Ketone NO COLOUR CHANGE

12. Chemical Tests on Carbonyl Compounds To detect the presence of a carbonyl compound: > 2,4-DNP A solution of 2,4-DNP in a mixture of methanol & sulphuric acid is known as BRADY’S REAGENT. Brady’s Reagent + Aldehyde/Ketone Yellow/Orange Precipitate. H H + H₂O = C C₂H₅ + O = C C₂H₅ Filter and recrystallise the 2,4-DNP derivative and record its melting point. This is compared against a database. Difficult to distinguish between Heptan-2-one and Cyclohexanone as the b.ps are very similar. The 2,4-DNP derivatives have different m.ps thus allowing easy identification.

13. Chemical Tests on Carbonyl Compounds Aldehyde or Ketone? Tollens’ Reagent is a weak OXIDISING agent that distinguishes between aldehydes and ketones. Aldehyde + [O] Carboxylic Acid SILVER MIRROR FORMED Ketones are NOT OXIDISED by Tollens’ reagent. Making Tollens’ Reagent: NaOH + AgNO₃ until a brown precipitate of Silver Oxide is formed. Dilute NH₃ added until precp. dissolves. The colourless solution is aka Ammonical Silver Nitrate.

14. Carboxylic Acids O = - C - Functional Group = COOH OH Solubility: As the carbon chain INCREASES, SOLUBILITY DECREASES. Molecules become more NON-POLAR. Carboxylic Acid + Metal SALT + H₂(g) Carboxylic Acid + Base SALT + H₂O Carboxylic Acid + Carbonate SALT + CO₂ + H₂O COOH’s are WEAK ACIDS and react with metals, bases and carbonates. First part of carboxylate comes from metal, base or carbonate. Salts formed from COOHs aka CARBOXYLATES. O Suffix ‘...oate’ = - C - O⁻

15. Esters Making esters is known as ESTERIFICATION. Carboxylic Acid + Alcohol COOHS + OHs produces an ESTER & H₂O Acid Catalyst: H₂SO₄ e.g. Ethanol + Propanoic Acid = Ethyl Propanoate + Water Acid Anhydride Ester Hydrolysis: This is the reverse of esterification. It ADDS WATER. Acid Hydrolysis = reflux + aqueous acid, reversible Alkaline Hydrolysis = reflux + aqueous alkali, makes sodium salt, non-reversible An acid anhydride is formed by the removal of H₂O from 2 molecules of carboxylic acids. This produces an ESTER and a CARBOXYLIC ACID. This process requires gentle heating. Esters from ACID ANHYDRIDES produce a GREATER YIELD. Esters are used in perfumes and as flavourings.

16. Fats and Oils Building Triglycerides 18:2(9,10) • Fats are used for: • Insulation • As an energy store • To protect organs Naming Fatty Acids: Unsaturated fats with MULTIPLE DOUBLE BONDS Double bond commonly between C9 and C10. Fats and Oils are esters of a long chained carboxylic acid. No. C atoms No. of = bonds Position of = bonds Fats m/p ABOVE room temp. Oils m/p BELOW room temp. Forming Triglycerides: Simple triglyceride derived from 2/3 of the SAME fatty acids. Natural/Mixed triglycerides derived from 2/3 DIFFERENT fatty acids. Triglycerides: Triglycerides are triesters of: Propane-1,2,3-triol (glycerol) 3 fatty acid molecules. HDLs – carry cholesterol FROM ARTERIES back TO LIVER LDLs – carry cholesterol FROM LIVER TO TISSUES. Fatty acids can also be used to make BIODIESEL.

17. Amines Amines are derivatives of ammonia. Amines are WEAK BASES and ACCEPT protons. Each lone pair on the N atom accepts a H⁺ Ammonia NH₃ Primary Amine RNH₂ Secondary Amine R ₂NH Tertiary Amine R₃N :NH₃ + H⁺ NH₄⁺ Adrenaline, Amphetamine, Phenylephrine A DATIVEBOND forms between the lone pair of the N atom and the H⁺. Base + Acid Salt Naming Amines Secondary Amines Alkylamine + Acid forms an ALKYLAMMONIUM SALT. e.g. Methylamine + Sulphuric Acid Methylammonium Sulphate N-methylpropylamine

18. Reactions of Amines Preparing Amines: Conditions: Excess NH₃ Solvent = Ethanal This is a NUCELOPHILIC SUBSTITUTION REACTION. CH₃CH₂CH₂Cl + NH₃ CH₃CH ₂CH₂NH₂ + HCl NH₃ + HCl NH₄⁺Cl⁻ Excess NH₃ is added so that it all reacts. Preparing Aromatic Amines: Nitroarenes are reduced using a mixture of TIN AND CONC. HCl Synthesis of Dyes from Phenylamines: Diazotisation Coupling Reaction Formation of HNO₂: NaNO₂ +HCl HNO₂ +NaCl + HNO₂ + 2HCl (<10°C) + H₂O

19. Br Cl Bromobenzene Benzene Chlorobenzene Br₂/FeBr₃ Cl₂/AlCl₃ Conc. HNO₃ Conc. H₂SO₄ 50°C N NaNO₂/HCl (aq) <10°C Phenol, NaOH N⁺ Cl⁻ Azo Dye NO₂ NH₂ Sn/ Conc. HCl Reflux Nitrobenzene Phenylamine Benzenediazonium chloride N N OH

20. O⁻ Na⁺ Sodium Phenoxide + H₂O NaOH O⁻ Na⁺ OH Sodium Phenoxide Na + H₂ Phenol 3Br₂ (aq) OH Br Br + 3HBr 2,4,6-tribromophenol Br

21. Amino Acids Amino acids make PEPTIDES and PROTEINS. 20 different amino acids in body. They are α-amino acids and have a BASIC AMINE GROUP and an ACIDIC CARBOXYL GROUP. Soluble in both acids and bases. R-group is usually –OH, -SH, -COOH or –NH₂ EXCEPT GLYCINE which has H as the R group (simplest). Carboxyl and Basic group can react to form a ZWITTERION (internal salt). Carboxyl donates proton to basic group. There is no overall charge as they cancel out. The ISOELECTRIC POINT is the pH at which there is no net charge. The zwitterion exists in this pH. H⁺ pH 1 OH⁻ pH 13 + H₂O H Amino acid acts as an ACID and DONATES a proton to the hydroxide ion. Amino acid acts as a BASE and ACCEPTS a proton from the acid.

22. Polypeptides and Proteins Amino acids join together to from PEPTIDES Acid Hydrolysis of Polypeptides and Proteins: Heated under reflux 6 mol dm⁻₃ 24 hours Acid Solution Peptide is separated and both become positive ions. Amino acids join together in a CONDENSATION REACTION and so eliminate H₂O. Alkaline Hydrolysis of Polypeptides and Proteins: Alkaline Solution Above 100°C. Peptide is separated into 2 original peptides and both become sodium salts. Proteins are long polypeptides with more than 50 amino acids.

23. Optical Isomerism OPTICAL ISOMERS= stereoisomers that are NON-SUPERIMPOSABLE mirror images. STEREOISOMERS = same structural formula, DIFFERENT ARRANGEMENT in space. CHIRAL CARBON= Carbon attached to 4 DIFFERENT atoms Optical isomers rotate the light CLOCKWISE & ANTICLOCKWISE. A mixture of equal amounts of optical isomers is known as a RACEMIC mixture – the rotations cancel each other out. Optical isomers exist in all amino acids (expt. Glycine). Only 1 of the isomers is synthesised naturally & only 1 will react with an enzyme.

24. Chirality in Pharmaceutical Synthesis • Advantage of single isomer: • Risk from undesirable side effects reduced • Drug doses reduced Separating optical isomers is difficult as they have the same properties. How to separate optical isomers: Use ENZYMES as biological catalysts Chiral Pool Synthesis Transition Element Complexes Ibuprofen has 2 optical isomers. It is sold as a mixture of both isomers.

25. Ester Linkage Condensation PolymerisationPolyesters O = To from an ester: A Carboxylic Acid An Alcohol (either on 1 or 2 molecules) -H lost from Alcohol, -OH lost from Carboxylic Acid Elimination of a by-product – usually H₂O. Ester Linkages e.g. Terylene and Poly(lactic) Acid - C - O • Uses of Polyesters: • Machine-washable • Machine-dryable • Resistant to stretching, shrinking and chemical attack • Burns easily

26. Amide Linkage Polyamides O = To form a polyamide: A Carboxylic Acid An Amine (can have 2 different monomers or just 1 with both functional groups.) Amide Linkage -OH lost from Carboxylic Acid, -H lost from Amine to form H₂O e.g. Nylon 6,6 and Kevlar _ _ _ C N _ H Nylon 6,6: Used widely in Clothing Kevlar: -fire resistant -stronger than steel -fire fighter clothing and bullet-proof vests.

27. PolyamideNylon 6,6 O O H = = _ H0 – C – (CH₂)₄- C – OH + H – N – (CH₂)₆ - N – H _ H Hexane-1,4-dioic Acid 1,6-diaminohexane O O H = = _ + 2n-1 H₂O C – (CH₂)₄- C – N – (CH₂)₆ - N _ H

28. PolyamidesNylon 6 O H = _ H-N-(CH₂)₅-C-OH O H = _ + n-1 H₂O N-(CH₂)₅-C

29. KevlarPolyamide H O O _ + = = N H N H C HO C OH _ H H O O _ = = + (2n-1) H₂O N C N C _ H

30. TerylenePolyester O O H H = = + HO C C COO⁻Na⁺ Na⁺COO⁻ HO C C OH OH H H O O H H = = + (2n-1) H₂O O C C C O C H H

31. Addition and Condensation Polymerisation Addition polymers have 1 MONOMER and there is NO BY-PRODUCT. Addition polymers contain a DOUBLE BOND.

32. Breaking down condensation polymers Hydrolysis of Polyesters H⁺/H₂O NaOH/H₂O

33. Breaking down condensation polymers Hydrolysis of Polyamides H⁺/H₂O NaOH/H₂O O O = = ⁺ ⁺ Na⁺OOC⁻– C – (CH₂)₄- C –COO⁻Na⁺

34. Degradable Plastics Degradable plastics break down into smaller fragments when exposed to HEAT, LIGHT OR MOISTURE. A biodegradable plastic breaks down COMPLETELY into CO₂ and H₂O Biodegradable polymers have bonds than undergo hydrolysis. Poly(lactic acid) – derived from corn starch Poly(glycolic acid) – isolated from sugar cane and unripe grapes. Photodegradable plastics are synthetic polymers designed to become weak and brittle when exposed to light for prolonged periods.

35. Separation by Chromatography ADSORPTION: The solid holds gas or liquid molecules ON THE SURFACE of a solid. Chromatography is used to separate components in a mixture. A mobile phase sweeps over a stationary phase. THIN LAYER CHROMATOGRAPHY Stationary Phase: Solid Mobile Phase: Liquid Different components have different AFFINITIES for the phases. GAS CHROMATOGRAPHY Stationary Phase: Solid/Liquid (silica) Mobile Phase: Gas The stationary phase slows the components down. They pass at different speeds thus separating the compound. A solid stationary phase separated by ADSORPTION.

36. Thin Layer Chromatography Rᶠ = distance moved by component distance moved by solvent front Limitations: Similar compounds have similar Rᶠ values. Difficult to find solvent that separates all of the compounds in a mixture. A chromatogram is a visible record showing the results of separation of a mixture.

37. Gas Chromatography Limitations: Similar retention times Not all substances will be detected No reference for unknown compounds. Stationary Phase:Thin layer solid/liquid coated on inside of capillary tube. Mobile Phase:Inert carrier gas e.g. helium Each component leaves the column at different times and is detected as it leaves the column. Gas Chromatography Mass Spectrometry Retention time in gas chromatography is the time for a component to pass from the column inlet to the detector. The area under the peak is proportional to the amount of compound in a sample. Components are separated by GC Detected by MS against a reference. GC-MS used in... Forensics Environmental Analysis Airport Security Space Probes

38. NMR Nuclear Magnetic Resonance Chemical Shift ᵟ is a scale that compares the frequency of an NMR adsorption with the frequency of the reference peak of TMS at 0ppm. TMS is added so that the spectrometer can be calibrated against the TMS reference peak. TMS is chemically UNREACTIVE and is removed from the sample after running the NMR. Low Resolution NMR Spectroscopy Solvent: D₂O Isotope of Hydrogen Produces no signal in spectrum as it has even number of nucleons. Other solvents: CDCl₃ (C peak usually removed from spectrum via evaporation) High Resolution NMR Spectroscopy Won’t absorb radio waves C₂H₅O-H +D-O-D C₂H₅O-D + D-O-H

39. NMR in Medicine An MRI scanner is like a large spectrometer in which the patient is the sample. MRI scanners detect SOFT TISSUE DAMAGE and in the diagnosis of tumours because they contain a high % of water. MRI scans are NON-IVASIVE so do NOT HARM the body’s cells. Disadvantages: Expensive, High training required

40. NMR Spectroscopy No. of peaks = No. of Chemical Environments Mass Spectrum = molecular ion (M) peak furthest right Identifies fragments