Chemistry AS Textbook

NES/Chemistry/AS  Example 29: The 7 isomers in dichloropropene Structural Display Skeletal Name Formula Formula Formula 1,1-dichloropropene H Cl Cl Cl CH3CH=CCl2 H C C C HH Cl CH3CCl=CHCl Cl Cl Cl Cl CH3CCl=CHCl C C Cl cis-1,2- CH2ClCH=CHCl H3C H Cl dichloropropene Cl H trans-1,2- C C dichloropropene Cl H3C Cl cis-1,3- H H Cl dichloropropene C C ClH2C Cl H Cl Cl Cl trans-1,3- CH2ClCH=CHCl CC Cl dichloropropene CH2ClCCl=CH2 Cl CHCl2CH=CH2 ClH2C H 2,3-dichloropropene Cl Cl H Cl 3,3-dichloropropene Cl HCCC H H Cl H H HCCC H Cl 200

NES/Chemistry/AS Effects of Isomers on Physical and Chemical Properties Melting Point and Boiling Point Any isomer that has more branches, or side chains will have a reduced melting and boiling point as the molecules are further apart from each other and this reduces the van der Waals' forces of attraction between the molecules. Chemical Reactions In functional group isomerism, if there is a different function group present then the isomers will have different chemical reactions. Optical Activity Optical isomers only affect the rotation of plane polarised monochromatic light. 201

NES/Chemistry/AS Topic 15 - Hydrocarbons Compounds containing only carbon and hydrogen are called hydrocarbons. This class of compound can be sub-divided into alkanes and alkenes. 15.1 Alkanes a) understand the general unreactivity of alkanes, including towards polar reagents b) describe the chemistry of alkanes as exemplified by the following reactions of ethane: (i) combustion (ii) substitution by chlorine and by bromine c) describe the mechanism of free-radical substitution at methyl groups with particular reference to the initiation, propagation and termination reactions d) explain the use of crude oil as a source of both aliphatic and aromatic hydrocarbons e) suggest how cracking can be used to obtain more useful alkanes and alkenes of lower Mr from larger hydrocarbon molecules 15.2 Alkenes a) describe the chemistry of alkenes as exemplified, where relevant, by the following reactions of ethene and propene (including the Markovnikov addition of asymmetric electrophiles to alkenes using propene as an example): (i) addition of hydrogen, steam, hydrogen halides and halogens (ii) oxidation by cold, dilute, acidified manganate(VII) ions to form the diol (iii) oxidation by hot, concentrated, acidified manganate(VII) ions leading to the rupture of the carbon–carbon double bond in order to determine the position of alkene linkages in larger molecules 202

NES/Chemistry/AS (iv) polymerisation (see also Section 21) b) describe the mechanism of electrophilic addition in alkenes, using bromine/ethene and hydrogen bromide/propene as examples c) describe and explain the inductive effects of alkyl groups on the stability of cations formed during electrophilic addition d) describe the characteristics of addition polymerisation as exemplified by poly(ethene) and PVC e) deduce the repeat unit of an addition polymer obtained from a given monomer f) identify the monomer(s) present in a given section of an addition polymer molecule g) recognise the difficulty of the disposal of poly(alkene)s, i.e. nonbiodegradability and harmful combustion products 15.3 Hydrocarbons as fuels a) describe and explain how the combustion reactions of alkanes led to their use as fuels in industry, in the home and in transport b) recognise the environmental consequences of: (i) carbon monoxide, oxides of nitrogen and unburnt hydrocarbons arising from the internal combustion engine and of their catalytic removal (ii) gases that contribute to the enhanced greenhouse effect c) outline the use of infra-red spectroscopy in monitoring air pollution (see also Section 22.2) 15.4 --A2 only-- 203

NES/Chemistry/AS 15.1 Alkanes This is a homologous series of saturated hydrocarbons with the general formula CnH2n+2 Alkanes are found in the fractions of petroleum. The carbon chain length ranges from CH4 in refinery gas to C70H142 in bitumen. Both aliphatic and aromatic compounds can be extracted from crude oil. Aliphatic compounds are open chain molecules (alkanes) and cyclic molecules (cycloalkanes). Aromatic compounds are benzene, C6H6 and its derivatives - these are covered at A2 level. They have a distinctive smell. Each successive alkane has an increase in molecular mass by 14. This is because the molecules get bigger by a CH2 each time. Alkanes are described as saturated because all the bonds between carbon atoms are single bonds. Every available space is filled (with hydrogen atoms). Chemical Properties of Alkanes Alkanes are quite unreactive. This is because the C-H bond has little polarity as there is only a small difference between the electronegativity of carbon and hydrogen. Types of reaction: 1. Combustion 2. Homolytic Free Radical Substitution 204

NES/Chemistry/AS 1. Combustion Like all hydrocarbons, alkanes burn in excess oxygen to produce carbon dioxide and water. Type of Reaction Combustion General Equation alkane + oxygen  carbon dioxide + water Conditions ignite in air / oxygen Example 1 C3H8 + 5O2  3CO2 + 4H2O If the supply of oxygen is limited, then incomplete combustion occurs. This results in carbon monoxide and even carbon (black solid) being formed as well. 2. Homolytic Free Radical Substitution with Halogens A substitution reaction is when an atom, or group of atoms, in one organic molecule is replaced by another atom or group of atoms. At the end of the reaction there are two products. An alkane can react with fluorine, chlorine, bromine, or iodine. One hydrogen atom leaves and one halogen atom joins. A hydrogen halide will also be made. Type of Reaction Homolytic Free Radical Substitution General Equation alkane + halogen  halogenoalkane + hydrogen halide Conditions UV light Example 2 CH4 + Cl2  CH3Cl + HCl The substitution is quite random and the chlorine can join at any position. It is also possible for a second, third, and subsequent substitutions to happen if more halogen is added. So in Example 2 it would be possible to produce CCl4 and 4 lots of HCl as well. 205

NES/Chemistry/AS How to Draw Mechanisms Mechanisms are a specific way of drawing what happens during a reaction. There are certain styles that must be used:  Bonds being broken must be shown in reagents  Bonds being made must be shown in products  Electrons involved in a mechanism step are shown as dots on the symbol  Any movement is always of electrons, shown by curly arrows. Electron Dots Reactions involving alkanes always have free radicals, which are shown by one dot on the symbol.  Example 3: A chlorine radical ●Cl This represents the chlorine free radical after the bond has broken (homolytic fission). The dot represents one of the electrons from the broken bond. Reactions with all other functional groups have nucleophiles, or electrophiles.  Example 4: A chloride nucleophile :Cl- This represents a chloride anion that has taken both electrons from the bond (heterolytic fission).  Example 5: A chlorine electrophile Cl+ This represents a chlorine cation that has not taken any electrons from the bond (heterolytic fission) 206

NES/Chemistry/AS Curly Arrows These show where the electrons move from. The start of the arrow shows where the electrons start from and the end of the arrow shows where they end up. The arrows are curly so they are not confused with normal straight arrows in a chemical equation. As we are showing bonds breaking, or being made the maximum number of electrons shown moving is two. A curly arrow with half an arrow head represents one electron moving.  Example 6: Homolytic fission of chlorine Cl Cl ● Cl + ● Cl A curly arrow with a full arrow head represents two electrons moving.  Example 7: Heterolytic fission of chlorine Cl Cl Cl+ + :Cl- 207

NES/Chemistry/AS The Mechanism for a Halogen Reacting with an Alkane Homolytic Free Radical Substitution This mechanism is the only mechanism studied at AS level that involves homolytic fission, which creates free radicals.  Example 8: The homolytic free radical substitution reaction of bromine with ethane. Step 1 - Initiation The halogen splits. Br Br ● Br + ● Br During an initiation step, two free radicals are always made. Step 2 - Propagation During a propagation step, one radical will form a bond making a molecule, but another radical is always made propagating the reaction. H5C2 H ● Br H5C2● + H Br Step 3 - Termination Two free radicals combine together to form a molecule. H5C2● ● Br H5C2 Br 208

NES/Chemistry/AS Cracking Alkanes Alkanes can be cracked, broken into smaller pieces, by heat and a catalyst (aluminium oxide). This produces:  Alkenes Plus:  Shorter chain Alkanes, which are more useful/easier to ignite  Hydrogen Hydrogen is usually made when cracking shorter alkanes  Example 9: Cracking decane C10H22 2C2H4 + C6H14  Example 10: Cracking ethane C2H6 C2H4 + H2 209

NES/Chemistry/AS 15.2 Alkenes This is a homologous series of unsaturated hydrocarbons with the general formula CnH2n Alkenes have an empirical formula of CH2 Each successive alkene has an increase in molecular mass by 14. This is because the molecules get bigger by a CH2 each time. Alkenes are described as unsaturated because there is at least one carbon-carbon double bond (C=C). Alkenes are made by cracking alkanes from crude oil. Chemical Properties of Alkenes Alkenes are more reactive than alkanes. This is because the C=C bond attracts electrophiles. Types of reaction: 1. Heterolytic Electrophilic Addition 2. Oxidation by cold/dilute acidified manganate(VII) ions 3. Oxidation by hot/concentrated acidified manganate(VII) ions 4. Addition Polymerisation 210

NES/Chemistry/AS 1. Heterolytic Electrophilic Addition An addition reaction is when an atom, or groups of atoms, add to the organic molecule to make one product only. An alkene can react with: a. Hydrogen b. Halogen c. Hydrogen halide d. Steam Type of Reaction Heterolytic Electrophilic Addition / Hydrogenation / Reduction General Equation alkene + hydrogen  alkane Conditions 180oC - Ni catalyst / Pt catalyst Example 11 C4H8 + H2  C4H10 Type of Reaction Heterolytic Electrophilic Addition / Halogenation General Equation alkene + halogen  dihalogenoalkane Conditions none Example 12 C2H4 + Br2  C2H4Br2 Type of Reaction Heterolytic Electrophilic Addition General Equation alkene + hydrogen halide  halogenoalkane Conditions none Example 13 C6H12 + HCl  C6H13Cl Type of Reaction Heterolytic Electrophilic Addition / Hydrolysis General Equation alkene + steam  alcohol Conditions 300oC - 60atm - H3PO4 Example 14 C3H6 + H2O  C3H7OH 211

NES/Chemistry/AS The Mechanism for Heterolytic Electrophilic Addition  Step 1 The non-organic molecule splits heterolytically.  Step 2 The electrophile adds to the organic molecule first creating a carbocation intermediate.  Step 3 The nucleophile then adds to the carbocation to form the product. A carbocation is an organic intermediate in which one of the carbon atoms has three σ bonds and a positive charge.  Example 15: The mechanism for the addition of ethene with hydrogen. Step 1 δ+ δ- H+ + :H- Step 2 HH HH HH HCCH HCCH H+ H carbocation Step 3 HH HH HCCH HCCH H HH :H- 212

NES/Chemistry/AS  Example 16: The mechanism for the addition of ethene with bromine liquid. Step 1 δ+ δ- Br+ + :Br- Step 2 Br Br HH HH HCCH HCCH Br+ Br Step 3 HH HH HCCH HCCH Br Br Br :Br-  If aqueous bromine is added to an alkene then the H+ ion and the OH- ion are also present. This can change the final step: Step 3 HH HH HCCH HCCH Br Br OH :OH- 213

NES/Chemistry/AS Markovnikov's Rule This applies to adding hydrogen halides, or steam, to alkenes. Essentially the hydrogen electrophile in step 2 adds to the carbon that already has the most hydrogen atoms more often. This product is called the major product.  Example 17: The mechanism for the addition of but-1-ene with hydrogen chloride. Step 1 δ+ δ- H+ + :Cl- H Cl Step 2 HHHH HHHH (Major) H C=C C C H H C C+ C C H HH H HH H+ secondary carbocation The hydrogen electrophile (H+) joins the first carbon atom as this has two hydrogens already, whereas the second carbon only has one hydrogen already. This is the major carbocation formed. Step 2 HHHH HHHH (Minor) H C=C C C H H CCCC H Step 3 HH + H+ HHH primary carbocation The hydrogen electrophile (H+) joins the second carbon atom less often, forming the minor carbocation. The chloride ion joins like the previous example, forming both:  2-chlorobutane (major product)  1-chlorobutane (minor product) 214

NES/Chemistry/AS Alkyl Group Inductive Effect This is the explanation why major and minor products form during addition reactions. Alkyl - a carbon chain Carbocation - a carbon ion with a positive charge Primary - a carbon atom/ion with 1 alkyl group (end of chain) Secondary - a carbon atom/ion with 2 alkyl groups (middle of chain) Tertiary - a carbon atom/ion with 3 alkyl groups (middle of chain with side chain) Carbocations are electron deficient and alkyl groups are electron-donating, so they tend to push electrons towards the carbocation. This spreads out the charge on the carbocation, making it more stable. The more alkyl groups on a carbocation, the lower the charge density there is on the carbon ion. Stability of carbocation intermediates: Tertiary > Secondary > Primary So tertiary and secondary carbocations (with lower activation energy) form in preference to primary carbocations, making it preferable for the halide ion, or hydroxide ion to join at position -2, rather than position -1.  Example 18: The addition of steam with propene Major product is propan-2-ol Minor product is propan-1-ol  Example 19: The addition of hydrogen bromide to but-2-ene Product is 2-bromobutane. The double bond is in the middle of a chain and only 1 product is possible.  Example 20: The addition of hydrogen chloride with 2-methylpent-2-ene Major product is 2-chloro-2-methylpentane Minor product is 3-chloro-2-methylpentane 215

NES/Chemistry/AS 2. Oxidation by cold/dilute acidified manganate(VII) ions Acidified manganate (VII) ions cause the π bond of the carbon-carbon double bond to break. The organic molecule is then oxidised to a diol. Type of Reaction Oxidation General Equation alkene + H2O + [O]  diol Conditions cold - dilute Example 21 CH3CH=CH2 + H2O + [O]  CH3CH(OH)CH2OH Note - the [O] is used to represent an oxidising agent, in this case acidified manganate (VII) ions. 3. Oxidation by hot/concentrated acidified manganate(VII) ions When the acidified manganate (VII) ions are hot and concentrated, both bonds in the carbon-carbon double bond break forming two separate molecules. The products can be carboxylic acids, ketones, or carbon dioxide and water depending on where the double bond is. Type of Reaction Oxidation alkene + [O]  carboxylic acid General Equation or ketone or carbon dioxide and water Conditions hot - concentrated Example 22 CH3CH=CH2 + 4[O]  CH3COOH + CO2 + H2O Example 23 CH2=CH2 + 6[O]  2CO2 + 2H2O Example 24 CH3CH2C(CH3)=CHCH3 + 3[O]  CH3CH2COCH3 + CH3COOH Note - in examples 22 and 23 the methanoic acid formed will be further oxidised to carbon dioxide and water. Note - in example 24 a ketone is formed because the double bond is on a carbon with a side chain. 216

NES/Chemistry/AS 4. Addition Polymerisation Addition polymers are formed from monomers of alkenes. Monomer Name Monomer Polymer Polymer Name ethene HH low density HH CC poly(ethene) CC H Hn HH high density HH CC poly(ethene) H Hn ethene HH poly(propene) HH CC CC poly(chloroethene) HH H CH3 n poly(vinyl chloride) propene HH HH P.V.C. H CC poly(tetrafluoroethene) chloroethene H Cl n (vinyl chloride) CCCH FF P.T.F.E. H CC F Fn H HH CC H Cl tetrafluoroethene FF CC FF Polymers vary in length, depending on how many monomers join together (up to 10000 monomers per polymer). This mixture of chain lengths results in varying strengths of van der Waals' forces of attraction thus addition polymers can 'soften'. 217

NES/Chemistry/AS Poly(ethene) - Low Density and High Density Poly(ethene) Low Density High Density Consists of branched chain Consists of straight chain molecules so the polymer cannot molecules so polymer can pack Properties pack closely to one another closer together Uses Softening point 120oC Softening point 130°C Plastic bags and packaging Water tanks and pipes, crates Electrical cable insulation and bottles Hospital equipment - can be heat sterilised Disposal of Addition Polymers Addition polymers do not readily biodegrade so there are few ways to dispose of them. 1. Bury in landfill sites - but take up a lot of space as they do not biodegrade. 2. Burn - but this releases toxic gases. 3. Recycle - but they are difficult to separate and process. 218

NES/Chemistry/AS 15.3 Hydrocarbons as fuels Crude oil is a mixture of alkanes, cycloalkanes and benzene. These are all hydrocarbon compounds. Each 'fraction' in crude oil can be separated by fractional distillation as each fraction has a different boiling point range. The shorter chain hydrocarbon fractions from crude oil are usually used as fuels, although this does cause pollution of the air. Pollutant Source Effect CO2 Complete combustion Climate change Toxic gas - causes CO Incomplete combustion carboxyhaemaglobin Carcinogens Unburnt hydrocarbons Incomplete combustion Acid rain NOx Car engines - N2 from air Acid rain SO2 Coal burning power stations Oxides of nitrogen and carbon monoxide levels produced by cars can be reduced by using a catalytic converter, using a platinum/rhodium catalyst. The following reactions take place: 2CO + 2NO  2CO2 + N2 unburnt hydrocarbons  CO2 + H2O 219

NES/Chemistry/AS Topic 16 - Halogen Derivatives The inclusion of a halogen atom within an organic molecule affects its reactivity. 16.1 Halogenoalkanes a) recall the chemistry of halogenoalkanes as exemplified by: (i) the following nucleophilic substitution reactions of bromoethane: hydrolysis, formation of nitriles, formation of primary amines by reaction with ammonia (ii) the elimination of hydrogen bromide from 2-bromopropane b) describe the SN1 and SN2 mechanisms of nucleophilic substitution in halogenoalkanes including the inductive effects of alkyl groups (see Section 15.2(c)) c) recall that primary halogenoalkanes tend to react via the SN2 mechanism; tertiary halogenoalkanes via the SN1 mechanism and secondary halogenoalkanes by a mixture of the two, depending on structure 16.2 Relative strength of the C-Hal bond a) interpret the different reactivities of halogenoalkanes (with particular reference to hydrolysis and to the relative strengths of the C-Hal bonds) b) explain the uses of fluoroalkanes and fluorohalogenoalkanes in terms of their relative chemical inertness c) recognise the concern about the effect of chlorofluoroalkanes on the ozone layer 220

NES/Chemistry/AS 16.1 Halogenoalkanes This is an homologous series containing carbon, hydrogen and at least one halogen atom. Remember, the halogens are Group 17 atoms: fluorine, chlorine, bromine and iodine. Chemical Properties of Halogenoalkanes Halogens have a higher electronegativity than carbon, so the C-Hal bond will have more polarity that a C-H bond. This gives rise to the increased reactivity of the halogenoalkanes, compared to the alkanes. HHH δ+ δ- H C C C Br HHH Types of reaction: 1. Heterolytic nucleophilic substitution 2. Elimination 221

NES/Chemistry/AS 1. Heterolytic Nucleophilic Substitution The joining nucleophile swaps places with the halide ion in the organic molecule. There are three different nucleophiles that can react with a halogenoalkane: Reagent Nucleophile Organic Product Non-Organic Product Sodium hydroxide :OH- Potassium cyanide :CN- Alcohol Sodium halide Ammonia :NH3 Nitrile Potassium halide Amine Hydrogen halide a. Sodium Hydroxide Type of Reaction Heterolytic Nucleophilic Substitution General Equation halogenoalkane + NaOH  alcohol + sodium halide Conditions aqueous - heat under reflux Example 1 C3H7Br + NaOH  C3H7OH + NaBr b. Potassium Cyanide Type of Reaction Heterolytic Nucleophilic Substitution General Equation halogenoalkane + KCN  nitrile + potassium halide Conditions ethanol solvent - heat under reflux Example 2 C3H7Cl + KCN  C3H7CN + KCl c. Ammonia Type of Reaction Heterolytic Nucleophilic Substitution General Equation halogenoalkane + NH3  amine + hydrogen halide Conditions conc. ammonia - ethanol solvent - heat under reflux Example 3 C2H5Br + NH3  C2H5NH2 + HBr 222

NES/Chemistry/AS 2. Elimination Type of Reaction Elimination General Equation halogenoalkane + NaOH  alkene + H2O + Na-halide Conditions ethanolic - heat under reflux Example 4 C3H7Cl + NaOH  C3H6 + H2O + NaCl In this reaction, the halide is removed with a hydrogen from a neighbouring carbon. This allows the formation of a carbon-carbon double bond. There will be major products and minor products formed - use the exact reverse of Markovnikov's rule.  Example 5: Reaction of 2-bromopentane with ethanolic sodium hydroxide Major Product: pent-2-ene CH3CHBrCH2CH2CH3 + NaOH  CH3CH=CHCH2CH3 + NaCl + H2O The halide is removed, followed by an adjacent hydrogen from the carbon atom with the least number of hydrogen atoms bonded to it. Minor Product: pent-1-ene CH3CHBrCH2CH2CH3 + NaOH  CH2=CHCHCH2CH3 + NaCl + H2O Here, after the halide is removed, a hydrogen is taken from the adjacent carbon with the most hydrogen atoms. 223

NES/Chemistry/AS Mechanism for Heterolytic Nucleophilic Substitution There are two possible mechanisms for this reaction, depending on the nature of the halogenoalkane:  Primary halogenoalkanes - usually SN2  Secondary halogenoalkanes - both SN1 and SN2  Tertiary halogenoalkanes - usually SN1 SN1 means that the nucleophilic substitution depends on just one reagent, in this case the halogenoalkane. There will be two steps to this mechanism. SN2 means that the nucleophilic substitution depends on two reagents, in this case the halogenoalkane and the substituting nucleophile. There will be only one step to this mechanism. 224

 Step 1 NES/Chemistry/AS  Step 2 SN1 Mechanism The halogenoalkane will split heterolytically producing a carbocation and a halide ion. The nucleophile will join the carbocation.  Example 6 The mechanism for the reaction of aqueous sodium hydroxide with 2-bromo-2-methylpropane. Step 1 CH3 CH3 H3C C + H3C δ+ δ- + :Br - C Br CH3 CH3 carbocation intermediate CH3 CH3 Step 2 H3C C+ H3C C OH CH3 CH3 :OH- The SN1 mechanism usually happens because a tertiary carbocation is more stable than a primary (or secondary) carbocation. This is due to:  3 electron-releasing alkyl groups dispersing the positive charge on the positive carbocation.  causing the activation energy to be lower than a primary halogenoalkane  forming the carbocation at a higher rate than a primary halogenoalkane This gives rise to a two step reaction mechanism, where each step is only dependant on one reagent - hence SN1. 225

NES/Chemistry/AS  Single step only SN2 Mechanism The halogenoalkane splits heterolytically at the same time as the nucleophile joins.  Example 7 The mechanism for the reaction of aqueous sodium hydroxide with bromoethane. Single Step H3C H δ- - H H Br H3C C OH + :Br - δ+ Br H3C C H OH H C transition state H :OH- Partial Bond The transitional state shows how the C-Br bond is breaking at the same time as the C-OH bond is forming. At no stage does carbon actually have 5 bonds, as it cannot expand its octet as it is in Period 2 and does not have vacant d-orbitals. The SN2 mechanism usually happens with primary halogenoalkanes as the activation energy is usually much higher than tertiary halogenoalkanes and the reaction requires the energy released from the formation of the C-OH bond. This gives rise to a single step mechanism, which is dependent on both reagents - hence SN2. 226

NES/Chemistry/AS 16.2 Relative Strength of the C-Hal Bond The strength of the C-Hal bond depends on the size of the halogen. Halogen Atomic Bond Bond Reactivity Radius Strength Increase Fluorine C-F Chlorine Increase C-Cl 467 Bromine C-Br C-I 346 Iodine 290 228 As the halogen's atomic radius increases, the bond length increases, causing the bond to get weaker. This means that the bond gets easier to break and the halogenoalkanes become more reactive (and have a faster rate of reaction) down the table. The usual way of identifying halogenoalkanes is by hydrolysis (similar reaction to example 1), followed by the addition of silver nitrate. Hydrolysis Chloroalkane Bromoalkane Iodoalkane Silver nitrate White Faster Yellow Precipitate Cream Precipitate Precipitate 227

NES/Chemistry/AS CFCs CFCs are chlorofluorocarbons, or chlorofluoroalkanes. Chemical Properties  Inert (high activation energy required)  Non-toxic  Non-flammable Uses  Anaesthetic  Non-stick linings (Teflon)  Aerosol propellants  Refrigerants As the C-Hal bonds have a high activation energy, they tend not react with most substances and many of their uses have been based on this. However, when CFCs are released into the atmosphere, sunlight can break the C-Cl bond forming chlorine radicals that can damage the ozone layer. CFCs are now being replaced by HFCs (hydrofluorocarbons) and HFEs (hydrofluoroethers) which do not contain chlorine, so cannot form chlorine radicals. 228

NES/Chemistry/AS Topic 17 - Hydroxyl Compounds This topic introduces the chemistry of a versatile class of organic compounds, hydroxyl compounds, which contain an –OH group. 17.1 Alcohols a) recall the chemistry of alcohols, exemplified by ethanol, in the following reactions: (i) combustion (ii) substitution to give halogenoalkanes (iii) reaction with sodium (iv) oxidation to carbonyl compounds and carboxylic acids (v) dehydration to alkenes (vi) formation of esters by esterification with carboxylic acids (vii) --A2 only-- b) (i) classify hydroxy compounds into primary, secondary and tertiary alcohols (ii) suggest characteristic distinguishing reactions, e.g. mild oxidation c) deduce the presence of a CH3CH(OH)– group in an alcohol from its reaction with alkaline aqueous iodine to form tri- iodomethane 17.2 Phenol --A2 only-- 229

NES/Chemistry/AS 17.1 Alcohols This is an homologous series containing carbon, hydrogen and at least one -OH group. Carbohydrates are compounds that contain carbon, hydrogen and oxygen, where the hydrogen:oxygen ratio is 2:1 Physical Properties of Alcohols As alcohols have a polar O-H functional group, they tend to have different properties compared to alkanes: Property Reason Higher melting point Hydrogen bonding between alcohol molecules Soluble in water Hydrogen bonding between alcohol and water molecules Alcohols are often used as solvent as the polar O-H group dissolves ionic compounds and the non-polar alkyl chain dissolves organic molecules. Alcohols are often used to allow ionic and organic reagents to mix and react. Chemical Properties of Alcohols Types of reaction: 1. Combustion 2. Oxidation 3. Reaction with sodium 4. Heterolytic nucleophilic substitution 5. Dehydration 6. Esterification 230

NES/Chemistry/AS 1. Combustion Alcohols burn in excess oxygen to produce carbon dioxide and water. Type of Reaction Combustion General Equation alcohol + oxygen  carbon dioxide + water Conditions ignite in air / oxygen Example 1 C3H7OH + O2  3CO2 + 4H2O If the supply of oxygen is limited, then incomplete combustion occurs. This results in carbon monoxide and unburnt carbon/hydrocarbon (black solid) being formed as well. 2. Oxidation Oxidation (and reduction) reactions vary, depending on whether the alcohol is primary, secondary, or tertiary. The oxidising agent is usually acidified potassium dichromate(VI) or acidified potassium manganate(VII). However, oxygen and air (slow) are also oxidising agents. The symbol [O] is used to show oxygen from an oxidising agent. Primary Aldehyde Carboxylic* Carbon Alcohol Ketone Acid Dioxide Secondary Alcohol Tertiary No Alcohol Reaction *Carboxylic acids - only methanoic acid and ethandioic acid oxidise, to carbon dioxide and water. 231

NES/Chemistry/AS Primary Alcohols Primary alcohols can be oxidised in steps: aldehyde, then carboxylic acid. Each step can be considered to happen separately one after another. Type of Reaction Oxidation General Equation primary alcohol + [O]  aldehyde + water Conditions O2 / air / oxidising agent Example 2a CH3CH2CH2OH + [O]  CH3CH2CHO + H2O Type of Reaction Oxidation General Equation aldehyde + [O]  carboxylic acid Conditions O2 / air / oxidising agent Example 2b CH3CH2CHO + [O]  CH3CH2COOH Primary alcohols can be considered oxidised directly to carboxylic acids if the oxidising agent is in excess. Type of Reaction Oxidation General Equation primary alcohol + [O]  carboxylic acid + water Conditions O2 / air / oxidising agent Example 2c CH3CH2CH2OH + 2[O]  CH3CH2COOH + H2O The following reaction only happens under certain conditions: Type of Reaction Oxidation General Equation carboxylic acid* + [O]  carbon dioxide + water Conditions oxidising agent Example 2d HCOOH + [O]  CO2 + H2O *Only methanoic acid and ethandioic acid 232

NES/Chemistry/AS Secondary Alcohols Secondary alcohols can only be oxidised to ketones. Type of Reaction Oxidation General Equation secondary alcohol + [O]  ketone + water Conditions O2 / air / oxidising agent Example 3 CH3CH(OH)CH3 + [O]  CH3COCH3 + H2O Further oxidation to a carboxylic acid is not possible as an acid functional group in the middle of a chain would have to have five bonds, which is not possible for carbon. Tertiary Alcohols Tertiary alcohols cannot be oxidised as forming a carbonyl (C=O) would require a tertiary carbon to form five bonds, which is not possible. 3. Reaction with Sodium Alcohol will react like an acid with dry sodium metal, substituting the hydrogen from the -OH alcohol functional group. Type of Reaction Redox General Equation alcohol + sodium  sodium alkoxide + hydrogen Conditions none Example 4 CH3CH2OH + Na  CH3CH2ONa + H2 sodium ethoxide This can be used as a test for alcohols as hydrogen gas is made, although it will give the same result with carboxylic acids as well. Note - other Group I metals can also react with alcohols. 233

NES/Chemistry/AS 4. Heterolytic Nucleophilic Substitution Alcohols can react with hydrogen halides to substitute the :OH- nucleophile with a halide nucleophile. This reaction can also happen with other compounds that contain a halide ion. Substituting Hydrogen SOCl2 PHal3 PHal5 Halide Halide √ √ Cl- √ Br- √* √* I- √* √* *The non-organic reagent is made in situ during the reaction. This is done by adding two inorganic reagents together that will form the hydrogen halide, which can then go on to react with the alcohol. Making HCl: NaCl + conc. H2SO4  HCl + NaHSO4 Making HBr: NaBr + dil. H2SO4  HBr + NaHSO4 Dilute sulphuric acid has to be used for NaBr as the bromide ion would act as a reducing agent. Iodide is an even stronger reducing agent and would reduce the sulphuric acid, rather than form hydrogen iodide. Making PBr3 in situ: 2P + 3Br2  2PBr3 Making PI3 in situ: 2P + 3I2  2PI3 234

NES/Chemistry/AS Type of Reaction Heterolytic Nucleophilic Substitution General Equation alcohol + hydrogen halide  halogenoalkane + water Conditions heat under reflux Example 5 CH3CH2CH2OH + HCl  CH3CH2CH2Cl + H2O Type of Reaction Heterolytic Nucleophilic Substitution General Equation alcohol + thionyl chloride  Conditions halogenoalkane + hydrogen chloride + sulphur dioxide Example 6 none CH3CH2CH2OH + SOCl2  CH3CH2CH2Cl + HCl + SO2 Type of Reaction Heterolytic Nucleophilic Substitution General Equation alcohol + phosphorus halide  Conditions halogenoalkane + phosphorus acid Example 7 heat under reflux 3CH3CH2CH2OH + PI3  3CH3CH2CH2I + H3PO3 Type of Reaction Heterolytic Nucleophilic Substitution General Equation alcohol + phosphorus pentachloride  halogenoalkane + hydrogen chloride + phosphorus Conditions Example 8 oxychloride none CH3CH2CH2OH + PCl5  CH3CH2CH2Cl + HCl + POCl3 Any reaction that produces gaseous hydrogen halide will have the observation 'misty white fumes'. This means that SOCl2 and PCl5 can also be used to test for alcohols. 235

NES/Chemistry/AS 5. Dehydration Concentrated sulphuric acid acts as a dehydrating agent and can remove hydroxide and a neighbouring hydrogen from alcohol. A carbon-carbon double bond forms as a result. Alternatively alcohols can be heated with a hot Al2O3 catalyst to form alkene and water. Type of Reaction Dehydration/Elimination General Equation Conditions alcohol  alkene + water Example 9 heat under reflux - conc H2SO4 or heat - Al2O3 catalyst CH3CH2CH(OH)CH3  CH3CH=CHCH3 + H2O The process is the reverse of Markovnikov's rule, so a hydrogen is removed from the carbon with the least number of hydrogen atoms to form the major product, as in example 9 - major product but-2-ene, minor product but-1-ene. 236

NES/Chemistry/AS 6. Esterification When concentrated sulphuric acid is added to a mixture of an alcohol and a carboxylic acid, then hydrogen and oxygen are removed to form water. This also produces an ester. Type of Reaction Dehydration/Elimination General Equation alcohol + carboxylic acid ⇌ ester + water Conditions heat under reflux - conc H2SO4 Example 10 CH3CH(OH)CH3 + HCOOH ⇌ HCOOCH(CH3)CH3 + H2O methylethyl methanoate Note - esterification is a reversible reaction, so the Yield is less than 100% Esters can be hydolysed (split up by reacting with water) Type of Reaction Acid Hydrolysis General Equation ester + water ⇌ alcohol + carboxylic acid Conditions heat under reflux - dilute strong acid (named) Example 11 CH3CH2COOCH2CH3 + H2O ⇌ CH3CH2OH + CH3CH2COOH Note - acid hydrolysis of an ester is a reversible reaction, so the Yield is less than 100% Type of Reaction Alkaline Hydrolysis General Equation ester + water  alcohol + soap* Conditions heat under reflux - NaOH Example 12 CH3CH2COOCH2CH3 + H2O  CH3CH2OH + CH3CH2COONa *Soap is the sodium salt of a carboxylic acid. 237

NES/Chemistry/AS Test for Alcohols Alcohols can be tested by several methods: Type of Sodium SOCl2 PCl5 H+/Cr2O72- Carboxylic Alcohol Acid Effervescence Misty white Misty white Orange to Primary (H2) fumes fumes Green Fruity smell (HCl) (HCl) Secondary Effervescence Orange to Fruity smell (H2) Misty white Misty white Green fumes fumes Fruity smell Tertiary Effervescence (HCl) (HCl) No (H2) Reaction Misty white Misty white fumes fumes (HCl) (HCl) Primary and secondary alcohols can be distinguished by oxidising with acidified potassium chromate(VI); followed by testing the carbonyl (aldehyde, or ketone) formed with Tollen's - see Topic 18. Iodoform Test The iodoform test is the reaction of alcohol with alkaline aqueous iodine to form tri- iodomethane. The difficult part of this test is what it actually tests positive for:  The presence of a -OH functional group (alcohol)  The presence of a =O functional group (aldehyde, or ketone)  BUT the functional group must be next to a -CH3, that means second from the end of the chain. Type of Reaction Halogenation followed by Alkaline Hydrolysis General Equation alcohol + I2 + NaOH  CHI3 + soap Conditions heat Example 13 CH3CH2OH + I2 + NaOH  CHI3 + HCOONa Note - example 13 is not complete, nor balanced. You only need to know the two organic products. The yellow precipitate of CHI3 can be identified by its melting point of 119oC. 238

NES/Chemistry/AS Topic 18 - Carbonyl Compounds This topic introduces the chemistry of the carbonyl compounds, aldehydes (RCHO) and ketones (RCOR'). 18.1 Aldehydes and ketones a) describe: (i) the formation of aldehydes and ketones from primary and secondary alcohols respectively using Cr2O72–/H+ (ii) the reduction of aldehydes and ketones, e.g. using NaBH4 or LiAlH4 (iii) the reaction of aldehydes and ketones with HCN and NaCN b) describe the mechanism of the nucleophilic addition reactions of hydrogen cyanide with aldehydes and ketones c) describe the use of 2,4-dinitrophenylhydrazine (2,4-DNPH) reagent to detect the presence of carbonyl compounds d) deduce the nature (aldehyde or ketone) of an unknown carbonyl compound from the results of simple tests (Fehling’s and Tollens’ reagents; ease of oxidation) e) describe the reaction of CH3CO– compounds with alkaline aqueous iodine to give tri-iodomethane 239

NES/Chemistry/AS 18.1 Aldehydes and Ketones Aldehydes and ketones together are known as carbonyls. Carbonyl - This is an homologous series containing carbon, hydrogen and at least one carbon-oxygen double bond. In the double bond, one bond is a σ bond and the second bond is a π bond. Aldehyde - This is an homologous series containing carbon, hydrogen and at least one RCHO group. This is a carbon-oxygen double bond at the end of an alkyl chain.  Example 1 - Butanal CH3CH2CH2CHO Ketone - This is an homologous series containing carbon, hydrogen and at least one RCOR' group. This is a carbon-oxygen double bond in the middle of an alkyl chain. '  Example 2 - Propanone CH3COCH3 240

NES/Chemistry/AS Reactivity of Carbonyls As oxygen has a higher electronegativity than carbon, the carbon-oxygen double bond is polar. The polar carbonyl group is unsaturated and takes part in addition reactions. Chemical Properties of Carbonyls Types of reaction: 1. Reduction 2. Oxidation 3. Heterolytic Nucleophilic Addition 241

NES/Chemistry/AS 1. Reduction Aldehydes are reduced to primary alcohols by reducing agents. Ketones are reduced to secondary alcohols by reducing agents. Reducing Agents:  H2 - Pt catalyst  H2 - Ni catalyst - 180oC  NaBH4 - alkaline solution - warm  LiAlH4 - dry ether - room temperature Type of Reaction Reduction General Equation aldehyde + [H]  primary alcohol Conditions see above Example 1 CH3CH2CHO + 2[H]  CH3CH2CH2OH Type of Reaction Reduction General Equation ketone + [H]  secondary alcohol Conditions see above Example 2 CH3COCH3 + 2[H]  CH3CH(OH)CH3 Note - [H] is used to represent hydrogen from a reducing agent. 242

NES/Chemistry/AS 2. Oxidation Aldehydes can be oxidised to carboxylic acids by oxidising agents. Ketones cannot be oxidised further. Type of Reaction Oxidation General Equation aldehyde + [O]  carboxylic acid Conditions O2 / air / oxidising agent - reflux Example 3 CH3CH2CHO + [O]  CH3CH2COOH This is the same reaction as Topic 17, example 2b. 243

NES/Chemistry/AS 3. Heterolytic Nucleophilic Addition Both aldehydes and ketones undergo heterolytic nucleophilic addition reactions with hydrogen cyanide. The hydrogen cyanide is made in situ from sodium cyanide and dilute sulphuric acid. Type of Reaction Heterolytic Nucleophilic Addition General Equation aldehyde + HCN  2-hydroxynitrile Conditions NaCN catalyst - dilute H2SO4 - room temperature Example 4 CH3CH2CHO + HCN  CH3CH2CH(CN)OH 2-hydroxybutanenitrile Note - this reaction lengthens the carbon chain by one extra carbon atom. Note - the nitrile carbon becomes carbon number 1 in naming, so the -OH is always on carbon number 2. Note - the -OH functional group becomes a substituent (as the nitrile functional group is more important for naming) so gets called hydroxy-. Type of Reaction Heterolytic Nucleophilic Addition General Equation ketone + HCN  2-hydroxynitrile Conditions NaCN catalyst - dilute H2SO4 - room temperature Example 5 CH3CH2COCH3 + HCN  CH3CH2C(CN)(OH)CH3 2-methyl-2-hydroxybutanenitrile 244

NES/Chemistry/AS Mechanism for Heterolytic Nucleophilic Addition The nucleophile :CN- attacks the δ+ carbonyl carbon. This is easier in aldehydes for two reasons. 1. There is only 1 alkyl group on the carbonyl carbon in aldehydes, so the δ+ charge is reduced less. 2. There is only 1 alkyl group on the carbonyl carbon in aldehydes, so the nucleophile is not hindered as much in trying to reach the carbonyl carbon. (It is physically easier for the nucleophile to get to the carbonyl carbon.)  Step 1 The HCN splits heterolytically  Step 2 The nucleophile (:CN-) adds to the carbonyl carbon as the carbon-oxygen π bond breaks  Step 3 The negatively charged intermediate reacts with H+ to form the product  Example 6: The addition of hydrogen cyanide with propanal. Step 1 δ+ δ- H+ + :CN- H CN Step 2 HH δ- H H :O- HCC H C C C CN O :CN- HH C δ+ HH H H H H :O- δ+ δ- H H OH H CN Step 3 H C C C CN H C C C CN + :CN- HH H HHH Note - you must show the :CN- nucleophile as a product in step 3. 245

NES/Chemistry/AS Testing for Aldehydes and Ketones There are three different reactions that work as tests for aldehydes and ketones. 2,4-DNPH Fehling's Tollens' Aldehyde yellow-orange red precipitate silver mirror Ketone precipitate x or yellow-orange grey/black precipitate precipitate x 2,4-Dinitrophenylhydrazine 2,4-DNPH is a yellow solution. It forms a yellow-orange precipitate with aldehydes, or ketones. The reaction looks complicated, but essentially it is a condensation reaction (removing water). H O+  C H3C 2,4-dinitrophenylhydrazine 2,4-dinitrophenylhydrazone (yellow solution) (yellow-orange precipitate) Note - water is also made as a product. Fehling's Solution This is a blue coloured solution of Cu2+ ions that acts as an oxidising agent with aldehydes to form a carboxylic acid and a red Cu2O precipitate. Tollens' Solution Tollens' solution contains Ag+ ions that act as an oxidising agent with aldehydes to form a carboxylic acid and a silver, or black Ag metal precipitate. 246

NES/Chemistry/AS Iodoform Test The iodoform test is the reaction of aldehydes, or ketones with alkaline aqueous iodine to form tri-iodomethane. It is the same as in Topic 17. The difficult part of this test is what it actually tests positive for:  The presence of a -OH functional group (alcohol)  The presence of a =O functional group (aldehyde, or ketone)  BUT the functional group must be next to a -CH3, that means second from the end of the chain. Type of Reaction Halogenation followed by Alkaline Hydrolysis General Equation aldehyde + I2 + NaOH  CHI3 + soap Conditions heat Example 6 CH3CHO + I2 + NaOH  CHI3 + HCOONa Type of Reaction Halogenation followed by Alkaline Hydrolysis General Equation ketone + I2 + NaOH  CHI3 + soap Conditions heat Example 7 CH3COCH3 + I2 + NaOH  CHI3 + CH3COONa Note - examples 6 and 7 are not complete, nor balanced. You only need to know the two organic products. The yellow precipitate of CHI3 can be identified by its melting point of 119oC. 247

NES/Chemistry/AS Nitriles Although nitriles are not a separate topic studied at AS Level, some of the reactions of nitriles are mentioned in Topics 18 and 19. Chemical Properties of Carbonyls Types of reaction: 1. Acid Hydrolysis 2. Alkaline Hydrolysis 3. Reduction 1. Acid Hydrolysis Nitriles can be broken down using dilute, strong acids. Type of Reaction Acid Hydrolysis General Equation nitrile + acid  carboxylic acid + ammonium salt Conditions reflux Example 8 CH3CH2CN + HCl + 2H2O  CH3CH2COOH + NH4Cl 2. Alkaline Hydrolysis Nitriles can be broken down using dilute, strong alkalis. Type of Reaction Alkaline Hydrolysis General Equation nitrile + alkali  soap* + ammonia Conditions reflux Example 9 CH3CH2CN + NaOH + H2O  CH3CH2COONa + NH3 *soap is the sodium salt of a carboxylic acid. In example 9 it is sodium propanoate. 248

NES/Chemistry/AS 3. Reduction Nitriles can be reduced to amines using a reducing agent, or sodium metal and ethanol. Type of Reaction Reduction General Equation nitrile + [H]  amine Conditions Example 10 LiAlH4 - dry ether or Na - ethanol CH3CH2CN + 4[H]  CH3CH2CH2NH2 249