The alkanes is an homologous series where all members fit the general formula C n H 2n+2. They have trends in physical properties e.g. density and m.p. and b.p. all increase with M r. They all undergo similar chemical reactions. Alkanes are SATURATED HYDROCARBONS. i.e. they contain only single C to C bonds and are made up of C and H atoms only.

Alkanes are obtained from crude oil by fractional distillation. They are mainly used as fuels. The large M r alkanes do not ignite easily so there is little demand for them as fuels so they are CRACKED to make smaller more useful alkanes and alkenes. Apart from combustion alkanes undergo few chemical reactions. This is for two main reasons: The bonds in alkanes are relatively strong. The bonds have a relatively low polarity as the electronegativity of C and H is similar. As a consequence alkanes can be used as lubricating oils, although they do degrade over time.

1. Similar for all alkanes 2. Not very reactive Reasons: 1. Contains the same single C-H and C-C covalent bonds. 2. Strong C-C and C-H bonds require a lot of energy to break.

 Smaller alkanes burn easily in air when ignited  If the combustion is complete, products formed are CO 2 and H 2 O. CH 4 + 2O 2  CO 2 + H 2 O H = -890kJ/mol  Reaction is exothermic - a result of the high relative strength of the C=O in CO 2 and O-H in H 2 O molecules. The large amount of heat energy released in making these bonds means the reaction is strongly exothermic. Write equations for the complete combustion of butane and octane.

 If the combustion is incomplete, products formed are CO, C (soot) and H 2 O. 2C H 4 + 3O 2  2CO + 4H 2 O CH 4 + O 2  C + 2H 2 O What problems do the gases released on combustion of alkanes cause?

 In the presence of ultraviolet light, methane can combine with chlorine to give a mixture of products (known as chloroalkanes)  Light energy is used to start the reaction (by providing the energy to break the covalent bond between the chlorine atoms in Cl 2.  The chlorine atom(radical) produced then reacts with alkane by substituting the hydrogen atom.

Free radical = species with an unpaired electron. Free radicals are formed by homolytic fission of bonds. In homolytic fission one electron from the shared pair goes to each atom. So Cl

+ or Cl 2  2Cl. unpaired electron Heterolytic fission of Cl – Cl would result in the formation of Cl + and Cl -. There are three steps in the mechanism: initiation, propagation and termination

CH 4 + Cl 2 CH 3 Cl + HCl Overall reaction equation Conditions ultra violet light (breaks weakest bond) excess methane to reduce further substitution i.e. homolytic breaking of covalent bonds Free radical substitution chlorination of methane

CH 4 + Cl CH 3 + HCl Cl 2 Cl + Cl CH 3 + Cl 2 CH 3 Cl + Cl CH 3 Cl CH 3 + Cl initiation step two propagation steps termination step UV Light CH 3 CH 3 CH 3 + CH 3 minor termination step Free radical substitution mechanism Also get reverse of initiation step occurring as a termination step.

CH 3 Cl + Cl 2 CH 2 Cl 2 + HCl Overall reaction equations Conditions CH 2 Cl 2 + Cl 2 CHCl 3 + HCl CHCl 3 + Cl 2 CCl 4 + HCl ultra-violet light excess chlorine Further free radical substitutions

All fit the general formula C n H 2n. Are unsaturated hydrocarbons as they contain a C = C. Much more reactive than alkanes. Industrial importance of alkenes: 1. Making polymers (plastics) 2. Hydrogenation of vegetable oils to make margarine 3. Hydration of ethene to make ethanol.

When naming alkenes have to include position of double bond, for example: CH 3 CH=CHCH 3 is but ene and CH 3 CH 2 CH=CH 2 is but -1- ene Alkenes undergo ADDITION reactions. Two substances combine to form one new substance. Unsaturated molecules are converted to saturated molecules.

1. Similar for all alkenes 2. More reactive than alkanes Reasons: 1. Contains the same C=C covalent bonds. 2. One of the C=C can break easily – allow other atoms to join to the carbon atoms resulting in an addition reaction.

 Alkenes burn in an excess air to form CO 2 and H 2 O. C 2 H 4 + 3O 2  2CO 2 +2H 2 O H = -1322kJ/mol  Reaction is exothermic.  If the combustion is incomplete, products formed are CO, C (soot) and H 2 O. C 2 H 4 + O 2  2C + 2H 2 O C 2 H 4 + 2O 2  2CO + 2H 2 O More soot (C) is produced compared to the corresponding alkane, due to higher percentage by mass of carbon. How can you tell whether the reaction is complete or incomplete?

1. To make margarine 2. To make alcohols, used as antifreeze and solvents 3. To make plastic (polymers) like poly(ethene), PVC 4. Used in agriculture (in low concentration) to help hasten the ripening of fruits like bananas.

 Addition of hydrogen (hydrogenation) Alkenes react with hydrogen in the presence of a nickel catalyst at 180 °C to form an alkane. e.g. H – C – C – H C =C H HH H + H 2 HH HH C 2 H 4 + H 2  C 2 H 6 ethene ethane

 Hydrogenation is used in margarine industry to convert oils containing unsaturated hyrocarbon chains into saturated compounds with higher melting points.  Margarine is a solid at room temperature

 Addition of halogens (halogenation)  Halogens react with alkenes at room temperature and pressure in a non-polar solvent to form a dihalogenoalkane. C =C H HH H + Br 2 H – C – C – H Br HH C 2 H 4 + Br 2  C 2 H 4 Br 2 ethene1,2-dibromoethane Write the equation when propene reacts with chlorine.

Bromine water is used as a test for unsaturation. In the presence of an alkene, bromine water turns from red brown to colourless. Alkanes do not react with bromine water.

C =C – C – C – Br + Br 2 (aq) colorlessambercolorless Test for Unsaturation – bromine water test for alkenes

C =C H HH H + HBrH – C – C – H HBr HH C 2 H 4 + HBr  CH 3 CH 2 Br ethene bromoethane Further addition reactions  Alkenes react with hydrogen halides (HCl, HBr etc.) to solvent to form a dihalogenoalkane.  The reaction occurs at room temperature and pressure. e.g.

C =C H HH H + H 2 SO 4 OSO 3 H H – C – C – H H H H Hydration (reaction with water)  Hydration reaction can be done in 2 ways:  addition of concentrated sulfuric acid to form alkyl hydrogensulfate. Water is then added to hydrolyse the product and produce an alcohol. The sulfuric acid is regenrated.  The reaction occurs at room temperature and pressure. e.g.

OSO 3 H H – C – C – H H HH + H 2 O OH H – C – C – H H HH + H 2 SO 4 C 2 H 4 + H 2 SO 4  CH 3 CH 2 OSO 3 H CH 3 CH 2 OSO 3 H + H 2 O  C 2 H 5 OH + H 2 SO 4 ethanol

C =C H HH H + H 2 O OH H – C – C – H H H H What advantages and disadvantages does this method have over production of ethanol by fermentation?  Alkenes can also undergo direct hydration to form an alcohol. Ethene can be converted to ethanol by reaction with steam in the presence of a phosphoric(V) acid (H 3 PO 4 ) catalyst at a pressure of 60 – 70 atm and a temperature of 300 °C.

nCH 2 = CH 2  CH 2 CH 2 n n = about 100 to CH 2 is the repeating unit  The formation of polymers involves reacting reacting with themselves to form a long chain molecule called a polymer. The individual molecules used to make the polymer are called monomers. Ethene is polymerised to form poly(ethene)

monomerrepeating unitpolymertypical uses CH 2 =CH 2 - CH 2 – CH 2 -poly(ethene) polythene film, bags poly(propene) polypropylene Moulded plastic, fibres poly(phenylethene) polystyrene packaging, insulation CH 2 =CHCH 3 - CH 2 – CH - CH 3 CH 2 =CHC 6 H 5 - CH 2 – CH - C6H5C6H5

monomerrepeating unit polymertypical uses pipes, flooring CH 2 =CHCl- CH 2 – CH - Cl poly(chloroethene) polyvinylchloride PVC Poly (tetrafluoroethene) PTFE CF 2 =CF 2 - CF 2 – CF 2 -Non-stick coating (Teflon)

 Alcohols are the homologous series with the general formula C n H 2n+1 OH.  They all contain the functional group, OH, which is called the hydroxyl group.  Alcohols can be classified as primary, secondary or tertiary, depending on the carbon skeleton to which the hydroxyl group is attached.

Draw out the structure, name and classify all the alcohols with the formula C 4 H 9 OH. RCH 2 OH 1 alkyl group on C next to OH so primary alcohol, 1° R 2 CHOH 2 alkyl groups on C next to OH so secondary alcohol, 2° R 3 COH 3 alkyl groups on C next to OH so tertiary alcohol, 3°

OHH H C H H C H H C H H C H Butan-1-ol primary OH H H C H H C H H C H H C H Butan-2-ol secondary

OHH H C H H C H H C CH 3 OH H H C H H C H HC CH 3 2-methylpropan-1-ol primary 2-methylpropan-2-ol tertiary

2CH 3 OH(l) + 3O 2 (g)  2CO 2 (g) + 4H 2 O(l) Δ H = -726kJ/mol C 2 H 5 OH(l) + O 2 (g)  2CO 2 (g) + 3H 2 O(l) Δ H = -1371kJ/mol  In countries such as Brazil, ethanol is mixed with petrol and used to power cars. Ethanol is less efficient as a fuel than petrol as it is already partially oxidised but does make the country less reliant on supply of petrol. As it can be produced by fermentation of sugar beet, many consider ethanol a carbon neutral fuel.

OHH H C H H C H +H2OH2O O H H C H C H ethanol ethanal Cr 2 O 7 /H +  Primary alcohols are oxidised first to aldehydes, such as ethanal. A suitable oxidising agent is acidified potassium dichromate(VI)

An aldehyde still has one hydrogen atom attached to the carbonyl carbon, so it can be oxidised one step further to a carboxylic acid. O H H C H C H O H H C H C OH ethanalethanoic acid Cr 2 O 7 /H +

 In practice, a primary alcohol such as ethanol is dripped into a warm solution of acidified potassium dichromate(VI).  The aldehyde, ethanal, is formed and immediately distils off, thereby preventing further oxidation to ethanoic acid, because the boiling point of ethanal (23 °C) is much lower than that of either the original alcohol ethanol (78 °C) or of ethanoic acid (118 °C). Both the alcohol and the acid have higher boiling points because of hydrogen bonding.  If oxidation of ethanol to ethanoic acid is required, the reagents must be heated together under reflux to prevent escape of the aldehyde before it can be oxidised further.

H H C HOH H C H H C H propan-2-ol H H C H O H C H C H propanone Cr 2 O 7 /H +  Secondary alcohols are oxidised to ketones. These have no hydrogen atoms attached to the carbonyl carbon and so cannot easily be oxidised further.

 When orange acidified potassium dichromate(VI) acts as an oxidising agent, it is reduced to green chromium(III) ions.  Primary and secondary alcohols both turn acidified dichromate(VI) solution from orange to green when they are oxidised, and this colour change can be used to distinguish them from tertiary alcohols.  Tertiary alcohols are not oxidised by acidified dichromate(VI) ions, so they have no effect on its colour, which remains orange.

 Named by using name of the alkane from which they are derived with the prefix chloro-, bromo- or iodo-.  For example: CH 3 CH 2 Br is bromoethane (CH 3 ) 2 CHCH 2 Cl is 1-chloro-2-methylpropane

Remember the position of the halogen atom must be indicated using the appropriate number so CH 3 CH 2 CH 2 Cl is 1-chloropropane and CH 3 CHClCH 3 is 2-chloropropane Halogenoalkanes can be classified in the same way as alcohols.

Key feature of halogenoalkanes is C X where X = Cl, Br or I What is notable about this bond compared with say, C – C and C – H? The halogen atom is more electronegative than C so the bond is polarised: C X ++++ ----

C Cl ++++ ---- C I ++++ ---- ORDER OF BOND POLARITIES: C Br ++++ ---- >> So is order of reactivity: chloroalkane > bromoalkanes > iodoalkanes? Is there another factor that ought to be considered before reaching a conclusion? BOND ENERGIES

Bond energies: BondBond energy in kJmol -1 C - Cl C - Br C - I This suggests that the order of reactivity is: iodoalkane > bromoalkanes > chloroalkanes

There’s only one way to find out which is best! Do an experiment, not fight! But what do halogenoalkanes react with? The  + carbon atom is susceptible to attack by NUCLEOPHILES. A nucleophile is a species with a lone pair of electrons. E.g. OH -, NH 3, CN -. When attack by a nucleophile occurs, the carbon – halogen bond breaks releasing a halide ion. A suitable nucleophile for experimentation is OH - from an aqueous solution of an alkali such as sodium hydroxide.

CH 3 CH 2 X + OH -  CH 3 CH 2 OH + X - OH has replaced the X so overall we have NUCLEOPHILIC SUBSTITUTION X = Cl, Br or I. How can we follow the rate of this reaction so that we can determine the order of reactivity? Halide ions  coloured precipitates when silver nitrate is added. So we can measure how long it takes for a precipitate to form.

S N 1 = unimolecular nucleophilic substitution (only one species in the slow step of the mechanism, rate determining step) S N 2 = bimolecular nucleophilic substitution (two species in the slow step of the mechanism, rate determining step) Use of curly arrows: Curly arrows are used in reaction mechanisms to show the movement of electron pairs.

to form a lone pair of electrons X either a bond pair of electrons or a lone pair of electrons either next to an atom TAILS come from HEADS point to form a bond pair of electrons or at an atom X

- OH CH 3 H OH C H Br - CH 3 H Br C H C CH 3 H H + + C H H + slow fast SN1SN1 Intermediate carbocation Heterolytic fission of C – Br bond

-OH HOH CH 3 C H Br- Br C HH + C H H HOBr- SN2SN2 Transition state

Which is best? S N 1 or S N 2? For primary halogenoalkanes – S N 2 For tertiary halogenoalkanes – S N 1 3° halogenoalkanes cannot undergo the S N 2 mechanism as 5 bulky groups would not fit around the C in the transition state - steric hindrance. 1° halogenoalknes are less likely to undergo S N 1 as this would involve the formation of a primary carbocation as an intermediate. Alkyl groups push electron density to the C atom they are attached to (positive inductive effect) which stabilises the positive charge. More alkyl groups mean a more stable carbocation.

2° halogenoalkanes react via a mixture of S N 1 and S N 2. The mechanism predominating depends upon the nature of the alkyl groups and the nature of the solvent.

alkane halogenoalkane alcohol ketone dihalogenoalkane alkene aldehyde poly(alkene) carboxylic acid trihalogenoalkane tetrahalogenoalkane M1 M2 M1 and M2 : You should know the mechanisms for these reactions The flow chart above enables you to convert 2-bromobutane into butanone using a two-step synthetic route 2-bromobutane butan-2-ol butanone reflux with NaOH reflux with H + /Cr 2 O 7

Devise two-step syntheses of the following products from the starting material. Include any experimental conditions. (a) Ethanoic acid from ethene (b) Butan-1-ol from butane (c) Propanal from 1-bromopropane