1. CONTENTS Structure of halogenoalkanes Naming of Halogenoalkanes Preperation of haloalkanes by alcohols Stability Physical properties of halogenoalkanes SN1 with mechanism SN2 with mechanism Reactions :  with aquous NaOH / KOH (nucleophillic substitution)  with alcoholic NaOH/KOH (elimination)  with ammonia (nucleophillic substitution)  with KCN (nucleophillic substitution)  with water (nucleophillic substitution) Tests for haloalkanes Uses of haloalkanes CFC’s THE CHEMISTRY OF HALOGENOALKANES

2. Before you start it would be helpful to… Recall the definition of a covalent bond Be able to balance simple equations Be able to write out structures for hydrocarbons and their derivatives Understand the different types of bond fission Recall the chemical properties of alkanes, alkenes and alcohols Intermolecular forces THE CHEMISTRY OF HALOGENOALKANES

3. STRUCTURE OF HALOGENOALKANES Format Contain the functional group C-X where X is a halogen (F,Cl,Br or I) Halogenoalkanes - halogen is attached to an aliphatic skeleton - alkyl group Structural difference Halogenoalkanes are classified according to the environment of the halogen Names Based on original alkane with a prefix indicating halogens and position. CH 3 CH 2 CH 2 Cl 1-chloropropaneCH 3 CHClCH 3 2-chloropropane CH 2 ClCHClCH 3 1,2-dichloropropane CH 3 CBr(CH 3 )CH 3 2-bromo-2-methylpropane PRIMARY 1° SECONDARY 2° TERTIARY 3°

4. Reactivity of halogenoalkanes Primary Secondary Tertiary + + + carbocation carbocation carbocation

5. Thirtiary carbocation Trigonal planar molecule to minimize repulsion between electron pairs bond angles of 120º Contains 3 bond pairs (no lone pairs) of Electrons * Formed by hetrolytic fission

6.  Naming haloalkanes are similar to those for naming alkanes  The halogens are written as prefixes: fluoro- (F), chloro- (Cl), bromo- (Br) and iodo- (I) e.g.

7.  When the parent chain has both a halogen and an alkyl substituent, the chain is numbered from the end nearer the first substituent regardless of what substituents are e.g.

9.  The compound has the systematic name A 2-chlorobutane B 3-chlorobutane C 1-chloro-1-methylpropane D 1-chloro-2-methylbutane

10. Structural formula Shows the atoms carbon by carbon, with the hydrogen and functional groups attached. CH 3 CH 2 CH 2 CH 2 OH Displayed formula Shows how all atoms are arranged, and all the bonds between them. H H H H H C C C C OH H H H H Skeletal formula Shows the bonds of the carbon skeleton only, with any functional groups. The hydrogen and carbon atoms aren’t shown. This is handy for drawing large complicated structures, like cyclic hydrocarbons. OH

11. STRUCTURAL ISOMERISM IN HALOGENOALKANES Different structures are possible due to... Different positions for the halogen and branching of the carbon chain 2-chlorobutane 2-chloro-2-methylpropane 1-chlorobutane 1-chloro-2-methylpropane

12. PHYSICAL PROPERTIES Boiling point Increases with molecular size due to increased van der Waals’ forces M r bp / °C chloroethane 64.5 13 1- chloropropane78.5 47 1-bromopropane124 71 Boiling point also increases for “straight” chain isomers. Greater branching = less relative surface area = lower inter-molecular forces bp / °C 1-bromobutaneCH 3 CH 2 CH 2 CH 2 Br 101 2-bromobutaneCH 3 CH 2 CHBrCH 3 91 2-bromo -2-methylpropane(CH 3 ) 3 CBr 73 Remember: the only methyl halide which is a liquid is iodomethane; chloroethane is a gas.

13. van der Waals dispersion forces (review) These attractions get stronger as the molecules get longer and have more electrons. That increases the sizes of the temporary dipoles that are set up. This is why the boiling points increase as the number of carbon atoms in the chains increases. Dispersion forces get stronger as you go from 1 to 2 to 3 carbons in the chain. It takes more energy to overcome them, and so the boiling points rise. The increase in boiling point as you go from a chloride to a bromide to an iodide (for a given number of carbon atoms) is also because of the increase in number of electrons leading to larger van der Waals dispersion forces. There are lots more electrons in, for example, iodomethane than there are in chloromethane - count them! Halogen atoms have grater number of elcetrons than hydrogen atom, increacing Van der Waals forces than alkenes

14. The carbon-halogen bonds (apart from the carbon-iodine bond) are polar, because the electron pair is pulled closer to the halogen atom than the carbon. This is because (apart from iodine) the halogens are more electronegative than carbon. The electronegativity values are: C2.5F4.0 Cl3.0 Br2.8 I2.5 forces due to the attractions between the permanent dipoles (except in the iodide case). The size of those dipole-dipole attractions will fall as the bonds get less polar (as you go from chloride to bromide to iodide, for example). Nevertheless, the boiling points rise! This shows that the effect of the permanent dipole-dipole attractions is much less important than that of the temporary dipoles which cause the dispersion forces. The large increase in number of electrons by the time you get to the iodide completely outweighs (more significant than) the loss of any permanent dipoles in the molecules.

15. Boiling temperature comparision Halagenoalkane  C-F bonds stronger than C-C Alkenes/alkanes  only london forces Boiling Temp. No. of carbon atoms 1 23

16. Solubility The halogenoalkanes are at best only very slightly soluble in water. (As they have more hydrocarbon part than polar part) In order for a halogenoalkane to dissolve in water you have to break attractions between the halogenoalkane molecules (van der Waals dispersion and dipole-dipole interactions) and break the hydrogen bonds between water molecules. Both of these cost energy. Halogenoalkanes tend to dissolve in organic solvents because the new intermolecular attractions have much the same strength as the ones being broken in the separate halogenoalkane and solvent.

17. Preperation oh halo-alkane

18. PREPERATION OF HALAGENOALKANES FROM ALKENE ReagentHydrogen bromide... it is electrophilic as the H is slightly positive ConditionRoom temperature. EquationC 2 H 4 (g) + HBr(g) ———> C 2 H 5 Br(l) bromoethane Mechanism Step 1As the HBr nears the alkene, one of the carbon-carbon bonds breaks The pair of electrons attaches to the slightly positive H end of H-Br. The HBr bond breaks to form a bromide ion. A carbocation (positively charged carbon species) is formed. Step 2The bromide ion behaves as a nucleophile and attacks the carbocation. Overall there has been addition of HBr across the double bond. ELECTROPHILIC ADDITION OF HYDROGEN BROMIDE

19. ADDITION TO UNSYMMETRICAL ALKENES Problem addition of HBr to propene gives two isomeric brominated compounds HBr is unsymmetrical and can add in two ways products are not formed to the same extent the problem doesn't arise in ethene because it is symmetrical. Mechanism Two possibilities ELECTROPHILIC ADDITION TO PROPENE

20. The general reaction looks like this:  Preperation oh halo-alkane from alcohols ROH + HX RX + H 2 O

21. CHLORINATION OF ALCOHOLS When PCl 5 is added to dry alcohol, clods of hydrogen cloride fumes are produced CH 3 CH 2 OH + PCl 5 CH 3 CH 2 Cl + POCl 3 + HCL (g) Hydrogen chloride test Hydrogen chloride gas forms a white smoke with ammonia. BROMINATION OF ALCOHOLS C 2 H 5 OH + HBr C 2 H 5 Br + H 2 O Dry conditions Room temp NaBr / KBr + 50% CONC. H 2 SO 4 Heat under reflux 3C 2 H 5 OH + PBr 3 3C 2 H 5 Br + H 3 PO 3 Moist red Phosperous + Br 2 KBr + H 2 SO 4 ---> KHSO 4 + HBr Heat under refux

22. IODINATION OF ALCOHOLS 3C 2 H 5 OH + PI 3 3C 2 H 5 I + H 3 PO 3 Moist red Phosperous + I 2 Heat under refux In this case the alcohol is reacted with a mixture of sodium or potassium iodide and concentrated phosphoric(V) acid, H 3 PO 4, and the iodoalkane is distilled off. The mixture of the iodide and phosphoric(V) acid produces hydrogen iodide which reacts with the alcohol. Phosphoric(V) acid is used instead of concentrated sulphuric acid because sulphuric acid oxidises iodide ions to iodine and produces hardly any hydrogen iodide. A similar thing happens to some extent with bromide ions in the preparation of bromoalkanes, (but not enough to get in the way of the main reaction). There is no reason why you couldn't use phosphoric(V) acid in the bromide case instead of sulphuric acid if you wanted to. C 2 H 5 OH + HI C 2 H 5 I + H 2 O NaI / KI + CONC. H 3 PO 4 H 3 PO 4 + KI ----> KH 2 PO 4 + HI

23. Instead of using phosphorus(III) bromide or iodide, the alcohol is heated under reflux with a mixture of red phosphorus and either bromine or iodine.

24. SECONDARY ALCOHOLS WITH HALOGENS + PCl 5 Butane-2-ol + PCl 5  2-cloro-butane

25. Tertiary alcohols react reasonably rapidly with concentrated hydrochloric acid, but for primary or secondary alcohols the reaction rates are too slow for the reaction to be of much importance. A tertiary alcohol reacts if it is shaken with with concentrated hydrochloric acid at room temperature. A tertiary halogenoalkane (haloalkane or alkyl halide) is formed TERTIARY ALCOHOLS WITH HALOGENS

26. Haloalkanes can be prepared from the vigorous reaction between cold alcohols and phosphorus(III) halides Preperation of halo-alkanes

27. Primary Secondary Tertiary H CH 3 CH 3 H 3 C C+ H C+ H 3 C C+ H CH 3 CH 3 Primary Secondary Tertiary Stability of carbocation increases

28.  Which of these compounds is a secondary halogenoalkane? A CH 3 CH(OH)CH 3 B CH 3 CCl(CH 3 )CH 3 C CH 3 CHClCH 3 D CH 3 CH 2 CH 2 Cl

29.  The formation of a carbocation from a halogenoalkane is an example of A homolytic fission. B heterolytic fission. C an initiation reaction. D a propagation reaction.  When a chloroalkane is heated with aqueous sodium hydroxide A no reaction occurs with primary, secondary or tertiary chloroalkanes. B a reaction occurs with primary and secondary chloroalkanes but not with tertiary chloroalkanes. C a reaction occurs with tertiary chloroalkanes but not with primary and secondary chloroalkanes. D a reaction occurs with primary, secondary and tertiary chloroalkanes.

30. NUCLEOPHILIC SUBSTITUTION Theory halogens have a greater electronegativity than carbon electronegativity is the ability to attract the shared pair in a covalent bond a dipole is induced in the C-X bond and it becomes polar the carbon is thus open to attack by nucleophiles nucleophile means ‘liking positive’ the greater electronegativity of the halogen attracts the shared pair of electrons so it becomes slightly negative; the bond is now polar.

31. OH¯ CN¯ NH 3 H 2 O NUCLEOPHILES : ELECTRON PAIR DONORS possess at least one LONE PAIR of electrons don’t have to possess a negative charge are attracted to the slightly positive (electron deficient) carbon examples are OH¯, CN¯, NH 3 and H 2 O (water is a poor nucleophile)

32. NUCLEOPHILIC SUBSTITUTION - MECHANISM the nucleophile uses its lone pair to provide the electrons for a new bond the halogen is displaced - carbon can only have 8 electrons in its outer shell the result is substitution following attack by a nucleophile the mechanism is therefore known as - NUCLEOPHILIC SUBSTITUTION

33. NUCLEOPHILIC SUBSTITUTION - MECHANISM Note the nucleophile has a lone pair of electrons the carbon-halogen bond is polar a ‘curly arrow’ is drawn from the lone pair to the slightly positive carbon atom a ‘curly arrow’ is used to show the movement of a pair of electrons carbon is restricted to 8 electrons in its outer shell - a bond must be broken the polar carbon-halogen bond breaks heterolytically (unevenly) the second ‘curly arrow’ shows the shared pair moving onto the halogen the halogen now has its own electron back plus that from the carbon atom it now becomes a negatively charged halide ion a halide ion (the leaving group) is displaced

34. The "electron pushing effect" of alkyl groups ?? You are probably familiar with the idea that bromine is more electronegative than hydrogen, so that in a H-Br bond the electrons are held closer to the bromine than the hydrogen. A bromine atom attached to a carbon atom would have precisely the same effect - the electrons being pulled towards the bromine end of the bond. The bromine has a negative inductive effect. Alkyl groups do precisely the opposite and, rather than draw electrons towards themselves, tend to "push" electrons away. The stability of the various carbocations (for understanding only )

35. This means that the alkyl group becomes slightly positive ( + ) and the carbon they are attached to becomes slightly negative ( - ) The alkyl group has a positive inductive effect. This is sometimes shown as, for example: The arrow shows the electrons being "pushed" away from the CH 3 group. The plus sign on the left-hand end of it shows that the CH 3 group is becoming positive. The symbols + and - simply reinforce that idea. (for understanding only )

36. Order of stability of carbocations primary < secondary < tertiary The importance of spreading charge around in making ions stable The general rule-of-thumb is that  If a charge is very localised (all concentrated on one atom) the ion is much less stable than if the charge is spread out over several atoms. Applying that to carbocations of various sorts... You will see that the electron pushing effect of the CH 3 group is placing more and more negative charge on the positive carbon as you go from primary to secondary to tertiary carbocations. The effect of this, of course, is to cut down that positive charge. At the same time, the region around the various CH 3 groups is becoming somewhat positive. The net effect, then, is that the positive charge is being spread out over more and more atoms as you go from primary to secondary to tertiary ions. The more you can spread the charge around, the more stable the ion becomes. (for understanding only )

37. When we talk about secondary carbocations being more stable than primary ones, what exactly do we mean? This means that it is going to take more energy to make a primary carbocation than a secondary one. If there is a choice between making a secondary ion or a primary one, it will be much easier to make the secondary one. Similarly, if there is a choice between making a tertiary ion or a secondary one, it will be easier to make the tertiary one.

38. Steric hindrance * It is simply as Prevention or retardation of reaction * Since the SN2 proceeds through a backside attack, the reaction will only proceed if the empty orbital is accessible. The more groups that are present around the area surrounding the leaving group, the slower the reaction will be. For the SN2, since steric hindrance increases as we go from primary to secondary to tertiary, the rate of reaction proceeds From primary (fastest) > secondary >> tertiary (slowest). * In another way, If the approach by the nucleophile to the carbon is made difficult by crowding by neighboring groups the transition state is more difficult to form, and the rate of the reaction slows. The blocking of access to a reactive site by nearby groups is referred to as steric hindrance. (for understanding only )

39. Primary carbocation (SN2 takes place)  Primary and tertiary carbocation intermediates have different stabilities  Because of inductive effects of alkyl groups stabilize thirtiary carbocation  Steric hindrance differs for attack on primary and tertiary carbon (in the molecule) / less space available for attack by OH on tertiary carbon / more space for attack by OH on primary carbon  As bulky / three alkyl groups obstruct attack by nucleophile  Inhibits formation of transition state Tertiary carbocation (SN1 takes place) Secondary carbocation ( SN1 or SN2 takes place ) Why the difference in mechanism??

40. NUCLEOPHILIC SUBSTITUTION - MECHANISM SN1SN1 Why S N 1 for thirtiary halagenoalkane ??  Tertiary carbocation is more stable  As inductive effects of alkyl groups stabilize tertiary carbocation  More Steric Hindrance / less space available for attack by OH¯on thirtiary carbocation As 3 alkyl groups obstruct attack by nucleophile with tertiary compound Br (:)Br

41. NUCLEOPHILIC SUBSTITUTION - MECHANISM SN2SN2 OH¯ CH 3 H H Br CH 3 H H OH Br - CH 3 H H Br (:)Br¯ Why S N 2 for primary halagenoalkane ??  Primary carbocation is less stable  As inductive effects of alkyl groups are less than that of secondary and tertiary carbocation  Less Steric Hindrance / more space available for attack by OH¯on primary carbocation  As central carbon atom is not surrounded by many alkyl / bulky groups which obstruct attack by the nucleophile

42. NUCLEOPHILIC SUBSTITUTION - MECHANISM ANIMATION SHOWING THE S N 2 MECHANISM

43. NUCLEOPHILIC SUBSTITUTION - RATE OF REACTION BasicsAn important reaction step is the breaking of the carbon-halogen (C-X) bond The rate of reaction depends on the strength of the C-X bond C-I 238 kJmol -1 weakest - easiest to break C-Br 276 kJmol -1 C-Cl 338 kJmol -1 C-F484 kJmol -1 strongest - hardest to break Experiment Water is a poor nucleophile but it can slowly displace halide ions C 2 H 5 Br(l) + H 2 O(l) ——> C 2 H 5 OH(l) + H + (aq) + Br¯(aq) If aqueous silver nitrate is shaken with a halogenoalkane (they are immiscible) the displaced halide combines with a silver ion to form a precipitate of a silver halide. The weaker the C-X bond the quicker the precipitate appears. Ag+ (aq) + X¯(aq) ——> AgX(s) AgCl white pptAgBr cream pptAgI yellow ppt

44. NUCLEOPHILIC SUBSTITUTION Reaction with Hydroxide ions

46. AQUEOUS SODIUM HYDROXIDE ReagentAqueous* sodium (or potassium) hydroxide ConditionsReflux/Heat in aqueous solution (SOLVENT IS IMPORTANT) ProductAlcohol Nucleophilehydroxide ion (OH¯) Equation e.g. C 2 H 5 Br(l) + NaOH(aq) ——> C 2 H 5 OH(l) + NaBr(aq) Mechanism * WARNING It is important to quote the solvent when answering questions. Elimination takes place when ethanol is the solvent - SEE LATER The reaction (and the one with water) is known as HYDROLYSIS NUCLEOPHILIC SUBSTITUTION OH¯ CH 3 H H Br CH 3 H H OH Br - CH 3 H H Br (:)Br¯

47. ELIMINATION ReagentAlcoholic sodium (or potassium) hydroxide ConditionsReflux/Heat in alcoholic solution ProductAlkene MechanismElimination EquationC 3 H 7 Br + NaOH(alc) ——> C 3 H 6 + H 2 O + NaBr Mechanism the OH¯ ion acts as a base and picks up a proton the proton comes from a carbon atom next to that bonded to the halogen the electron pair left moves to form a second bond between the carbon atoms the halogen is displaced overall there is ELIMINATION of HBr.  Generally elimination happend more readily with secondary or thirtiary halagenoalkane Complication With unsymmetrical halogenoalkanes, you can get mixture of products

48. ELIMINATION Complication The OH¯ removes a proton from a carbon atom adjacent the C bearing the halogen. If there had been another carbon atom on the other side of the C-Halogen bond, its hydrogen(s) would also be open to attack. If the haloalkane is unsymmetrical (e.g. 2- bromobutane) a mixture of isomeric alkene products is obtained. but-1-ene but-2-ene can exist as cis and trans isomers

49. Also make sure you know how to name thirtiary halo-alkanes And the alkenes after elimination

50. ELIMINATION ANIMATED MECHANISM With Alcoholic sodium (or potassium) hydroxide

51. NUCLEOPHILIC SUBSTITUTION AMMONIA ReagentAqueous, alcoholic ammonia (in EXCESS) ConditionsReflux in aqueous, alcoholic solution under pressure ProductAmine NucleophileAmmonia (NH 3 ) Equation e.g.C 2 H 5 Br + 2NH 3 (aq / alc) ——> C 2 H 5 NH 2 + NH 4 Br (i) C 2 H 5 Br + NH 3 (aq / alc) ——> C 2 H 5 NH 2 + HBr (ii) HBr + NH 3 (aq / alc) ——> NH 4 Br Mechanism NotesThe equation shows two ammonia molecules. The second one ensures that a salt is not formed.

52. NUCLEOPHILIC SUBSTITUTION AMMONIA Why excess ammonia? The second ammonia molecule ensures the removal of HBr which would lead to the formation of a salt. A large excess ammonia ensures that further substitution doesn’t take place - see below Problem Amines are also nucleophiles (lone pair on N) and can attack another molecule of halogenoalkane to produce a 2° amine. This too is a nucleophile and can react further producing a 3° amine and, eventually an ionic quarternary ammonium salt. C 2 H 5 NH 2 + C 2 H 5 Br ——> HBr + (C 2 H 5 ) 2 NH diethylamine, a 2° amine (C 2 H 5 ) 2 NH + C 2 H 5 Br ——> HBr + (C 2 H 5 ) 3 N triethylamine, a 3° amine (C 2 H 5 ) 3 N + C 2 H 5 Br ——> (C 2 H 5 ) 4 N + Br¯ tetraethylammonium bromide a quaternary (4°) salt

53. POTASSIUM CYANIDE ReagentAqueous, alcoholic potassium (or sodium) cyanide ConditionsReflux in aqueous, alcoholic solution ProductNitrile (cyanide) Nucleophilecyanide ion (CN¯) Equation e.g. C 2 H 5 Br + KCN (aq/alc) ——> C 2 H 5 CN + KBr(aq) Mechanism NUCLEOPHILIC SUBSTITUTION

54. POTASSIUM CYANIDE ReagentAqueous, alcoholic potassium (or sodium) cyanide ConditionsReflux in aqueous, alcoholic solution ProductNitrile (cyanide) Nucleophilecyanide ion (CN¯) Equation e.g. C 2 H 5 Br + KCN (aq/alc) ——> C 2 H 5 CN + KBr(aq) Mechanism Importanceextends the carbon chain by one carbon atom the CN group can be converted to carboxylic acids or amines. Hydrolysis C 2 H 5 CN + 2H 2 O ———> C 2 H 5 COOH + NH 3 Reduction C 2 H 5 CN + 4[H] ———> C 2 H 5 CH 2 NH 2 NUCLEOPHILIC SUBSTITUTION

55. POTASSIUM CYANIDE ANIMATED MECHANISM NUCLEOPHILIC SUBSTITUTION

56. Amine Lone pair

57. NUCLEOPHILIC SUBSTITUTION WATER Details A similar reaction to that with OH¯ takes place with water. It is slower as water is a poor nucleophile. EquationC 2 H 5 Br(l) + H 2 O(l) ——> C 2 H 5 OH(l) + HBr(aq)

59. Testing for halogenoalkanes

60. ion presentobservation Cl - white precipitate Br - very pale cream precipitate I-I- very pale yellow precipitate Testing for halogenoalkanes Add aqueous acidified silver nitrate Note : It is reacted with NaOH to hydrolyse the halogenoalkane to an alcohol and release the halogen as a halide ion. It is heated to make the reaction faster. The test for a halide ion is done using silver nitrate. The solution needs to be acidic to avoid interference by other ions, and nitric acid contains no ions (unlike hydrochloric acid) that would interfere

61. Halogenoalkanes and water and alakli

62. ELIMINATION v. SUBSTITUTION The products of reactions between haloalkanes and OH¯ are influenced by the solvent SOLVENT ROLE OF OH – MECHANISMPRODUCT WATERNUCLEOPHILESUBSTITUTIONALCOHOL BASEELIMINATIONALKENE Modes of attack Aqueous solnOH¯ attacks the slightly positive carbon bonded to the halogen. OH¯ acts as a nucleophile Alcoholic solnOH¯ attacks one of the hydrogen atoms on a carbon atom adjacent the carbon bonded to the halogen. OH¯ acts as a base (A BASE IS A PROTON ACCEPTOR) Both reactions take place at the same time but by varying the solvent you can influence which mechanism dominates.

63. USES OF HALOGENOALKANES SyntheticThe reactivity of the C-X bond means that halogenoalkanes play an important part in synthetic organic chemistry. The halogen can be replaced by a variety of groups via nucleophilic substitution. PolymersMany useful polymers are formed from halogeno hydrocarbons Monomer PolymerRepeating unit chloroethene poly(chloroethene) PVC - (CH 2 - CHCl) n – USED FOR PACKAGING tetrafluoroethene poly(tetrafluoroethene) PTFE - (CF 2 - CF 2 ) n - USED FOR NON-STICK SURFACES

64. USES OF HALOGENOALKANES Chlorofluorocarbons - CFC’s – REFRIGERENT dichlorofluoromethane CHFCl 2 refrigerant trichlorofluoromethane CF 3 Cl aerosol propellant, blowing agent bromochlorodifluoromethane CBrClF 2 fire extinguishers CCl 2 FCClF 2 dry cleaning solvent, degreasing agent All are/were chosen because of their  LOW REACTIVITY  HAVE HIGH ENTHALPY OF VAPORIZATION  NON-TOXICITY  NON-FLAMMABLE  NON-CORROSSIVE  HAS A MODERATE DENSITY IN LIQUID FORM AND RELATIVELY HIGH DENSITY IN VAPOUR FORM

65. CH 3 CH 2 CH 2 CH 2 OH + HBr CH 3 CH 2 CH 2 CH 2 Br + H 2 O NaBr + H 2 SO 4 ---> NaHSO 4 + HBr Heat under reflux Prepearing Halagenoalkanes in the lab Sodium bromide, butane-1-ol and water are placed in the flask. The flask is put into a beaker of cold water and conc. Sulfuric acid is added slowly from the flask. This flask is cooled because the reaction at this stage is exothermic. The dropping funnel is removed and the apparatus refluxed on water bath for 30 minutes

66. The apparatus shown below is set up, the mixture is distilled and is collected. The distillate will be collected as two layers - a lower organic layer upper aqueous layer Continuation 1-bromobutane distilled into the organic layer

67. Organic layerAqueous layer Unreacted butane-1-ol BromineOxides of sulfure But-1-eneHbr Water Impurities that maybe in the distillate Continuation  Aqueous layer is discarded. The organic layer can be purified and finally redistilled to produce pure 1-bromobutane

68. PURIFYING BROMOALKANE BROM WATER  transfer the contents of the collection flask to a separating funnel.

69.  To get rid of any remaining acidic impurities (including the bromine and sulphur dioxide), return the bromoethane to the separating funnel and shake it with either sodium carbonate or sodium hydrogencarbonate solution.  Add some anhydrous calcium chloride to the tube, shake well and leave to stand. The anhydrous calcium chloride is a drying agent and removes any remaining water. (It also absorbs ethanol, and so any remaining ethanol may be removed as well (depending on how much calcium chloride you use)  Transfer the dry bromoethane to a distillation flask and fractionally distil it, collecting what distils over at between 35 and 40°C.

70. 1-bromo butane from butane-1 -ol

71. ReagentAlcoholic sodium (or potassium) hydroxide ConditionsHeat in alcoholic solution ProductAlkene Nucleophilehydroxide ion (OH¯) Equation e.g. C 3 H 7 Br + NaOH(alc) ——> C 3 H 6 + H 2 O + NaBr LABORATORY PREPARATION OF ALKENE FROM HALAGENOALKANE C 3 H 7 Br + NaOH(alc) C3H6C3H6

72. NaOH/KOH inethanol/alcohol NaOH/KOH in water/ aqueous * NaBr/KBr & (50% or moderately conc) H2SO4 / * P & Br2 / PBr 3 /PBr5 / * NaBr /KBr & H3PO4 / * HBr NH3 (in alcohol /in a sealed tube /at high pressure)

73. REVISION CHECK What should you be able to do? Recall and explain the physical properties of halogenoalkanes Recall and explain the chemical properties of halogenoalkanes based on their structure Recall and explain the properties of nucleophiles Write balanced equations for reactions involving substitution and elimination Understand how the properties of a hydroxide ion are influenced by the choice of solvent Recall the effect of CFC’s on the ozone layer CAN YOU DO ALL OF THESE? YES NO

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