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AN INTRODUCTION TO SYNTHETIC ORGANIC CHEMISTRY KNOCKHARDY PUBLISHING.

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Presentation on theme: "AN INTRODUCTION TO SYNTHETIC ORGANIC CHEMISTRY KNOCKHARDY PUBLISHING."— Presentation transcript:

1 AN INTRODUCTION TO SYNTHETIC ORGANIC CHEMISTRY KNOCKHARDY PUBLISHING

2 CONVERSIONS INTRODUCTION This Powerpoint show is one of several produced to help students understand selected topics at AS and A2 level Chemistry. It is based on the requirements of the AQA and OCR specifications but is suitable for other examination boards. Individual students may use the material at home for revision purposes or it may be used for classroom teaching if an interactive white board is available. Accompanying notes on this, and the full range of AS and A2 topics, are available from the KNOCKHARDY SCIENCE WEBSITE at... Navigation is achieved by... either clicking on the grey arrows at the foot of each page orusing the left and right arrow keys on the keyboard KNOCKHARDY PUBLISHING ORGANIC REACTION SEQUENCES

3 ESTERS REACTIONS OF ORGANIC COMPOUNDS ALKANES ALKENES HALOGENOALKANES ALCOHOLS AMINES ALDEHYDES KETONES CARBOXYLIC ACIDS POLYMERS NITRILES DIBROMOALKANES CONVERSIONS

4 K K ESTERS REACTIONS OF ORGANIC COMPOUNDS ALKANES ALKENES HALOGENOALKANES ALCOHOLS AMINES ALDEHYDES KETONES CARBOXYLIC ACIDS A A P P S S T T G G T T N N R R POLYMERS E E NITRILES H H J J DIBROMOALKANES U U U U I I B B L L D D M M Q Q O O F F C C V V

5 CHLORINATION OF METHANE Initiation Cl 2 ——> 2Cl radicals created Propagation Cl + CH 4 ——> CH 3 + HClradicals used and Cl 2 + CH 3 ——> CH 3 Cl + Cl then re-generated Termination Cl + Cl ——> Cl 2 radicals removed Cl + CH 3 ——> CH 3 Cl CH 3 + CH 3——> C 2 H 6 Summary Due to the lack of reactivity of alkanes you need a very reactive species to persuade them to react Free radicals need to be formed by homolytic fission of covalent bonds This is done by shining UV light on the mixture (heat could be used) Chlorine radicals are produced because the Cl-Cl bond is the weakest You only need one chlorine radical to start things off With excess chlorine you will get further substitution and a mixture of chlorinated products A A CONVERSIONS

6 ELECTROPHILIC ADDITION OF HBr Reagent Hydrogen bromide... it is electrophilic as the H is slightly positive Condition Room temperature. Equation C 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. B B CONVERSIONS

7 Reagent Bromine. (Neat liquid or dissolved in tetrachloromethane, CCl 4 ) Conditions Room temperature. No catalyst or UV light required! Equation C 2 H 4 (g) + Br 2 (l) ——> CH 2 BrCH 2 Br(l) 1,2 - dibromoethane Mechanism It is surprising that bromine should act as an electrophile as it is non-polar. C C ELECTROPHILIC ADDITION OF BROMINE CONVERSIONS

8 DIRECT HYDRATION OF ALKENES Reagent steam Conditions high pressure Catalyst phosphoric acid Product alcohol Equation C 2 H 4 (g) + H 2 O(g) C 2 H 5 OH(g) ethanol Use ethanol manufacture Comments It may be surprising that water needs such vigorous conditions to react with ethene. It is a highly polar molecule and you would expect it to be a good electrophile. However, the O-H bonds are very strong so require a great deal of energy to be broken. This necessitates the need for a catalyst. D D CONVERSIONS

9 HYDROGENATION E E Reagent hydrogen Conditions nickel catalyst - finely divided Product alkanes Equation C 2 H 4 (g) + H 2 (g) ———> C 2 H 6 (g) ethane Use margarine manufacture CONVERSIONS

10 POLYMERISATION OF ALKENES ETHENE EXAMPLES OF ADDITION POLYMERISATION PROPENE TETRAFLUOROETHENE CHLOROETHENE POLY(ETHENE) POLY(PROPENE) POLY(CHLOROETHENE) POLYVINYLCHLORIDE PVC POLY(TETRAFLUOROETHENE) PTFE “Teflon” F F CONVERSIONS

11 AQUEOUS SODIUM HYDROXIDE ReagentAqueous* sodium (or potassium) hydroxide ConditionsReflux 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 The reaction (and the one with water) is known as HYDROLYSIS NUCLEOPHILIC SUBSTITUTION G G CONVERSIONS

12 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. Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE) H H CONVERSIONS

13 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 The amine produced is also a nucleophile (lone pair on N) and can attack another molecule of halogenoalkane to produce a 2° amine. This in turn is a nucleophile and reacts further producing a 3° amine and, eventually a 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 4° salt H H CONVERSIONS

14 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 Importanceit extends the carbon chain by one carbon atom the CN group can then be converted to carboxylic acids or amines. ReductionC 2 H 5 CN + 4[H] ——> C 2 H 5 CH 2 NH 2 HydrolysisC 2 H 5 CN + 2H 2 O ——> C 2 H 5 COOH + NH 3 NUCLEOPHILIC SUBSTITUTION I I J J K K CONVERSIONS

15 ELIMINATION ReagentAlcoholic sodium (or potassium) hydroxide ConditionsReflux 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 C atom next to the one bonded to the halogen the electron pair moves to form a second bond between the carbon atoms the halogen is displaced; overall there is ELIMINATION of HBr. With unsymmetrical halogenoalkanes, a mixture of products may be formed. L L CONVERSIONS

16 ELIMINATION OF WATER (DEHYDRATION) Reagent/catalystconc. sulphuric acid (H 2 SO 4 ) or conc. phosphoric acid (H 3 PO 4 ) Conditionsreflux at 180°C Productalkene Equation e.g. C 2 H 5 OH(l) ————> CH 2 = CH 2 (g) + H 2 O(l) Mechanism Step 1protonation of the alcohol using a lone pair on oxygen Step 2loss of a water molecule to generate a carbocation Step 3loss of a proton (H + ) to give the alkene NoteAlcohols with the OH in the middle of a chain can have two ways of losing water. In Step 3 of the mechanism, a proton can be lost from either side of the carbocation. This gives a mixture of alkenes from unsymmetrical alcohols... M M CONVERSIONS

17 OXIDATION OF PRIMARY ALCOHOLS Primary alcohols are easily oxidised to aldehydes e.g. CH 3 CH 2 OH(l) + [O] ———> CH 3 CHO(l) + H 2 O(l) it is essential to distil off the aldehyde before it gets oxidised to the acid CH 3 CHO(l) + [O] ———> CH 3 COOH(l) N N Aldehyde has a lower boiling point so distils off before being oxidised further OXIDATION TO ALDEHYDES DISTILLATION OXIDATION TO CARBOXYLIC ACIDS REFLUX Aldehyde condenses back into the mixture and gets oxidised to the acid CONVERSIONS

18 OXIDATION OF ALDEHYDES Aldehydes are easily oxidised to carboxylic acids e.g. CH 3 CHO(l) + [O] ———> CH 3 COOH(l) one way to tell an aldehyde from a ketone is to see how it reacts to mild oxidation ALDEHYES are EASILY OXIDISED KETONES are RESISTANT TO MILD OXIDATION reagents includeTOLLENS’ REAGENTand FEHLING’S SOLUTION TOLLENS’ REAGENT Reagentammoniacal silver nitrate solution Observationa silver mirror is formed on the inside of the test tube Productssilver + carboxylic acid EquationAg + + e - ——> Ag FEHLING’S SOLUTION Reagenta solution of a copper(II) complex Observationa red precipitate forms in the blue solution Productscopper(I) oxide + carboxylic acid EquationCu 2+ + e - ——> Cu + O O CONVERSIONS

19 OXIDATION OF SECONDARY ALCOHOLS Secondary alcohols are easily oxidised to ketones e.g. CH 3 CHOHCH 3 (l) + [O] ———> CH 3 COCH 3 (l) + H 2 O(l) The alcohol is refluxed with acidified K 2 Cr 2 O 7. However, on prolonged treatment with a powerful oxidising agent they can be further oxidised to a mixture of acids with fewer carbon atoms than the original alcohol. P P CONVERSIONS

20 REDUCTION OF CARBOXYLIC ACIDS Q Q Reagent/catalystlithium tetrahydridoaluminate(III) LiAlH 4 Conditionsreflux in ethoxyethane Productaldehyde Equation e.g. CH 3 COOH(l) + 2[H] ———> CH 3 CHO(l) + H 2 O(l) CONVERSIONS

21 REDUCTION OF ALDEHYDES R R Reagentsodium tetrahydridoborate(III) NaBH 4 Conditionswarm in water or ethanol Productprimary alcohol Equation e.g. C 2 H 5 CHO(l) + 2[H] ———> C 3 H 7 OH(l) CONVERSIONS

22 REDUCTION OF KETONES S S Reagentsodium tetrahydridoborate(III) NaBH 4 Conditionswarm in water or ethanol Productsecondary alcohol Equation e.g. CH 3 COCH 3 (l) + 2[H] ———> CH 3 CH(OH)CH 3 (l) CONVERSIONS

23 ESTERIFICATION Reagent(s)carboxylic acid + strong acid catalyst (e.g conc. H 2 SO 4 ) Conditionsreflux Productester Equation e.g.CH 3 CH 2 OH(l) + CH 3 COOH(l) CH 3 COOC 2 H 5 (l) + H 2 O(l) NotesConcentrated H 2 SO 4 is also a dehydrating agent, it removes water as it is formed causing the equilibrium to move to the right and thus increasing the yield of ester. Uses of estersEsters are fairly unreactive but that doesn’t make them useless Used as flavourings Naming estersNamed from the alcohol and carboxylic acid which made them... CH 3 OH + CH 3 COOH CH 3 COOCH 3 + H 2 O from ethanoic acid CH 3 COOCH 3 from methanol METHYL ETHANOATE T T CONVERSIONS

24 HYDROLYSIS OF ESTERS U U Reagent(s)dilute acid or dilute alkali Conditionsreflux Productcarboxylic acid and an alcohol Equation e.g.CH 3 COOC 2 H 5 (l) + H 2 O(l) CH 3 CH 2 OH(l) + CH 3 COOH(l) NotesIf alkali is used for the hydrolysis the salt of the acid is formed CH 3 COOC 2 H 5 (l) + NaOH(aq) ———> CH 3 CH 2 OH(l) + CH 3 COO - Na + (aq) CONVERSIONS

25 BROMINATION OF ALCOHOLS Reagent(s) conc. hydrobromic acid HBr(aq) or sodium (or potassium) bromide and concentrated sulphuric acid Conditionsreflux Producthaloalkane EquationC 2 H 5 OH(l) + conc. HBr(aq) ———> C 2 H 5 Br(l) + H 2 O(l) MechanismThe mechanism starts off in a similar way to dehydration (protonation of the alcohol and loss of water) but the carbocation (carbonium ion) is attacked by a nucleophilic bromide ion in step 3. Step 1protonation of the alcohol using a lone pair on oxygen Step 2loss of a water molecule to generate a carbocation (carbonium ion) Step 3a bromide ion behaves as a nucleophile and attacks the carbocation V V CONVERSIONS

26 © 2003 JONATHAN HOPTON & KNOCKHARDY PUBLISHING THE THE END AN INTRODUCTION TO SYNTHETIC ORGANIC CHEMISTRY


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