Upload
lianchen251110
View
90
Download
5
Tags:
Embed Size (px)
Citation preview
SYNTHETIC SYNTHETIC ORGANIC ORGANIC
CHEMISTRYCHEMISTRY
ESTERSESTERS
REACTIONS OF ORGANIC COMPOUNDSREACTIONS OF ORGANIC COMPOUNDS
ALKANESALKANES ALKENESALKENES
HALOGENOALKANESHALOGENOALKANES
ALCOHOLSALCOHOLS
AMINESAMINES
ALDEHYDESALDEHYDES
KETONESKETONES
CARBOXYLIC ACIDSCARBOXYLIC ACIDS
POLYMERSPOLYMERS
NITRILESNITRILES
DIBROMOALKANESDIBROMOALKANES
CONVERSIONSCONVERSIONS
AMIDESAMIDES
HYDROXYNITRILESHYDROXYNITRILES
ALKYL HYDROGEN SULPHATES
ALKYL HYDROGEN SULPHATES
KK
ESTERSESTERS
REACTIONS OF ORGANIC COMPOUNDSREACTIONS OF ORGANIC COMPOUNDS
ALKANESALKANES ALKENESALKENES
HALOGENOALKANESHALOGENOALKANES
ALCOHOLSALCOHOLS
AMINESAMINES
ALDEHYDESALDEHYDES
KETONESKETONES
CARBOXYLIC ACIDSCARBOXYLIC ACIDS
AA
PP
SS
TTGG
TT
NN
RR
POLYMERSPOLYMERS
EE
NITRILESNITRILES
HH
JJ
UU
UUII
BB LL
DD
MM
QQOO
FF
CC
VV
ESTERSESTERS
REACTIONS OF ORGANIC COMPOUNDSREACTIONS OF ORGANIC COMPOUNDS
ALKENESALKENES
HALOGENOALKANESHALOGENOALKANES
ALCOHOLSALCOHOLS
AMINESAMINES
ALDEHYDESALDEHYDES
KETONESKETONES
CARBOXYLIC ACIDSCARBOXYLIC ACIDSNITRILESNITRILES
DIBROMOALKANESDIBROMOALKANES
CONVERSIONSCONVERSIONS
HYDROXYNITRILESHYDROXYNITRILES
ALKYL HYDROGEN SULPHATES
ALKYL HYDROGEN SULPHATES
AMIDESAMIDES
WWWW
XX
XX
YY
CHLORINATION OF METHANECHLORINATION OF METHANE
Mechanism free radical substitution
Initiation Cl2 2Cl• radicals created
Propagation Cl• + CH4 CH3• + HCl radicals used and
Cl2 + CH3• CH3Cl + Cl• then re-generated
Termination Cl• + Cl• Cl2 radicals removed
Cl• + CH3• CH3Cl
CH3• + CH3• C2H6 don’t use this as a termination
step in the examSummaryDue 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.
AA
CONVERSIONSCONVERSIONS
ELECTROPHILIC ADDITION OF HBrELECTROPHILIC ADDITION OF HBr
Reagent hydrogen bromide... it is electrophilic as the H is slightly positive
Condition room temperature
Equation C2H4(g) + HBr(g) C2H5Br(l) bromoethane
Mechanism electrophilic addition
Step 1 As 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 H-Br bond breaks to form a bromide ion.A carbocation (positively charged carbon species) is formed.
Step 2 The bromide ion behaves as a nucleophile and attacks the carbocation.Overall there has been addition of HBr across the double bond.Note that unsymmetrical alkenes will give a mixture of isomeric products.
BB
CONVERSIONSCONVERSIONS
Reagent bromine (liquid or dissolved in tetrachloromethane, CCl4 )
Conditions room temperature. No catalyst or UV light required.
Equation C2H4(g) + Br2(l) CH2BrCH2Br(l) 1,2 - dibromoethane
Mechanism electrophilic addition
It is surprising that bromineshould act as an electrophileas it is non-polar. The Br2 molecule is polarised by the region of high electron density around the double bond.
CC ELECTROPHILIC ADDITION OF BROMINEELECTROPHILIC ADDITION OF BROMINE
CONVERSIONSCONVERSIONS
DIRECT HYDRATION OF ALKENESDIRECT HYDRATION OF ALKENES
Reagent steam
Conditions high pressure
Catalyst c. H3PO4 (phosphoric acid)
Product alcohol
Equation C2H4(g) + H2O(g) C2H5OH(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.
DD
CONVERSIONSCONVERSIONS
HYDROGENATION (REDUCTION)HYDROGENATION (REDUCTION)EE
Reagent hydrogen
Conditions nickel catalyst - finely divided
Product alkanes
Equation C2H4(g) + H2(g) C2H6(g) ethane
Use margarine manufacture
CONVERSIONSCONVERSIONS
POLYMERISATION OF ALKENESPOLYMERISATION OF ALKENES
ETHENE
EXAMPLES OF ADDITION POLYMERISATION
PROPENE
TETRAFLUOROETHENE
CHLOROETHENE
POLY(ETHENE)
POLY(PROPENE)
POLY(CHLOROETHENE)
POLYVINYLCHLORIDE PVC
POLY(TETRAFLUOROETHENE)
PTFE “Teflon”
FF
CONVERSIONSCONVERSIONS
AQUEOUS SODIUM HYDROXIDE
Reagent aqueous* sodium (or potassium) hydroxideConditions reflux in aqueous solution (SOLVENT IS IMPORTANT)Product alcoholNucleophile hydroxide ion (:OH¯)
Equation C2H5Br(l) + NaOH(aq) C2H5OH(l) + NaBr(aq)
Mechanism nucleophilic substitution
*WARNING It is important to quote the solvent when answering questions.A different reaction (elimination) takes place when ethanol is the solvent.
NUCLEOPHILIC SUBSTITUTIONNUCLEOPHILIC SUBSTITUTIONGG
CONVERSIONSCONVERSIONS
NUCLEOPHILIC SUBSTITUTIONNUCLEOPHILIC SUBSTITUTION
AMMONIA
Reagent aqueous, alcoholic ammonia (in EXCESS)Conditions reflux in aqueous, alcoholic solution under pressureProduct amineNucleophile ammonia (:NH3)
Equation C2H5Br + 2NH3 C2H5NH2 + NH4Br
Mechanism nucleophilic substitution You need to show the removal of the proton here by another ammonia molecule to form NH4Br
Notes The equation shows two ammonia molecules.Excess ammonia is used to prevent further substitution (SEE NEXT SLIDE)
HH
CONVERSIONSCONVERSIONS
NUCLEOPHILIC SUBSTITUTIONNUCLEOPHILIC SUBSTITUTION
AMMONIA
Why excess ammonia?The second ammonia molecule ensures the removal of HBr, effectively as NH4Br.
A large excess ammonia ensures that further substitution doesn’t take place - see below.
ProblemThe amine produced is also a nucleophile (lone pair on N) and can attack another molecule of the original 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.
C2H5NH2 + C2H5Br HBr + (C2H5)2NH diethylamine, a 2° amine
(C2H5)2NH + C2H5Br HBr + (C2H5)3N triethylamine, a 3° amine
(C2H5)3N + C2H5Br (C2H5)4N+ Br¯ tetraethylammonium bromide, a 4° salt
HH
CONVERSIONSCONVERSIONS
POTASSIUM CYANIDE
Reagent aqueous, alcoholic potassium (or sodium) cyanideConditions reflux in aqueous , alcoholic solutionProduct nitrile (cyanide)Nucleophile cyanide ion (:CN¯)
Equation C2H5Br + KCN C2H5CN + KBr(aq)
Mechanism nucleophilic substitution
Importance It extends the carbon chain by one carbon atom.The CN group can then be converted to carboxylic acids or amines.
Hydrolysis C2H5CN + 2H2O C2H5COOH + NH3
Reduction C2H5CN + 4[H] C2H5CH2NH2
NUCLEOPHILIC SUBSTITUTIONNUCLEOPHILIC SUBSTITUTIONII
CONVERSIONSCONVERSIONS
REDUCTION OF NITRILESREDUCTION OF NITRILESJJ
CONVERSIONSCONVERSIONS
Reagent nickel catalyst and H2
Product amine
Equation C2H5CN + 4[H] C2H5CH2NH2 propylamine
Note that lithium tetrahydridoaluminate(III), LiAlH4, would achieve exactly the same effect. This reagent is not on your exam however.
HYDROLYSIS OF NITRILESHYDROLYSIS OF NITRILESKK
CONVERSIONSCONVERSIONS
Reagent waterConditions reflux in acidic or alkaline conditionsProduct carboxylic acid
Equation C2H5CN + 2H2O C2H5COOH + NH3
ELIMINATIONELIMINATION
Reagent alcoholic sodium (or potassium) hydroxide
Conditions reflux in alcoholic solution (SOLVENT IS IMPORTANT)
Product alkene
Equation C3H7Br + NaOH(alc) C3H6 + H2O + NaBr
Mechanism elimination
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 and overall there is ELIMINATION of HBr.
With some unsymmetrical haloalkanes, a mixture of isomeric products may be formed.
LL
CONVERSIONSCONVERSIONS
ELIMINATION OF WATER (DEHYDRATION)ELIMINATION OF WATER (DEHYDRATION)
Reagent/catalyst conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)
Conditions reflux at 180°C
Product alkene
Equation C2H5OH(l) CH2 = CH2(g) + H2O(l)
Mechanism (NO LONGER NEEDED FOR YOUR EXAM)
Step 1 Protonation of the alcohol using a lone pair on oxygen.Step 2 Loss of a water molecule to generate a carbocation.Step 3 Loss 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 some unsymmetrical alcohols.
LL
CONVERSIONSCONVERSIONS
OXIDATION OF PRIMARY ALCOHOLSOXIDATION OF PRIMARY ALCOHOLS
Primary alcohols are easily oxidised to aldehydes by acidified (with H2SO4) K2Cr2O7 (turns green from orange)
e.g. CH3CH2OH(l) + [O] CH3CHO(l) + H2O(l)
It is essential to distil off the aldehyde before it gets oxidised to the acid
CH3CHO(l) + [O] CH3COOH(l)
NN
The aldehyde has a lower boiling point so distils off before being oxidised further.
OXIDATION TOALDEHYDES
DISTILLATION
OXIDATION TOCARBOXYLIC ACIDS
REFLUX
The aldehyde condenses back into the mixture and gets oxidised to the acid.
CONVERSIONSCONVERSIONS
OXIDATION OF ALDEHYDESOXIDATION OF ALDEHYDES
Aldehydes are easily oxidised to carboxylic acids
e.g. CH3CHO(l) + [O] CH3COOH(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 include TOLLENS’ REAGENT and FEHLING’S SOLUTION
TOLLENS’ REAGENTReagent ammoniacal silver nitrate solutionObservation a silver mirror is formed on the inside of the test tubeProducts silver + carboxylic acidEquation Ag+ + e- Ag
FEHLING’S SOLUTIONReagent a solution of a copper(II) complex Observation a red precipitate forms in the blue solution Products copper(I) oxide + carboxylic acidEquation Cu2+ + e- Cu+
OO
CONVERSIONSCONVERSIONS
OXIDATION OF SECONDARY ALCOHOLSOXIDATION OF SECONDARY ALCOHOLS
Secondary alcohols are easily oxidised to ketones
e.g. CH3CHOHCH3(l) + [O] CH3COCH3(l) + H2O(l)
The alcohol is refluxed with acidified (with H2SO4) K2Cr2O7
However, on prolonged treatment with a more powerful oxidising agent they can be further oxidised to a mixture of acids with fewer carbon atoms than the original alcohol.
As far as your exam is required, they cannot be oxidised past ketones with acidified K2Cr2O7.
PP
CONVERSIONSCONVERSIONS
REDUCTION OF CARBOXYLIC ACIDS REDUCTION OF CARBOXYLIC ACIDS (NOT NEEDED FOR YOUR EXAM)(NOT NEEDED FOR YOUR EXAM)
Reagent/catalyst lithium tetrahydridoaluminate(III), LiAlH4
Conditions reflux in ethoxyethane
Product aldehyde
Equation CH3COOH(l) + 2[H] CH3CHO(l) + H2O(l)
CONVERSIONSCONVERSIONS
REDUCTION OF ALDEHYDESREDUCTION OF ALDEHYDESRR
Reagent sodium tetrahydridoborate(III), NaBH4
Conditions warm in water (aqueous conditions)
Product primary alcohol
Mechanism nucleophilic addition
Equation C2H5CHO(l) + 2[H] C3H7OH(l)
Mechanism only need to show the H+ here
CONVERSIONSCONVERSIONS
REDUCTION OF KETONESREDUCTION OF KETONESSS
Reagent sodium tetrahydridoborate(III), NaBH4
Conditions warm in water
Product secondary alcohol
Mechanism nucleophilic addition (see aldehydes for an example)
Equation CH3COCH3(l) + 2[H] CH3CH(OH)CH3(l)
CONVERSIONSCONVERSIONS
ESTERIFICATIONESTERIFICATION
Reagent(s) carboxylic acid, alcohol, strong acid catalyst (eg. conc. H2SO4)
Conditions heat under reflux
Product ester
Equation CH3CH2OH(l) + CH3COOH(l) CH3COOC2H5(l) + H2O(l)
Notes Concentrated H2SO4 is also a dehydrating agent. It removes
water as it is formed causing the equilibrium to move to the rightand thus increasing the yield of ester.
Uses of esters Esters are fairly unreactive but that doesn’t make them useless.Used as flavourings, plasticisers and solvents.
Naming esters Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH CH3COOCH3 + H2O
from ethanoic acid CH3COOCH3 from methanol
METHYL ETHANOATE
TT
CONVERSIONSCONVERSIONS
HYDROLYSIS OF ESTERSHYDROLYSIS OF ESTERSUU
Reagent(s) dilute acid or dilute alkali
Conditions heat under reflux
Product carboxylic acid and an alcohol
Equation CH3COOC2H5(l) + H2O(l) CH3CH2OH(l) + CH3COOH(l)
Notes If alkali is used for the hydrolysis the salt of the acid is formed.
CH3COOC2H5(l) + NaOH(aq) CH3CH2OH(l) + CH3COO-Na+(aq)
CONVERSIONSCONVERSIONS
BROMINATION OF ALCOHOLS BROMINATION OF ALCOHOLS (NOT NEEDED FOR YOUR EXAM)(NOT NEEDED FOR YOUR EXAM)
Reagent(s) conc. hydrobromic acid HBr or sodium (or potassium) bromide and concentrated sulphuric acid
Conditions reflux
Product haloalkane
Equation C2H5OH(l) + conc. HBr(aq) C2H5Br(l) + H2O(l)
Mechanism The 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 1 Protonation of the alcohol using a lone pair on oxygen.
Step 2 Loss of a water molecule to generate a carbocation (carbonium ion).
Step 3 The bromide ion behaves as a nucleophile and attacks the carbocation.
VV
CONVERSIONSCONVERSIONS
Reagent conc. H2SO4
Equation C2H4 + H2SO4 CH3CH2OSO3H
Mechanism electrophilic addition
WW ELECTROPHILIC ADDITION OF SULPHURIC ELECTROPHILIC ADDITION OF SULPHURIC ACIDACID
CONVERSIONSCONVERSIONS
Note that if the resulting ethyl hydrogen sulphate is heated in water, HYDROLYSIS takes place producing an alcohol (ethanol) and regenerating the H2SO4. Hence the sulphuric acid has acted as a catalyst overall.
C2H5OSO3H + H2O C2H5OH + H2SO4
Reagent hydrogen cyanide, HCN
Conditions heat under reflux (alkaline conditions)
Product haloalkane
Equation CH3 CHO + HCN CH3CH(OH)CN 2-hydroxypropanenitrile
Mechanism
XX NUCLEOPHILIC ADDITION IN ALDEHYDES NUCLEOPHILIC ADDITION IN ALDEHYDES AND KETONESAND KETONES
CONVERSIONSCONVERSIONS
Watch out for the possibility of optical isomerism in hydroxynitriles.
Conditions anhydrous conditions here (depends on the nucleophile used)
Equation eg. CH3 COCl + CH3NH2 CH3COONHCH3 + HCl
N-methylethanamide
Mechanism nucleophilic addition-elimination need to show H+ being lost also (this mechanism doesn’t, but should)
YY NUCLEOPHILIC ADDITION-ELIMINATIONNUCLEOPHILIC ADDITION-ELIMINATION
CONVERSIONSCONVERSIONS
Remember that in industry, acid anhydrides are the preferred method for acylation reactions like this, not acyl chlorides as shown here. This is for three reasons:
• secondary product is a carboxylic acid, not HCl (and so is less corrosive)
• anhydrides are cheaper to use
• anhydrides are less susceptible to hydrolysis