ESTERSESTERS

REACTIONS OF ORGANIC COMPOUNDSREACTIONS OF ORGANIC COMPOUNDS

ALKANESALKANES ALKENESALKENES

HALOGENOALKANESHALOGENOALKANES

ALCOHOLSALCOHOLS

AMINESAMINES

ALDEHYDESALDEHYDES

KETONESKETONES

CARBOXYLIC ACIDSCARBOXYLIC ACIDS

POLYMERSPOLYMERS

NITRILESNITRILES

DIBROMOALKANESDIBROMOALKANES

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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

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DIBROMOALKANESDIBROMOALKANES

UU

UUII

BB LL

DD

MM

QQOO

FF CC

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CHLORINATION OF METHANECHLORINATION OF METHANE

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

SummaryDue to the lack of reactivity of alkanes you need a very reactive species to persuade them to reactFree radicals need to be formed by homolytic fission of covalent bondsThis is done by shining UV light on the mixture (heat could be used)Chlorine radicals are produced because the Cl-Cl bond is the weakestYou only need one chlorine radical to start things offWith excess chlorine you will get further substitution and a mixture of chlorinated products

AA

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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

Step 1 As the HBr nears the alkene, one of the carbon-carbon bonds breaksThe 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 2 The bromide ion behaves as a nucleophile and attacks the carbocation.Overall there has been addition of HBr across the double bond.

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Reagent Bromine. (Neat 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

It is surprising that bromineshould act as an electrophileas it is non-polar.

CC ELECTROPHILIC ADDITION OF BROMINEELECTROPHILIC ADDITION OF BROMINE

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DIRECT HYDRATION OF ALKENESDIRECT HYDRATION OF ALKENES

Reagent steam

Conditions high pressure

Catalyst 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

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HYDROGENATIONHYDROGENATIONEE

Reagent hydrogen

Conditions nickel catalyst - finely divided

Product alkanes

Equation C2H4(g) + H2(g) ———> C2H6(g) ethane

Use margarine manufacture

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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”

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AQUEOUS SODIUM HYDROXIDE

Reagent Aqueous* sodium (or potassium) hydroxideConditions Reflux in aqueous solution (SOLVENT IS IMPORTANT)Product AlcoholNucleophile hydroxide ion (OH¯)

Equation e.g. C2H5Br(l) + NaOH(aq) ———> C2H5OH(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 SUBSTITUTIONNUCLEOPHILIC SUBSTITUTIONGG

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NUCLEOPHILIC SUBSTITUTIONNUCLEOPHILIC SUBSTITUTION

AMMONIA

Reagent Aqueous, alcoholic ammonia (in EXCESS)Conditions Reflux in aqueous, alcoholic solution under pressureProduct AmineNucleophile Ammonia (NH3)

Equation e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br

(i) C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr

(ii) HBr + NH3 (aq / alc) ——> NH4Br

Mechanism

Notes The 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)

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NUCLEOPHILIC SUBSTITUTIONNUCLEOPHILIC 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

ProblemThe 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.

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

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POTASSIUM CYANIDE

Reagent Aqueous, alcoholic potassium (or sodium) cyanideConditions Reflux in aqueous , alcoholic solutionProduct Nitrile (cyanide)Nucleophile cyanide ion (CN¯)

Equation e.g. C2H5Br + KCN (aq/alc) ———> C2H5CN + KBr(aq)

Mechanism

Importance it extends the carbon chain by one carbon atomthe CN group can then be converted to carboxylic acids or amines.

Hydrolysis C2H5CN + 2H2O ———> C2H5COOH + NH3

Reduction C2H5CN + 4[H] ———> C2H5CH2NH2

NUCLEOPHILIC SUBSTITUTIONNUCLEOPHILIC SUBSTITUTIONII

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ELIMINATIONELIMINATION

Reagent Alcoholic sodium (or potassium) hydroxide

Conditions Reflux in alcoholic solution

Product Alkene

Mechanism Elimination

Equation C3H7Br + NaOH(alc) ———> C3H6 + H2O + NaBr

Mechanism

the OH¯ ion acts as a base and picks up a protonthe proton comes from a C atom next to the one bonded to the halogenthe electron pair moves to form a second bond between the carbon atomsthe halogen is displaced; overall there is ELIMINATION of HBr.

With unsymmetrical halogenoalkanes, a mixture of products may be formed.

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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 e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)

Mechanism

Step 1 protonation of the alcohol using a lone pair on oxygenStep 2 loss of a water molecule to generate a carbocationStep 3 loss of a proton (H+) to give the alkene

Note Alcohols with the OH in the middle of a chain can havetwo 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...

LL

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OXIDATION OF PRIMARY ALCOHOLSOXIDATION OF PRIMARY ALCOHOLS

Primary alcohols are easily oxidised to aldehydes

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)

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Aldehyde has a lower boiling point so distils off before being oxidised further

OXIDATION TOALDEHYDES

DISTILLATION

OXIDATION TOCARBOXYLIC ACIDS

REFLUX

Aldehyde condenses back into the mixture and gets oxidised to the acid

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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+

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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 K2Cr2O7. 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.

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REDUCTION OF CARBOXYLIC ACIDSREDUCTION OF CARBOXYLIC ACIDSQQ

Reagent/catalyst lithium tetrahydridoaluminate(III) LiAlH4

Conditions reflux in ethoxyethane

Product aldehyde

Equation e.g. CH3COOH(l) + 2[H] ———> CH3CHO(l) + H2O(l)

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REDUCTION OF ALDEHYDESREDUCTION OF ALDEHYDESRR

Reagent sodium tetrahydridoborate(III) NaBH4

Conditions warm in water or ethanol

Product primary alcohol

Equation e.g. C2H5CHO(l) + 2[H] ———> C3H7OH(l)

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REDUCTION OF KETONESREDUCTION OF KETONESSS

Reagent sodium tetrahydridoborate(III) NaBH4

Conditions warm in water or ethanol

Product secondary alcohol

Equation e.g. CH3COCH3(l) + 2[H] ———> CH3CH(OH)CH3(l)

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ESTERIFICATIONESTERIFICATION

Reagent(s) carboxylic acid + strong acid catalyst (e.g conc. H2SO4 )

Conditions reflux

Product ester

Equation e.g. 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 uselessUsed as flavourings

Naming esters Named from the alcohol and carboxylic acid which made them...

CH3OH + CH3COOH CH3COOCH3 + H2O

from ethanoic acid CH3COOCH3 from methanol

METHYL ETHANOATE

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HYDROLYSIS OF ESTERSHYDROLYSIS OF ESTERSUU

Reagent(s) dilute acid or dilute alkali

Conditions reflux

Product carboxylic acid and an alcohol

Equation e.g. 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)

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BROMINATION OF ALCOHOLSBROMINATION OF ALCOHOLS

Reagent(s) conc. hydrobromic acid HBr(aq) 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 a bromide ion behaves as a nucleophile and attacks the carbocation

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