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

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

3. CHLORINATION OF METHANE
Initiation Cl2 ——> 2Cl• radicals created
Propagation Cl• CH ——> CH3• HCl radicals used and
Cl CH3• ——> CH3Cl Cl• then re-generated
Termination Cl• + Cl• ——> Cl2 radicals removed
Cl• + CH3• ——> CH3Cl
CH3• + CH3• ——> C2H6
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
CONVERSIONS

4. ELECTROPHILIC 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 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 2 The bromide ion behaves as a nucleophile and attacks the carbocation.
Overall there has been addition of HBr across the double bond.
CONVERSIONS

5. ELECTROPHILIC ADDITION OF BROMINE
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) ,2 - dibromoethane
Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.
CONVERSIONS

6. DIRECT 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.
CONVERSIONS

7. HYDROGENATION Reagent hydrogen
Conditions nickel catalyst - finely divided
Product alkanes
Equation C2H4(g) H2(g) ———> C2H6(g) ethane
Use margarine manufacture
CONVERSIONS

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

9. NUCLEOPHILIC SUBSTITUTION AQUEOUS SODIUM HYDROXIDEG
NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE
Reagent Aqueous* sodium (or potassium) hydroxide
Conditions Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Product Alcohol
Nucleophile 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
CONVERSIONS

10. NUCLEOPHILIC SUBSTITUTION
AMMONIA
Reagent Aqueous, alcoholic ammonia (in EXCESS)
Conditions Reflux in aqueous, alcoholic solution under pressure
Product Amine
Nucleophile Ammonia (NH3)
Equation e.g. C2H5Br NH3 (aq / alc) ——> C2H5NH NH4Br
(i) C2H5Br NH3 (aq / alc) ——> C2H5NH 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)
CONVERSIONS

11. 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.
C2H5NH 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
CONVERSIONS

12. NUCLEOPHILIC SUBSTITUTION
POTASSIUM CYANIDE
Reagent Aqueous, alcoholic potassium (or sodium) cyanide
Conditions Reflux in aqueous , alcoholic solution
Product 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 atom
the CN group can then be converted to carboxylic acids or amines.
Hydrolysis C2H5CN H2O ———> C2H5COOH NH3
Reduction C2H5CN [H] ———> C2H5CH2NH2 J K
CONVERSIONS

13. ELIMINATION Reagent Alcoholic sodium (or potassium) hydroxide
Conditions Reflux in alcoholic solution
Product Alkene
Mechanism Elimination
Equation C3H7Br NaOH(alc) ———> C3H H2O 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.
CONVERSIONS

14. 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 oxygen
Step 2 loss of a water molecule to generate a carbocation
Step 3 loss of a proton (H+) to give the alkene
Note Alcohols 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...
CONVERSIONS

15. OXIDATION 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)
OXIDATION TO
ALDEHYDES
DISTILLATION
OXIDATION TO
CARBOXYLIC ACIDS
REFLUX
Aldehyde has a lower boiling point so distils off before being oxidised further
Aldehyde condenses back into the mixture and gets oxidised to the acid
CONVERSIONS

16. OXIDATION 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’ REAGENT
Reagent ammoniacal silver nitrate solution
Observation a silver mirror is formed on the inside of the test tube
Products silver + carboxylic acid
Equation Ag e- ——> Ag
FEHLING’S SOLUTION
Reagent a solution of a copper(II) complex
Observation a red precipitate forms in the blue solution
Products copper(I) oxide + carboxylic acid
Equation Cu e- ——> Cu+
CONVERSIONS

17. OXIDATION OF SECONDARY ALCOHOLSP
OXIDATION 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.
CONVERSIONS

18. REDUCTION OF CARBOXYLIC ACIDSQ
REDUCTION OF CARBOXYLIC ACIDS
Reagent/catalyst lithium tetrahydridoaluminate(III) LiAlH4
Conditions reflux in ethoxyethane
Product aldehyde
Equation e.g. CH3COOH(l) [H] ———> CH3CHO(l) H2O(l)
CONVERSIONS

19. REDUCTION OF ALDEHYDES
Reagent sodium tetrahydridoborate(III) NaBH4
Conditions warm in water or ethanol
Product primary alcohol
Equation e.g. C2H5CHO(l) [H] ———> C3H7OH(l)
CONVERSIONS

20. REDUCTION OF KETONES Reagent sodium tetrahydridoborate(III) NaBH4
Conditions warm in water or ethanol
Product secondary alcohol
Equation e.g. CH3COCH3(l) [H] ———> CH3CH(OH)CH3(l)
CONVERSIONS

21. T
ESTERIFICATION
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 right
and thus increasing the yield of ester.
Uses of esters Esters are fairly unreactive but that doesn’t make them useless
Used as flavourings
Naming esters Named from the alcohol and carboxylic acid which made them...
CH3OH + CH3COOH CH3COOCH H2O
from ethanoic acid CH3COOCH from methanol
METHYL ETHANOATE
CONVERSIONS

22. HYDROLYSIS OF ESTERS 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)
CONVERSIONS

23. BROMINATION OF ALCOHOLSV
BROMINATION 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
CONVERSIONS