Download m4 organic reaction pathways

REACTIONS OF ORGANIC COMPOUNDS
POLYMERS
ALKANES
DIBROMOALKANES
ALKENES
KETONES
ALCOHOLS
ALDEHYDES
HALOGENOALKANES
AMINES
ESTERS
NITRILES
CONVERSIONS
CARBOXYLIC ACIDS
REACTIONS OF ORGANIC COMPOUNDS
POLYMERS
DIBROMOALKANES
KETONES
P
F
C
S
D
ALKANES
ALKENES
E
ALCOHOLS
M
N
A
B
G
L
T
U
R
V
HALOGENOALKANES
ALDEHYDES
ESTERS
U
H
O
Q
I
T
AMINES
J
NITRILES
K
CARBOXYLIC ACIDS
A
Initiation
CHLORINATION OF METHANE
Cl2
Propagation
Cl• + CH4
Cl2 + CH3•
Termination
Cl• + Cl•
Cl• + CH3•
CH3• + CH3•
——> 2Cl•
radicals created
——> CH3• + HCl
——> CH3Cl + Cl•
radicals used and
then re-generated
——>
——>
——>
radicals removed
Cl2
CH3Cl
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
ELECTROPHILIC ADDITION OF HBr
B
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
C
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)
1,2 - dibromoethane
Mechanism
It is surprising that bromine
should act as an electrophile
as it is non-polar.
CONVERSIONS
D
Reagent
DIRECT HYDRATION OF ALKENES
steam
Conditions high pressure
Catalyst
phosphoric acid
Product
alcohol
Equation
C2H4(g) +
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.
H2O(g)
C2H5OH(g)
ethanol
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
HYDROGENATION
E
Reagent
Conditions
hydrogen
nickel catalyst - finely divided
Product
alkanes
Equation
C2H4(g) + H2(g)
Use
margarine manufacture
———>
C2H6(g)
CONVERSIONS
ethane
POLYMERISATION OF ALKENES
F
EXAMPLES OF ADDITION POLYMERISATION
ETHENE
POLY(ETHENE)
PROPENE
POLY(PROPENE)
CHLOROETHENE
POLY(CHLOROETHENE)
POLYVINYLCHLORIDE
TETRAFLUOROETHENE
PVC
POLY(TETRAFLUOROETHENE)
PTFE
CONVERSIONS
“Teflon”
G
NUCLEOPHILIC SUBSTITUTION
AQUEOUS SODIUM HYDROXIDE
Reagent
Conditions
Product
Nucleophile
Equation
Aqueous* sodium (or potassium) hydroxide
Reflux in aqueous solution (SOLVENT IS IMPORTANT)
Alcohol
hydroxide ion (OH¯)
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
NUCLEOPHILIC SUBSTITUTION
H
AMMONIA
Reagent
Conditions
Product
Nucleophile
Aqueous, alcoholic ammonia (in EXCESS)
Reflux in aqueous, alcoholic solution under pressure
Amine
Ammonia (NH3)
e.g. C2H5Br + 2NH3 (aq / alc) ——> C2H5NH2 + NH4Br
Equation
(i)
(ii)
C2H5Br + NH3 (aq / alc) ——> C2H5NH2 + HBr
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
NUCLEOPHILIC SUBSTITUTION
H
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.
C2H5Br
——>
HBr
+
(C2H5)2NH
diethylamine, a 2° amine
(C2H5)2NH + C2H5Br
——>
HBr
+
(C2H5)3N
triethylamine, a 3° amine
(C2H5)3N
——>
(C2H5)4N+ Br¯
C2H5NH2
+
+
C2H5Br
tetraethylammonium bromide, a 4° salt
CONVERSIONS
NUCLEOPHILIC SUBSTITUTION
I
POTASSIUM CYANIDE
Reagent
Conditions
Product
Nucleophile
Equation
Aqueous, alcoholic potassium (or sodium) cyanide
Reflux in aqueous , alcoholic solution
Nitrile (cyanide)
cyanide ion (CN¯)
e.g. C2H5Br +
KCN (aq/alc) ———> C2H5CN + KBr(aq)
Mechanism
Importance
J
K
it extends the carbon chain by one carbon atom
the CN group can then be converted to carboxylic acids or amines.
Hydrolysis
Reduction
C2H5CN + 2H2O ———> C2H5COOH +
C2H5CN + 4[H] ———> C2H5CH2NH2
CONVERSIONS
NH3
ELIMINATION
L
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 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
ELIMINATION OF WATER (DEHYDRATION)
L
Reagent/catalyst
conc. sulphuric acid (H2SO4) or conc. phosphoric acid (H3PO4)
Conditions
reflux at 180°C
Product
alkene
e.g. C2H5OH(l) ————> CH2 = CH2(g) + H2O(l)
Equation
Mechanism
Step 1
Step 2
Step 3
Note
protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation
loss of a proton (H+) to give the alkene
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
OXIDATION OF PRIMARY ALCOHOLS
N
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]
———>
OXIDATION TO
ALDEHYDES
DISTILLATION
CH3COOH(l)
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
O
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
Cu2+ + e- ——> Cu+
CONVERSIONS
OXIDATION OF SECONDARY ALCOHOLS
P
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
Q
REDUCTION OF CARBOXYLIC ACIDS
Reagent/catalyst
lithium tetrahydridoaluminate(III) LiAlH4
Conditions
reflux in ethoxyethane
Product
aldehyde
Equation
e.g.
CH3COOH(l)
+ 2[H]
———>
CONVERSIONS
CH3CHO(l) + H2O(l)
REDUCTION OF ALDEHYDES
R
Reagent
sodium tetrahydridoborate(III) NaBH4
Conditions
warm in water or ethanol
Product
primary alcohol
Equation
e.g.
C2H5CHO(l) + 2[H]
———>
CONVERSIONS
C3H7OH(l)
S
REDUCTION OF KETONES
Reagent
sodium tetrahydridoborate(III) NaBH4
Conditions
warm in water or ethanol
Product
secondary alcohol
Equation
e.g. CH3COCH3(l) + 2[H]
———>
CONVERSIONS
CH3CH(OH)CH3(l)
ESTERIFICATION
T
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
from ethanoic acid
CH3COOCH3 + H2O
CH3COOCH3
METHYL ETHANOATE
CONVERSIONS
from methanol
U
HYDROLYSIS OF ESTERS
Reagent(s)
dilute acid or dilute alkali
Conditions
reflux
Product
carboxylic acid and an alcohol
Equation
Notes
e.g. CH3COOC2H5(l) + H2O(l)
CH3CH2OH(l) + CH3COOH(l)
If alkali is used for the hydrolysis the salt of the acid is formed
CH3COOC2H5(l) + NaOH(aq) ———> CH3CH2OH(l) + CH3COO-Na+(aq)
CONVERSIONS
BROMINATION OF ALCOHOLS
V
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)
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
———>
C2H5Br(l) + H2O(l)
3.
Step 1
Step 2
Step 3
protonation of the alcohol using a lone pair on oxygen
loss of a water molecule to generate a carbocation (carbonium ion)
a bromide ion behaves as a nucleophile and attacks the carbocation
CONVERSIONS