benzene
TRANSCRIPT
Hassan Subhie GCE A-level Chemistry Unit 5 1
Benzene the basis of aromatic compounds:
o The study of organic chemistry is divided into 2 main sections:
1. Aliphatic compounds: compounds based on carbon chains.
2. Aromatic compounds: compounds based on benzene ring structure.
Structure of benzene:
o The molecular formula of benzene is C6H6.
o The molecule of benzene has a regular hexagonal shape and is planar.
o The benzene ring doesn’t consist of alternate double and single bonds.
o All the bond lengths are the same are between single C-C and and double C=C.
o Each C-C bond in benzene is therefore neither a single nor a double bond but and
intermediate type lying between the 2 forms.
o Each C-atom in benzene forms 2σ bonds with 2 carbon atoms and 1 σ bond with
hydrogen atom and will still have one p orbital with single electron.
o These p orbitals overlap below and above the plane of the molecule forming a
continuous or deloecallised π system.
o This gives stability to the benzene structure which is why it undergoes substitution
reaction rather than addition reactions.
o Benzene can be represented either by:
Thermo chemical evidence for the structure of benzene:
o The enthalpy change of hydrogenation of cyclohexene is KJmol120
H2+
o The enthalpy of hydrogenation of benzene if it has 3 C=C double bonds is
KJmol360)120(3
Organic Chemistry III
Hassan Subhie GCE A-level Chemistry Unit 5 2
3 H
2+ KJmol120
o The actual of hydrogenation of benzene is KJmol208 .
Benzene can’t have cyclohexatriene structure. It has a more stable structure
which is the delocalized structure.
o The extra stability of benzene gained by delocalization is equal to
360-208 =152 KJmol- which is called delocalization enthalpy or stabilization
enthalpy.
152 KJ
-360 KJ
-208 KJ
Cyclohexane
Naming compounds of benzene:
NO2
Cl
CH3
Benzene Nitro Benzene Chloro benzene methyl benzene
(Toluene)
Hassan Subhie GCE A-level Chemistry Unit 5 3
C2H
5
NH2
OH
CH2Cl
Ethyl benzene phenyl amine Phenol Chloro methylbenzene
SO3H
CO2H
CHO
NO2
NO2
Benzene sulphonic acid Benzoic acid Benzyldehyde 1,4 dinitrobenzene
Benzene carboxylic acid
NO2
NO2
NO
2
NO2
CH2CO
2H
1,2 dinitrobenzene 1,3 dinitrobenzene Phenyl ethanoic acid
CO2H
C2H
5
NH2
CHO
Br
BrBr
2-ethyl benzoic acid 2-amino benzyldehyde 1,3,5 tri bromo benzene
Hassan Subhie GCE A-level Chemistry Unit 5 4
COCH3
OCOCH3
Br
BrBr
OH phenyl ethanone Phenyl ethanoate 2,4,6 tri bromo phenol
Physical properties of benzene:
o Colorless liquid with boiling point 80
0C and melting point 5.5
0C
o Immiscible with water.
o It burns with a luminous smoky flame.
o It is toxic and its inhalation over a period of time leads to anaemia and leukemia.
Types of reactions of benzene:
o Benzene makes electrophilic substitution reaction rather than addition reaction
although the benzene ring is unsaturated.
Reason:
Benzene ring is thermodynamically very stable due to the delocalization of
the pi molecular orbitals.
Addition reactions to benzene would result in disruption of this
delocalization and so reduces the stability of the benzene ring.
Substitution reactions can occur with only temporary disruption and so the
stability of the benzene ring is maintained.
Reactions of Benzene:
Benzene reacts mainly by electrophilic substitution reactions.
1. Nitration: Benzene reacts with a mixture of nitric acid and sulphuric acid to form
nitro benzene.
Reaction: OHNOHCHNOHCSOH
225636642
HNO3
H2SO
4
NO2
H 2O
Nitro benzene
+ +
Hassan Subhie GCE A-level Chemistry Unit 5 5
Reagent: concentrated HNO3
Condition: mix with concentrated H2SO4 at 50 0C
2. Bromination: Benzene reacts with liquid Bromine in the presence of a catalyst of
anhydrous iron(III) bromide.
Reaction: HBrBrHCBrHC l 56)(266
Br2
Br
H Br
Bromo benzene
+ +
Reagent: Br2 (l)
Condition: Liquid Br2 with iron catalyst.
3. Friedel-Craft alkylation: Benzene will react with halogeno alkanes in the
presence of a catalyst of anhydrous aluminium
chloride.
Reaction: HClRHCRClHC 5666
Reagent: Halogeno alkane RCl
Condition: Anhydrous AlCl3 catalyst.
E.g. HClHCHCClHCHC 52565266
C2H
5Cl
C2H
5
H Cl
Ethyl benzene
+ +
4. Friedel-Craft acylation: Benzene will react with acid chlorides in the presence
of a catalyst of anhydrous aluminium chloride.
Reaction: HClCORHCRCOClHC 5666
Reagent: Acyl chloride RCOCl
Condition: Anhydrous AlCl3 catalyst.
Hassan Subhie GCE A-level Chemistry Unit 5 6
C
H ClRC//O
__Cl
//O
__R
+ +
E.g. HClCOCHHCCOClCHHC 356366
Phenyl Ethan one
C
H Cl//O
__Cl
//O
__CH
3
CH3C+ +
Reactions of compounds with a carbon containing side chain:
o Compounds such as methylbenzene C6H5CH3 and ethyl benzene C6H5C2H5 can be
oxidized, and the product will contain a COO- group on the benzene ring,
regardless of the number of carbon atoms in the chain. On acidification benzoic
acid is produced.
Reaction: 225656 3][6 COOHCOOHCOHORHC
R
6 [O] OH
_
_
COO
3 H
2O CO
2+ + + +
Reagent: Potassium manganate (VII) + NaOH
Condition: heat under reflux
E.g. 22565256 3][6 COOHCOOHCOHOHCHC
Hassan Subhie GCE A-level Chemistry Unit 5 7
Phenols:
o These contain the –OH group connected directly to the benzene ring.
OH
Phenol
Naming derivatives of Phenols:
OH|
|Cl
OH|
CH3
OH|
|NO
2 4-chloro phenol 2-methyl phenol 4-nitro phenol
Br
BrBr
OH 2,4,6 tri bromo phenol
Physical properties of phenols:
o Colorless solid
o Its vapor is toxic
o Burns the skin
o Although it forms some hydrogen bonds with water, its only partially soluble and
forms two layers when added to water at room temperature.
o It is a weaker acid than carbonic acid, and so it will not liberate CO2 from sodium
hydrogen carbonate, unlike carboxylic acid which will.
Hassan Subhie GCE A-level Chemistry Unit 5 8
o Acidity of Phenol
Phenol is weakly acidic dissociating in water
C6H5OH C6H5O- + H
+ Ka=1.3 x 10
-10 mol dm
-3
Phenol is too weak an acid to affect litmus paper or liberate carbon dioxide from
carbonates and hydrogencarbonates, but it dissolves readily in alkalis.
o It is soluble in alkali as NaOH, as it is a weak acid and reacts with NaOH to form
colorless solution.
)(2)(56)()(56 laqaql OHOHCOHOHHC
H 2O
OH
OH
_
_
O
+ +
Phenate ion
Reactions of Phenol:
o The presence of the OH group makes substitution into the benzene ring easier.
No catalyst is required.
1. With Bromine water:
Reaction: HBrBrHOHCBrOHHCaq
33 326)(256
OH| |
+3 Br
2+
|
OH
Br
Br
Br
3 HBr
2,4,6- tribromo phenol
Reagent: Bromine.
Condition: aqueous
Observation: When red-brown bromine in H2O is added, a white ppt. rapidly forms and
bromine decolorizes .
Hassan Subhie GCE A-level Chemistry Unit 5 9
2. With acid chlorides: Phenol reacts as an alcohol, and an ester is formed.
Reaction: HClHRCOOCRCOClOHHC 5656
OH
RCOCl
|OOCR
H Cl+ +
Reagent: Acyl chloride.
E.g. HClHCOOCCHCOClCHOHHC 563356
Phenyl ethanoate
OH|OOCCH
3
H ClCH3COCl+ +
Phenyl amine :( Aniline)
o Phenyl amine has a NH2 group attached to the benzene ring.
|NH
2
Preparation:
o Nitrobenzene is reduced by tin and concentrated hydrochloric acid when
heated under reflux.
o Then sodium hydroxide is added to liberate the phenyl amine, which is
removed by steam distillation.
Reaction: Nitrobenzene→ Phenyl amine
OHNHHCHNOHC 2256256 2][6
Reagent: Tin and concentrated hydrochloric acid.
Hassan Subhie GCE A-level Chemistry Unit 5 10
Condition: heat under reflux, and then add sodium hydroxide
Reactions of Phenyl amine:
1. Its weakly basic and reacts with acids to form salts.
ClNHHCHClNHHC 356256
2. Reacts with acyl chlorides and and benzoyl chlorides
HClNHCOCHHCCOClCHNHHC 3563256
Phenylethanamide
HClHNHCOCHCCOClHCNHHC 565656256
o These substituted amides are sometimes used as pharmaceuticals,
e.g. paracetamol.
NHCOCH
3
|OH
|
3. Reacts with nitrous acid to form a diazonium salt.
Reaction: Phenyl amine → diazonium salt.
Reagents: nitrous acid at 5 0C
Conditions: add dilute hydrochloric to phenylamine and cool the solution to
5 0C.Sodium nitrite solution is then added, keeping the temperature close
to 5 0C and above 0 0C
OHNHCNOHNHHC 22562256 22
|NH
2
+ +
|
______N N
2 H
2O+NO
2-H
2 +
+
Diazonium salt
Hassan Subhie GCE A-level Chemistry Unit 5 11
o The solutions of diazonium ions react with phenol. A yellow precipitate of diazo
compound is obtained. This reaction is called coupling.
N+
OH__+N __N=N__ __OH
Azo dye
O Na+
N2Cl H
+
__N=N
__
HO
aromatic diazonium chloride
red precipitate
+
+
o The main use of this reaction is to produce synthetic dyes some acid indicator.
Questions:
1. State the structural formulae of the organic product obtained by the reaction of:
a) Propylbenzene with alkaline potassium manganate (VII).
b) Phenol with sodium hydroxide solution.
c) Phenol with propanoyl chloride
2. What would you observe if bromine water were added to aqueous phenol?
3. State the conditions for the conversion of nitrobenzene to phenylamine.
4. (a) Describe the mechanism for the reaction of bromine with benzene .
(b) Why does benzene undergo a substitution reaction with bromine rather than an
addition reaction?
5. In the reaction between propanone and hydrogen cyanide no reaction occurs unless a
small amount of a base such as sodium hydroxide is added. Explain these observations.
6. When benzenediazonium chloride is prepared, the reaction is carried out at about 5 0C.
(a) Why are these conditions chosen?
(b) What is the formula of the product obtained by the reaction of the benzene
diazonium chloride solution with phenol in alkaline solution?
Hassan Subhie GCE A-level Chemistry Unit 5 12
7. Robert Lister was the first to use phenol as an antiseptic, saving the life of a young
boy who had been in a traffic accident. At that time phenol was known as “carbolic
acid”.
(a) (i) Draw the structural formula of phenol.
[1]
(ii) When phenol is added to water, a weakly acidic solution is formed. Write an
equation explaining this observation.
[2]
(b) Phenol reacts with bromine to make tribromophenol, which is analogous to the
commercially available antiseptic trichlorophenol (TCP). Write the equation for the
reaction between bromine and phenol.
[2]
(c) Vinegar (ethanoic acid solution) also has medicinal properties. Explain how a
solution of sodium hydrogencarbonate may be used to distinguish between a solution
of vinegar and one of carbolic acid.
[2]
8. (i) What is observed when phenol reacts with bromine water
[2]
(ii) Write a balanced equation for the reaction and name the organic product.
[3]
Hassan Subhie GCE A-level Chemistry Unit 5 13
Mechanisms of Reactions:
Reaction mechanisms tell us:
1. Which bonds are broken and how they are broken.
2. Which bonds are made and in what order they are made.
A curly arrow ( ) represents the movement of a pair of electrons, either from a
bond or a lone pair.
A half-headed arrow ( ) represents the movement of a single electron.
Types of bond fission:
o There are 2 types of bond fission:
1. Homolytic bond fission: When a bond breaks evenly and one electron
moves to each atom forming a free radical.
A__
B A B +
2. Heterolytic bond fission: When a bond breaks unevenly and the 2
electrons go to one atom forming a nucleophile
and an electrophile.
A B
__A
+B:+
Note: Ensure that the curly arrow starts from a bond or an atom with alone pair of
electrons.
Ensure that the curly arrow points towards an atom either forming a negative ion or
a new covalent bond.
Types of mechanisms:
I. Homolytic free radical substitution
1. Reaction between an alkane and chlorine or bromine.
e.g. reaction of CH4 and Cl2.
There are three stages in this type of mechanism.
Hassan Subhie GCE A-level Chemistry Unit 5 14
1st
Initiation: Light energy causes homolytic fission of Cl2.
Cl
2 Cl
2nd
Propagation: each propagation step involves a radical reacting with a molecule to
produce a new molecule.
.CH4
Cl CH3 H Cl+ +
. Cl CH
3then : Cl2
CH3Cl+ +
3rd
Termination: Involves two radicals joining with no radicals being produced.
. Cl CH
3CH
3Cl+
Cl Cl
2Cl +
.CH
3.CH
3CH
3CH
3+
2. Homolytic free radical addition: ( Homolytic polymerization)
o The conditions are very high pressure (100 atm) and a trace of O2.
o The conditions produce radicals R , which attack the ethene molecule.
Initiation: .CH
2=CH
2R
R CH
2CH
2+
Propagation:
.CH2=CH
2 R CH2
CH2
.R CH2
CH2 CH
2CH
2__+
etc….
Note: Another way of polymerization of ethene is by using titanium tetrachloride and
aluminium tri ethyl, but this is not a hemolytic mechanism.
II. Heterolytic electrophylic addition.
Reaction of alkenes with halogens or hydrogen halides.
The reaction takes place in 2 steps:
Hassan Subhie GCE A-level Chemistry Unit 5 15
1. The electrophile accepts the electrons to form a new bond with one of
the carbon atoms and the same time π the Br-Br bond breaks.
2. The intermediate carbon cation then bonds with Br- ion.
C C
BrBrBr
Br
CC
Br:Br-
CC
BrBr
+
With HBr:
C C
Br
CC
:Br-
CC
Br
H
H H
+
Note: With the addition of a hydrogen halide to unsymmetrical alkene, the hydrogen atom adds
to the carbon which already has more hydrogen atoms directly bonded to it.
This is because:
30 carbocation is more stable then 20 carbocation then 10 carbocation.
C C C
C____|
C C C C C C
most stable least stable
> >+ +
+
Hassan Subhie GCE A-level Chemistry Unit 5 16
III. Heterolytic, electrophilic substitution:
In general
E+
+
HE
E
+ H+
+
E+: is the electrophile which could be NO2
+, Br
+ C2H5
+ or RCO
1. Nitration of benzene:
i. H2SO4 is a stronger acid than nitric acid and so protonates it
432342 HSONOHHNOSOH
2232 NOOHNOH
ii. The NO2+ is the electrophile and attacks the benzene ring.
iii. The HSO4- ion pulls off a H
+ and reforms H2SO4
+
H
+
NO2+NO2+ NO2
NO2
HSO4_
H2SO4
The loss of H+ results in the reforming of the benzene ring and the gain in
stability associated with the ring.
2. Bromination of benzene:
i. The catalyst FeBr3 reacts with Br2 to form the electrophile
Br+
ii. Br+ attacks the benzene ring
iii. FeBr4- pulls off a H
+ and reforms FeBr3
BrFeBrFeBrBrBr 43
Hassan Subhie GCE A-level Chemistry Unit 5 17
+
H
+
Br+ Br Br
FeBr4-
HBr FeBr3+
3. Friedel’s craft alkylation:
452352 AlClHCAlClClHC
+
H
+ +
C2H5+ C2H5
C2H5
AlCl4-
HCl AlCl3
IV. Heterolytic, nucleophilic substitution:
Reaction between the halogeno alkanes and hydroxide or cyanide or cyanide ions
There are two mechanisms depending on the type of halogenoalkanes.
With primary (10) halogenoalkanes, SN2 mechanism is dominant.
Reaction proceeds via a transition state when the lone pair of electrons on the
nucleophile attacks the halogeno alkanes.
CHO:
_
Br
CH3
HH
H
CHO Br
H
CH3
_
C
HH
CH3
HO
With tertiary (30) halogenoalkanes the SN1 mechanism is dominant.
This happens in 2 steps:
1. The halogeno alkane ionizes in the relatively slow rate-determining step to
form an intermediate carbocation.
Hassan Subhie GCE A-level Chemistry Unit 5 18
C___ ___
|
|CH
3
CH3
CH3
Br C___
|
|CH
3
CH3
CH3
:Br_
+ +
2. This rapidly forms a bond with the nucleophile OH-
C___
|
|CH
3
CH3
CH3
OH.. _
C___
|
|CH
3
CH3
CH3
___OH Br:+ +
V. Heterolytic, nucleophilic addition:
Reaction between carbonyl compound and hydrogen cyanide.
Reaction between carbonyl compound and hydrogen cyanide.
Although the reaction is the addition of HCN, the 1st step is the nucleophilic
attack by the CN- ion the carbon atom in the carbonyl compound.
The negatively charged oxygen in the anion then removes an H+ from the HCN
molecule. This produces another CN- ion, which reacts with another carbonyl
group and so
C//
R ___
H
O
CN:_
CR
O
H______
|
|CN
_:
H___
CN
CR
O
H______
|
|CN
___H
CN_
+
The conditions of this reaction are either HCN and trace of base or KCN and
small amount of dilute H2SO4.
Note: -Any optical activity is maintained in a SN2 reaction
-If the mechanism is SN1, a reactant which is an optical isomer will produce a
racemic mixture as carbocation can be attacked from either side.
Hassan Subhie GCE A-level Chemistry Unit 5 19
Summary of the reactions of aromatic substances:
benzene
C6H
6
concHNO3
conc H2SO
4
C6H
5NO
2
Br2
Fe catalyst
C6H
5Br
C2H
5Cl
AlCl3 catalyst
C6H
5C
2H
5
CH3COCl
AlCl3 catalyst
C6H
5COCH
3
Phenol
C6H
5OH
OH_
(aq)
Br2 (aq)
CH3COCl
C6H
5O
_
C6H
2Br
3OH
CH3COOC
6H
5
H+
(aq)
HNO2
PhyenylamineC
6H
5NH
2
C6H
5NH
3+
C6H
5N
2+
At5 CO
Hassan Subhie GCE A-level Chemistry Unit 5 20
Tests for alkenes or unsaturation: (C=C) Test: Shake the alkene with red-brown bromine solution
Observation: it gets decolorized.
Test: Shake with cold alkaline purple potassium manganate (VII),
Observation: it rapidly goes green
Test for Halogeno alkanes:
Test: Warm the halogeno alkane under reflux with aqueous sodium
hydroxide then acidify with dilute nitric acid to remove excess of OH- ions
then add aqueous silver nitrate.
Observation: If white ppt. forms which is soluble in dilute ammonia Chloro
alkane
If creamy ppt. forms which is partially soluble in dilute ammonia
Bromo alkane
If yellow ppt. forms which is insoluble in concentrated ammonia
Iodo alkane
Test for OH group:( in alcohols and acids)
Test: Add dry PCl5
Observation: Steamy acidic fumes of hydrogen chloride are produced (HCl)
Test for the type of alcohol:
Test: Warm with dilute sulphuric acid and potassium dichromate (VII)
solution.
Organic Tests
Hassan Subhie GCE A-level Chemistry Unit 5 21
Observation: -Primary (10) and secondary (20) alcohols will reduce the orange
dichromate ions to green.
-Tertiary (30) alcohols don’t affect the color
To distinguish between 10and 20 repeat the experiment and distil the product
into ammonical silver nitrate solution:
10 alcohols are oxidized to aldehydes, which give silver mirror
20 alcohols are oxidized to ketones, which don’t react.
Test for Carbonyl group C=O: (Aldehyde and Ketone)
Test: Add solution of 2,4-dinitro phenyl hydrazine (Brady’s Reagent).
Observation: An orange or red precipitate is seen.
Test for CHO group aldehydes:
There are two different tests to differentiate between aldehydes and ketones:
Test 1: Add ammonical silver nitrate solution and warm.
Observation: silver metal is precipitated in the form of a mirror on the side of a
test tube with aldehydes , with ketones no reaction.
Test 2: Add Fehling’s or Benedict’s solution and warm.
Observation: A red precipitate of copper (I) oxide is formed with aldehydes, no
reaction with ketones.
Iodoform reaction:
Test: an organic compound is added to a mixture of iodine and dilute NaOH
(or a mixture of NaI and NaOCl).
Observation: a pale-yellow precipitate of iodoform, CHI3
This test works with compounds containing the CH3C=O and
alcohols containing the CH3CH(OH) group.
Hassan Subhie GCE A-level Chemistry Unit 5 22
Test for the carboxylic acids, COOH group:
Test: Add Na2CO3 or NaHCO3 solution.
Observation: Colorless gas is evolved which turns limewater milky.
In synthesizing organic compounds, chemists normally chose the shortest possible
route.
Organic reactions never give 100% yield.
The more the number of stages in the synthesis of a compound, the lower is the %
yield.
E.g. To prepare butanol from propanol 2 routes are possible.
Route 1:
propanol
OHKHSOBrHCSOHKBrOHHC 24734273
OHHCNHHCCNHCBrHCHClNaNOLiAlHKCN
94273737324
toxic reagent Expensive reagent Very low yield
Route 2:
propanol
OHKHSOBrHCSOHKBrOHHC 24734273
C3H
7Br step
1Mg/dry ether
C3H
7MgBr
step 2
methanol followed by hydrolysisC
4H
9OH
Step 1 gives excellent yield
Step 2 gives good yield.
Route 2 is better as it gives better yield.
Ex. 1. How can you convert propan-2-ol into 2-methyl propan-2-ol ?
2. How would you make hexan-3-ol from propan-1-ol using no other
reagents except solvents?
Organic Synthesis
Hassan Subhie GCE A-level Chemistry Unit 5 23
Answer:
1.
propan-2-ol
[O]propanone
CH3MgBr
CH3C CH
3
CH3
OMgBr
H+
CH3C CH
3
CH3
OH
2.
CH3CH
2CH
2OH
Propan-2-ol
K2Cr
2O
7 H+
/
CH3CH
2CHO
KBr H2SO
4/CH
3CH
2CH
2Br
Mg/dryether
CH3CH
2CH
2MgBr
H+
(aq)
CH3CH
2CH(OH)CH
2CH
2CH
3
Hexan-3-ol
Main groups of synthetic reactions:
1. No change in carbon skeleton.
2. If number of carbon atoms in the chain increases by one carbon
The step could be:
-Halogeno alkanes +KCN
-Carbonyl+HCN
-Gringard reagent CH3MgBr with a carbonyl compound
-Gringard reagent with CO2 or with methanal.
3. If number of carbon atoms in the chain increases by more than one carbon
-Gringard reagent with more than one carbon with carbonyl compounds.
-Friedel- craft reaction for aromatic substances.
4. If the number of carbon atoms is decreases by one.
-Hoffman’s degradation of amides with NaOH and Br2
222 CONaBrRNHNaOBrRCONH
Hassan Subhie GCE A-level Chemistry Unit 5 24
-The iodoform reaction of methyl ketone and methyl secondary alcohols
RCOCH
3
X2 / NaOH
RCOOH
-The oxidation of the aromatic side chain in alkaline KMnO4
C
6H
5C
2H
5
KMnO4 / OH
-
C6H
5COOH
No Change in the carbon Skeleton:
Reactions of this type include
1. Simple substitution reactions involving non-carbon containing groups.
e.g. OH , Br , I , NH2 ………..
2. Oxidation of alcohols, aldehydes, reduction of aldehydes, ketones, nitriles and
carboxylic acids.
3. Elimination yielding C=C double bonds, amides and nitrilels, esters formation
and hydrolysis.
Example:
RCH2CH
2OH
PCl5
RCH2CH
2X
NH3 Ethanol/
RCH2CH
2NH
2
K2Cr
2O
7/ H
+
RCH2CHO
K2Cr
2O
7/ H
+
RCH2COOH
NaNO2 / H Cl
Hassan Subhie GCE A-level Chemistry Unit 5 25
RCH2CH
2OH
K2Cr
2O
7 / H+
conc. H2SO
4 RCH=CH2
H2SO
4/H
2O
RCH(OH)CH3
RCOCH3
LiAlH4
/dry ether
RCHXCH3
HX
KOH/ethanol
RCH(NH2)CH
3
NH3/ethanol
KOH(aq)
NaNO2/HCl
conc. H2SO
4
Exercise: How would you convert?
1- bromo propane into 2- bromo propane
Number of carbon atoms in the skeleton is increased by one or more than one:
RCH2X RCH
2MgX
RCH2COH|
|R
R,,
,
RCOR
,,,
HCHO
RCH2CH
2OH
CO2
RCH2COOHRCH
2CN
KCN
H+
(aq)
LiAlH4 /ether
RCH2CH
2NH
2
HNO2
Hassan Subhie GCE A-level Chemistry Unit 5 26
Problems: 1. How would you convert proapn-2-ol into?
a. 2,3 dimethyl butan-2-ol
b. Propene.
2. How would you convert phenyl ethanone into
a. Benzoic acid
b. Phenyl amine
c. 2,4,6 tribromo phenol
Polymerization:
There are two types of polymers:
1. Addition polymers: The monomers contain one or more C=C groups. In the
formation of these polymers the pi bond in the
monomer is broken.
E.g. poly (ethene), Poly (propene), poly (vinychloride) and poly (tetra fluorine)
o The monomer and the polymer have the same empirical formula.
2. Condensation Polymerization: Both the monomers have two functional
groups, one at each end. Polymerization
involves the loss of a simple molecule (usually
H2O or HCl) as each link forms.
o Examples are nylon (a polyamide) and terylene (a polyester)
Nylon: It consists of two types of monomers
(i) di carboxylic acid HOOC-----COOH and diamine H2N -----NH2
(ii) di acid di chloride ClOC-----COCl and diamine H2N -----NH2
C
O
___OH -----------C\ \
O
___ OH//
C__ C------- ____N H
|H
N
|__
H
H
C
O
___-----------C\ \
O
___//
C__ C------- ____N
|H
N
|H
+
C
O
___-----------C\ \
O
___Cl
//C__ C------- ____N H
|H
N
|__
H
H
C
O
___-----------C\ \
O
___//
C__ C------- ____N
|H
N
|H
Cl +
Hassan Subhie GCE A-level Chemistry Unit 5 27
(CH2)4
______CC
ClCl
OO
(CH2)6________ NN
H
HH
H
di acid chloride di amine
H Cl
(CH)4
___ ___C||
C||
______N|
(CH2)6
___N|
n
OO H H
__________
+
Tyrelene: It consists of two types of monomers
(i) di carboxylic acid HOOC-----COOH and diol HO -----OH
(ii) di carboxylic acid dichloride ClOC-----COCl and diol HO -----OH
C
O
___ O
H
-----------C\ \
O
___Cl
//C__ C------- ____ H|H
|__
H
H
C
O
___O
H
-----------C\ \
O
___//
C__ C------- ____
H
O
H|
H
OCl|
|| ||H
+
C
O
___ O
H
-----------C\ \
O
___//
C__ C------- ____ H|H
|__
H
H
C
O
___O
H
-----------C\ \
O
___OH
//C__ C------- ____
H
O
H|
H
O|
|| ||H
HO +
____COOHHOOC HOCH
2CH
2OHN N
______
||C||
C___
O___
CH2CH
2
___O
n
O O
________
+
Hassan Subhie GCE A-level Chemistry Unit 5 28
Synthetic polymers are not easily biodegraded and can cause an environmental
problem of disposal. One answer is for them to be recycled.
Polymers usually soften over a range of temperature rather than have a sharp
melting temperature .The reason for this is that the polymer is a mixture of
molecules of different chain lengths.
Practical techniques:
1. Recrystallisation: It is a technique used to purify solids.
Steps: - Dissolve the solid in a minimum of hot solvent.
- Filter the hot solution through a preheated funnel using a fluted
filter paper.
- Allow to cool.
- Filter under reduced pressure using Buchner funnel.
- Wash with a little cold solvent and allow to dry.
- The solid should have a sharp melting point.
2. Mixing:
Reactions of organic chemistry often use immiscible liquid reactants and its thus
necessary to shake or stir the mixture.
Shaking is done in a stoppered flask for reactions that don’t build up pressure or
in flask fitted with condenser for reflux.
When there is difference in densities of immiscible liquids as in nitration of
benzene or reduction of nitrobenzene, vigorous shaking is the only method of
bringing the reactants together.
3. Temperature control:
Most organic reactions are relatively slow and therefore heating is needed for the
reaction to occur. This is done on a steam bath or by using an electric heating
mantle.
Since many organic reactants are volatile and heating causes them to escape , thus
heating is done under reflux.
4. Heating under reflux:
This process boiling under reflux is used in organic reactions where the reactions
are often slow and reagents are volatile.
Hassan Subhie GCE A-level Chemistry Unit 5 29
The process consists of a condenser mounted vertically on top of the reaction
vessel so that any vapors escaping during the heating process will condense to
liquid and run back into the flask.
By this way the heating process can be carried out for a longer period.
1
2
3
45 6
7
8
9
1 10
2
3
45 6
7
8
9
11
5. Safety precaution:
No naked flame when chemicals are flammable i.e. use water bath or electric
heating mantle.
Do the reactions in a fume cupboard when a reactant or a product is toxic ,irritant
or carcinogenic.
6. Simple distillation:
Used to separate chemicals that have big difference in their boiling points.
E.g. bromoethane has boiling point 38 oC and ethanol boiling point is 78
oC.
The liquid with the lower boiling point distils first.
Hassan Subhie GCE A-level Chemistry Unit 5 30
1
2
3
45 6
7
8
9
1 10
2
3
45 6
7
8
9
11
7. Fractional distillation:
This is used to separate a mixture of 2 liquids.
This mixture can usually be separated into pure samples.
But if the boiling points are too close, a good separation will be difficult.
Hassan Subhie GCE A-level Chemistry Unit 5 31
Temperature /composition diagram:
If a liquid of composition X is heated, it will boil at T1
oC to give a vapor of
composition Y. If this is condensed and reboiled, it will boil at a temperature T2 oC
giving a vapor of composition Z. Eventually pure B will distill off the top and pure A
will be left in the flask.
8. Measuring melting point:
The solid is crushed on a clean porous tile in order to put it in a capillary tube.
Capillary tube is attached to a thermometer and placed in water or oil.
The liquid is heated gently and stirred continuously.
The temperature when the solid starts to melt is recorded.
Heating is continued and temperature at which the solid is completely melted is
recorded.
The difference between the temperatures should be ± 3 oC or less.
Hassan Subhie GCE A-level Chemistry Unit 5 32
Errors in the method:
- Solid may be impure/damp.
- Heating is too rapid.
- Water is a poor conductor and therefore temperature of solid is different to
that on the thermometer.
- Insufficient stirring.
- Difficult to see solid and thermometer together.
- Difficult to decide when melting starts.
- Melting point not sharp.
Improvement for the method:
- Purify the solid.
- Use a heating block (Thiele tube)
- Stir more or heat more slowly.
- Use a magnifying glass.
- Repeat and take the average temperature.
9. Measuring boiling point:
- The liquid is placed in an ignition tube and the tube is fixed to the thermometer.
- Place a capillary tube with a closed end inside the ignition tube with the opened end
inside the liquid.
- Clamp the thermometer in a beaker of water or oil.
- Heat the water gently with stirring until rapid stream of continuous are produced.
Record the temperature.
- Cool the water bath and record the temperature at which bubbles stop escaping.
- The difference between the 2 temperatures should be less than or equal to 5 oC.
Hassan Subhie GCE A-level Chemistry Unit 5 33
Errors in the method:
- Can’t see the temperature together.
- Heating is too rapid.
- No enough stirring.
- Water is a poor conductor and therefore temperature of the liquid is different to that
on the thermometer
Improvement for the method:
- Use fractional distillation to observe the steady boiling point when the boiling occurs.
Applied Organic chemistry:
Targeting pharmaceutical compounds:
Pharmaceutical compounds which are ionic or have several groups that can form
hydrogen bonds with water, will tend to be retained in aqueous (non fatty) tissues.
These water-soluble compounds will be excreted from the body.
Compounds with long hydrocarbon chains and with few hydrogen-bonding
groups will be retained in fatty tissues. These compounds will be stored in the
body.
Chemists chemically modify the structure of pharmaceuticals, in order to change
the solubility and retention time of the material in the body.
Nitrogenous fertilizers:
There are three types of fertilizers:
1. Quick release, such as those containing NO3
- and NH4
+
2. Slow release, such as urea, NH2CONH2
3. Natural, such as slurry, compost and manure.
The first two are soluble and can be leached out. The last is bulky and contains
little nitrogen.
Esters, oils and fats:
Simple esters are used in food flavorings, perfumes and as solvents for glues,
varnishes and spray paints.
Fats are esters of propan-1, 2,3-triol and saturated acids such as stearic acid
C17H35COOH.
Hassan Subhie GCE A-level Chemistry Unit 5 34
When fats are hydrolyzed by boiling them with aqueous sodium hydroxide, they
produce soap which is the sodium salt of the organic acid.
Vegetable oils are liquids, which are esters of propan-1, 2,3-triol and
unsaturated acids.
Margarine is made from polyunsaturated vegetable oils by partial
hydrogenation (addition of hydrogen) so that only a few double bonds remain
Questions:
1. How would you distinguish between:
a. Pentan-3-one and pent anal
b. 2-bromo propane and 2-chloro propane
c. Methyl propan-2-ol, methyl propan-1-ol and butan-2- ol?
2. How would you prepare from propene:
i. 2-bromopropne
ii. propanone
iii. 2,3-dimethyl butn-2-ol?
3. How would you make
i. Phenol
ii. Benzoic acid
iii. Phenyl benzoate,
Using benzene and methylbenzene as only organic materials?
4. How would you convert oleic acid CH3 (CH2) 7CH=CH (CH2) 7CO2H into
a. Stearic acid, CH3 (CH2) 16CO2H
b. Plamitic acid, CH3 (CH2) 14CO2H
c. Octadeca-8, 10-dienoic acid CH3 (CH2) 6 CH=CH (CH2) 6 CO2H?
5. Starting from butan-1-ol and methanol as only organic materials, how would you
make
i. Butanone
ii. Propanoic acid
iii. Methyl butanoate
iv. Propylamine
v. Butene
vi. 2-methylbutne?
Hassan Subhie GCE A-level Chemistry Unit 5 35
Interpret simple fragmentation patterns from a mass spectrometer
An organic compound produces ions in a mass spectrometer. The ions generate
pulses of electric current which are sent as signals to a computer (or chart plotter)
to be displayed as a mass spectrum.
On the spectrum the large peak on the right is the parent molecular ion and this
indicates the relative molecular mass of the compound.
In the spectrometer the molecules are fragmented into positive ions which form a
pattern which depends on the structure of the molecule. e.g. ether CH3OCH3
Example 1: Another compound of relative molecular mass 46 also contains carbon,
hydrogen and oxygen has the mass spectrum below. Identify each fragment
and the structural formula.
Hassan Subhie GCE A-level Chemistry Unit 5 36
Problem 2: Sketch the mass spectrum of propanal showing the masses of each fragment
Interpret simple infrared spectra
The bonds in organic molecule absorb infra-red radiation. This happens when the
frequency of the radiation matches the natural frequency of vibrations in the bonds.
These might be stretching, or bending vibrations.
A spectrometer shines infra-red light at a sample of an organic material and measures
how much of the light is absorbed. A measure of the frequency (wave number) is
displayed in the spectrum. Each bond has its own frequency (wave number) and this
can be used to identify the bonds present in a compound.
bond wavenumber/cm-1 seen on spectrum
C-H 2840 - 3095
C-C 1610 - 1680
C=O 1680 - 1750
C-O 1000 - 1300
C-Cl 700 - 800
O-H 3233 - 3550
2500 - 3300
N-H 3100 - 3500
Infra-red spectrum for propanone
o Is substance A in the infra-red spectrum below most likely to be ethyl
ethanoate or butane?
Hassan Subhie GCE A-level Chemistry Unit 5 37
5.5a(iii)(c) Low resolution nuclear magnetic resonance spectra (NMR) Hydrogen atoms can be detected using this sort of spectrometry. The nucleus
of a hydrogen atom, the proton, spins and so has a magnetic moment. This can
be aligned or not aligned with a magnetic field . When electromagnetic
radiation of the right frequency is applied resonance occurs and the protons flip
from one state to the other and absorb energy. This absorption of energy is
used to detect protons in organic compounds. The exact resonance frequency
for a proton (hydrogen atom) depends on its environment. For example the
frequency is different for hydrogen atoms in CH3, CH2 , C6H5- and in O-H.
Trimethylsilane, TMS, is used as a standard. The distance in the spectrum
from the TMS peak is called the chemical shift.
Type of proton Chemical Shift (ppm)
R-CH3 0.9
R-CH2 1.3
R-CH2-O- 4.0
C6H5- 7.5
-O-H 5.0
-CHO 9.5
Hassan Subhie GCE A-level Chemistry Unit 5 38
To which functional groups do the protons in the following NMR spectrum
belong? Identify the compound.
Sketch an NMR spectrum for propanal.
An excellent guide to low resolution nmr.
5.5a (iii) (d) The interpretation of simple ultra-violet/visible spectra.
Some chemical structures absorb electromagnetic radiation in the ultra violet
part of the spectrum. These include conjugated (contain alternate double and
single bonds) dienes. E.g. 1,3-butadiene. The ultraviolet absorption spectrum
Hassan Subhie GCE A-level Chemistry Unit 5 39
for 2,5-dimethyl-2,4-hexadiene is shown below.
Ultra-violet wavelengths are from about 200nm to about 400nm. Visible light
has wavelength between 400nm and 800nm. -carotene, which gives carrots
their orange colour absorbs at 497nm. Lycopene, which gives tomatoes their
red colour, absorbs at 505nm. Both of these compounds have 11 conjugated
double bonds.