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CHAPTER 1
INTRODUCTION TO ORGANIC CHEMISRY
Organic chemistry can be defined as the study of organic compounds.
Organic compounds are compounds of carbon (with the exception of
oxides of carbon, carbonates, bicarbonates and metallic carbides). Before
1828, organic compounds are believed to be from plants and animals until
when urea was artificially
made from ammoniumthiocyanate by a German
Ag
NH4CN
NH2 – C – NH2
scientist called Freidrich Wohler.
Ammoniumthiocyanate
Urea
Organic compound play major roles in the chemistry of living things and
some areas like food, dyes, perfumes, soaps, Nano-technology e.t.c.
Organic compound contain the following elements:
i.
Carbon (main element)
ii.
Hydrogen and oxygen
iii.
Halogen and nitrogen
iv.
Phosphorus, sulphur and metals
1.1
EXCEPTIONAL ABILITIES OF CARBON
The presence of many organic compounds in the universe is due to the
following properties of carbon.
Ø
CATENATION
This is the ability of carbon to combine (link – up) with itself to form
straight, branched or ring compound.
C−C
C
−C−C−C−
C−C
C
C
C
Straight chain
Ø
branched chain
ring
TERAVALENCY
The tetravalence nature of carbon makes it easily for carbon form single,
double or triple bond with one another.
−C−C
Single bond
Ø
C=C
C≡C
double bond
triple bond
REACTIVITY
Carbon has the ability to read easily with both metals and non – metals to
form different compounds.
1.2
GENERAL FEATURES OF ORGANIC COMPOUNDS
i.
COVALENCY
Organic compounds are covalent in nature, hence they do not ionize in
solution and are non – conductor of electricity. Also they possess low
melting and boiling points.
ii.
FLAMMABILITY
Most organic compounds are highly flammable and they burn
enothermicallty in a plentiful supply of air to produce Co2 and H2o and in
limited supply of air produce Co and H2o. The energy obtained from the
organic compound can be used as source of fuel.
iii.
SOLUBILITY
Most organic compounds are insoluble in water with the exception of
those that contain hydroxyl group (-OH) like phenol, methanol e.t.c.
iv.
REACTIVITY
Reactions involving organic compounds are generally slow because of the
absence of mobile ions they usually require heat, catalyst or though mixing
to speed up the reaction rate.
v.
THERMAL INSTABILITY
Most organic compounds are unstable as they decompose to simpler
organic molecules at a temperature above 500oC. This process is known as
thermal cracking and it is important in petroleum industry.
vi.
Presence of weak intermolecular forces of attraction
Organic compounds are mainly gases, volatile liquids or volatile solids as
the forces of attraction between them include hydrogen bond,
vandarwaal’s forces (London dispersion forces and dipole – dipole
attraction forces).
1.3
PURIFICATION AND ISOLATION OF ORGANIC COMPOUNDS
Most organic compounds can be obtained in a swampy environment,
plants, animals, air beneath the earth crust. The purification of the organic
compounds depends on the physical nature of the compound.
Gaseous organic compounds can be purified by removing agents (such as
CuSO4for removing ethyne); other crystallization, distillation, fractional
crystallization,
fractional
distillation,
steam
extraction
and
chromatography.
1.4
TERMS IN ORGANIC CHEMISTRY
Ø
Homologous series
This is the family of organic compounds that follows a regular structural
pattern in which each successive member differs by molecular mass of 14
(i.e – ch2): organic compounds that are related to one another are grouped
into family known as homologous series and each member is called
homologue.
Common examples include alkane, alkene, alkanol, alkanal, alkanone,
alkanoate etc.
CHARACTERISTIES OF HOMOLOGOUS SERIES
1.
All members of the series conform to a general molecular formular.
2.
All members of the series have the same methods of preparation
though condition may vary.
3.
Each successive member differ with a molecular formular – ch2 and
molecular mass 14.
4.
Chemical properties of members are similar chemical reaction.
5.
The physical properly depend on the carbon atoms per molecule and
if varies down the series. For example, in the family of alkane, the
boiling point, melting point and density increases down the series but
their solubility in water decreases down the series.
Ø
Alkyl group
This includes all groups derived from alkane by the loss of one hydrogen
atom. They are named by replacing – ane from alkanes by – yl.
PARENT ALKANE
NAME
ALKYL GROUP
NAME
CH4
Methane
CH3
Methyl
C 2H 6
Ethane
C 2H 5
Ethyl
C 3H 8
Propane
C 3H 7
Propyl
C6H14
Hexane
C6H13
Hexyl
C12H26
Dodecan
e
C12H25
Dodecyl
CHARACTERISTICS OF ALKYL GROUP
i.
They have general molecular formular CnH2nH where n≥l.
ii.
They are usually represented by R.
iii.
They determine the physical properties of organic compounds.
iv.
They do not exist alone (they are usually attached to a parent organic
compound).
Ø
Functional group
This can be defined as on atom or radical or bond common to homologous
series which determine the chemical properties of the series.
The table shows common homologous series with functional group.
CLASS OF COMPUND
GENERAL STRUCTURE
FUNCTIONAL GROUP
Alkane
R–H
−
Alkyl
(haloalkanes)
halide R – X
−X(X = F, B, d or I)
Alkenes
R – CH = CH – R1
=
Alkenes
R – C≡C – R1
≡
Aromatic compounds
Benzene ring
Alcohols (Alkanol)
R − OH
−OH
Thiols
R – SH
−SH (sulfhydryl group)
Alkanal (Aldehyde)
−COH
R−C−H
Alkanone (katone)
−CO−
1
R−C−R
Alkanoic
(carboxylic acids)
Ethers
Estars (Alkanoates)
acid
−CO2H
group)
R−C−OH
(carbonyl
−O− (alkoxy group)
1
R−O−R
R−C−O−R
− CO2R (carboalkoxy
group)
Primary Amines
R−NH2
−NH2 (amino group)
Secondary Amines
R−NH− R1
NH
Tertiary Amines
R3−N
N−
Amides
R−C−NH2
−CONH2(carboxamide
s group)
Nitrile
R−C≡N
−CN (cyano group)
Nitroalkanes
R−NO2
−NO2(nitrogroup)
1.5
1
SATURATED AND UNSATURATED COMPOUNDS
A compound is saturated if single cadent bond exist between the carbon
atoms. Alkane and cycloalkane are saturated compounds.
A compound is unsaturated if double or triple bond exist between carbon
to carbon atoms (as in alkene and alkyne) and also if double or triple bond
exist between carbon atom to another element(s).
−C−C−
−C−C−X
−C−C−OH
−C≡C−
C=C
H−C−C≡N
H−C−C−H
Saturated
Unsaturated
The degree of unsaturated increases with the multiplicity of the bond.
i.e
i.
Aromatic > Alkyne > Alkene
Decreasing unsaturation
ii.
Alkene < diene < triene
Increasing unsaturation
iii.
Alkyne < diyne < triyne
Increasing unsaturation
In summary
Organic compounds
Saturated
[generally alkenes
and cycloalkanes]
Aromatics
unsaturated
Alkene
Alkyne
CHAPTER 2
DETERMINATION OF ELEMENTS & FORMULA
Elements in an organic compound can be known by the analysis of the
sample qualitatively and the amount of the elements presents can be
determined qualitatively.
2.1
QUALITATIVE ANALYSIS
1.
Detection of carbon and hydrogen
This is done by
-
Heating sample organic compound with dry CuO
(Compound) + CuO (CU(s) + CO2(g) + H2O 10
-
Evolved gas is passed through line water and it turns it milky.
Ca(OH)2 + CO2(g)(CaCO3(s) + H2O(1)
-
H2O is passed through anhydrous Cus04 and turns it blue.
2.
Detection of N, S and X
For this to be done, the following steps must be taken
-
Take an organic sample and mix with sodium pellet in a test tube.
-
Heat the content to red hot and allow to cool.
-
Add methanol and water, and then filter.
-
If N is present, it’s converted to NaCN, if sulphur is present, if is
converted to Na2s and if halogen is present. It’s converted to NaX.
(compound) + Na ( Na+x-+ Na+CN-+Na+s2.
-
The salts are then dissolved in water to form salt solution and
filtered.
-
Divide the filtrate into four.
-
To filtrate I, add BaSO4, boil and add fecl3 and dil.HCl. if nitrogen is
present, the solution turns perussian blue.
-
To filtrate 2, add Na3 (Fe NO (CN) 5) which is blue, the solution turns
purple if sulphur is present.
-
To filtrate 3, add HNO3 then boil (HCN and H2S are formed and
escape with air), after that add AgNO3 (aq) a precipitate is formed. If
precipitate is
a.
White, chlorine is present
b.
Green, bromine is present
c.
Yellow, iodine is present
To filtrate 4, add benzene and chlorine water, then shake. Two layers
are observed (top layer being the organic layer). If the organic layer is
a.
Colourless, chlorine is present
b.
Yellow, bromine is present
c.
Purple, iodine is present
2.2
QUANTITATIVE ANALYSIS
1.
Combustion analysis
A known amount of organic compound is heated in a stream of dry oxygen.
If there is carbon it is converted to CO2 and if there is hydrogen, it is
converted to water.
Co2 is then absorbed into a weighed testtube containing soda lime.
NaOH(aq)+Co2(g) ( NaHCO3
Calculation
Initial weight = weight of test tube + NaOH
Final weight = weight of test tube + NaOH + absorbed CO2.
(Weight of Co2 = final weight – Initial weight.
Weight of C = atomic weight of C x weight of Co2
obtained
Molecular weight of Co
H2o is absorbed into a weighed test tube containing anhydrous magnesium
chlorate (MgClO4).
Calculation
Initial weight = weight of test tube + MgClO4
Final weight = weight of test tube + MgClO4 + absorbed H2O
(Weight of H2O = final weight – initial weight
Weight of H2 = 2 x atomic weight of hydrogen x weight of H2o obtained
Molecular weight of H2o
2.
Dumas method
A known amount of organic sample is taken and mixed with CUO which is
then heated in a steam of co2; nitrogen and its oxides are produced, the
oxides are then converted to nitrogen by passing them through a hot
copper the total volume of nitrogen present is known by using nitro meter
Sample + CuO → N2(g)↑ + (NO2NO2 etc.) cu→Na↑
Weight of N = molecular weight of nitrogen x volume (cm3)
22400
Weight of nitrogen = ρ v
ρ → density of nitrogen
v → volume of nitrogen
3.
Carius method
A known mass of organic sample is treated with fuming HNO3 in a sealed
test tube, all the present sulphur are oxidized in form of H2so4 after which
BaCl2(ae) is added in the test tube
Sample + HNO3(aq) → H2SO4
H2SO4(aq) + BaCl2(aq) → BaSO4ppt(s) + HCl(aq)
Filter the precipitate, wash with distilled water and allow to dry; then weigh.
Weight of S = atomic weight of S x weight of Baso4
Molecular weight of Baso4
A known organic sample is treated with turning HNO3 and HX is formed,
add AgNO3 solution to the hydrogenhalide
Sample + HNO3(aq) → HX
HX + AgNO3(aq)→ AgX + HNO3(aq)
Filter the precipitate and wash with disfilled water, allow to dry and weigh.
Weight of X = atomic weight of X
x weight of AgX
Molecular weight of AgX
2.3
Formular determination
Empirical formula gives the simple ratio of each element in a molecule of a
compound. If can be calculated if:
i.
percentage composition of each element is known
ii.
atomic ratio of each element is known
NOTE:
Atomic ratio =% composition of element
Atomic weight of element
Molecular formula gives the exact number of element in a compound.
Before this can be done, the molecular mass and empirical formula of the
compound must be known.
Structural formula indicates how atoms are arranged within a molecule of
an organic compound.
Examples
1.
A compound known to contain carbon, hydrogen and oxygen was
found to contain 40% carbon 6.67% hydrogen and 53.33% oxygen. If the
compound has a vapour density of 45. Calculate the molecular formula of
the compound.
Solution
C
H
O
40
6.67
53.33
12
1
16
3.33
6.67
3.33
3.33
3.33
3.33
1
2
1
Empirical formula = CH2O
Molecular mass of the compound = 2 x v.d
= 2 x 45
= 90
Mass of CH2 O = 30
n = Molecular formula
Empirical formula
= 90
=3
30
Molecular formula = (empirical formula)n
= (CH2O)3
Molecular formula = C3H6O3
2.
On the combustion analysis of 20g of an organic sample, it produces
28g of H2O and 56g of Co2. Calculate the percentage composition of
carbon and hydrogen in the sample.
Solution
Weight of H2 = 2 x atomic weight of hydrogen x weight of H2O obtained
Molecular weight H2O
= 2 x 1 x 28
18
= 56 3.1g
18
% composition of hydrogen = mass of hydrogen x 100%
Mass of sample
= 3.1 x 100%
20
= 15.5%
Weight of C = atomic weight of carbon x weight of Co2 obtained
Molecular weight of Co2
= 12 x 56
44
= 15.3g
% composition of carbon = weight of C x 100%
Weight of sample
= 15.3 x 100%
20
= 76.5%
3.
Combustion analysis of 0.176g of an organic sample is known to
contain carbon, hydrogen and possibly oxygen produces 0.196g of H20 and
0.242g 0f Co2. Calculate the empirical formula of the compound.
Solution
Weight of carbon = atomic weight of carbon x weight of Co2 obtained
Molecular weight of Co2
= 12 x 0.242
44
= 0.066g
% composition of C = weight of C in sample x 100%
Weight of sample
= 0.066 x 100%
0.176
= 37.5%
Weight of hydrogen = 2 x atomic weight of hydrogen x weight of H2O
obtained
Molecular weight of water
= 2 x 1 x 0.196
18
= 0.022g
% composition of hydrogen = weight of hydrogen in sample x 100%
Weight of sample
= 0.022 x 100%
0.176
= 12.5%
% composition of oxygen = (100 – (12.5 + 37.5))%
= 50%
C
H
O
37.5
12.5
50
12
1
16
3.125
12.5
3.125
3.125
3.125
3.125
1
4
1
Empirical formula = CH4O
A compound containing C, H and N as a molar mass of 45g.When 0.1g of
the compound was subjected to Dumas method of analysis, 24.8cm3 of
N2(g) was produced, if the compound contains 53.33% of carbon. Calculate
the molecular formula of the compound (density of nitrogen =
0.00125gem-3).
Solution
Weight of nitrogen = density x volume
= 0.00125 x 24.8
= 0.031g
% composition of nitrogen = 0.031 x 100%
0.1
= 31%
% composition of carbon = 53.33%
% composition of hydrogen = 100% - (53.33% + 31%)
= 15.67%
C
N
H
53.33
31
15.67
12
14
1
4.44
2.21
15.67
2.21
2.21
2.21
2
1
7
Empirical formula = C2NH7
Molar mass of C2NH2 = (2 x 12) + (14) + (1 x 7)
= 45g
n = Molecular formula mass = 45
Empirical formula mass
45
n=1
since n = 1
Molecular formula = Empirical formula = C2H7N.
CHAPTER 3
HYB RIDIZATION AND BONDING
3.0
Bond formation
The formation of bonds is as a result of the tendency of unpaired
electron(s) in the orbital’sto pair up with another single unpaired
electron(s). The mode and manner of these overlaps lead to the formation
of either sigma bond or pi – bond.
1.
Sigma bond (δ – bond ).
This is the overlap of atomic orbitals to form a single bond. There are three
possibilities in which s orbitals and p orbitals can overlap to form this
i.
S and S overlap
ii.
P and P overlap
iii.
S and P overlap
2.
Pi bond (π – bond)
This is formed as a result of lateral overlap of two parallel unhybridized P
orbitals of adjacent atoms.
Bonding electrons in δ – bond lie along same line with the nuclei of the
atoms involved, as such the degree of overlap is higher than that of π –
bond, hence δ – bond is stranger than π – bond.
π – bond is always the centre of reaction and reaction with easy breakage
of this bond is addition reaction.
3.1
Hybridization
This is the mixing of atomic orbitals (outer shells) of same energy level to
form an equivalent orbital in shape, energy and size. This can be of
different types, only three types will be the area of concentration in organic
chemistry as the bonding in organic compounds involves only S and P
orbitals.
1.
SP3 hybridization
This occurs when carbon atom is bonding with four atoms to make four
single bond. The mixing of S and P orbitals is of the ration 1 :3, as such SP3
hybrid orbitals are formed.
+
2s
sp3
2p
sp3
sp3
sp3
All compounds of alkane except methane possess overlap of sp3 orbitals.
sp3
sp3
in methane and other higher alkanes, there is overlap between carbon and
other elements; there by producing a δ – bond.
Alkyl group of a compound is sp3hybridized the bond angle between
carbon and other elements, molecule in sp3 – hybridization as a bond
angle of 109.5o.
Co-H bond length is 1.10Ao while C – C bond length is 1.54Ao. The
dissociation energy is 88 kcalmol-1.
2.
sp3hybridization
One S – orbital mixes with two P –orbitals leaving one P – orbital
unhybridized. sp3orbitals are arranged in a trigonal arrangement and has a
bond angle of 120o.
2p
Sp2
2s
The three hybrid orbitals formed are used to form a single bond while the
unhybridized P – orbitals overlap to form the other bond which is II in
nature. The three sp2 – orbitals lie on a plane in which the unhybridized P
orbital lies perpendicular to such plane
C
C
δ - bond
C
C
C=C
The bond energy of C – C bond is 146keallmol and the bond angle for C =
C is 1200, its bond length is 1.34Ao.
3.
Sp hybridization
In this one S – orbital mixes with one P – orbital leaving two P – orbitals
unhybridized cylindrically along the δ – bond produced by C – C bond.
Sp hybridized orbital is linear in shape and the bond of C ≡ C is 1.20Ao.
2p
2s
sp
sp
π - bond
C
δ - bond
C≡C
C
π - bond
CHAPTER 4
HYDROCARBONS
These are organic compounds that contain carbon and hydrogen only. The
main source of hydrocarbon is petroleum (a dark, sticky viscous liquid).
They are numerous hydrocarbons and can be group into aliphatic and
aromatic.
Hydrocarbons
Aliphatic
Aromatic
Cyclo alkane e.g cyclopropane
Acyclic
cyclic
Cyclo alkene e.g cyclopentene
Alkane
4.1
Alkene
Alkyne
Petroleum (Crude Oil)
This is a dark sticky viscous liquid found in underground deposit.
Petroleum is a mixture of gases, liquids, solid alkanes, alkanes,
cycloalkanes as well as aromatic hydrocarbon.
Petroleum is originated from the chemical decomposition of organic
compounds over billions of years. Since petroleum can also be produced
when enzyme (organic catalyst) is acted upon by chemical substance. The
constituents of crude oil can be separated by fractional distillation and the
process of removing its impurities is known as straight run process or
refining.
4.2
Petrol
This is also known as motor spirit. It is used as fuel in motor vehicles and
other machineries. Petrol usually vapourize in the engine during
combustion. The rate of petrol dependson the arrangement of the atom in
the petrol. Branched chain carbon atom such as 2,2,4 – trimethylpentane
has slow rate of combustion which indicates high performance.
4.3
knocking
This causes wearing and tearing of engine part (i.e it reduces the efficiency
of engine). It’s caused when there is large straight chain hydrogen in
petrol.
Knocking can be reduced by adding petro; additive (anti-knock agent) such
as lead tetraethyl (pb (C2H4) 4) or 1,2 – dibromoethene. Pb (C2H4)4causes air
pollution due to the incomplete combustion of the lead in petrol.
4.4
Octane Rating
This is a scale numbered 0 – 100 that is used to measure the grade of
quality of petrol with branched chain hydrocarbon is of higher performance
and its octane rating almost 100. High quality petrol is known as super
petrol, extra petrol or premium motor spirit (PMS) while low quality petrol is
known as regular petrol or ordinary petrol.
Octane can be defined mathematically as
Octane number = Amount of 2,2,4 – trimethyl pentane in petrol sample
Amount of petrol sample
Petrols of low octane number can be made to have octane number by:
1.
Cracking
This is the process of braking down larger organic compound with lower
volatility into smaller organic compound with low volatility. Cracking can
either by thermal or catalytic.
Catalytic cracking is better than cracking because
i.
the reaction can be controlled
ii.
quality product will be obtained
C15H32 catalyst C8H18 + 2C2H4 + C3H6
Petrol
2.
ethane propene
Catalytic reforming
This is the conversion of alkane or cycloalkane to aromatic compound so
as to improve the quality and the rate of combustion of fuel by
dehydration.
Ni
180oC, 5 atom
3.
+ Hcl
Catalytic Isomerization
This is the process of converting alkanes with straight chain into branch
chain hydrocarbons so as to improve the quality of petrol.
Alcl3
Octane
4.5
180oC
2,2,4 -
petroleum fractions
Name of
fraction
Boiling point range
(oC)
Carbon atoms
in molecules
uses
i. petroleum
gas
Below 40
1–4
Used as fuel and
manufacture of other
organic compounds.
ii. petrol
(gasoline)
40 – 200
4 – 12
Used as fuels in
aeroplane and
vehicles
iii. kerosene
200 – 250
12 – 18
Used as fuel for
lighting, heating and
jet engines
iv. diesel (gas
oil)
250 – 350
18 – 25
Used as fuel for
heating and diesel
engine at is also used
as rans – metal in
cracking process
v. lubricating
oil
350 – 500
More than 20
Used as in lubricating
moving parts of
machines. Making
candles, cream and
hair care product.
vi. bitumen
(petroleum
coke)
Above 50
More than 35
Used in surfacing
roads, binding agent
for roofing materials.
4.6
Petro chemicals
These are the process form of petroleum, the product are known as value
added product because after processing, their market value increases
thirteen times the original value of petroleum common examples of
petrochemicals are ethane, aromatic compounds, synthetic gases and
natural gases.
CHAPTER 5
ALKANES
These are saturated hydrocarbons with single bond between their carbon
atoms and with a general formula CnH2n+2. Where n ≥ 1. The simplest
member being methane (called marshgas and dampfire gas).
Alkanes have free rotation about C – C, as such they possess S bond, are
Sp3 hybridized and bond angle is 109.5o. Atoms can be converted into one
another by their free rotation (conformation) and when this occurs
torsional energy is required.
Alkanes are generally stable and they undergo molecular reaction under
special condition Alkanes are generally inert towards reagents hence they
are called paraffins.
Example: Determine the molecular formula and the name of the alkane if
CnH10 is a member of the alkane?
Solution
CnH10 = CnH2n+2
H10 = H2n + 2
10 = 2n + 2
2n == 10 – 2
2n = 8
2
2
n=4
C4H10 = CnH10
The alkane is butane.
5.1
Nomenclature
Alkanes generally ends with ‘ane’.
No of C
prefix
1 carbon
- meth
2 carbons - Eth
3c
- prop
4c
- but
5c
- pent
No of C
prefix
11 c
- undec
12 c
- dodec
13 c
- tridec
14 c
- tetradec
15 c
-pentadec
16 c
- hexadec
The following rules should be employed naming alkane.
i.
Identify the longest chain as it will be the parent chain
ii.
The un-numbered carbon atoms are named substituent (alky group)
iii.
Consider the substituent halogen group in the compound and it must
end with o fluorine as fluoro, chlorine as chloro, bromine as bromo,
iodine as iodo.
iv.
Substituent with multiple numbers must start with prefixes such as
di, tri, tetra, etc.
v.
Some of the position of the substituent must be minimal
vi.
The position of the substituent must proceed their name.
vii.
Substituent must be named in alphabetical order.
viii. Numbers are separated by comma while a number and a name are
separated by hyphen.
Examples
H
H
H
H
H
H
H
H
1.
H–C–C–C–C-H
butane
H H
H
H
H
H
H
H
H
H
H
2.
C3 H 4 – C – C – C – C2 H 5 → H – C8 – C7 – C6 – C5 – C4 – C3 – C2 – C1
H H H
H H H
H H H H H
–H
→ 6 – methyloctane ×
← 3 – methyloctane √
3.
CH3CH(Cl)
CH (Cl) CH3
H
Cl
H
H
H
H – C 14 – C 23 – C 32 – C 41 – H
H
Cl
H
→
←
1
2, 3 – dichlorobutane
C
H3
CH3 - 2C - CH3
4.
H
H(CH3) 3 C
H3C – C – CH3
H
H
H3C - C - CH3
H – C – 3C
H
H H
C–
C-H
H
H H 5C 2 – C –
H – C7 – C6 – C5
H
H
H
H
H
H
CH3 H
9
8
7
6
H 3C 1 – C 2 – C 3 – C 4 – C 55 – C 46 - 3C 7 – 2C 8 – 1C 9 – H
H H
H
H H
H
H
H
H
H–C–
H
H H
4
C – H
– CH3
CH3
H – 3C 1 – H
H 3C – 2C 2 – H
1
C 3H 3
2 – methyl – 5- (2 – methylpropyl) nonane
H
6.
CH2CH3
CHCH3
H
C2– 1CH3
H
C3
H
C4
C5
H
CH CH
H
H
3 - methylpentane
5.2
Physical Properties of Alkanes
i.
They are non – polar compounds.
ii.
They possess symmetrical directional bond.
iii.
Boiling points and melting points increases as the number of carbon
atom increases as a result of increment in the intermolecular forces.
5.3
Preparations
→ Decarboxylation of sodium alkanoates
Heating sodium alkanoates with sodalime
RCOONa + NaOH → R – H + Na2CO3
From sodalime
e.g
CH3COONa + NaOH → CH4 + Na2 CO3
C5H11COONa + NaOH → C5 H12 + Na2CO3
Sodalime (isa mixture of Cao and caustic sodium) is preferred to caustic
soda because it is not deliquescent and does not attack glass easily.
→
Hydrogenation of alkanes and alkynes.
CnH2n + H2 pd/ni/pt CnH2n+2
pd → palladium
CnH2n + H2 pd/ni/pt CnH2n+2
ni → Nickel
pt → platinum
→
Reduction of Alkyhalides
i.
by hydrolysis of Grignard reagent.
R – MgX + H2O → R – H + Mg (OH) x
e.g
C2H5mgBr + H2o → C2H6 + Mg (OH) Br
Grignard reagent is considered as the magnesium salt of extremely weak
acid and it is produced when a solution of alkylhalide (RX) in dry ether is
allowed to stand over a turning of metallic magnesium.
A vigorous reaction takes place and the solution becomes cloudy.
R – X + mg dry Et2o R – MgX
Grignard
reagent
RMgX + HOH → R – H + Mg (OH)X
RMgX + NH3 → R – H + Mg (NH2)X
Hence alkanes are weak acids.
ii.
by metals
(a) 2R – X + Zn H+ 2R – H + ZnX2
Example.
2C2H5 Br + Zn H+ 2C2 H6 + Zn Br2
(b) 2R – x + Na Et2O 2R - + 2NaX (wurtz reaction).
Example
2CH3Cl + 2Na Et2O C2H6 + 2Nacl.
Reactions
→ Combustion reaction
CnH2n+2 + () H2 O2→ n Co2 + (n + 1) H2o
Example
C2H6 + o2 → 2co2 + 3H2O
→ Substitution reaction.
i.
Nitration
CnH2n+2 +HNO3 → CnH2n+1 No2+ H2o
Example
C3H8 + HNO3 → C3H7NO2 + H2o
ii.
Sulphonation
CnH2n+2 + H2S207 → CnH2n+1SO3H + H2SO4
Example
CH4 + H2S2o7 → CH3SO3H + H2SO4
iii.
Chlorination
Chlorine is usually reacted with alkane (iodine does not react and bromine
react less readily). Chlorination is a photo catalytic reaction.
The reaction follows free radical mechanisms which are
a.
Initiation
Homolytic cleavage of a molecule of chlorine yield two chlorine molecules
Cl2
b.
u
2Cl.
Propagation
A Chlorine radical combines with an alkane molecule, abstract a hydrogen
atom.
C l. + CH4 → CH3 + HCl
The alkyl radical collides with cl2, abstract chlorine atom
CH3 + Cl2 → CH3 Cl + Cl2
CH3 Cl + Cl → CH2 Cl + HCl
CH2 Cl + Cl2 → CH2Cl2 + Cl
CH2Cl2 + Cl →CHCl2 + HCl
CH2 Cl + Cl2 → CHCl3+ Cl
CHCl3 + Cl → CCl3 + HCl
CCl3 + Cl2 → CCl4 + Cl
c.
Termination
The radicals are converted to stable molecules and this stops the reaction.
Cl. + Cl. → Cl2
CH3 + Cl → CH3Cl
CH3 + CH3 → C2H6
Chlorination of higher alkanes
CH3 CH CH2 CH3
CH3 CH2 CH2 CH3 + Cl2
u
Cl
major product
The following are deduced from chlorination
CH3 CH2 CHreaction
2 CH2Cl
i.
Alkanes react with chlorine in dark room at temperature more than
250ooc or under the influence of ultra – violet light.
ii.
The presence of small amount of oxygen slows down the reaction for
a period of time after which the reaction proceeds normally.
5.5
Uses
i.
Methane is used to produce hydrogen from process.
ii.
Alkane (especially methane) is used to produce ccl4
(tetrachloromethane) which is an important organic solvent.
iii.
Alkanes are used as fuel either alone or combine with other gases.
iv.
Production of chloroform, phormalane (trichloromethane CHCl3)
which is used in surgical operation.
CHAPTER 6
ALKENES
These are unsaturated hydrocarbons with general molecular formula C2H2n
where n ≥ 2 i.e CH2 (n = 1) does not exist. Alkenes are oily liquids and this
is because when there derivative react with halogen, they produce oily
compounds hence are called olefins.
In an alkene molecule, there exist a double bond between its carbon atoms
and the carbon atoms are sp2 hybridized, trigonal in shape and as such
have bond angle of 120o. the availability of electrons in the double bond(s)
make alkenes chemically more reactive than alkanes; the double bond
contains a sigma bond and a pi – bond.
6.1
Nomenclature
Alkenes generally end with in ‘ene’. The following rules must be applied
when naming alkenes.
i.
In naming a straight chain alkenes, the carbon atoms are numbered
giving the lowest number to the carbon atom bearing the functional
group (c = c).
ii.
Choose the longest chain as the parent hydrocarbon.
iii.
Consider the position of the double bond and give it the lowest
number ( - C = C - )
iv.
Consider the substituent alkyl group.
v.
Consider the substituent halogen group.
vi.
When a particular substituent group or atom is more than one, the
prefix di, tri, tetra for 2,3,4 respectively is used before its name.
vii.
When there exist more than a substituent group or atom, name them
alphabetically.
viii.
The number of carbon atoms carrying the substituent or functional
group is written before their respective names.
Example I
H H H H H
H – C 15 – C 24 – C 33 = C 42 – C 51 – H
H H
H
→ pent – 3 – ene ×
← pent – 2 – ene √
Example II
H
H
CH3 H
H
H – C 16 – C 25 = C 34 – C 43 = C 52 – 6C 1H 3
H
→ – 3 – methylhexa – 2,4 - diene √
← – 4 – methylhexa – 2,4 - diene ×
H
H
CH2CH3
H
H
H–C–C=C–C–H
H
H
H
Example III
→
H H
1
3
H–C–
H
H
C 2 – 3C 1
H
C3
H
= 4C 2 – 15C
H
3 – methylpent – 3 – ene ×
3 – methylpent – 2 – ene √
Example IV
Cl Cl Br Br Cl
H5C2 – C = C – C – C – C – Cl
Cl Cl
C 2H 5
H Cl
H
H
H
H
Br
Br
→ CHI7 –
– C 26 – C 35 = C 44 – C 53 – C 62 – C 71
H
H
H C2 H H
H
C
H
H
→
Cl
5,6 – dibromo – 7,7,7 – trichloro – 5 – ethythept – 3 – ene √
Cl
Cl
←
1,1,1 – trichloro – 5, 6 – dibromo – 3 – ethylhept – 4 – ene ×
6.2
Physical properties
i.
They are non – polar
ii.
Boiling point and melting point increases with increases carbon
atoms.
iii.
They are colourless.
iv.
They are less dense than water.
6.3
Preparation
→
Cracking of higher alkanes
C15H32 12 atom
C8H18 + C3H6 + 2C2 H4
520oC
petrol
propane ethane
Diesel oil
→
Dehydrogenation of alkanes
CnH2n+2
Al2O3
CnH2n
E.g.
H H H
H
H – C – C – C – C – HAl2O3
H H H H
H H H
H
+H – C = C – C = C – H
Con H2SO3
H H H
H
H H H
H
C=C–C–C–H +H–C–C=C–C–H
H
H H
H
H
→ Dehydration of alcohols
CnH2n+1 OH
CnH2n
Example
C2H5OH
C 2H 4
CH3
CH3
H+
OH
The impurities such as So2 and Co2 can be removed using caustic soda
(NaOH).
→Dehalogenation of vicinal dihalide.
–C–C–
x
Zn/acetone
C=C
x
Example
CH3 – CH – CH – CH3 Zn/acetone CH3 CH = CH CH3
Cl
Cl
Cl
Zn/acetone
Cl
→
Reduction of alkyne
Lindlar catalyst
R1
R
H2,pd/caco3
C=C
H
H
cis - alkane
R – C ≡ C – R1
H
R
Na,NH3
C=C
H
Trans -
6.4
Reactions
→
Addition reaction
i.
Hydrogenation (sabatier’s reaction)
Cn H2n+ H2
Ni
Cn H2n+2
E.g.
Cn H2n+ H2
ii.
Ni
C3 H 8
Halogenation
cc1
C = C + X2
Example
–C – C –
x x
Cl
+ Cl2
cc1
Cl
R1
H H
H H
H H H
H – C – C = C – C – H + Cl2
H–C–C–C–C–H
H
H
iii.
cc1
H
H Cl Cl H
Hydration
CnH2n + H2O
H+
CnH2n+1 OH
E.g.
H 2O
H+
CH3 – CH2 – CH = CH2
CH3 – CH2 – CH – CH3
OH
iv.
Halohydrin formation
H 2O
1
R – C = C – R + X2
R – CH – CH R1 + HX
Example
H 2O
CH3 CH = CH2 + Cl2
CH3 – CH – CH2
Cl
*
OH
Glycol formation
Kmn04
HCo2OH
CnH2n
CnH2nOH
Kmn04
HCo2OH
CH3 CH = CH2
CH3 – CH – CH2
OH
vi.
OH
Alkylation
C=C+R–H
H+
–C–C–
H
Example
R
H
H
H+
a.
CH3 – C = CH2 + CH4
CH3
b.
CH3 – C – CH2 – CH3
H
CH3
H+
CH3 – C = CH2 + CH3 – C – H
CH3
CH3 – C – CH2 – C – CH3
CH3
vii.
CH3
CH3
H
Epoxidation
C=C
Per
acid
C–C
Example
CH3 CH = CHCH3
viii.
Per
acid
CH3 CH2 – CH2 CH3
Ozonolysis
Riii
Ri
ii
R
C=C
iv
R
O3
H2o/zn/H+
Ri
Riii
C=o+o=C
ii
R
Riv
Mechanism
Riii
Ri
ii
R
C=C
iv
R
O3
Ri
Riii
C
ii
R
C
Riv
H2O/H
Zn
+
Ri
Riii
C=O=C
ii
R
Ri
Riv
Riii
C=O +O=C
Rii
Riv
Example
CH3CH = CH2
O3
H2O/H+/zn
CH3CH =O + O = CH2
O
O
– C – CH3 + H – C – H
ethanal
ix.
methanone
Hydrohalogenation
CnH2n + HX
CnH2n+1 x
Example
CH3CH CH3 (markonikorv)
CH3CH = CH2
X
H2O2CH3CH2 CH2 (Anti – markonikorv)
X
It should be noted that
i.
Addition reaction is a reaction in which two molecules combine to
yield a single product.
ii.
The affecting reagent is known as addendum (pl. addenda).
→
Combustion
Alkene combust in air to produce CO2 and H2O like other
hydrocarbons.
→
Polymerization
This is a process by which two or more simple molecules
(monomers) link up to from giant molecules (polymers) with or
without the elimination of small particles, is eliminated, it is called
Addition polymerization
Examples of polymers include polyethene, PVC etc.
6.5
Uses of alkenes
i.
Derivatives such as 1,2 – dibromoethene is used as petrol additive to
prevent knocking.
ii.
Use as starting materials for compounds such as alkanol, alkane et.
iii.
Alkanes are used to produce important materials such as plastic,
rubber etc.
CHAPTER 7
ALKYNES
These are highly unsaturated compounds with triple bonds there carbon
atoms with general formula CnH2n-2 where n ≥ 2 (the simplest member is
ethyne). They possess one sigma bond and two pi-bonds hence there is
availability of electrons across the triple bond so they are highly reactive
than the corresponding alkanes and alkenes.
7.1
Classification of Alkynes
On broad basis, alkynes can be classified into two which are
→
Terminal Alkynes
These are alkynes in the atom. Atom carrying the triple bond is attached to
one or two hydrogen atom(s). The triple bond is usually located at the
extreme end of the alkynes; the hydrogen(s) attached to the last carbon
H H
H H
atom is acidic.
Example
H C≡CH
→
H–C–C–C≡C–H
H C≡C–C≡C H
Non – terminal Alkynes
These are the alkynes in which the carbon carrying the triple bond is
attached to by many alkyl groups.
H
H
H–C–C≡C–C–H
H
7.2
C 2H 5
H3C – C ≡ C – CH3
H
Nomenclature
The rules governing the naming of alkynes are similar to the rules in
alkenes’ nomenclature.
Example
H
1.
H – C 13 – C 22 ≡ C 31 - H
H
→
Prop – 2 – ene ×
←
Prop – 1 – ene (propene) √
CH3
CH3
C 16 ≡ C 25 – 3C 2
C 3H 7
CH3
H
4
C3
H
H
5
C2
H
2.
H – C ≡ C – C – CH3 →
H–
3, 2 – dimethylhexyne
4,4 – dimethylhex – 5 – yne ×
H
3.
H
H
H – C 15 – C 24 = C 33 – C 42 ≡ C 51 – H
H
→
pent-2-ene-4-yne ×
←
pent-4-ene-1-yne √
7.2
Physical properties
i.
They are colourless and odourless
ii.
They are non – polar
iii.
Ethyne is less dense than air
7.3
preparations
→
Action of cold water on calcium carbide
CaC2 + H2O → C2H2 + Ca (OH)2
The calcium carbide can be gotten from coal and/or lime stone.
Coal → coke
20000c CaC2
Limestone → CaO
→
Dehalogenation of tetrahalide
x
x
R – C – C – R1
x
Zn
R – C ≡ C – R1 + 2Zn X2
x
Br Br
E.g.
CH3 – C – C –H
Zn
CH3 C ≡ CH + 2 Zn Br2
Br Br
→
Dehydrohalogenation of vicinal dihalide
alc.
koH
1
R – CH – CH – RNaNH
3
X
R – C ≡ C – R1
X
Example
alc. koH
CH3CH – CH2 NaNH3 CH3C ≡ CH
Br
Br
7.4
Reactions
→
Combustion
Alkynes burn with luminous smoky flame.Ethyne combust in air at a
temperature above 3000oc to produce oxyacetylene flame used for cutting
and welding metals.
→
Substitution reaction
This reaction is only undergone by terminal alkyne.
i.
with solution
HC ≡ CH + Na
liq. NH3
2HCCNa + H2
sodiumacetylid
ii.
with solution of Cu2 Cl2 in NH3
HC ≡ CH + Cu2 Cl2 NH4 OH
CuCCu
Copper
acetylide
iii.
with ammonical solution of AgNo3
AgNO3
HC ≡ CH NH4OH AgCC Ag + 2HNO3
Silveracetylide
→
Reaction as acids
(white)
Since the terminal hydrogen of a terminal alkyne is labile (i.e acidic), hence
they can react with bases.
ether
i.
R – C ≡ CH + LiNH2
R – C ≡ C- Li ++ NH3
Lithium alkyl acetylide
E.g.
ether
CH3 – CH – C ≡ C – H + LiNH2
CH3
ii.
with Grignard reagent
CH3 – CH – C ≡ C- Li ++ NH3
CH3
Lithium isopropyl
R – C ≡ CH + R1 Mg Brether
R – C ≡ C Mg Br + R – H
Alkynyl magnesiumAlkane
bromide
E.g.
ether
HC ≡ CH + C2 H5 Mg Br
→
Addition reaction
i.
Hydrogenation
CnH2n-2 + 2H2
Ni
150
HC ≡ C Mg Br + C2 H6
Ethynyl
magnesium
CnH2n+2
o
R
Na or Li
aqeous NH3
H
H
C=C
R1
Trans - alkene
R – C ≡ C – R1
Lindlar
R
H
R1
C=C
cis - alkene
E.g.
CH3 – C ≡ C – CH3 + 2H2
Ni
Na
aq. NH
CH3 – C ≡ C – CH3
CH3CH2 CH3CH2
150o
CH3
3
H
H
C=C
CH3
H
CH3 – C ≡ C – CH3
Lindla
CH3
Catalyst
H
ii.
Halogenation
a.
CnH2n-2 + 2X2 → R – C – C – R
x
x
E.g.
x
CH3
C=C
H
x
Br Br
CH3C ≡ CH + 2Br2 → CH3C – CH
Br
b.
Br x
x
R – C ≡ C – R1 + X2 → R – C = C – R1
Br Br
E.g.
CH3C ≡ CH + Br2 → H3C – C = C – 11
iii.
Nitrile formation
Alumna
R–C≡N
573k
Hydrohalogenation
H X
Hx
1 Hx
1
R – C ≡ C – R1 + NH3
iv.
R–C≡C–R
R–C=C–R
H X
R – C – C – R1
H
The second stage of the reaction is in accordance with the markonikov’s
rule which states “if an attacking reagent is added to an unsymmetric
alkene, the highly electronegative element from the agent should be added
to the carbon with fewer number of hydrogen atom”.
Cl H
Cl
E.g.
HCl
HCl
Cl H
CH3C ≡ CH
v.
CH3 – C = C H2
CH3 -C – C – H
Keto-enol
O
tautomeis
dil. H2SO4
1
1
R –– CH2 –C
R – C ≡ C – R + H 2O
R–C=C–R
o
m
HgSO4, 60 c H OH
Hydration
If the alkyne is terminal, aldehyde is produced. If not is ketone is produced.
H O
E.g.
CH3C ≡ CH + H2O
HgSO4,
H2 SO4
CH3 – C – C – H
H
Propan
H H O
H
H2SO4
o
CH3C – C ≡ C – CH3 HgSO4, 60 c CH3– C – C – C – CH3
H H
Penta – 2 vi.
Polymerization
The product obtained by polymerizing ethyne depends on the condition
3C2H2
400o
C 6H 6
7.5
Uses
i.
Ethyne is used as fuel in hunter’s lamp.
ii.
Alkynes can be used to produce compounds such as alkenes,
alkanes, alkanals, alkanones etc.
iii.
Oxyacetylene flame produced from the combustion of ethyne is used
for cutting and welding metals
CHAPTER 8
AROMATIC HYDROCARBONS
These are compounds with general formula Cn H2n-6 where n ≥ 6. They
contain (4n + 2) π electrons and obey Huckel’s rule (it states “conjugated
system containing complete and uninterrupted cycle of p – orbital are
stabilized relative to their cyclic counterpart if they contain (4n+2) π
electrons”).
According to August Kekule in 1865, the structure of benzene is based on
the concept of resonance and it proposed structure include.
H
C
H
H
C
C
H
C
C
H
C
H
Benzene can be obtained from destructive distillation of coal from the coal
tar (a mixture of different organic solvent which include benzene). Benzene
can also be obtained from Naphta (ore of the light fraction of petroleum).
8.1
Nomenclature
The following rules are considered in naming aromatic compounds.
i.
In naming monosubstituted benzene, we prefix the name of the
substituent group to the word benzene.
Example
Cl
1.
4.
OH
Chloro
2.
Hydroxy benzene (phenol)
CH3
5.
NO2
Methyl benzene (Tolurne)
3.
Nitrobenzene
NH3
Amino benzene (aniline)
ii.
If 2/3 groups are present in benzene ring, they are name by numbers
or ortho, para and meta.
Example
Cl
1
1.
6
5
4
NO2
1
2.
6
5
2
3
1,3 – dichloro
benzene
Cl
3.
or
4.
3 – bromonitro benzene
or
3
meta – bromonitro
Br
benzene
2
4
OH
1 Br
2
6
2 – bromo – 4 3
5
Iodophenol
OH
1 No2
2 2 – nitro phenol
6
3 Ortho – nitro phenol
5
4
CH3
1
6
5
4
2
3
4 – chlorotoluene
Para – chlorotoluene
Cl
COOH No
2
1
2
6
3
4
5.
6.
o2N
5
Other aromatic compounds include
CH3
CH3
2,6 – dinitrobenzoic acid
OH
xylen
Anthracen
Naphtol
COOH
CHO
et.c.
8.2
banzoic
Physical properties
banzaldehyde
i.
Benzene is insoluble in water.
ii.
They are colourless liquid with sweet smell.
iii.
Boiling point of not less than 80oc.
8.3
Preparation
→ Action of quicklime on benzoic acid.
COOH
+ CaO
→
Polymerization of ethyne.
3C2H2
8.4
Reactions
+ CaCO3
naphthlene
→
Addition reaction
i.
Hydrogenation
+ 3H2
ii.
Ni
200oc
→
Halogenation in the presence of sunlight
Cl
Cl
Cl
u.v
+ 3cl2
Cl
Cl
Cl
Substituent reaction
i.
Halogenation
+ x2
x
Lewis
I
+ Cl
Nitration 2
No+ 2Hcl
AlCl3
+ HNO3 Conc.
iii.
+ H 2o
Sulponation
So 2H
+ SO3
iv.
where X = Cl, Br
Ch
E.g.
ii.
+ XH
H2SO4
+ H 2o
Friedel – craft alkylation
R
+ R – X Lewis
E.g.
+ HX
CH3
+ CH3Cl
AlCl3
+ HCl
:
+ I2
HNO3
8.5
Uses of substituted Benzene
i.
Halobenzene
x
NaoH
OH
NH3
ii.
Nitrobenzene
NO2
Phenol
NH2
HN2
Zn/Hcl
Anilin
iii.
Alkylated benzene
R
COOH
KMno4
8.6
Uses of Benzene and its compound
i.
It is used as fuel (Benzol) in car.
ii.
It is used as solvent in industry and dry cleaning.
iii.
Benzene styrene used as plastic, phenol used as disinfectant,
toluene used as explosives, aniline used in drug manufacture.
CHAPTER 9
ALKANOLS (ALCOHOLS)
They possess a general formula CnH2n+1OH (R – OH) and are of the
homologous series with functional group OH. Alcohols may possess single
bond, double bond, halogen or additional hydroxyl group.
9.1
Classification
Alkanols can be classified into four categories based on the number of OH
per molecule.
→
Monohydric alkanols
These contain one OH group in a molecule. Monohydric alkanols can
be divided into
i.
Primary alkanols (1o)
These are the alkanols which the carbon atom carrying the – OH is
attached to an alkyl group.
R
H – C – OH
OH
E.g. CH3 – C – H
H
ii.
H
Secondary alkanols (2o)
These are the alkanols in which the carbon atom having the – OH is
attached to two alkyl groups.
R
R1 – C – OH
H
CH3
E.g.
H3C – C - OH
H
Propan – 2 – 01
Tertiary alkanols (3o)
R
CH3
These are the alkanols in which the carbon atom having the – OH is
1
R
– OHalkyl groups.
E.g.
H 5C 2 – C attached –
toCthree
3 – methylhexan – 3 OH Rii
C 3H 7
→
Dihydric alkanols
iii.
These contain two OH – groups per molecules. They are known as
diols.
→
Trihydric alkanols
These contain three OH – group per molecules. They are known as
triols.
→
Polyhdric alkanols
These contain more than three OH – group per molecules. They are
polyols.
9.2
Nomenclature
1.
CH3OH
CH2 Methanol
OH
OH
Example
2.
CH3 – CH – CH3 3.
Propan – 2 – 01
(methyl
cyclopentenol
OH
OH
4. CH2 –
1,2 – ethanadol
(athylene glycol
Br Br Br Br
5.
C Br2 OH (C Br OH)2 CBr3 → Br – C – C – C – C – Br
Br Br Br Br
1,1,2,3,4,4,4,4 – heptabromo but – 1,2,3 – triol
9.4
Physical properties
Alkanols contain lipophilic (alkane – like group) and hydrophilic
(water – like group). The properties of alcohols depend on the OH – group
which is modified based on the size of the alkyl group.
The boiling point of alkanols increases with increasing number of carbon
atom, also the more branched an alcohol is, the lower the boiling points.
Generally, other properties such as melting point, solubility relative density
increases in the same trend as boiling point.
9.5
Preparation
→
Hydrolysis of alkanes
Conc. H2SO4
CnH2n + H2O
CnH2n+1 OH
E.g.
C 2H 4 + H 2O
→
Conc. H2SO4
C2H5OH
Hydroboration
CnH2n + B2H6
H 2O 4
CnH2n+1 OH
HO
E.g.
CnH2n + B2 H6
→
i.
H 2O 4
C2H5OH
HO
Reduction of Aldehydes or ketones (carbonyl compounds)
Using Grignard reagent
H
diethylether
R MgX + R1CHO
R – C– R1
H+
aldehyd
OH
diethylether
R MgX + R1CORII
R – C– R1
+
H
II
R
Generally, formaldehyde produce 1o – alcohol, other aldehyde produce 2o –
alcohol while ketones produce 3o alcohol.
E.g.
O
OH
E + 2O
H+
CH3 MgCl + H – C – H
H 3C – C – H
H
O
E + 2O
H+
OH
CH3 MgCl + CH3 CH2 – C - H
O
E + 2O
CH3 MgCl + CH3 C – CH3
ii.
H3C – C –CH3CH2
H+
CH3 – C –CH3
CH3
Using NaBH4 (sodium tetrahydrido borate III)
i.
RCHO + NaBH4
Alcohol
ii. H+
OH
i.
RCHO + NaBH4
E.g.
ii. H+
R – C – R1
EtOH/H2O
CH3 C – C2 H5 + Na BH4
iii.
OH
CH3 C – C2 H5
H+
Using Li AlH4 (Lithium tetra hydrido aluminate III).
Dry either
Aldehyde/ketone + Li AlH4
Alcohol
H+
This reaction proceed under dry condition because it produces Al(OH)3 and
LiOH under wet condition.
LiAlH4 + 4H2O → 4H2 ↑ + LiOH + Al (OH) 3
iv.
Catalytic hydrogenation
Adehyde/ketones + H2
9.6
Ni
200o
Alcohol
FERMENTATION
This is the slow decomposition of large organic compound (such as
starch) into smaller organic compound (such as ethanol) by micro –
organisms. This method is used to produce ethanol in a large scale from
starchy food.
Starchy foods are cruched and grounded into powered form, malt (which
contains diastase to convert starch into maltase) is added at (50 – 60) oc
for an hour. Yeast (which contains maltase to glucose and glucose of
ethanol respectively) is added at room temperature.
diastase
2 (C6 H10 O5)n + nH2O
nC12H22O11
C12H12O11 + H2O
C6H12O6
maltase
zymase
2C6 H12O6
C2H5OH + 2Co2
9.7
Reactions
→
Williamson reason
Alcohols react with active metals to librate hydrogen gas and form
alkoxide
R – OH + M(s) → R – O – M + ½ H2↑
E.g.
R – OH + Na → R – O – Na + ½ H2
→
i.
ii.
Dehydration
Access
R – OH
Alkene
180o
Conc. H2 SO4
R – OH
ether
145o
E.g.
C2H5OH
Access H2SO4
180o c
–C=C
Conc. H2 SO4
C 2H 5 – O – C 2H 5
145o c
→
Formation of alkylhalide
This is achieved using Socl2, Pcl3, Pcl5 or Hcl.
R – OH + Pcl3/Pcl5/SOCl2
→ R – Cl + Hcl
Zncl
R – OH + HCl
R – Cl H2O.
E.g.
C2H5OH + pcl3 → C2H5 + Hcl
C2H5OH +Hcl
Zncl
C2H5cl H2O
If should be noted that reaction of alcohol with thionylchloride give a
stereospecitic product of isomers depending on the chiral centre.
→
Oxidation of alcohol
Primary alkanol undergoes partial oxidation using a selective oxidizing
agent called pyrimidium chloro-chromate (PCC) with the formula
C5H5NHCrO3cl to produce alkanal and complete oxidation using KMnO4,
K2Cr2O7 to give Alkanoic acid. Secondary alkanol undergo oxidization to
produce alkanone.
H
R – C – OH
H
H
R – – OH
H
9.8
Uses
(O)
(O)
O
R–C–H
O
R – C – R1
(O)
R – C – OH
i.
Sometime use as anti – freeze agent.
ii.
Used to produce important such as alkane, alkenes, alkanal etc.
iii.
Used as fuel.
iv.
Used as solvent for variety substances.
CHAPTER 10
ETHERS
These are related structurally to alcohol but possess general formula R – O
– R1 where R & R1 are alkyl or aryl groups.
10.1 Physical properties
i.
Possess low melting points due to their dipole moment.
ii.
Very soluble in water due to formation of hydrogen bonding.
O
R1
H
H
R
O
¨
10.2 Preparation
→
Williamson synthesis
R – X + R1 O M → R – O R1 + M X
Alkoxide
E.g.
CH3 Br + C2H5 ONa → C2 H5 O CH3
→
Form alkenes
H2SO4
CnH2n
ether + R – OH
80OC
E.g.
H2SO4
C2H2 80OC C2H4 OC2H5 + C2H5OH
10.3 Epoxides
These are cyclic ethers of 3 – membered ring. They can be prepared
by
→
→
Catalytic oxidation of ethylene
Ag
C
– C = C– + O2
O
250 C
Reduction of hylohydrin
OH X
C
C
–C–C– →
O
E.g.
CH3CH CH2 → CH3 – CH – CH2
O
C
O
10.4 Nomenclature
Ethers can be named by
i.
H
using their trivial name
O– C – C – H
CH3 – O – CH3 dimethylether
H H
phenylethylether
CH3 – O – C2H5 methylethther
ii.
H
from the corresponding alkane derivative
CH3
Phenoxy
iii.
Methoxy
cycle ethers are names by adding prefix ‘oxa’ to the corresponding
name of the cyclic hydrocarbon
Oxacyclubutane (cyclobutoxide
Oxacyclopentane
Dioxacyclohenane (1,4 –
→
Reaction
Ethers are generally unreactive and for this they are used as reaction
solvents.
CHAPTER 11
CARBONYL COMPOUND (ALKANALS AND ALKANONES)
O
Alkanalas are organic compounds with functional formula RCHO (R – C –
H) where R is an alkyl or aryl group. They are formally called Aldehydes
and we called terminal carbonyls. They possess an unpleasant smell.
O
Alkanones are organic compounds with functional formula RCOR1 (R – C –
R1). Alkanones are formally called ketones and they possess a sweet
smell. Alkanones sometimes have double bond(s) between their carbon
atom (s) and these are referred to us as alkenones.
11.1 Nomenclature of alkanals
This is general from corresponding name of carboxylic acid by replacing
its ‘ic’ with aldehyde, or ‘oic’ with ‘al’.
O
H
O
–C–H
C H
Formaldehyde from formic
acid
O
C
benzaldehyde from benzoic
O
CH3 C – H
H
Ethanal from ethanoic acid
Cyclohexane
carbaldehyde
Acetaldehyde from acetic
acid
Hence the following rules need to be applied when naming aldehydes.
i.
Choose the longest chain as the parent chain.
ii.
Consider the substituent alkyl and halo groups.
iii.
When there is more than a substituent group or atom present, name
alphabetically.
iv.
The compound is named as a derivate of the longest chain but end
the suffix – al is used instead of the last letter ‘e’ of the
corresponding alkane.
H
H
H
H
H–C–C–C–C–H
–C–H
H H H H
Butana
Cyclohexene
O
C–H
H O
H H
H
O
H–C–C=C–C–H
H
but – 2 -
11.2 Nomenclature of alkanones
Rule (i – iv) in naming aldehyles remain. The last letter of ‘e’ of
alkane is change to ‘one’ similarly the ‘cc’ (or ‘occ’) ending of carbolic
acid is ‘one’.
e.g.
O
CH3 C – CH3
Propanone from propanoic
Example 1
O H H H
– C 1– C 2 – C 3 – C 4 –
H H H
Butyrophenone (1 – pheny ibutan – 1 –
Example
O
C
11.3
Preparations
1,1 depheny
O H
– C– C – H
H
Acetophenone (L Example 2
H
O
H
C 14 – C 23 – C 33
H
C 41
4 – pheylbut – 3 – one
×
H
H3
→
Oxidation of alcohol
→
Friedel craft acylation
O
R – C – Cl + Ar – H
Fecl
ArCo – R
acyl
Example
O
C
O
CH3 – C – Cl +
Fecl
CH3
pheylethanon
→
Reduction of Nitriles with Grignard reagent.
O
ethe
1
R – C ≡ N + R Mgx H+ R – C – R1
Example
O
ethe
– C – C ≡ N + CH3 Mg Br H+ H3C – C – CH3
11.4 Reactions
→
Formation of alcohol
i.
Using Grignard reagent.
R – Mgx + HCHO → 1o – alcohol
R – Mgx + Aldehyde → 2o – alcohol
R – Mgx + ketone → 3o – alcohol
ii.
Using metallic hydride complex
NaBH4
LiAlH4
Aldehyde/ketone
→
alcohol
Cyanohydrin formation
Carbonyl compounds + HC ≡ B
O
C
R
→
R1 HC ≡ N
(H)
cynahydrin
OH
R – C – CN
R1
Formation of diols (hydration)
O
R – C – H + H 2O →
O
R – C –R1 + H2O →
OH
R–C–H
OH
OH
R – C – R1
OH
Example
OH
O
H 3C – C – C 2 H 5 + H 2O → H 3C – C – C 2 H 5
OH
→
Witting reaction
Carbonyl compound + trialkylylide → Alkenes + trialky phosphine
oxide.
O
R –O
C – H + R”3 – P = RI →
C=C
R – C – H + R”3 – P = RIII →
C=C
Example
O
H – C – H + (CH3)3P = CH24 →
H
H
(CH3)3P = CHCH3 + H3C – C – C2H5 →
→
H 3C
H 5C 2
H
C=C
+ (CH3)3 P = O
H
C = CHCH3
Reaction with alcohol (Alcoholysis)
If a mole of alcohol is reacted with a mole of alkanal, a hemi – acetal
is formed, but if 2 moles of alcohol is used an acetal is formed.
Similarly if a mole of alcohol is reacted with a mole of ketone, a hemi –
ketal is formed while if 2 mole of alcohol is used a ketal is formed.
OH
O
H+
I
R – C – ORI
H
Hemi -
R – C – H + R OH
O
H
H+
R – C – H + RIOH
R – C –I ORI
OR
Aceta
ORII
H+
R – C – RI + RIIOH
R – OR
C –II ORI
E.g.
O
OH
H+
CH3C – H + C2H5OH
CH3C – OC2H5
OH
CH3C – H + 2CH3OH
H+
CH3C – OCH3
OCH3
11.5 Uses
i.
Formalin (a solution of methanol in water) is used to preserve
biological specimens.
ii.
Propanone is a useful solvent for many organic materials.
iii.
Methanal is useful in plastic production.
11.6 Physical properties
i.
Carbonyl compounds possess high melting and boiling due to the
presence of hydrogen bonding.
ii.
Majority of their compounds are liquid at room temperature (HCHO
and CH3CHO are gases).
iii.
Solubility in water decreases with decreasing number of carbon
atoms.
CHAPTER 12
ALKANOIC ACIDS
These are compounds with the molecular formula RCOOH where R is an
alkylgroup or arylgroup. They are formally called
carbonylic acids or
O
organic acids. They have a carbonyl functional group (R – C) and alkanol
group ( - OH) which are responsible for their chemical properties.
Carboxylic acids with two or more carbon atoms undergo functional group
isomerism with esters. The derivatives of alkanoic acids include esters
amides, acylchlorides.
12.1 Classification of alkanoic acids
Carboxylic acids can be generally classified into four namely:i. monocarboxyle
acid
ii. dicarboxylic acid
iii. tricarboxylic acid
→
Monocarboxylic acids
These are the organic acids that possess single – COOH per
molecule. They are responsible for the sour taste in most unripe
agricultural product. They have general formula CnH2n+1 COOH where n ≥ o.
E.g H
O
H – C – C – OH
H
→
Ethanoic
Dicarboxylic acids
These are alkanoic acids with two carboxylic groups per molecule.
They have general formulaCnH2n (COOH)2
E.g. COOH
COOH
Ethanedioic acid
→
Tricarboxylic acids
These are alkanoic acids with these – COOH per molecule. E.g
COOH
COOH
COOH
Propan – 1,2,3 – trioic
→
Aromatic carboxylic acids
These are alkanoic acids in which the carboxylic group is attached
COOH
directly to an aromatic compound e.g
12.2 Physical properties
i.
They are polar in nature (only the first four aliphatic members are
soluble in water, carbon five member is partially soluble in water and
members with carbons are insoluble).
ii.
Boiling points increases with increasing carbon atoms
iii.
They are soluble in less polar solvent (e.g ether, alcohol, benzene
etc.)
iv.
They turn blue litmus red.
12.3 Degree of Acidity
The acidity of carboxylic acid depends on
→
The molecular mass of alkylgroup
The higher the molecular mass of the alkyl group (attached to the
COOH), the lower the acidity of a carboxylic acid.
H COOH
H–1
a
CH3 COOH
C2H5 COOH
CH3 = 15
C2H5 = 29
b
c
hence acidity decrease from a – c.
→
Electronegativity of elements
If an electronegative element is present in a molecule of alkanoic acid, the
acidity of such molecule will be high, to the acidity of a carboxylic acid can
also be determine by
i.
The nature of electro negative element
E.g.
Cl CH2COOH
Br CH2 COOH
F CH2COOH
a
b
c
Acidity increases from a – c because Cl has the lowest electro negativity
value.
ii.
The number of electronegative elements per molecule; as the
number of electro negativity elements increases, acidity increases.
iii.
distance between the electronegative element and the COOH: The
closer the electronegative element to the carboxylic group (COOH), the
higher the acidity.
Generally, acidity of an acid determines its polarity.
12.4 Preparation
→
Oxidation of primary alcohol
→
Oxidation of alkylated benzene.
→
Hydrolysis of Nitriles
R – C ≡ N+H2O
Example
H
+
O
RC – OH + NH4
O
CH3C – OH
+
H
CH3 ≡ N+ H2O
→
Carbonation of Grignard reagent
H+
RMgX + CO2
R – COOH
E.g.
CH3MgCl + CO2
H+
O
CH3 C – OH
12.5 Nomenclature
The naming of carboxylic acid follows same rule as that in previous
chapter only that the suffix ‘oic’ is added.
H H O
H O
O
CH3 C – OH
H – C – C – C – OH
H H
Ethanoic
acid
H
H
H 3C
C
C2
H
C1
2 - hydroxylpropanoic
acid O
O
C4
C3
CH3
OH
H
C
H
C – C – OH
H
Phenyl ethanoic
acid
Cl Cl O
H – C = C – C - OH
H
H
2,3 – dihloro pro 2 – enoic
O
2,2,2 – trimethylbut – 1,4 – dioic
12.6 Reactions
→
Formula of salts
Since it is an acid, on reacting with a salt is formed.
OH
RCOOH
RCOO
H+
E.g.
CH3COH + NaOH
→
CH3COONa + H2O
Formation of alkylhalide (Hunsdieker reaction)
Alkanoic acid on reacting into halogen loss a carbon atom to from
alkyl halide in the presence of mercury oxide.
O
R – C – OH + X2
E.g. O
CH3C – OH + Br2
→
HgO
R – X + CO2 + HgX
HgO
R – Br + CO2 + HgBr
Hell – Volhard – Zelinskii reaction
Aliphatic carboxylic acids reacts with chlorine or bromine in the presence
of phosphorus to substitute only the – hydrogen atoms (i.e hydrogen
atoms on the carbon directly before COOH).
H O
R–C–C–H
Br2P
H
Br O
R – C – C – OH
H
→
Formation of its derivatives
i.
acid chlorides formation
O
R – C – OH + SOCl3
O
CH3C – OH + SOCl3
Br2P
Br
R – C – C – OH
Br
O
R – C – Cl
O
CH3C – Cl
ii.
Esters formation(Alcoholysis of carboxylic acid).
O
O
1
R – C – OH + R – OH → R – C – OR1 –H2O
iii.
Amide formation
O
O
O
R – C – OH + R1 – NH2 → R – C – NHR1
Amide
iv.
Acid anhydride
O
O
O
O
R – C – OH + R1 – C – Cl → R – C – O – C – R1
→
Formation of alkane (decarboxylation)
This is done by carboxylic acid with soda lime (Cao slated with
NaOH) O
R – C – OH
R – H + CO2
Example
O
CH3C – OH
CH4 + CO2
12.7 Uses
i.
They are used for making dynes
ii.
Ethanoic acid is used in coagulating rubber latex
iii.
Ethanoic acid is used in preserving and flavouring food
iv.
Carboxylic acids are used in making other compounds like alcohol,
alkane etc.
CHAPTER 13
ALKANOATES (ESTEIR)
These are derivative of alkanoic acids and they are very important in
nature because they have sweet and inviting smell and are responsible for
flavor in most fruits and smell in flowers. They have the general formula
RCOOR1 where R and R1 are alkyl group.
13.1 The rule follows that of the previous but the R1 is named before R;
and a suffix ‘oate’ is added.
R1 → C5H11
1
C2H5COOCH3
C2H5COOC5H11
R → CH3
R → CH
Methylpropanoate R → C H
Pentylethanoate (found in
O H H
C–O–C–C–
H H
C2H5COOC5H11
R1 → C5H11
R→CH
Pentylpentanoate (found in
Ethyl
benzoate
H H
H H O
H–C=C–C–C–C–
H H
R1 → C2H5
R → C= C –
Ethyl propen – 2 – oate (ethyl – 2
propenoate)
13.2 Preparation
→
Alcoholysis of carboxylic acid (fischer esterification)
O
O
H+
O
O
R – C – OH + R1 – OH
E.g.
CH3C – OH + C2H5OH
→
R – C – O R1 + H2O
H+
CH3C – O – C2H5
Reaction of carboxylic and with diazomethane
O
R – C – OH + CH2N2 → R – C – OCH3 + N2(g) +
diazometha
13.3 Reactions
→
Reduction to form tertiary alcohol
O
OH
I
II
R – C – OR + R MgX → R – C - RII
RII
E.g.
O
OH
CH3C – OCH3 + C2H5Mgx → CH3 – C – C2H5
C 2H 5
→
O
Hydrolysis
O
R – C – O – R + H2O → R – C – OH + R1 – OH
1
→
E.g. O
Aminolysis (Amide formation)
O
O
1
R – C – O – R + NH3 → R – C NH2 + R1 – OH
O
CH3C– CH3 + NH3→ CH3C – NH3 + CH3OH
→
Alcoholysis (trans esterification)
O
O
I
II
R – C – O – R + R – OH → R – C - ORII
E.g.
O
O
CH3C – OC3H7 + C6H11OH → CH3C – OC5H11
13.4 Uses
i.
Esters are mainly used as solvents for cellulose nitrate and quick
drying substances like paints, adhesives and nails – vanishes.
ii.
They are used in perfumes and cosmetics.
iii.
They are used as artificial flavouring for foods.
CHAPTER 14
ALKANOYL CHLORIDES
These are also carboxylic acid derivate which the general formula RCOCl
where R is alkyl they are generally called acyl chlorides. They are prepared
by reacting thionylchloride with carboxylic acid.
14.1 Nomenclature
The suffix ‘oyl’ is substituted for ‘oic’ in alkanoic acid.
E.g.
O
CH3C – Cl
Ethanoyl chloride
14.2 Reactions
→
Hydrolysis
Acyl chloride hydrolysesOto form carboxylic acid
O
O
O
R – C – Cl + H2O → R – C – OH
E.g.
R CH3C – Cl + H2O → CH3C – OH
→
Alcoholysis (ester formation)
O
O
1
R – C – Cl + R – OH → R – C – OR1
E.g.
O
O
R CH3C – Cl + C3H7OH → CH3C – C3H7
→
Friedel craft acylation
O
C
O
R
+ R – C– Cl →
→
Formation of acid anhydride
O
O
O
O
R – C – Cl + R1 – C – OH → R – C – O – C – R1
E.g.
O
O
O
O
CH3C – Cl + H3C – C – OH →CH3C – O – C + CH3
→
Reduction
i.
With LiAlH4
O
R – C – Cl + LiAlH4
dry
H+
OH
R–C–H
H
1 OH
o
E.g.
O
dry
C2H5C – Cl + LiAlH4
H+
C 2H 5C – C – H
H
ii.
with alkyLithium (R – Li)
O
H+
R - C – Cl + RI – Li
OH
R – C – RI
RI
E.g.
iii.
O
CH3C – Cl + C2H5Li
OH
H+
H 3B – C – C 2H 5
C 2H 5
with Li (AlH (OCCH3)3)
O
R – C – Cl + Li(AlH (OCCH3) 3) → R – C – H
→
Aminolysis
O
O
R – C – Cl + NH3 → R – C – NH2
E.g.
O
O
CH3C – Cl + NH3 → CH3C – NH2
CHAPTER 15
AMIDES
These are also derivative of Alkanoic acids by replacing – OH of the
Alkanoic acid by an amino group – NH2.
O
O
R – C – OH → R – C – NH2
Amides can be represented with the functional group R – C – N. hence
apart from being Alkanoic acid derivative they are also ammonia
derivatives.
Amides are classified into three namely: i. 1o amide
ii. 2o amide
iii. 3o amide
Since amides are from Alkanoic acid, on replacing the – H group with
amino group – NH2, a 1o amide is formed, if one of the hydrogen atom is
placed by an alkyl (or aryl group), a secondary (2o) amide is formed; if the
two hydrogen atoms are replaced by alkyl (aryl groups), we have a tertiary
amide.
15.1 Nomenclature
The ending ‘oic’ of alkanoic acid (parent acid) is replace by amide
H H
O
H – C – C – C – NH2
H H
H
H
H
H C 2H 5
O
H – C – C – C – C – NH2
propanamid
2 – ethyl
H
H
O CH3
H – C – C – N – C 2H 5
H
H 3C – C – C – N – C 2H 5
CH3
N – ethyl, N – methyl
N – ethyl methy
15.2 Preparation
→
Aminolysis of acylchloride and ester.
→
Reacting carboxylic acid with amine.
15.3 Reactions
→
Hydrolysis (formation of carboxylic acid)
O
O H
H+
H+
O
O
R – C – NH2 + H2O
E.g.
R – C – OH
O
CH3C – N – H3 + H2O
→
CH3C – OH
Formation of amine (Hoofman degradation)
O
R–C–N
Base
R – NH2 + CO23
E.g.
O
CH3C – NH2
CH3NH2
O
CH3C – NCH3
CH3NHCH2
CHAPTER 16
AMINES
These are ammonia derivative in which at least one of the hydrogen is
replaced by an alkyl group or an aryl group.
Amines have tendency to form double bond and are trigonal pyramidal in
structure.
R2NH
NH3
R - NH2
R3N
Generally amines can be aromatic, aliphatic or alicyclic. They are used to
manufacture nylon.
16.1 Classification of Amines
Amines can be classified based on the number of replaceable hydrogen in
ammonia (lie the number of alkyl/aryl group(s) present) into three, these
are:
→
Primary Amines (1o – amine)
These are amines that have only hydrogen atom of ammonia being
replaced by an alkyl or aryl group. They are represented
by R. –
. N– or Ar –
..
N–
Example
H
..
H 5C 2 – N – H
Ethylamine
→
N–
Phenylamine
2
– C –NH
C–
C–C–
2 – amino
Secondary Amines (2o – amine)
These are amines in which two hydrogen atoms of ammonia are
replaced by alkyl or aryl groups. They can be generally represented by
H
H
H
Ar – N – R or R – N – R or Ar – N – Ar
Examples
H
H
N
H 3C – N – C 2H 5
Diphenylamine
→
H
Methyl
N – CH3
Methyl aniline
Tertiary Amines (3o – amine)
These are amines in which all three hydrogen atoms of ammonia are
replaced with alkyl or aryl groups. They can be generally represented as
R
R – N – R or R – N – R or Ar – N – Ar or Ar – N – Ar
Ar
Ar
R
Examples
CH3
H3C – N – CH3
N – C 2H 5
CH3
Trimethylamin
N
Triphenylamin
N - ethyl, N – methyl
16.2 Physical properties
i.
All amines are basic in nature.
ii.
Amines have strong odour characteristics of dead fish.
iii.
First two members of amine (methylamine and ethylamine) are
gases at room temperature while other members are liquid.
iv.
Smaller members are soluble in water.
v.
High melting and boiling point.
16.3 Degree of Basicity in Amine
The degree of basicity of amines are strongly affected by
→
Increasing number of Aryl group or Alkyl group
The greater the molar (the more the alkyl group) of the alkyl group,
the more increase in the basicity of the amine while as the aryl group
increases, basicity decrease.
Hence aliphatic amines are more basic than there aromatic
counterpart.
→
Presence of electronegative elements
Generally, amines are electron rich, the presence of electronegative
elements such as Cl, F, Br, and I decreases the basicity by withdrawing the
electron (lone pairs) available.
→
Hybridization state
Since nitrogen can exist in three hybridization state, it implies such
state will affect the basicity of amine this is because electrons are most
tightly held in sp hybridization. Hence basicity of amine decreases from
sp3 state towards sp2 state to sp state.
R – CH2 – N – , R – CH = N
Decreasing basicity
16.4 Solubility of Amines
R–C≡N
The solubility of amines is strongly determined by the basicity of the
amine, the more basic an amine the lower its solubility (i.e solubility of
amines decreases with increasing basicity). This is due to the presence of
hydrogen bonding between the hydrogen molecules and the water
molecules and the nitrogen atoms of the amine groups in the amine.
R
R
R
H – N …. H – N … H – N
H
H
H
.
R
.
.
R – N – H …. N – H
H
H
16.5 Nomenclature
The system of naming amines depends on the class of amine.
Primary and secondary amines are named by considering the alkyl or aryl
(groups) attached to the nitrogen atom followed by the ending name
‘amines’ or it is possible to name the amine as hydrocarbon (i.e
considering the amino (NH2) group as substitutent). Tertiary amines are
named using nitrogen atom of the amino.
The alkyl or aryl group with the greatest number of carbon atoms is used
as the parent chain.
H
Examples
H 7C 3
H N-H H
NH2
H – C1 – C2 – C3 – C4 – H
H
Propylamine (amino
H
Aminobenzen
H
H
2 - aminobutane
H
H
H 3C – N – C 2H 5
Methylethylamine
N – C 2H 5
N – ethylaniline
(N –
(N –
16.6 Preparations
→
Reduction of Nitro compounds
i.
R – NO2
Fe/Hcl
R – NH2
E.g.
ii.
Fe/Hcl
CH3 CH2 NO2
CH3 CH2 NH2
H2/pt
Ar – NO2
Ar – NH2
Example
NO2
NH2
H2/pt
→
Ammonolysis of alkyhalides
R – X + NH3 → R – NH2 + HX
1O – amine
R–X
R2 – NH + HX
2O – amine
R3 – X
R3N + HX
3O – amine
CH3
H 7C 3 – N – C 2H 5
(N – ethyl – N – methyl
amino propane)
→
Reduction of Nitriles. Using LiAlH4
LiAlH4
R – C = N Ether/H+ R – C – NH2
Example
H+
Ether CH3CH2C H2NH2
CH3CH2 = N + LiAlH4
→
Ammonolysis of primary alcohol
Al203
R – OH + NH3 350o
R – NH2 + H2O
16.7 Reactions
→
Alkylation (reactions with haloalkanes)
H
R – NH2 + R1X → R – N – R1 + H X
R2 NH + R1 X → R2 – N – R1 + HX
The limitation of this is the isolating of the intermediate amine, hence
quaternary amine is formed.
→
Acylation (amide formation)
Amide + acylhalide → amide
O
O
R – NH2 + R – C – X base R – C – NHR
E.g.
O
CH3NH2 + CH3 – C – Cl
CH3 CH2
OH-
O
CH3N – C – CH3
H
O
OH-
O
CH3 – N – H + CH3 – C – Cl
CH3– C – N – CH2CH3
CH3
→
E.g.
Salt formation
R – NH2 + H+
(+)
R – NH3
NH2
(+) (–)
NH3 Cl
+ HCl
→ Diazotization
Primary aromatic amines react with Nitrous acid to produce diazonium
salts. Primary aliphatic amines also give diazonium salt but the resulting
salt is unstable as it reacts rapidly with water to give mixture of products.
NaNO2, HCl
+
Ar – NH2
Ar – N2
HNO2
NaNO2, HCl
R – NH2
E.g.
HNO2
+
R – N2
+
N=N
NH2
NaNO2 HCl
+ HNO2
CHAPTER 17
AMINO – ACIDS
These are derivatives of carboxylic acids in which an hydrogen atom of the
alkyl group has been replaced by an amino – group. Amino acids are the
R
basis structural units of proteins with the general
formula
H – C – COOH
NH2
The carbon atom to which the amino – group is attached is known as –
carbon. Since amino acid have two functional groups (i.e COOH and NH2)
the can exist as acids, bases or salts based on the PH of the medium.
17.1 Physical properties
i.
Amino acids are crystalline solids with melting and boiling points.
ii.
They are conic in nature.
iii.
They have low solubility in non – polar solvent.
iv.
They are more soluble in water than in less polar solvent such as
ether.
17.2 Nomenclature
Although amino acids have common names but they are named as
the derivative of carboxylic acid (using IUPAC system of naming).
H
O
H2N – CH2 – CH2 – CH2 – CH2 – C – OH
5
4
3
2
CH3 – C – COOH
3
1
H – C – COOH
NH2
3 – amino propanoic acid
O
O
NH2
(Alanine)
O
HO – C – CH2 – CH2 – CH – C – OH
2
CH3NH
– CH
– CH – C – OH
Aminoethanoic acid
OH
1
NH2
5aminopentanoic
H
2
5
4
3
2
1
2 – amino – 1, 5 – pentadioic
acid
2 – amino – 3- hydroxyl butanoic
acid
17.3 Preparation
Amino acids can be prepared by the action of concentrated ammonia
solution on 2- cholo carboxylix acids (produced from Hell – rolhard
zelinskii reaction)
redphosphor
Cl
R1 – CH2 – C – OH + Cl2
R – CH – COOH
2 – chloro carboxylic
Cl
R – CH – COOH + 2NH3
R – C – COOH + NH4Cl
H
CHAPTER 18
FATS AND OILS
Fats and oils belong to a group of compound known as lipids and are hiher
homologous of eaters. Fats are solid and of animal while oils are liquid
mainly from plants at room temperature.
Fats are saturated whiles oils are unsaturated.
18.1 Physical properties
i.
Fats have high melting points and oil have lower melting points.
ii.
Fats and oils are soluble in water
iii.
They decompose at temperature above 300oc.
18.2 Reactions
→
Hydrogenation of oils
This is also known as hardening of oil because if produces solid fats.
The process involves passing hydrogen into unsaturated oil at 200oc at
satn in the presence of finely divided nickel as catalyst.
Ni
CH2 CH2
CH
+H2
200oc, 5
Hardened oil
→
Saponification
This is the alkaline hydrolysis of esters to produce soap and gyceol.
This method is used to manufacture soap in large quantity. The bases
used are NaOH (to produce hard soap) and KOH (to produce soft soap).
Fats or oil + caustic alkali → soap + glycerol.
Concentrated sodiumchloride is ussally use to decrease solubility of
the soap and to separate the glycerol from the solution. This process is
know as salting out process.
18.3 Soaps and Detergents
Detergents are commercially packaged powders or liquids used as
cleansing agents. They can soapy or soapless. Soupy detergents are the
sodium or potassium salt of long chain organic acid. Hence the have the
general formular RSO4 -Na+ or RSO4-kt and they are acidic in nature.
Soaps are biodegradable product of saponification which do not
form scum with hard water.
18.4 Uses of fats and oils
i.
They are essential food ingredients.
ii.
They are used in making soaps.
iii.
They are used in making candles and glycerols.
iv.
They are raw materials in paints and varnishes industry.
CHAPTER 19
CARBOHYDRATES
Carbohydrates are naturally occurring organic compounds that contain
carbon, hydrogen and oxygen with the general formula Cx (H2O)3.
Carbohydrate can be classified into simple sugars and complex sugars.
Simple sugars are crystalline, soluble in water and have sweet smell.
Structurally simple sugars can be divided into monosaccharides (e.g
glucose, fructose, galactose), and disaccharides (e.g maltose and
sucrose).
Complex sugars are non – crystalline, insoluble and tasteless. They are
mainly polysaccharides e.g cellucose and starch.
Carbohydrates with the structure of alkanal – CHO are known as aldoese
while those with structure of alkanone are known as ketoses
Carbohydrate
Simple sugar
Monosaccharide
complex sugar
Disaccharides
polysachardaries
Glucose fructose galactose sucrose maltose lactose starch glycogen
cellulose
Uses
i.
Glucose is used in the manufacture of jam and sweets
ii.
Glucose is used as intermediate source of energy.
iii.
Sucrose is used in preserving food.
iv.
Ceramella (a sucrose) is used for flavoring and in confectimary.
v.
Starch is used mainly as food.
vi.
It is used as stiffening agent.
vii. Cellulose is used mainly in the manufacture of paper, cellophane,
ropes, textiles, gum, cotton, and explosives.
CHAPTER 20
POLYMERS
Polymers are macromolecules formed from simple molecules. The
process of forming a large molecule (polymer) from simple units
(monomers) is known as polymerization. Polymer consist of repeating unit
and its molecular size is not fixed.
20.1 Types of polymerization
→
Addition polymerization
In this, all atoms in the monomer are present in the polymer.
→
Condensation polymerization
In, this there is elimination of small molecules such as water, caron
(iv) oxide, methanol etc.
20.1 Types of polymers
→
Natural polymers
These are polymeric substances that occur in nature (i.e
livingthings). Examples of natural polymers include certain carbohydrates
(such as starch and cellulose) and all proteins. If should be noted that fats
and oils are not polymers as they are not large enough to be considered as
macromolecules.
Natural polymers can be used to produce artificial polymers of desired
qualities when subjected to chemical treatment. This led to the production
of very useful synthetic polymers.
→
Synthetic polymers
These are commonly called plastics which can be soften by heat or
pressure and than moulded into any desired shape.
Plastics can be classified into two namely:
i.
Themoplastics: These are synthetic materials which can be soften
repentedly by heat and remolded for example polythene, polystyrene,
nylon, polypropene, terylene and perspex.
ii.
Thermosets: These are synthetic materials that cannot be soften or
melted by that or remoulded ones they are formed or set. For
example Bakelite and urea – methanol.
PROPERTIES OF PLASTICS
i.
They have resistant to fungal and bacterial attack.
ii.
They are good insulators of heart electricity.
iii.
Their raw material are cheaply and readily available from crude oil
refining.
iv.
They are strong but light, inert to air, water and other chemical.
v.
They are manufactured at a low costs using mould and automated
manufacturing process.
EFFECTS OF PLASTICS
i.
Plastics cause land air pollution because on burning toxic gases are
released into the atmosphere.
ii.
Heavy dependence of crude oil.
iii.
Plastic materials are non – biodegradable.
20.2 Types of polymerization
There are basically two types of polymerization (process of making
polymers) these are
→
Addition polymerization
This is the process whereby two or more of monomers join together
without the elimination of any small molecules: The monomers must
be simple unsaturated molecules 9such as alkenes and alkymes).
Polymers made from ethane or its derivatives are called Vinyl-type
polymers. The
physical properties of varies as the condition(s) is are altered (i.e
temperature,
pressure or catalyst).
Poly (Ethene) – polythene
This is polymerized ethane formed when ethane is heated to a temperature
of about 250oc and pressure above 500atm together with traces of oxygen.
It can be represented by (CH2 – CH2)n where n = 100 to 1000.
In the polymerization process, the double bond of ethane becomes
converted to a single bond leaving free electrons which can link up other
electrons (similar) in other molecules to form the chain.
n(H2C = CH2) → CH2 - CH2 – (CH2 - CH2)n - CH2 -CH2.
Poly(ethane) can be of low density (as the one obtained above) and low
density polyethene is used in making plastic bags, bottles, kitchen wears
and transparent films for packing stuffs; it is also used as wire and cable
insulators because of its insulating features.
There is also high density polythene (prepared at a lower
temperature and pressure in the presence of catalyst) and are used in
making large bones, crates, detergent bottles, dustbins and hander plastic
bags.
Perspex
This is formed by the polymerization of methyl – 2 – methy propenoate in
the presence of Lauroyl peroxide (organic catalyst). Perspex is a strong,
hard, transparent thermoplastic with a glossy surface (therefore used as
replacement for glass).
It is used in making car rear lights, windshields, contact lens, street – lamp,
fish tanks et.c.
COOCH3 COOCH3 COOCH3
n CH2 = CCOOCH3 → . . . CH2 – C – CH2 – C – CH2 – C . . . .
n
CH3
CH3
CH3
CH3
Methyl – 2 – methyl
Poly (chloro ethene)
Perspe
This is also known as polyvinyl chloride (PVC). It is manufactured
from ethene or ethyne. Chloroethene is polymerized at about 60oc in he
presence of hydrogen peroxide (as catalyst).
n CH2 = CH Cl → CH2 – CHCl – CH2 – CHCl2 n – CH2 – CHCl2
PVC
PVC is used in making tough plastic pipes, cable covering, artificial leather
clohs, table clothes, rain coats, handgloves, tiles and records.
Polystyrene
This is made from phenylethene (or styrene) by refluxing in the presence of
benzoyl peroxide solution (as catalyst). Polystyrene is a clear transparent
plastic which is hard but brittle. Styrofoam (expanded polystyrene) is
formed when air with a foaming agenct is blown into heated polystyrene.
Styrofoam is a light, white, opaque solid which floats in water; and hence
are used as ceiling tiles and lamp shades.
Polystyrene is used as packing material or as shock absorbers for fragile
objects, also used for making disposables cups and food containers.
nCH2 = CH → . . . CH2 – CH – CH2 –CH –CH2 – CH . . .
n
phenylethen
polystyren
Poly (propenonitrile)
This is prepared by the polymerization of propenonitrile in the presence of
organic peroxides. Propenonitrile is derived from ethane by sybstituting
one of the hydrogen atom with a – CN group.
nCH2 = CHCN → . . . CH2 – CHCN – CH2 n– CHCN . . .
Propenonitrile
Polypropenonitrile is used in making textile and wovlly garments.
→
Condensation polymerization
This is the process where by two or more monomers link together to
form the polymer with the elimination of a small particle (like watter, CO2
et.c).
If two condensing monomers are different, they produce a copolymer but if
they are same a homopolymer is formed.
nA
nA
nB
. . . A – A – A – A – A – A . . . . + X (homopolymer)
Where X is a by product
Like H O, HCl or NH
. . . A – B – A – B – A – B . . . . + X (copolymer))
There are two main types of condensation polymer which are polyester
(e.g terylene) and polyamide (e.g nylon).
Terylene
This is made by heating benzene – 1, 4 – carboxylic acid with ethane – 1,2
– diol in the presence of an acid (as a catalyst). Terylene is a soft,
synthetic fibre capable of raining an almost permanent crease.
nHOOC
COOH + nHO(CH2)2 OH
H+
nH2O
. . . OC
COO (CH2)2O – OC
COO (CH2)2O – CO
...
Terylene
Terylene is used in the manufacture of synthetic textiles and for making
sails of boat.
Nylon
Various nylon can be made by using monomers. The most common
neing Nylon 6.6 is prepared by heating an aqeous solution of hexane – 1,6
– diamine with hexane – 1,6 – dioic acid in tetrachloromethane.
Nylon is used for making ropes, fishing lines, net, nylon has a tensile
strength as such it is used as a synthetic fibre and also to replace steel
part in machines like gear and bearing.
nH2N(CH2)6 NH2 + nHOOC (CH2)4 COOH
nH2O
. . . OC (CH2) CO – NH (CH2)6 NH – CO (CH2)4 CO – NH (CH2)6 NH . . .
Nylon 6.6.
20.3 Rubber
This a very important polymer and can be natural or synthetic.
→
Natural rubber
This is a polymer of isoprene (2 – methylbut – 1,3 – diene) which are
obtained from rubber tree as a white, sticky, liquid latext that comes out
when the bark of the tree is cut.
If the lated is heated, it changes into an elastic solid called rubber. The
quality of this natural rubber is improved by adding sulphur then heated in
order to produce a disulphide bond which will form a strong cross –
linkage to hold long rubber chains together. This process is known as
vulcanization.
Vulcanized rubber have a high tensile strength, long durability and
elasticity over a wide temperature range.
CH3
CH3
CH3
nCH2 = C – CH = CH3 → . . . (CH2 – C = CH – CH2)n CH 2 – C = CH – CH2
Natural rubber
n....M–M–M–M....+ns
polymeric chain of rubber sulphur
....A–A–A–A....
S
S
....A–A–A–A....
Where A = (CH2(CH3)C = CHCH2)n
S
S – cross link
....A–A–A–A....
→
Synthetic Rubber
A number of synthetic rubber have been made from busta – 1, 3 –
diene and (CH2 = CH – CH = CH2) or 2,3 – dimethylbuta – 1,3 – diene (CH2
= CH3 – CCH3 = CH3) alone or in combination. The first synthetic rubber
was neoprene (poly 2 – chlorobuta -, 3 – diene), later styrenebutadiene
rubber (SBR), Thiokol, poly (buta – 1,3 – diene) and poly (2 – methyl
propene) were invented.
SBR is obtained by the co – polymerization of styrene with buta – 1,3 –
diene, it is vulcanized by heating it with 3% by mass of sulphur. SBR is
used in making vehicle tyres and foot wear because of its high resistance
to abrasion.
CHAPTER 21
ISOMERISM
This is define as the existence of two or more compounds with the same
molecular formula but different molecular structure. Isomerism is an
important phenomenon in organic chemistry.
Isomerism can be divided into structural and stereo isomerism.
21.1 Structural Isomerism
This is divided into three groups
→
Chain Isomerism
This is the existence of two or more compounds with the same
molecular formular but different arrangement of atom(s) or carbon
skeleton. This is common to the alkane.
H H H H H
H H H H
H–C–C–C–C–C–H
H H H H H
2-
C5H12
H
H
H–C–C– C–C–H
H
H H
CH3
H
Example I
H
H
H
H
H
H CH3 H CH3 H
H–C–C–C–C–C–C–C–C–H H–C–C– C–C–C–H
H
H
H
H
H
H CH3 H
C8 H18
H
H
2,2,4 – trimethyl pentane
Example II
→
Functional group Isomerism
This is the existence of two or more compounds with the same
molecular formular but different functional group.
E.g.
H OH
H–C–C–H
H
–C–O–C–
H
dimethyleth
ethanol
C 2H 6O
→
Position Isomerism
This is the existence of two or more compounds with the same
molecular formula but different position of functional group.
H OH H
H H H
E.g.
H–C–C–C–H
H
H
H
Propan – 2 –
HO – C – C – C – H
H H
H
Propona
H
H H
H
H–C=C–C–C–H
H H
–C–C=C–C–
H
But – 2 - ene
But 21.2 Stereo Isomerism
steroIsomer are compounds having the same sequence of convalentlly
bonded atom but have different spatial orientation of these atoms in
space. There are three of stereo isomerism and these are
→
Geometric Isomerism
This is the isomerism of the group the double bond. When the
groupes are on the same side (up or down), they are cis but when opposite
each other, they are trans.
R1
R
C=C
H
H
ci
R1
H
H
C= C
trans
R
The cis and compounds are similar in chemical properties but slightly
differs in physical properties because they have some functional group but
different arrangement of carbon atom(s).
→
Optical Isomerism
Compound that can rotate on a plane polarized light is said to be
optically active. If the compound rotate to the right, it is called dextro
rotatary and if it rotates to the left it is called laevo rotatary.
A chiral compound (i.e a compound in which the centre carbon
(chiral carbon) is attached to different group) is optically active and their
mirror images are non – super imposable, hence the two isomers are
.
CH2OH
CH2OH
called enantiomers.
.
.
.
.
.
H 3C –
C2CH5– H
Cl
C
H
HO
H
C
C
C2HH5 – C – H3C
Cl
COOH
.
.
.
.
.
.
COOH
H
.
H
OH
HO
C
C
C
H
OH
H
Stereo isomers that contain chiral centre and one part of the molecular is a
mirror image of the other part and super imposable one another is known
as mesocompound.
E.g.
CH3
H
C
Cl
....................
.. H
Cl
mirror
C
CH3
Hence mesocompounds are not optically active.
CHAPTER 22
TESTS
22.1 Tests for saturation
→
Bromine (in methylene chloride) test
Br
C = C + Br2 → – C – C –
Br
Br
Br
Br Br
Br Br
C ≡ C + Br2 → C = C + Br2 → – C – C –
Procedure
1 drop of the unknown is added to 0.5ml of methylene chloride. Add a
fresh dilute solution of bromine in methylene chloxde dropwise with
shaking until bromine colour persist. If the bromine colour is discharged
without the evolution of hydrogenbromide test, then the unknown is an
unsaturated compound.
The solution should be placed in an appropriate waste container.
Complications
i.
This test should be employed in conjunction with Baeyer test (dil.
KMn04).
ii.
Electron – with drawing groups in the vinylic position can slow down
bromine addition to the point that a negative test is emoneously
produced.
iii.
tertiary amines (like pyridine) form perbromides upon treatment with
bromine and lead to false positive test.
iv.
Aliphatic and aromatic amines can discharge bromine colour without
the evolution of HBr.
→
Baeyer Test
OH OH
C = C + KMnO4
O 2H
–C–C–
OH OH
O 2H
OH OH
OH OH
C ≡ C + KMnO4
–C–C–
Procedure
Dissolve a drop or 0.02g of the unknown in 0.5ml reagent grade acetone.
Add 1% aqeous solution of potassium permanganate (KMn04) drop wise
with shaking. If more than one drop of reagent is required to give a purple
colour to solution, unsaturated or an easily oxidized functional group is
present.
The disappearance of the KMn04’s purple colour and the appearance
of a brown suspension of MnO2 is a positive test.
After the experiment, the solution should be diluted with water and
flush down the drain.
Complications
i.
water insoluble compounds should be dissolved in ethanol, methanol
or acetone.
ii.
Sometimes the brown precipitate fails to form and the solution turns
reddish – brown.
iii.
Easily oxidized compounds give a positive test. Most aldehydes and
ketones, formic acid and it ester give a positive test.
iv.
Alcohols with trace of impurities give a positive test.
v.
Phensls and aryl amines give a positive test.
vi.
Carboxyl compounds which decolorize bromine/methylene chloxide
usually give a negative test.
→
Iginition test
Procedure
Heat a small sample of on a spatula, first, hold the sample near the side of
a burnsea burner to see if it melts normally and then burns. Heat it in the
flame. Aromatic compounds often burns with a high smoky flame.
A sooty yellow flame is an indication of an aromatic ring or other centres of
unsaturation.
22.2 Tests for Alcohols
→
Lucas tests
ROH + HCl
ZnCl2
RCl + H2O
This test is based on the reactive of the different types of alcohol to form
corresponding carbocations easily. An equimolar mixture of ZnCl2 and HCl
is the reagent. The alcohol is protonated by this mixture and water group
attached to the carbon is replaced by Cl.
Procedure
Add 2ml of the lucas reagent to 0.2ml or 0.2g of unknown at room
temperature. Cover the tube with a stopper and shake vigorously, then
allow to stand note the time required for the formation of alkylchloxiden
which appears as in soluble layer or emulsion in
a.
Immediate to 2 – 3 mintes : 30 alcohol
b.
5 – 10 minutes: 20 alcohol
c.
No reaction: 10alcohol
Complication
The test applies only to alcohols soluble in the reagent (monofunctional
alcohol lower than hexyl and some polyfunctional alcohols)
→
Jones oxidation
OH
O
3R – C – H + HCrO4-+ 8H+ 4H2O → 4 [Cr (H2O)6]3+ + 3 (R – C – OH)
H
OH
O
R – C – R + HCrO4 + 8H+ 4H2O → 2 [Cr (H2O)6] + 3 (R – C – R1)
1
-
3+
H
Procedure
Dissolve 10mg or 2 drops of the unknown in 1ml of pure acetone in a
test tube and add to the solution 1o small drop of jones reagent
(chronicacid and sulfuric acid). A positive test is marked by the formation
of a green colour within 15 seconds upon addition of the orange – yellow
reagent to a primary or secondary alcohol.
Complications
i.
Enols may give a positive test.
ii.
phenols give a dark coloured solution which is not blue – green like a
positive test.
22.3 Test for carbonyl compounds
→
Tallen’s Test
O
R – C H + 2Ag (NH3) 2 OH → RCO2NH4 + Ag + 2NH3 + H2O
Preparation of Tollen’s Reagent
Into a test tube which has been rinsed with 3M NaOH, place 2ml of
0.2M silver nitrate (AgNO3)
, and add a drop of 3M NaOH. Add 3ml of dilute ammonia until all the
brown precipitate of silver oxide dissolves.
Procedure
Add a drop or a few crystals unknown to 1ml of a freshly prepared
Tolen’s reagent. Gentle heating can be employees if no reaction is
immediately observed. Formation of a silver mirror or a black precipitate is
a positive test and as such the unknown is Alkanal.
Complications
i.
The test tube must be clean and oil – free if silver mirror is to be
observed.
ii.
Easily oxidized compounds give a positive test. For example
aromatic amine and some phenols.
→
Jones oxidation
R – CHO
HCr04
H+
O
RC – OH
Complications
Aldexydes are better characterized in other ways. The colour usually
develops in 5 – 15 secs.
→
Brandy’s test
HNNH2
NO2
O
R – C – H/R1 +
O 2N
2,4 -
N
N = C – R1(H)
+ H 2O
H
2,4 –
dinitrophenylhydrazine
Procedure
Add a solution of 1 or 2 drops or 30mg of unknown in 2ml of ethanol
to 3ml of 2,4 – dinitrophenylhydrazine reagent. Shake vigorously and if no
precipitate is formed immediately, allow to stand for 15 minutes.
Formation of a precipitate is a positive test.
Complications
i.
Some ketones give oils which will not solidify
ii.
Some alcohols is not purified may contain aldehyde or ketone
impuries.
→
Fehling’s solution test (Benedict’s solution test)
O
2+
R – C – H + 2CO NaOH + H2O → R – C –O –Na+ + Cu2O + H+
Blue
Red
Procedure
Add a few drop of fehling’s solution to an unknown sample, then
heat. A positive test is shown when a brick – red ppt is formed (the
unknown sample is alkanal).
22.4 Test for carboxylic acid
O
O
R – C – OH + Na2CO2 → R – C – O – Na + + H2O + CO2
Procedure
A few drop or crystals of an unknown sample are dissolve in 1ml of
methanol and slowly added to 1ml of a saturated solution of
sodiumbicarbonate. Carbon(iv) oxide gas is evolved if the unknown is
carboxylic acid.
22.5 Tests for phenols and Nitro groups
→
Iron (III) chloride test
Pyridin
3ArOH + fecl3
Fe (OAr)3
e
Coloured complex
Procedure
Dissolve 15mg of the unknown compound in 0.5ml of water or water
– alcohol mixture and add 1 – 2 drops of 1% aqeous iron (III) chloride
solution A red, blue, green, or purple colour is a positive test.
Complications
The iron III chloxide test is not completely reliable for acidic phenols.
→
Iron (II) hydroxide test
RNO2 + 6 Fe (OH)2 + 4H2O → RNH2 + 6 Fe (OH)3
Procedure
Add about 10mg of the compound to 1ml of the ferrous ammonium
sulphate reagent in a test tube and then 0.7ml of the 2N alcoholic
potassium hydroxide reagent. Cover the tube with a stopper and shake
note the colour of the precipitate after 1 minute formation of a red – brown
precipitate of iron (III) hyoloxide is a positive test.
Complications
The red – brown precipitate of Fe(OH)3 is formed by the oxidation of
fe(OH)2 by the nitro compound which in turn is reduced to the primary
amine. A negative test is indicated by a greenish precipitate. In some
cases partial oxidation may cause a darking of the ferrous hydroxide
(Fe(OH)2).
Practically, all nitro compounds give a positive test in 30 seconds.
The speed with which the nitro compound is reduced depends on its
solubility.
22.6 Testfor Amines
Amines can be distinguished using Hinsberg Test
NaOH
R – NH2 +
SO2Cl
SO2NR- Na + + NaCl + 2H2O
Soluble
H+
SO2NRH
insolubl
R2 – NH +
SO2Cl
NaOH
SO2NR2 + NaCl + 2H2O
Insoluble
H+
R 3N +
SO2Cl
NaOH
no visible
SO2- Na+ + NR3 + NaCl + H2O
Insoluble
H+
R3NH+Cl
Procedure
soluble
Add 5ml of 10% NaOH solution and 0.4ml of
benzenesulphonylchloride to 0.3ml or 300mg of unknown in a test tube;
close the test tube with a stopper and shake the mixture vigorously. After
all the benzene sulphonyl chloxide has reacted, cool the solution and
separate the residue, if present from the solution. Test the residue for
solubility in 10% Hcl solution. If no resolve remains, then treat the solution
with 10% Hcl solution and observe whether a precipitate forms.
1o amine dissolves in base and precipitates from acid is a positive
test; 2o amine forms precipitate from base and no change from acid is a
positive test, 3o amine form precipitates from base and dissolves in acid is
a positive test.
Complications
i.
Amphoteric compounds give erroneous results.
ii.
Some sodium salts of benzene sulphoamides of primary amines are
insoluble in the Hinsberg solution and may appear to be secondary
amines
iii.
Some tertiary amine in hydrochloride salts are insoluble are insoluble
in dilute Hcl and water and may also appear to be secondary amines.
22.7 Tests of carbohydrates
→
The Molisch Test
O
H
C
O
H
C
OH
H2SO4
H
C
OH
–3H2
H
Procedure
C
OH
H
C
H
O
C
C
C
C
H
H
CH2OH
2ml of a sample solution is placed in a test tube. Two drops of the molisch
reagent (a solution of a – napthol in ethanol) is added. The solution is then
poured
slowly into a tube containing 2ml of concentrated sulphuric acid so that
two layers
are formed. A purple product at the interface of the two layers is a positive
test.
Monosaccharides give a rapid positive test, disaccharides and
polysaccharides react slower.
→
Benediets’s Test (Test for reducing sugar)
H
O
O
OH
2+
C + 2Cu + 5OH →
C
R
R
+ Cu2O + 3H2O
Procedure
1ml of a sample solution is placed in a test tube. 2ml of benedict’s
reagent (a solution of sodium citrate and sodium carbonate mixed with a
solution of copper sulphate) is added. The solution is then heated in a
boiling water both for three minutes.
The formation of a reddish precipitate within three minutes is a
positive test for reducing sugars.
→
Iodine / potassium Iodide Test (Test for starch)
Procedure
2ml of a sample solution is placed in a test tube. Two drops of iodine
/ potassium iodide solution and 1ml of water are added.
The formation of a blue – black complex is a positive test for starch
(polysaccharides may give other colours including blue or red).
→
Barfoed’s test (test for reducing monosaccharides)
H
O
O
HO
C – 2 Cu2+ + 2H2O → C + Cu2O + 4H+
R
R
Procedure
1ml of a sample solution is placed in a test tube. 3ml of Barfoed’s
reagent (a solution of cupric acetate and acetic acid) is added. The
solution is then heated in a boiling water both for three minutes. The
formation of a reddish precipitate within three minutes is a positive test for
reducing monosaccharides.
Reducing disaccharides also give a positive test (as they undergo
same reaction) but at a slower rate.
→
Seliwanoff’s test (test for ketoses)
H
C
O
C
C
C
C
COOH
Procedure
½ ml of a sample solution in attest 2ml of seliwanoff’s reagent (a solution
of resoreinol and Hcl) is added. The solution is then heated in water bath
for 2 minutes. The formation of a red product shows ketoses is present.
→ Bial’s test (test for pentoses)
H
C
O
H
C
OH
H
C
OH
H
C
OH
O
H
C
H
O
C
C
C
C
H
H
CH2OH
Procedure
2ml of a sample solution is placed in a test tube. 2ml of Bial’s reagent (a
solution of orcinol, Hcl and ferric chloride) is added. The solution is then
heated gently in a hot water bath. If the colour is not obvious, more water
can be added to the tube. The formation of a bluish product gives a
positive test for pentoses.
Hexoses generally react to form green, red or brown product.
REVISED QUESTIONS & ANSWER
1.
Which of the following compounds determines the octane rating of
petrol?
a.
1,2,3 – trimethylpentane
trimethylpentane
c.
2.
Alkane
b.
2,2,4 – trimethylpentane
Alkanol
c.
Alkanal
d.
Alkyne
The following substances are examples of addition polymer except
a.
Nylon
polychloroethane
4.
d.
2,3,5 –
Which of the following compounds would react with ethanoic acid to
give a sweet smelling liquid?
a.
3.
2,3,5 – trimethyloctane
b.
b. perspex
c. polyethane
d.
When bromine is added to ethane at room temperature, the
compound formed is
a.
1,1 – dibromoethane
b.
1,1 – dibromoethene
c.
5.
1,2 – dibromethane
a.
H
H
H–C–C
c.
d. CH3
H
C=O
O–C–H
C=O
C H2CH3
H
CH3
b.
Ethanol b.
ethanoic acid
methanoic acid
c.
methanol d.
the regent that can be used to ethane from ethyne is
ammoniacal silvertrioxonitrate (V) solution
b. benedict
c.
bromine water d.
fehling’s solution
Consider the following reaction equation: C16 H34 → C5H12 + C11H22 the
process represented by the equation is
a.
9.
O
The compound that makes palm wine sour after exposure to air for
few days is
a.
solution
8.
H
O
H–C–C
OH
H
a.
7.
1,2 – dibromoethene
Which of the following compounds would react with sodium
trioxocarbonate (IV) to librate carbon (IV) oxide?
H
6.
d.
cracking
b. fermentation c. polymerization d. reforming
Consider the following structures of organic compounds
CH3
CH3
C–C
H
and
H
CH3
H
C–C
CH3
H
When of the following statements about the structure is not correct?
a.
are geometric isomers b.
are saturated hydrocarbons
c.
have similar physical properties
same homologous series
d. are members of the
10.
Which of the following substances would not produce ethanol when
fermented?
a. cane sugar
b. glucose c. starch
d.
vinegar
11.
An alkanol can be prepared by the reaction of an alkene with
a.
concentrated tetraoxosulphate
tetrachloroethene
b.
bromine in
c.
aqueous potassium tetraoxomanganate (VII)
hydroxide solution
12.
which of the industrial process depends on the action of enzymes?
a.
of beer
liquefaction of air
d.
catalytic cracking
13.
14.
b. manufacture of soap
c. brewing
A tertiary alkanol has a molecular formular C4H10O. what is the
structural formula of the compound.
a.
(CH3) CHCH2OH b.
d.
CH3 CH2 CH2 CH2OH
CH3 CH2CH (OH) CH3
c.
(CH3)3 COH
A compound contains 7.75% hydrogrn, 37.21% carbon and 55.04%
chlorine. Determine the empirical formular of the compound. (H =
1.00; C = 12.0; Cl = 35.5)
a.
15.
d. sodium
C3H3Cl
b.
C2H5Cl
c.
C3H8Cl
d.
Catalytic hydrogenation of oils result in the production of
C 5H 2I
a.
16.
17.
soaps
b.
detergents c.
alkanes
d.
magarine
What is the molecular formula of a compound whose empirical
formula is CH2O and molar mass is 180? (H = 1, C = 12, 0 = 16).
a.
C 4H 8O 2
e.
C12H22OH
b.
C 4H 8O 3
c.
C6H10O5
d.
C6H12O6
The compound with the structure given below
H H
H
H
R–C–C–C–C–H
H H
H
H
Where R is an alkyl group is classified as an
a.
alkanoic acid
alkyhalide
b.
unsaturated compound
d.
alkane
e.
aromatic compound
c.
18.
Petrol is obtained from diesel by a.
c.
catalysis d.
polymerization
19.
The compound with the structural formula below is a product of
complete oxidation of
H H
H
distillation b.
cracking
e.
dehydrogenation
O
H–C–C–C–C
OH
H H
a.
butan
––
2H
– 01 b. 2 – methybutan – 1 – 01 c. propan – 2
H–C
– 01
H
d.
2 – methylpropan 2- 01
e.
3 – methybutan – 2 – 01
20.
21.
The product of the reaction ethanol and acidified K2cr2O7 is
a.
ethanal
b.
ethyl ethanoate c.
d.
ethyne
e.
ethanedioic acid
Examples of polmers include the following except
a.
22.
23.
starch
glass
b. nylon
c.
wool d.
perpex
e.
An alkene may be converted to an alkane by
a.
halogenations
b.
hydrolysis c.
d.
hydrogenation
e.
decomposition
dehydration
The following decolourizes bromine water except
a.
24.
ethanoic acid
C 2H 6
b. C2H4
c. C2H2
d. C3H4
e.
C 3H 6
Which of the following statement is not correct about the compound
represented below?
COOH
COOH
25.
a.
its basicity is 4
b.
it is a di—carboxylic acid
c.
its boiling point is higher than that of corresponding alkane
d.
it reacts with alkanols under suitable conditions
e.
it produces effervescence with saturated NaHCO3 solution
Alkanes are used mainly
a.
in the production of plastics b.
fuels
as domestic and industrial
26.
c.
in the textile industry
e.
as fine chemical
d.
29.
drugs
b. margarine c.
CH3 CH2 CH3
d.
CH3 CH2CHO
paraffin
d.
product
soapy detergents
b.
CH3CO CH3
c. CH3CH(OH) CH3
What is the product of the reaction between ethanol and excess
acidified KMnO4 solution?
CH2 = CH2
b. CH3COOH
c. CH3 - CH3
d. CH3O CH3
Hydrocarbons which react with ammoniacal copper (I) chloride
solution confirm to the general formula
CnHn b.
CnH2n
c. CnH2n+2
d.
CnH2n-2
Alkanes undergo the following reactions except
a.
32.
copper (I)
copper (II) hydroxide
a.
a.
31.
copper (II) oxide c.
Which of the following compounds readily reacts with sodium to
librate hydrogen?
a.
30.
copper (I) oxide b.
H2, Ni catalyst
Consider the reaction: vegetable oil High
the reaction is applied in the manufacture
of
temperature
a.
28.
in the hydrogenation of oils
Glucose reduces Fehling’s solution on warming to
a.
chloride
27.
d.
addition
b.
substitution
hydration c.
polymerization d.
Which of the molecular below are geometric isomers?
CH3
CH3
CH3
a.
CH3 – CH – CH2 – CH3 and CH3 – CH2 – CH – CH3
b.
CH3 – CH2 – CH – CH3 and CH3 – C – CH3
c.
CH3 – CH2 – CH2 – CH3 and CH3 – CH2 – CH – CH3
Br
d.
C=C
Benzene
C6H8 b.
CH2 b.
H
H
b.
ethanol
c.
turpentine d.
water
C6H10
c.
C6H12 d.
C6H14
CH3 c.
CH4 d.
C 2H 4
A hydrocarbon containing 88.9% carbon has the empirical formula
[H=1, C=12]
a.
37.
C=C
What is the empirical formula of a hydrocarbon containing 0.08
moles of carbon and 0.32 moles of hydrogen?
a.
36.
CH3
The complete hydrogenation of C6H6 in the presence of Nickel
catalyst at 200oc gives
a.
35.
Br
Which of the following substances is a suitable solvent for
perfumes?
a.
34.
and
Br
H
33.
Br
H
CH
b. CH2
c.
C 2H 3
d. C2H5
What is the IUPAC name of the compound with the follow structure?
H
H
H
H–C–
H–C–C=C–H
38.
a.
2-methybutene
b.
c.
2-methyprop-1-ene
2-methylprop-2-ene
d.
but-1-ene
Consider
H organic compound x with the following structure
H H the
H–C–C–C–H
H OH H
The complete oxidation of x gives
a.
39.
40.
b. propanoic acid
c. propanone d.
The hydrolysis of groundnut oil by potassium hydroxide is known as
a.
hydrogenation
d.
neutralization
b.
saponification
c.
esterification
When ethanol is heated with excess concentrated tefraoxosulphate
(VI) acid, the organic product formed is
a.
41.
proponal
propene
ethanal
b.
ethanoic acid
c.
ethane
d.ethane
Arrange the following in order of increasing boiling point
I.
CH3CH2 CH2 OH
II. CH3CH2 OH
IV.
CH3CH2 CH2 CH2 CH3
III. . CH3CH2 CH2 CH2OH
a.
42.
I<II<IV<III
b.
IV<III<II<I
c.
I<II<III<IV d. IV<II<II<III
Which of the following compounds would not give a precipitate with
ammoniacal AgNO3 solution
a.
CH
CH3C≡ CCH3
b. HC ≡ CH
c. CH3C≡ CH
d. CH3 CH2C≡
Use the structure of the compound below to answer questions 43 – 44.
H
H–C–H
H
H H
H – C – C – C – C – OH
H
43.
H
The name of the compound is
a.
c.
44.
H
H–C –
2,3-dimethybutan-1-0l
d.
3-methypentan-1-0l
The product of the complete oxidation of the compound will be an
c. alkanoic acid d. alkanone
Which of the following is an alkanoates
a.
46.
2,3-dimethybutan-4-0l
2-methypentan-1-0l
a. alkane b. alkanal
45.
b.
CH3COOH
b. CH3COO CH3 c. CH3 CH3OH d. CH3 CH2COOH
O
The compound H2N – CH2 – C – OH
a.
has no functional group b. is monofunctional
difunctional
d.
is trifunctional
c. is
47.
Both addition and substitution reactions can be undergone by
a.
48.
ethane
hexane
d.
propane
b.
substitution
c.
elimination d.
aliphatic b.
unsatutated
aromatic
c.
saturated d.
CnHb + 502 → 2CO2 + 4H2O. The hydrocarbon CaHb in the reaction is
a.
51.
c.
A hydrocarbon which contains the maximum number of hydrogen
atom of every carbon is said to be
a.
50.
benzene
The characteristic reaction of alkanal and alkanone is
a.
addition
nutralization
49.
b.
ethane
b.
alkane
c.
alkene
d.
aromatic
The product of the reaction between excess ethanol and conc. H2SO4
is
a.
ethane
b.
ethanoic acid
c.
ethylhydrogentetraoxosulphate VI d.
ethoxyethane
52.
The role played by coc.H2SO4 in the esterification of alkanoic acid is
53.
A drunkard breathe carried a significant level of alcohol. The colour
change observed when this breathe is passed into acidified K2Cr2o4
solution is
a.
orange to pink
d.
orange to blue
b.
orange to purple c.
54.
The roles of NaCl in the manufacture of soap are
55.
starch and cellulose are
orange to green
a.
56.
57.
sugar
b. polymer
c.
isomer
d.
hydrocarbon
If one mole of hydrocarbon containing 4g of hydrogen and its molar
mass is 40g. what is the homologous series of the compound? (C =
12, H = 1)
a.
Alkene
above
H O H
b.
Alkyne
c.
Alkane
d. none of the
H –HC – C –HC – H
The compound above is an
a.
58.
alkanal
c.
alkanol
d.
alkanone
A primary amide is generally represented by the formula
a.
59.
alkanoate b.
RCONH2
b.
RCONHR
c.
RCONR2
d.
RCOOR
The dehydration of ammonium of alkanoic acids produces a
compound with a general formular
O
O
a.
R – C – NH2
b.
R–NH2
c. R – C
R
R-C
O
d.
60.
Which of the
O following compounds in solution will turnOred litmes
paper blue?
R
a.
R – C –RN – R b.
RNH2
c.
R–C
d.
RIORII
61.
An organic compound contains 60% carbon, 13.3% hydrogen and
26.7% oxygen. Calculate the empirical formula
OR
a.
C 3H 8O
b.
C6H13O
e.
[C=12, H=1,O=15]
c.
C 4H 9O
d.
C5H12O
H
62.
CH3 – C – CH2 – CH = CH2
CH3
The IUPAC nomenclature for the compound above is
63.
64.
65.
a.
3 – methylpent – 2 – ene
ene
b.
2 – methypent – 1 –
c. 2 – methylpent – 4 – ene
ene
d.
4 – methylapent – 1 –
The compound that will react with sodium hydroxide to form salt and
water only is
a.
(CH3)3COH
d.
C6H12O6
b. CH3CH = CH2
c.
CH3 CH2COOH
Which of the following fractions is used as raw material for cracking
process?
a.
lubricating oil
d.
kerosene
b.
bitumen
c.
diesel oils
An organic compound with a pleasant smell is likely to have a
general formula
a.
CnH2n+1COOH
b.
CnH2n+1COO CnH2n+1
c.
CnH2n+1CO CnH2n+1
d.
CnH2n+1CHO
66.
The reaction between and ethylethanoate produces
a.
propanol and papanamide
b.
c.
ethanol and ethanomide
ethanamide
ethanol and propanamide
d.
propanol and
67. The decarboxylation of ethanoic acid will produce carbon (IV) oxide
and
a.
68.
ethane
b.
propane
c.
butane
d.
methane
2 – methylbutan – 2- 01 is an example of a
a.
alkanol
d.
primary alkanol b.
dihydric alkanol
secondary alkanol
c. tertiary
ESSAY
1a.
An organic compound x contains 40% carbon, 6.67% hydrogens, the
reat being oxygen. If x has a relative molecular mass of 60 determine
i.
empirical formula
ii.
Molecular formula
[H = 1, C = 12, O = 16]
b.
An alkanoic acid y has a relative molecular mass of 74.
i.
State the functional group of y.
ii.
What type of reaction is involved when y is converted an alkanoate.
iii.
Determine the structural formula of y.
iv.
Write an equation for the reaction & between y and sodium.
v.
If x in (a) above boils of 118oc and belongs to the some homologous
series as y, state with reason whether the boiling point of y will be
equal to, higher or lower than 118oc.
c i.
What is fermentation?
ii.
Write an equation for the fermentation of glucose.
iii.
What most be added to glucose solution to make it ferment?
iv.
Explain why a tightly corked glass bottle filled with to the brim with
fresh palmwine shatters on standing.
(WASSCE MAY/JUNE)
2003
2a.
Reaction of 2,2 – dimethyl – 1 – butane with HBr leads to an alkyl
bromide C6H13Br. On treatment of this with KOH in ethanol,
elimination of HBr to give an alkene occurs and a hydrocarbon that is
isomeric with the starting alkene is formed. Using chemical equation,
show all the steps involved in the transformation.
b.
product the major product of the following reaction.
i.
+ HBr
A
ii.
+ BrH2O2
B
c.
Show how the geometrical isomers of hen – 3 – ene could be
synthesized from a named alkyne
(BUK (2010)
3(a) Write the structure of the product for the observation in each of the
following reactions.
i.
A mixture of butanoic acid and ethanol warned in the presence of
concentrated H2SO4 gives off a fragrant odour.
ii.
Sodium dissolves in propon – 2 – 01 with effervescence to give a
solution which on evaporation to degree leaves white precipitate.
b.
consider the compound CH 3 CH 2COO CH 2 CH 3
i.
Name the compound
ii.
Write the structural formula of the compound.
iii.
State the reagents and conditions for the function of the compound
(WASSCE MAY/JUNE 2006)
4(a) What term is used to describe each of the following process?
i.
Alkaline hydrolysis of fats and oils.
ii.
The conversation of glucose into ethanol by enzymatic action.
iii.
Thermal decomposition of higher petroleum fraction into lower
molecular mass hydrocarbons in the presence of catalyst.
b i.
Write structure and IUPAC name for one alkanoic acid with the
molecular formula C4H8O2.
ii.
Arrange the following compounds in order of increasing boiling point.
Butane; Butanoic acid; methylpropane.
iii.
Give an explanation for your answer in 3 (b) (ii).
c i.
Ethanol was used for preparing a gas x which decolourized bromine
water. Identify x and describe briefly its laboratory preparation.
ii.
Write an equation to show how ethanol reacts with sodium.
iii.
Give the reagent and reaction conditions for the conversion of
ethanol into C2H5COOC2H5.
d.
State the type of reaction in each of the conversions below.
i.
C6H6 → C6H5CH3
ii.
nC2H4 → (CH2 – CH2)n
iii.
CH 3 CH 2CH(OH) CH 3 → CH 3 CH 2 C CH 3
iv.
(C6H10O5)n → C6H12O6
v.
CH 2
CH 2
H 2c
CH 2 →
H 2c
CH 2
H 2c
CH
H 2c
CH 2
(WASSCE MAY/JUNE 2002)
5.
2.32g of a hydrocarbon on ignition gave 7.3g of carbon (IV) and
2.98g of water. If the vapour density of the hydrocarbon is 28.
Calculate.
i.
The empirical formula
ii.
The molecular formula
iii.
The structural formula
iv.
Draw the structure of possible isomers and name them.
6.
A hydrocarbon y which decolorized bromine water and forms a silver
salt with amoniacal silver trioxonitrate solution y was found to have a
molar mass of 54gmol-1.
i.
deduce the molecular formula of y.
ii.
draw the structural formula of y.
iii.
draw the structures of possible isomers and name them.
7.
Draw the structure and name the product of the complete oxidation
of the following alkanols.
H
H CH3
[o]
H
H CH3
KMnO4/H+
?
i.
H–C–C–C–O–H
H
H CH3 H
Cro3
ii.
H–C–C–C–C–H
H
O CH3 H
[o]
?
H
Cl H H H
iii.
[o]
Cl – C – C – C – C – Cl K2Cr2O7/H+
Cl Cl O H
?
H
8.
A liquid C6H12O2 undergone hydrolysis to an acid Q and an alkanol z.
oxidation of z concerts it to Q. suggest a structure of the original
compounds and write equations for all reactions mentiones.
9.
Give equations to illustrate the method used to synthesize each of
the following alkanoic acid.
i.
Butanoic acid from butan – 1 – 01.
ii.
P – chlorobenzoic acid from p – chlorotoluene.
iii.
pentane - > 5 – dioic acid from cyclopentene.
iv.
2 – methylbutanoic and from 2 – methylbutanol.
10.
Using mechanisms and equation only, describe structurally the
synthesis of ethoxyethane.
11.
i.
Complete each of the following equations, giving the structure and
IUPAC of the main products.
COOH
H+
PCl5
?
?
+ HOCH2CH2OH
v.
CO2H
CH3
ii.
iii.
LiAlH4
C2H5COOH
?
vi.
?
CH3
Br2 + P
CH3CH2CO2H
[O]
?
vii.
?
CH3CH = CHCO2H
?
LiAlH4
iv.
CH3(CH2)2CO2H
12.
Complete each of the following equations, giving the structures and
names (IUPAC) of the organic products.
O
H
2, 200 C
Na2Cr2O7/H+
?
CH3(CH2)2CHO
iv.
CH2(CH2) CHO
Ni
i.
viii.
KMnO4/H+
?
?
CH2CH3
ii.
CH3(CH2)2CHO + 2C2H5OH
H+ or OH
-
? v.
CH3CHCH2CHO
HCN
?
HCl(n)
CH2CH2CH3
H+ / OH-
?
iii.
HCHO + C2H5MgCl
13.
Complete each of the following equations giving the structures and
IUPAC of the main product.
i.
O
O
Ni,H2, 200OC
LiAlH4
?
+
iv.
CH2CH2COCH2CH3
Mg Br
HCN
?
H+
?
NaBH4
?
?
ii.
iii.
(CH3)2CH (CH2)2COCH3
14.(a)
Write the structural formula of (i) 2,2,4 – trimethylpentane
(ii) ethylmethanoate (iii) tran, - 2, 3 – dimethylbut – 2 – ene
b.
Write the structure of the straight chain compound that is isomeric
with 2,2,4 – trimethylpentane.
c.
Write chemical equations to illustrate the oxidation of (i) secondary
alkanol (ii) adihydric alcohol
d.
When crushed cassava was warmed with dilute hydrochloric acid, a
sweet tasting compound D was obtained. When the mixture distilled,
a clear and colourless liquid E was obtained. When liquid E was
warmed with athanoic acid in the presence of a few drops of
concentrate tetraoxosulphate (VI) acid, a component F with fruity
smell was obtained.
i.
What class of compounds does D belong?
ii.
Name E and F
iii.
Write the functional group in F
iv.
Write the equation for the reaction between E and ethanoic acid in
the presence of concentrated tetraoxosulphate VI acid.
v.
Name the type of reaction that takes place between E and ethanoic
acid.
e.
Arrange the following compounds in their correct order of increasing
boiling points.
CH3 CH2 CH2OH, CH3 CH2 CH2 CH2 CH2 CH3 and CH3 CH2 CH2 CH3.
Explain the order.
WASSCE MAY/JUNE 1988
15(a) Write the structural formula and the name of compounds having the
formula C2H2Cl2
WASSCE MAY/JUNE 1988
(b)
Consider the compounds represented as A and B below:
H
A.
Cl
C=C
Cl
relationship
H
H
B.
Cl
C=C
H
Cl
:what is the
between A and B
WASSCE MAY/JUNE 1989
16(a) Write the chemical equation for the formation of a named alkanoate.
(b)(i) What are the monomers of protein called?
(ii)
Write the two functional groups present in the monomers named in
b(i) above.
(iii)
State the type of reaction that leads to the formation of proteins from
their monomers.
WASSCE MAY/JUNE 1989
17(a) List 3 characteristics of a homologous series.
(b)
Give one example of (i) alkanes, (ii) alkynes.
(c)
A hydrocarbon contains 7.7% by mass of hydrogen and 92.3% by
mass of carbon. The relative molar mass of the compound is 78.
(i)
Derive the empirical formula of the compound and hence its
molecular formula.
(ii)
Name the hydrocarbon and write its structural formula.
(d)
Two hydrocarbons, X and Y were treated separately with acidified
potassium tetraoxomaganate (VII). Solution X decolorized the
solution and Y did not. Which of X and Y will undergo
(i)
Substitution reaction only
reactions
(iii)
Polymerization?
(c)
If ethanol is converted into ethanoic acid.
(i)
What are the conditions required.
(ii)
Name the type of reaction that will be involved and write the
equation.
(ii) both addition and substitution
WASSCE MAY/JUNE 1989
18.
Three functional group isomers were found to have the same
molecula, weight of 60. Combustion analysis of 0.2g of these
compounds each produce 0.2933g of CO2 and 0.1201g of water.
(i)
Calculate the percentage composition, empirical and molecular
formula of these compounds.
(ii)
Draw all these isomers and name them.
19(a)(i)
(ii)
Define the term addition polymerization.
What type or organic compounds undergo addition polymerization.
(iii)
List two factors which effect the strength of polymers.
(b)
The diagram below shows some reaction pathways involving
ethanol.
X
H+
Catalyst CH3 CH2COOH
A
B
C2H4 ← C2H5OH → CH3COOH
(i)
Write the name and structural formula of the organic product X.
(ii)
State the reagent for the conversion indicated as A
(iii)
What type of reaction does ethanol undergo during the process of
conversion indicated as B.
(c) (i) Write a balanced equation for the complete combustion of ethanol in
oxygen.
(ii)
Write a balanced equation for the complete combustion of ethanol in
oxygen.
(iii)
Calculate the volume of oxygen required at S.T.P for the omplete
combustion of 2,3g of ethanol [H = 1; C = 12; O = 16; molar volume of
gases at S.T.P = 22.4dm3].
(WASSCE, MAY/JUNE) 1990)
20.
Benzene contains six carbon atoms and six hydrogen atoms
(a) (i) Draw two stable structures of benzene to show how these atoms are
arranged.
(ii)
What is the concept behind these structures?
(b)
Give (i) two uses of benzene
(ii) one industrial source of benzene.
(WASSCE, MAY/JUNE) 1990)
ANSWERS
OBJECTIVES
1.
D.
2.
B – Ethanoic acid + alkanol
3.
A – Nulon is a polyamide which is a condensation
4.
C-
esters + water sweet smelling liquid
C = C + Br2 → – C – C –
H
Br Br
O No2CO3
CH4 + CO2↑
5.
A– H–C–C
OH
H
6.
B – As ethanol oxidizes to ethanoic acid.
7.
A
8.
A
9.
B
10.
D – Vinegar is not a carbohydrate
11.
A
12.
C
13.
C – (CH3)3C(OH) → CH3 – C – CH3
OH
CH3
14.
H
C
Cl
7.75
37.21
55.04
1
12
35.5
7.75
3.10
1.55
1.59
1.55
1.55
5
2
1
:.
Empirical formular C2H5Cl.
15.
E
16.
D – (Empirical formular)n = molecular formular
(CH2O)n = 180
(12 + 2 + 16)n = 180
30n = 180
N=6
:.
Molecular formular = (CH2O)6
= C6H12O6
17.
D
18.
B
19.
B – only 1o alc undergoes complete oxidation to alkanoic acid.
20.
C
21.
22.
D – CnH2n – 2 + H2 Ni
23.
A – All unsaturated hydrocarbons decolorizes bromine water.
24.
A
25.
B
26.
A
27.
B
28.
C – CH3CHCH3 + Na → CH3 – C – CH3 + H2↑
29.
B
30.
D
31.
D
32.
D
OH
CnH2n + 2
ONa
33.
+ H2 →
(C6H6+3H2 → C6H12)
34.
C-
35.
C–C
H
0.08
0.32
0.08
0.08
1
=
4
Empirical formula = CH4
36.
% H = (100 – 88.9%)
= 11.1%
C
H
88.9
11.1
12
1
37.
C
38.
C
39.
B
40.
D
41.
D
42.
A – Only terminal alkyne give a precipitate with ammoniacal AgNO3
solution.
43.
A
44.
C
45.
B
46.
C
47.
B
48.
A
49.
C
50.
B – Comparing with
CaHb + (a + b/4) O2 → a CO2 + b/2 H2O
a + b/4 = 5
a=3
b/2 = 4 → b = 8
C3H8 → propane (alkane).
51.
D
52.
Catalyst
53.
C
54.
i. to reduce solubility
ii. to separate soap from glycerol
55.
B
56.
B
Cx H4 = 40
12x = 36
12x + 4 = 40
12
12x = 40 – 4
x=3
12
C3H4 → Alkyne
57.
D
58.
A
59.
B
60.
B
61.
A
62.
D
63.
C
64.
C
65.
B
66.
C
67.
D
68.
C
ESSAY
a.
% of carbon = 40%
% of hydrogen = 6.67%
% of oxygen = 100 – (40 + 6.67)
= 53.33%
i.
C
H
O
40
6.67
53.33
12
1
16
3.33
6.67
3.33
3.33
3.33
3.33
1
2
1
Empirical formular = CH2O
ii.
To determine molecular fomula
n (CH2O) = 60
n (12 + 2 + 16) = 60
30n = 60 → n = 60/30
Molecular formula = 2(CH2O)
= C 2H 4O 2
bi.
COOH
ii.
Esterification
iii.
CnH2n+1COOH = 74
12n + (2n + 1) + 12 + (2 x 16) +1 = 74
12n + 2n + 1 + 12 + 32 + 1 = 74
14n + 46 = 74
14n = 74 – 46
14n = 28
14
14
n=2
:.
The compound is C2H5 COOH (propanoic acid)
H H O
H – C – C – C – OH
H H
iv.
C2H5COOH + Na(s) → C2H5COONa + H2↑
v.
The boiling point of y will be higher than that of x because molecular
mass of the alkyl group in y is greater than that in x.
c i.
ii.
Fermentation is the decomposition of large organic molecules into
smaller molecules.
Zymas
C6H12O6
2C2HSOH(cl) + 2CO2(g)
iii.
zymase
iv.
This is because the palm wine exert much pressure on the bottle.
alc. KOH
+ H Br →
Br
2a.
b. i.
+ H Br →
Br
Br
+ H Br, → H2O2
ii.
H H
H H
c.
Since hex – 3 – ene is H – C – C – C ≡ C – C – C – H
H H
H H
i.e
H 5C 2 – C ≡ C – C 2 H 5
Lindar
catalyst
H 5C 2
C=C
H
H 2C 5
H
H 5C 2 – C ≡ C – C 2 H 5
Na, NH3
H 5C 2
C=C
H
3a. (i) CH3(CH2)2 COOH + C2H5OH
ae
Structure of product
H
H
H O
H
C3HCOOC2H5 + H2O
este
O
H
O
H–C–C–C–C–O–C–C–H
H H H
H H
ii.
CH3CH2(OH) CH3+Na(s) → CH3CH2C(ONa) + H2↑
Structure of product
H H
O
H – C – C – C – Na
H H
i.
ethylpropanoate
O
O
H
O
H
H 2C 5
H H
ii.
H
H–C–C–C–O–C–C–H
H H
iii.
H
H
or
H
(a) ethanol
(b) conc. H2SO4
(c) hot water bath
4(a). (i)
saponification
(ii)
fermentation
(iii)
cracking
H
b i.
H H
O
O
H – C – C – C – C – OH
H H H
Structure
or
OH
name is butanoic acid.
ii.
butane < methylpropane < butanoic acid.
iii.
methylpropane has a higher boiling point than its straight chain
isomer butane because the more branched on compound is, the
more the vander waal’s forces between its molecules and the higher
its boiling point.
Butanoic acid has the highest boiling point because of the presence
of hydrogen bond.
c.i.
x is ethane
laboratory preparation
ethanol is heated with excess conc. H2SO4 at a temperature of 170oc.
this occurs in two stages.
I.
The ethanol and the acid will be mixed in volume ratio of 1 : 2
respectively to form ethylhydrogen sulphate (VI)’
II.
On heating the ethylhydrogen sulphate (VI), it decomposes into
tetraoxosulphate (VI) acid to p and produce ethane in the presence of
excess H2SO4.
C2H5OH(ae)+ H2SO4 → C2H5HSO4 + H2O
C2H5HSO4 → C2H4 (g)+ H2SO4(ae)
iii.
OH + 2 Na (s) → 2
ONa + H2↑
sodiumethoxide
propanoic acid
d.
i.
alkyalation (substitution)
ii.
polymerization
iii.
oxidation
iv.
hydrolysis
v.
hydrogenation
ii.
5.
2
Weight of CO2 = 07.3g
Weight of water = 2.98g
Weight of cabon = atom weight of c x weight of CO2
Molecular weight of CO2
= 12 x 7.3g
44
= 1.99g
% of carbon = weight of carbon
x 100%
Weight of hydrocarbon
= 1.99
x 100%
2.22
85.8%
% of hydrogen = 100% - 85.8%
= 14.2%
i.
C
H
85.8
14.2
12
1
7.15
14.2
7.15
7.15
1
1.98 = 2
Empirical formula = CH2
ii.
Molecular weight = 2 x V.D
= 2 x 28
= 56
(Empirical formula)n = molecular formula
(CH2)n = 56
(12 + 2)n = 56
14n = 56
n = 56
14
=4
Molecular formula = (CH2)4
= C 4H 8
H
iii.
iv.
H H H
H–C–C=C–C–H
H
H
H
H H H
H–C=C–C–C–H
H H
H
but – 1 –
H
H H
H–C–C=C–C–H
H
H
6.
but – 2 –
Y is an alkyne
Cn+H2n – 2 = 04
(12 x n) + (2n – 2) (1) = 54
12n + 2n – 2 = 54
14n – 2 = 54
14n = 56
14
14
n=4
i.
Molecular formula is C4H6
H
ii.
H
H–C≡C–C–H
H
H
but – 1 –
H
iii.
H
H–C≡C–C–C–H
H
H
but – 2 –
7.
i.
no visible reaction (it’s 3o – alc)
H
H
Cl
O
ii.
Cl – C – C – C – C – Cl
H
Cl Cl
H
CH3 H
iii.
H –HC – C – C – C – H
O CH3 H
H H O
8.
H
H H
The liquid is H – C – C – C – O – C – C – C – H
H
H
H
H
H
H OH
H
H H
H H
The structure of Z is H – C – C – C – H
O
The structure of Q is H – C – C – C – OH
C2H12O2(1) + H2O → C3H6O2 + C3H6OH
↓ [O]
C 3H 6O 2
O
[O]
i.
OH
OH
COOH
CH3
[O]
ii.
Cl
O
Cl
OH
[O]
iii.
O
H+
OH
O
OH
iv.
[O]
OH
O
[O]
9i.
OH
OH
CH3
COOH
ii.
O
iii.
OH
[O]
O
H+
OH
O
[O]
iv.
OH
10.
Williamson ethe synthesis
OH
C2H5OH + Na → C2H5ONa
↓ C2H5Cl
C2H5O C2H5 + NaCl
Mechanism
OH + Na(s) →
+
O – Na →
(-)
H
←
*
Substitution reaction of alcohol
Cl
ONa(+)
C2H5OH + C2H5Cl → C2H5O C2H5 + HCl
Mechanism
..
O +
Cl
→
Cl
O
H
H
↓
O
11.
i.
COOCH2CH2OH
ii.
C2H5OH
iii.
CH3C(Br2CO2H
iv.
CH2(CH3)CH3
v.
COCl
COOH
vi.
COOH
vii.
O
OH
OH
O
viii.
CH3CH = CH CH2 OH
12.
i.
CH3(CH2)2 CO2H
OC2H5
ii.
CH3(CH2)2
C–H
OC2H5
iii.
CH2(OH) C2H5
iv.
CH2 (CH2) CH2 OH
CH 2 CH3
v.
CN
CH3CHCH2CH(CN) (OH)
HCl
Cl
CH 2CH2CH3
13.
OH
i.
OH
ii.
iii.
(CH3)2CH (CH2)2 C (CN) (OH) CH3
iv.
CH2CH2CH (OH) CH2CH3
CH2CH3
CH3
14.i. H3C – C – C – CH – CH3
CH3 H CH3
ii.
O
HC – O – C2H5
H CH3
H
O
or
H
H
or
H
O
H 3C
H 3C
CH3
CH3
iii.
H–C–C=C–C–H
or
C=C
CH3
H H
b.
H H H H
H–C–C–C–C–C–C–C–C–H
H H
H H H H
OH
I
(c)(i) R – C - R
OH OH
ii.
H H
[O]
[O]
H2C – CH2
or
H H
O
R – C - RI
O
O O
HO – C– C – OH
(d) (i) D – monosaccharide
(ii)
(iii)
E – ethanol, F ethylethanoate
O
F belongs to R – C – ORI
iv.
CH3COOH + C2H5OH
v.
Esterification
e.
CH3CH2CH2CH3<CH3CH2 CH2 CH2 CH2 CH3< CH3 CH2 CH2OH
Conc. H2SO4
CH2C – OC2H5 + H2O
increasing boiling point
Propanol (CH3 CH2 CH2OH) has the highest boiling points due to the
presence of hydrogen bond. Since both butane and heptanes are alkanes,
the boiling point of alkanes increases with increasing carbon atoms per
colecular in straight chain hence heptane is of higher boiling point than
butane. H Cl
Cl H
H Cl
Cl H
15(a) H – C – C – H
H–C–C–H
1,2 – dichloro ethane 1,1; dichloroethane
(b)
They are isomers
H
OH
O
16(a) H – C – C – OH + H – C – H
H+
O
H3C – C – O CH3
H
b.
Amino acids
ii.
Carboxylic group and amino group.
iii.
polymerization
17(c)(i)
C
H
92.3
7.7
12
1
7.7
7.7
7.7
7.7
1
i
Empirical formula = CH
Molecular formula = (empirical formula)n
R.MM = (CH) n
78 = (12 + 1)n
n = 78
13
= 6
Molecular formula = (CH)6
= C 6H 6
(ii)
Benzene
(d) (i) Y
(ii)
X
(iii)
X
(e) (i)
(ii)
Boiling with acidified KMnO4 or acidified K2Cr2O7
oxidation
C2H2OH
18.
[O]
H3C COOH
Data given
Weight of sample = 0.2g
Weight of CO2 obtained = 0.2933g
Weight of H2O produced = 0.1201g
Molar mass of CO2 = 44 molar mass of H2O = 18
(i)
Weight of carbon =
12
x weight of CO2 obtained
Molar mass of CO2
= 12 x 0.2933
14
= 0.089
Wright of hydrogen = 2 x 0.1201
18
= 0.013g
% composition of C = weight of carbon x 100%
Weight of sample
= 0.08 x 100%
0.2
= 40%
% composition of H = weight of hydrogen x 100%
Weight of sample
= 0.013 x 100%
0.2
= 6.5%
% composition of O = 100% - (40% + 6.5%)
= 100% - 46.5%
= 53.3%
Empirical formula
C
H
O
40
6.5
53.3
12
1
16
3.33
6.5
3.34
3.33
3.33
3.33
1
2
1
Empirical formula
Molecular formula = (Empirical formula)n
R.M.M = (CH2O)n
60 = (12 + 2 + 16)n
(30)n = 60
n = 60
30
n=2
molecular formula = (CH2O)2
= C 2H 4O 2
H O
ii.
H – C – C – OH
H
Ethanoic acid
H
O
H–C–O–C–H
H
Methylmethanoate
19(a)(i)
Addition polymerization is the combination of several
monomers to form a polymer without any loss or gain.
(ii)
unsaturated compounds
(iii)
I. number of cross – linkages
II. forces of attraction or repulsion between and within
molecules
(b) (i)
CH3CH2COOC2H5
Ethylpropanoate
(ii)
acidified K2Cr2O7
(iii)
dehydration
(c)(i)
C2H5OH + 3O2 → 2CO2 + 3H2O
(ii)
from the reaction
1 mole of ethanol requires 3 moles of oxygen
Molar mass of ethanol = (2 x 12) + (1 x 5) + (16 x 1) + (1 x 1)
= 24 + 5 + 16 + 1
= 46mol – 1
:.
46g of ethanol requires 3 x 22.4dm3 of oxygen at S.T.P
46g of ethanol requires 67.2dm3 of oxygen at S.T.P
2.3g of ethanol required xdm3 of oxygen at S.T.P
x = 2.3 x 67.2
46
= 3.36dm3
Hence 3.36dm3 of oxygen required at S.T.P for the complete combustion of
2.3g of ethanol.
20.(a)(i)
(ii)
concept of resonance
(b) (i) I. production of dynes
II. solvent for rubber
(iii)
destructive distillation of coal