Ministry of Higher Education
and Scientific Research
Ibn khaldun University College
ً‫وزارة التعليم العالً والبحث العلم‬
‫كليت ابه خلدون الجامعت‬
‫قسم تقىياث المختبراث الطبيت‬
General chemistry
‫الكيمياء العامت وظري‬
‫المرحلت االولى‬
Lecturer
Msc: Tamam AL-Jbori
Matter:
Matter is any thing that occupies space and has weight. States of Matter
are solid, liquid, and gas.
Classification of Matter:
Matter
Pure substances
Elements
-
-
Simplest
Mixtures
Compounds
- Composed of 2 or
pure
more components
substances
- Can be
Cannot be
decomposed
decomposed
(separated by
chemical means)
3
Solutions
(Homogeneous mixture)
Heterogeneous
mixtures
Composed of 2
- Composed of 2 or
-
or more elements
more components
-
- 2 or more phases
Single phase
Chemical Bonds: Chemical bonds can be divided into: 1. Ionic bond: An ionic bond occurs when one or more electrons are
transferred from one to another
e.g. Li (1s2 2s1) –e-
Li+ (1s2) + e-
F (1s2 2s2 2p5 ) + eF- (1s2 2s2 2p6 )
..
..
_
Li . + . F.. :
Li+ [ : F
:
]
LiF
..
e.g. CaCl2
2.
Covalent bond: A covalent bond results from sharing of a pair of
electrons between atoms.
.
.
e.g.
H +H
H2
H
. x
. HX C
. x
.
. C .+ 4HX
.
H
X
H.
H
C
H
CH4
methane
H
H
..
.
.
HX N
.X X H
H
..
N. . + . 3H X
e.g.
NH3
H
ammonia
CH2 = CH2
CH
3.
H N H
= CH
Coordinate covalent band: Where a pair of electrons from one atom is
shared by two atoms.
H
CL
H N: + B CL
H
CL
H CL
H
H N: B CL
H
4
H
N
B
H CL
CL
4.
Hydrogen bonding: A particularly strong dipole – dipole attraction
occurs when hydrogen is covalently bonded to a very small highly
electronegative element such as fluorine, oxygen or nitrogen.
δS+
δS-
H
H
O
--
δS O
H
δS--
H
δS+
O
H
H
Errors and statistics:
Accuracy:
The accuracy of a determination may be defined as the concordance
between it and the true or most probable value (relative error)
Precision:
Precision may be defined as the concordance a series of measurements
of the same quantity. The mean deviation or the relative mean deviation is a
measure of precision.
Example: A substance was known to contain 49.06 + 0.02 percent of given
constituents A. The result obtained by two observers using the
same substance and the same general technique were:
Observe (1). 49.01; 49.21; 49.08 Mean 49.10 percent.
Relative mean error = (49.10 – 49.06) / 49.06 = 0.08 percent.
Relative mean deviation = [(0.09 + 0.11 + 0.02) / 3 × 100 / 49.10 =
0.15 percent.
5
Observe (2). 49.40; 49.44; 49.42 Mean = 49.42 percent.
Relative mean error = (49.42 – 49.06) / 49.06 = 0.73 percent.
Relative mean deviation = [(0.02 + 0.02 + 0.00) / 3 × 100 / 49.42 =
0.03 percent.
Classification of errors:
The errors may be divided into:
A)
Determinate (systematic, constant) errors:
1. Instrumental errors:
(Impurities in reagents).
2. Methodic errors:
(Solubility of a precipitate, in complete reaction)
3. Operative errors
(Loss of precipitate during filtration)
4. Personal errors
(Color blindness)
B)
Indeterminate errors:
Indeterminate errors are frequently called accidental or random,
errors, such errors can be attributed to no known cause, nor can they be
predicted as to magnitude or direction for any single measurement in a
series or for any single measurement standing alone,
6
Express of concentration
The gram formula weight (gfw):
Is the summation of atomic weights, in grams, of all the atoms in the
chemical formula of a substance. Thus, the gram formula weight for H 2 is
2.016 (2 X 1.008) g; for NaCl it is 58.44 (35.45 + 22.99) g.
Formality:
Is defined as the number of gram-formula weights (gfw), of solute per
liter of solution.
F=
wt
gfw
X
1000
V(ml)
Gram molecular weight (gmw):
Is employed rather than gram formula weight when we are concerned
with a real chemical species. Thus, the gram molecular weight of H2 is its
gram formula weight 2.016g if we are dealing with the substance NaCl in
water; we will not assign to it a gram molecular weight because this species is
not found in aqueous media.
It is perfectly proper to assign gram molecular weights to Na + (22.99g)
-
and Cl (35.45g) since these are real chemical entities (strictly, these should
be called gramionic weights rather than gram molecular weights, although
this terminology is seldom encountered).
Molarity or molar concentration:
Is defined as the number of molecular weights or moles of a solute
per a liter of solution (or the number of millimoles per millimeter).
Wt
1000
M=
X
Gmw
V (ml)
7
Normality:
Is defined as the number of gram
* Equivalent weights (equivalents) of a substance dissolved in one liter
of solution (or number of milliequivalents per millimeter).
The equivalent weight of a substance depends upon the type of reaction it
undergoes.
Molecular weight
Gram-equivalent weight =
n (equivalency)
Example of equivalency
Solute
n
Oxidation state
NaCl
1 (where is of interest)
BaCl2
2 (where Ba2+ is of interest)
ALCl3
3 (where Al3+ is of interest)
Na2SO4
2 (where SO42¯ is of interest)
Replaceable Hydronium and Hydroxide Ions
HCl
1
HNO3
1
H2SO4
1 (where product is HSO4¯ )
2 (where product is SO42¯ )
NaOH
1
KOH
1
Ba(OH)2
2
Change in oxidation state
FeCl3
1 (where product is Fe 2+ )
MnO4¯
5 (where product is Mn +2 )
SnCl4
2 (where product is Sn +2)
* gram – equivalents =
Grams of solute
Gram – equivalent weight of solute
8
N=
Wt
Eqw
X
1000
V (m1)
Parts per million; part per billion:
For very dilute solutions, it is convenient to express concentrations in
term of:
Part per million (PPM) =
Weight of solute
X 10 6
Weight of solution
mg
PPM ≈
Liter solution
Part per billion (ppb) =
Weight of solute
X 10 9
Weight of solution
P – functions:
In some circumstances it is convenient to express the concentration of
an ion in terms of the negative logarithm of its molar concentration. The most
common p – function is:
P = - log [H3O+]
The term in brackets is the molar hydronium ion concentration.
Density and specific Gravity:
The density of a substance measure its mass per unit volume and has
units of Kg / liter or g / ml.
Specific gravity of a material is the ratio of its mass to that of an equal
volume of water at 4C º. Specific gravity is unit less because water at 4C º has
a density of exactly 1g / ml.
Percentage concentration:
Weight percent (W / W) =
Volume percent (V / V) =
Wt of solute
Wt of solution
X 100
Volume of solute
Volume of solution
9
X 100
Weight percent (W / V) =
Wt of solute, g
Vol. of solution, ml
X 100
Solution – Diluent Volume Ratios:
The composition of a dilute solution is sometimes specified in terms of
the volume of a more concentrated reagent and the volume of solvent to be
used in diluting it. Thus a 1:4 HCl solution contains four volumes of water for
each volume of concentrated Hydrochloric acid taken.
V1 C1 = V2 C2
N=
M=
D X % X 10
eq. wt
D X % X 10
M. Wt
Molal solution contains a gram – molecular weight of solute dissolved
in 1000 g of solvent.
10
Chemical Equilibrium
In a solution of acetic acid, two reactions occurring simultaneously:
CH3COOH + H2O
H3O + CH3COOO¯………(I)
H3O + + CH3COO¯
CH3COOH + H2O ………(II)
And
When the rate which the ions are formed by reaction (I) is equal to the
rate at which they disappear by reaction (II), there concentrations in the
solution will no change with time. In fact, the concentrations of all the species
will remain constant from this point on, this call equilibrium.
To indicate chemical equilibrium in a reacting system we use a set of
double arrow
in
the
chemical
equation.
Thus
the
equilibrium that we have been discussing is expressed as
CH3COOH + H2O
H3O + CH3COOO¯
The use of this notation implies that the forward reaction (the reaction
going from to right) is occurring at the same rate as the reverse reaction (the
reaction from right to left).
For strong electrolytes, the reverse reaction does not occare, for e.g.
NaCl would write simply
Na+(aq) + Cl¯(aq)
NaCl
Equilibrium constant:
The generalized equation for a chemical equilibrium
mM + nN
pP + qQ
The equilibrium – constant expression for this reaction is
K=
[p] p [Q] q
[M] m [N] n
11
Where the letters in brackets represent the molar concentration of
dissolved solutes or partial pressures (in atmospheres) if the reacting
substances are gases.
Common Types of Equilibrium – Constant Expressions:
1. Dissociation of water
Aqueous solutions always contain small amounts of hydronium and
hydroxide ions as a consequence of the dissociation reaction
H3O + + OH ¯
2H2O
The equilibrium constant for this reaction.
K=
[ H3O+] [OH ¯ ]
[H2O] 2
[H2O] in the equation can be taken as constant
K [H2O]2 = Kw = [H3O+] [OH ¯ ]
Where the new constant Kw is given the special name, the ion – product
-
constant for water. ( ≈ 1 X 10 14 at 25 C˚ )
pKw
= - log Kw
-
= - log (1 X 10 14)
= - (-14) = 14
log Kw = log [H+] + log [OH ¯ ]
- log Kw = (-log [H+]) + (-log [OH ¯ ])
pKw
= pH + pOH
= pH + pOH
12
Example 1:
What is the ph of a 0.002M HCl solution?
[H+] = 0.002M = 2 X 10– 3 M
pH = - log [H+]
= - log [2 X 10– 3 ]
= - (log 2 + log 10– 3)
= - {0.3 + (– 3)}
= - (- 2.7)
= 2.7
Example 2:
What is the pH of a 5 X 10– 4 M NaOH?
Method 1
Kw = [H+] [OH ¯ ]
+
[H ] =
1 X 10– 14
5 X 10
–4
= 0.2 X 10– 10 or 2 X 10 – 11 M
pH = - log (2 X 10 – 11) = 10.2
Method 2
pOH = - log [OH ¯ ]
pOH = - log (5 X 10– 4)
= 3.3
pH = pKw – pOH
= 14 – 3.3
= 10.7
13
2. Equilibrium involving slightly soluble ionic solids:
When an aqueous solution is saturated with a sparingly soluble salt
(e.g. silver chloride) one or more equilibrium will be established:
AgCl(s)
AgCl(aq)
AgCl(aq)
Ag + + Cl ¯
K=
[AgCl] aq
[AgCl] s
K [AgCl] s = K1 = [AgCl] aq
The concentration of AgCl in a saturated solution is constant at any
given temperature:
K2 =
[Ag +] [Cl ¯]
[AgCl] aq
K1 K2 = Ksp - [Ag +] [Cl ¯]
The product of these two constants, Ksp, is called the solubility –
product constant.
14
Buffer Solutions
Solutions containing both week acids and their salts or week bases and
their salts. They have the capacity of resisting changes of pH when either
acids or alkalies are added to them for example, suppose 1 ml of 0.1 M HCl is
added to 99 ml of pure water pH7. The [H+] of the mixture will be about 0.001
M and its pH about 3. Suppose, however, 1 ml of 0.1 M HCl is added to 99
ml of buffer solution containing 0.1M acetic acid and 0.1M sodium acetate.
The pH of the buffer solution to begin with is 4.73. When the HCl is added,
the pH becomes about 4.72.
A similar condition exists when alkali is added to the buffer mixture.
1ml of 0.1M NaOH added to 99 ml of pure water gives a pH about 11. Where
as if it is added to 99ml of the acetate buffer solution (pH 4.73), the pH
becomes 4.74.
The chemical mechanism according to which buffers function may be
illustrated by what happens when NaOH and HCl are added to an acetate
buffer solution are as follows.
CH3COO¯ + H+
CH3COOH
CH3COONa
weakly dissociated
CH3COO¯ + Na+ completely dissociated
Upon addition of NaOH
CH3COOH + NaOH
CH3COONa + H2O
CH3COONa
The addition of the alkali decreases the CH3COOH in the buffer and
increases the CH3COONa. The pH of the solution increases in proportion to
the change in ratio of salt to acid in the buffer solution
15
Upon addition of HCl
CH3COOH
CH3COONa + HCl
CH3COOH + NaCl
In this case the HCl reacts decrease CH3COONa and increase
CH3COOH in the buffer. The pH of the solution falls in proportion to the
change in ratio of salt to acid in the solution.
The Henderson – Hasselbalach equation for buffer solutions:
In the equilibrium reactions for the dissociation of HA and BA in a
buffer solution, we have.
+
HA
H + Aˉ, weakly dissociated
+
{CH3COOH
H + CH3COOˉ}
+
B + Aˉ, completely dissociated
Ba
+
{CH3COONa
CH3COOˉ + Na }
+
Ka =
[H ] [Aˉ]
[HA]
[HA]
+
[H ] = Ka
[Aˉ]
+
or [H ] = Ka
+
pH = - log [H ] = - log Ka
= - ( log Ka + log
= - log Ka - log
[HA]
[BA]
[HA]
[BA]
[HA]
[BA]
16
)
[HA]
[BA]
+
or [H ] = Ka
[Acid]
[Salt]
= - log Ka = pKa
pH = pKa + log
and
[BA]
[HA]
- log
[HA]
= log
[BA]
or pH = pKa + log
[BA]
[HA]
[salt]
[acid]
A similar equation that would apply to a basic buffer is:
pOH = pKb + log
[salt]
[base]
Table = salt – acid ratio and pH
concentration
Percent
CH3COONa
CH3COOH
ratio
dissociation
Molar
Normal
Salt / Acid
Of acid
pH
0.00
0.2
0.00
1.00
2.7
0.01
0.2
0.05
0.20
3.4
0.05
0.2
0.25
0.04
4.1
0.10
0.25
0.50
0.02
4.4
0.15
0.2
0.75
0.012
4.6
0.20
0.2
1.00
0.01
4.7
0.30
0.2
1.50
0.006
4.9
17
Capacity:
The larger the concentrations of the components of the buffer, the more
effective it is at resisting changes in pH.
pH = pKa + log
[salt]
[acid]
Buffered solutions are very frequently used in (quantitative) analytical
chemistry for example.
1.
Ensure the proper progress of a reaction by removing H 3O+ or
OH ˉ reaction products as the save formed.
2.
Control the level of pH during a precipitation process.
3.
Maintain a required level of pH in order to guarantee the proper
reaction of an indicator substance.
4.
Are encountered at certain stages of titration processes.
Buffered solutions are of particular importance in most life processes.
Many of the chemical mechanisms of body are pH – dependent; and as
a case in point, the pH of human blood ranges about ± 0.05 pH units
around a value of 7.40 – Any marked change of a few tenths above or
below this range can mean eventual death. Acid and base substances
introduced into the system are usually carried by the blood as buffered
form until the excesses (e.g., carbon dioxide and uric acid) can be
removed by the lungs or kidneys. Changes in the normal level of the
blood pH may provide for conditions, which prevent the buffering and
transporting of such substances.
18
Analytical chemistry : deals with identification , characterization and
measurement of the chemical species present in a sample.
Qualitative analysis : is to determine what the constituents are in the
sample.
Quantitative analysis: is to determine how much of each of constituent
is in the sample.
Thus the role of an analytical chemist is essential not only to chemical
science
but also to allied areas in biology , food science , medicine ,
biochemistry and engineering.
In pharmacology , accurate adjustment of a drug dose for an individual
may necessitate the determination of its levels in the blood stream and in the
pharmaceutical industry the quality of the manufactured drugs in tablet ,
solution , and emulsion form must be carefully controlled. Slight changes in
composition or in purity of the drug itself can effect the therapeutic value.
The analytical chemist and the analysis :The concept of measurement is basic to analytical chemistry. A simple
measurement may involve properties such as mass, current, voltage, volume,
or time. The measurement of such properties as absorption or emission of
energy, optical rotation, over voltage, refractive index, equilibrium constant,
rate constant, activation energy, or heat of reaction is more complex. Whether
simple or complex, the reliability, utility, accuracy and specifity of these
measurements are the responsibility of the analytical chemist. The analytical
chemist is not only concerned with carrying out the analysis, but also with the
how, why and where of using such measurements for analysis, separation, for
the elucidation of the fundamental chemistry involved in problem.
Volumetric Analysis:A quantitative analysis based upon the measurement of volume.
19
Titration:- Is the process by witch the quantity of analyte in a solution is
determined from the amount of a standard reagent it consumes.
Standard solution:- The reagent of exactly known concentration that is used
in a titration.
The concentration of a standard solution is established either
directly or indirectly:1. By dissolving a carefully weighed quantity of the pure reagent and diluting
to an exactly known volume.
2. By titrating a solution containing a weighed quantity of a pure compound
with the reagent solution.
Standardization:- the process whereby the concentration of a standard
solution is determined by titration of a primary standard is called a
standardization.
Indicator:- is the chemical compound used to detect the end-point in
volumetric analysis that exhibits a change in color as a result of
concentration changes occurring near the equivalence point.
Primary standards:Good primary standards include the following:1.
Highest purity.
2.
Stability.
3.
Absence of hydrate water.
4.
Ready availability at reasonable cost.
5.
Reasonably high equivalent weight.
Standard solution:An ideal standard solution for titrimetric analysis would have the
following properties:20
1. Its concentration should remain constant for months or years after
preparation to eliminate the need for restandardization.
2. Its reaction with the analyte should be rapid, complete and can be
described by balanced chemical equation.
3. A method must exist for detecting the equivalence point between the
reagent and the analyte ; that is a satisfactory end point is required.
End point in volumetric methods:End points are based upon a physical property which changes in
characteristic way at or near the equivalence point in the titration. The most
common end point involves as color changes due to the reagent, the analyte,
or an indicator substance. Other physical properties, such as electrical
potential, conductivity, temperature, and refractive index, have also been
employed to locate the equivalence point in titrations.
It is convenient to classify volumetric methods according to four
reaction types:1. Precipitation reactions:AgNO3 + NaBr
AgBr(s) + NaNO3
2. Neutralization reactions:NaOH +HCl
H2O + NaCl
3. Oxidation-reduction reaction:Sn+2 + Hg2+ + Cl -
H2Cl2 + Su4+
4. Complex-formation reaction:2CN - + 2Ag+
Ag[Ag(CN)2] (S)
21
Theory of indicator:Acid-base indicators are generally organic compounds which behave as
weak acids or bases. The dissociation or association reactions of indicators are
accompanied by internal structural rearrangements that are responsible for the
changes in color.
H3O+ + In-
H2O + HIn
(acid color)
(base color)
Or
In
(base color)
InH+ + OH-
+ H2O
(acid color)
Ka =
[H3O+] [In-]
[HIn]
Kb =
[InH +] [OH -]
[In]
And
The expressions can be rearranged to give :
[In-]
[HIn]
=
Ka
[H3O+]
And
[InH +]
[In]
=
Kb
[OH -]
=
Kb [H3O+]
Kw
22
A solution containing an indicator will show a continuous change in the
color with variations in pH. The human eye is not very sensitive to these
changes, however typically, a five-to tenfold excess of one form is required
before the color of that species appears predominant to the observer
[In-]
1
≤
[HIn]
[In-]
10
10
≥
[HIn]
1
For the full acid color:
[H3O+] [In-]
[HIn]
=
[H3O+] 1
10
= Ka
[H3O+] = 10 Ka
And similarly, for the full basic color:
[H3O+] 10
1
+
[H3O ] =
= Ka
1
10
Ka
To obtain the indicator range, we take the negative logarithms of the two
expressions. That is,
Indicator pH range = - log 10 Ka to – log
= - 1 + pKa to – ( - 1 ) + pKa
= pKa ± 1 .
23
Ka
10
Spectrophotometric Method:
Many determinations made in the clinical laboratory are based upon
measurements of radiant energy emitted, transmitted or absorbed.
Components of Instruments for Absorption Measurements:
1. power supply.
2. a source of radiant energy, usually containing a large group of wave
lengths, e.g. , a tungsten lamp.
3. a monochromator e.g. , a filter, prism, or diffraction gratting
4. a simple container to hold the sample and blank e.g., cuvette.
5. a detector system to detect the transmitted radiant energy and convert it
into electrical energy so that it may be measured, e.g., a photocell.
6. a signal indicator, e.g., a meter reading percent T or absorbance.
Light
source
Monochromator
Entrance
slit
Cuvet
Exit
slit
Detector
Meter
major components of a single beam spectrophotometer
Sample
Mirrors
Light
source
Reference
Monochromator
Entrance
slit
Exit
slit
Io
Io = Ia + It
24
Ia
Meter
Detector
It
Io = incident radiation power
A = log
It
Io
Io
T% = It
It = transmitted radiation power
A = absorbance
T = Transmittance
X 100
T% = percent Transmittance
ℓ = path length
Aαℓ
ε = molar absorptivity
AαC
C = concentration
A=εℓC
A = log T
25
26
ORGANIC CHEMISTRY
Organic chemistry is the chemistry of the compounds of carbon.
Certain organic compounds contain only two elements, hydrogen and
carbon are known as hydrocarbon which are divided as follow:
Hydrocarbons
Aliphatic
Alkanes
Aromaatic
Alkenes
Alkynes
Cyclic Aliphatic
Alkanes :
The general formula for alkanes is Cn H2n + 2 where n represents
the number of carbon atoms. That is, if a compound contains n carbon
atoms the number of hydrogen atoms is twice n plus two more.
27
Table 1 . Alkanes
Number of
carbon
atoms
Name
Molecular
formula
Compound Structural
formula
Structural formula
H
1
Methane
C H
H
CH4
CH4
H
2
3
Ethane
Propane
C 2 H6
H
C 3 H8
H
H
H
C
C
H
H
H
H H
C
C C
CH3 – CH3
H
CH3 – CH2 – CH3
H
H H H
H H H H
H
C
C
C
C
H
H
H
H
H
CH3 – CH2 – CH2 – CH3
or
or
4
Butane
CH3 – CH – CH3
H H H
C4 H10
H
C C C
H
H
H C H
H
5
Pentane
C5 H12
3 Isomers *
6
Hexane
C6 H14
5 Isomers
7
Heptane
C7 H16
9 Isomers
8
Octane
C8 H18
18 Isomers
9
Nonane
C9 H22
35 Isomers
10
Decane
C10 H24
75 Isomers
28
H
CH3
* The number of isomers increases rapidly as the compound
contains more carbon atoms. C40 H82 has 62,491,178,805,831
possible isomers.
Alkyl Groups:
When a hydrogen atom is removed from an alkane , an alkyl
group is formed. The names of the alkyl groups are obtained by
changing the eniting of the name from (ane) to (yl) .
Naming Hydrocarbons:
An
international
system
of
nomenclature
for
organic
compounds has been devised and is recognized and used by chemists
all over the world. This system was devised and approved by the
International Union of Pure and Applied Chemistry and is frequently
designated by the initials IUPAC. The rules of IUPAC system are:
1. Pick out the longest continuous chain of carbon atoms.
2. Identify that chain as an alkane.
3. Pick out the alkyl groups attached to that chain.
4. Number the carbons in the chain , starting at whichever end of the
chain will give the smallest numbers to the carbons to which the
alkyl groups are attached. Continue the numbering of this carbon
chain in the same direction from one end to the other.
5. List the numbers and the names of alkyl groups
6. Use commas between numbers and a dash between a number and a
letter.
29
Examples:
CH3
1
2
3
4
5
6
7
8
CH3 – CH – CH – CH2 – CH – CH – CH2 – CH3
CH3 CH3
C 2H5
. 2,3,6 – trimethyl – 5 – ethyl octane
CH3
5
4
3
2
1
CH3 – CH – CH – CH2 – CH – CH3
6CH2
CH3
7CH3
2,4,5 – trimethylheptane.
Cycloalkanes:
Alkanes also exist in the shape of a ring. Such structures are
called cycloalkanes. They are named by placing the perfix cyclo
before the name of the corresponding straight chain alkanes. Thus the
cyclic alkane : of three carbons is called cyclopropane. Its structure is
CH2
H 2C
CH2
or
Alkenes:
Alkenes have a double bond (two bonds) between two of the
carbon atoms. The general formula for alkenes is CnH2n
30
Number of
carbon
atoms
2
3
Name
Ethene
Propene
Molecular
formula
C 2 H6
H
H
C 3 H6
H
4
Butene
Compound Structural
formula
Structural formula
H
H
C
C
H
H
H
H H
C
C C
H
H H
CH2 ═ CH2
H
H
H
H
H
H
C
C
C
C
H
H
H
H
CH3 – CH ═ CH2
H
or
CH2 ═CH – CH2 – CH3
C 4 H8
H
H
H
H
H
C
C
C
C
H
H
H
H
5
Pentene
C5 H10
6 Isomers
6
Hexene
C6 H12
15 Isomers
7
Heptene
C7 H14
30 Isomers
8
Octene
C8 H16
66 Isomers
or
H CH3 – CH ═ CH – CH3
4
CH2 ═ CH – CH – CH2 – CH3
3
5
CH3
1
3 – methyl – 1 – pentene
CH3
2
3 – methyl cyclopentene
31
Alkyne:
Consider two carbon atoms connected by a triple bond.
– C ≡ C – { ethyne
H – C ≡ C – H acetylene }
The general formula for alkynes is CnH2n – 2
HC ≡ C – CH2 – CH2 – CH3
CH3 – C ≡ C – CH – CH3
1 – pentyne
CH3
4 – methyl – 2 – pentyne
Reactions:
Saturated hydrocarbons react by a process known as
substitution.
CH4 + Cl2
light
CH3Cl + HCl
Further chlorination may substitute additional chlorines for
hydrogens.
Alkanes may be oxidized to yield carbon dioxide, water, and
energy.
CH4 + 2O2
CH3
CO2 + 2H2O + energy
CH3
2CH3 – C – CH2 – CH – CH3 + 25 O2
16 CO2 + 18H2O + energy
CH3
2,2,4 – trimethylpentane
Incomplete oxidation (combustion) of alkanes yield carbon
monoxide in place of carbon dioxide:
32
2CH
. 4 + 3O2
2CO + 4H2O + energy
Carbon monoxide is poisonous because it combines with
hemoglobin and prevents that compound from carrying oxygen to the
cells.
Un saturated hydrocarbons react by a process known as addition.
CH2 ═ CH2 + H2
CH3 – CH3
ethene
ethane
CH2 ═ CH2 + Cl2
CH2Cl – CH2Cl
H – C ≡ C – H + 2Cl2
CHCl2 – CHCl2
CH3 – CH ═ CH2 + HI
CH3 – CH – CH3
I
Markownikoff rule:
when an asymmetrical reagent (e.g. HI) adds to an asymmetrical
ethylenic linkage (e.g. propylene) , the +ve part (hydrogen) will go to
the carbon atom holding the greatest number of hydrogen atoms.
CH ≡ CH
HBr
CH2 ═ CHBr
HBr
CH3CHBr2
Aromatic hydrocarbons:
A class of hydrocarbon which is characterized by certain
properties that differ from those of alkenes. The parent of all these
aromatic substances is benzene which has molecular formula C6H6
suggested as a cyclic structure containing three double bonds.
33
Kekule suggested that the position of the double and single
bonds could change , producing two structures that represent
benzene.
H
H
C
C
H
C
C
H
H
C
H
C
C
H
H
C
C
C
H
H
C
H
C
H
Organic chemists abbreviate these formulas to
or
or
or
benzene is a colorless, liquid , insoluble in water but is soluble
in alcohol and ether. Benzene is toxic when taken internally. Contact
with the skin is harmful and continued inhalation of benzene vapors
decreases red and white blood cell count. Benzene is now considered
to be mildly carcinogenic and care must be taken with it use.
CH3
CH2CH3
ethyl benzene
methyl benzene
If more than one hydrogen is substituted , a three possible
distributed benzenes are obtained.
x
x
x
x
x
1,2 – or ortho = o
1,3 – or meta
34= m
x
1,4 – para = p
CH3
CH3
CH3
CH3
CH3
o – xylene
1,2 – dimethyl benzene
anthracene
naphthalene
Reactions:
+ 3Cl2
CH3
p – xylene
1,4 – dimethyl benzene
m – xylene
1,3 – dimethyl benzene
Cl
Sun light
Cl
Cl
Cl
Cl
benzene hexachloride
Cl
Br
+ Br2
Fe
+ HBr
bromobenzene
catalyst known as halogen carrier
NO2
+ H 2O
+ HNO3
nitrobenzene
SO3H
+ HOSO3H
+ H 2O
benzene sulphonate
CH3
+ CH3Cl
AlCl3
+ HCl
COOH
CH3
(O)
KM4O4
Benzoic acid
35
methyl benzene
Alcohols and Phenols:
Alcohols and phenols are hydrocarbons in which one or more
hydrogen atoms are replaced by (-OH) hydroxy groups. If (-OH)
group is attached directly to an aromatic ring , the compound is
phenol. All other hydroxy compounds are alcohols.
CH3 – OH
OH
CH2 – OH
Ch3 OH
methyl alcohol
phenol
benzyl alcohol
Primary alcohol
CH3CH2OH
Secondary alcohol
CH3–CH–CH3
OH
O - cresol
CH3
CH3– C–CH3
OH
Tertiary alcohol
CH3OH
CH3CH2CH2OH
OH
CH3CH2CHCH3
methanol
propanol
2 – butanol
Reactions:
C12H22O11 + H2O
enzyme
4 C2H5OH + 4CO2
CH3COOH + C2H5OH
CH3COOC2H5 + H2O
ethylacetate ( an aster )
CH3OH + HCl
2C2H5OH
H2SO4
◦
140 C
CH3Cl + H2O
C2H5OC2H5 + H2O
diethyl ether
36
2CH3OH + 2Na
2CH3ONa + H2
CH3ONa + C2H5Br
CH3OC2H5 + NaBr
methyl ethyl ether
OH
ONa
+ NaOH
+ H2O
ONa
OCH3
+ CH3I
+ NaI
phenyl methyl ether
CH3CH2OH
Ethanol
KMnO4
(O)
OH
CH3–CH–CH3
2–propanol
CH3CHO
KMnO4
(O)
CH3 CH2 CH CH3
OH
ethanal
KMnO4
(O)
CH3COOH
ethaniac acid
O
CH3–C–CH3
Conc. H2SO4
HEAT
acetone
CH3–CH=CH CH3
According to Saytzoff’s rule which is state that the hydrogen
atom eliminated from the carbon which carries the least number of
hydrogen atom.
37
OH
OH
NO2
+
o–nitrophenol
OH
NO2
OH
NO2
NO2
p–nitrophenol
NO2
2, 4, 6 – trnitrophenol (picric acid)
Aldehydes and Ketons:
These compounds which contain the carbonyl group ( C = O). If the
H
carbonyl group is attached to one hydrogen atom at least ( R–C=O ) the
compound is classified as an aldehyde. If the carbonyl group is attached to
R
two alkyl group ( R–C =O )the compound is a ketone.
O
H–C–H
O
CH3–C–H
methanal
(formaldehyde)
ethanal
(acetaldehyde)
CHO
benzaldehyde
CH3 CO CH3 acetone
CH3 CO CH2 CH3 methyl ethyl ketone
38
COCH3
Methyl phenyl ketone (acetophenone)
CH3 CH CO CH CH2 CH3
CH3
CH3
2,4 – dimethyl – 3 – hexanone
Reactions:
CH3 CHO + H2
CH3 CH2 OH
O
CH3 – C – CH3 + H2
O
CH3 – CH – CH3
HCHO + CH3 MgI
CH3CH2OMgI
H2O
OMgI H O
2
CH3–C–CH3
CH3
O
CH3– C–CH3 + CH3 MgI
CH3
C=O + NH2OH
CH3
hydroxylamine
CH3
C=NOH + H2O
CH3 acetoxime
CH3
C=O + NH2NH2
CH3
hydrozine
CH3
C=N–NH2 + H2O
CH3 hydrazone
CH3 CHO
(O)
CH3CH2OH
OH
CH3–C–CH3
CH3
CH3COOH
acetaldehyde
CH3CHO + Cu++ (Fehling)
Acetic acid
(O)
CH3COOH + CuO
red
39
CH3CHO + Ag (NH3)2+ NO3ˉ (Tollen)
O
CH3–C –CH3
(O)
CH3COOH + Ag (mirror)
(O)
CH3COOH +CO2+ H2O
K2Cr2O7 / H2SO45
{Ketones are oxidized with powerful oxidizing agent}
Carboxylic Acids:
The carboxylic compounds are those containing carboxylic group
O
( – C – OH ).
According to the I.U.P.A.C system the acid is named after the alkane
from which it is derived by adding suffix – oic after dropping the ending – e .
HCOOH
methanoic acid
formic acid
CH3COOH
ethanoic acid
acetic acid
COOH
Benzoic acid
CH3 – CH – CH2 – CH – CH2 – CH – COOH
CH3
C 2H5
Cl
2 – chloro – 4 – ethyl – 6 – methyl heptanoic acid.
40
Reactions:
1. Salt formations:
CH3COOH + NaOH
CH3COONa + H2O
CH3COOH + NH3
CH3COONH4
2. Ester formation:
CH3COOH + C2H5OH
CH3COOC2H5 + H2O
C6H5COOH + CH3OH
C6H5COOCH3 + H2O
3. Formation of acid chloride:
CH3COOH + PCl5
CH3COCl + POCl3 + HCl
C6H5COOH + SOCl2
C6H5COCl + SO2 + HCl
4. Halogenation:
CH3CH2COOH
Br2/P
CH3CHBrCOOH
Br2/P
CH3CBr2COOH
Reaction of salts of acids:
1) Preparation of alkanes
CH3COONa + NaOH(CaO)
CH4 + Na2CO3
sodalime
2CH3COONa + 2H2O
hydrolysis
CH3CH3 + 2CO2 +2KOH + H2
2) Preparation of acid amide
CH3COONH4
ammonium acetate
heat
O
CH3– C – NH2 + H2O
acetamide
41
Amines:
Amines are organic bases, they can be considered as
substituted ammonia compounds. The amines are classified into
primary, secondary and tertiary amines.
R – NH2
R2NH
R 3N
Primary amine
secondary amine
tertiary amine
CH3NH2
methyl amine
(CH3)2NH
dimethyl amine
(CH3)3N
trimethyl amine
(C2H5)2NCH3
diethyl methyl amine
Reactions:
1) Salt formation:
CH3NH3+ Clˉ
CH3NH2 + HCl
2) Acylation (amine to amide)
O
CH3CH2NH2 + CH3C – Cl
Athylamine
acetylchloride
O
CH3CH2NH – C – CH3 + HCl
amide
3) Sulphonamide preparation:
SO2CL + CH3NH2
NaOH
SO2CL + (CH3)2NH
SO2CL + (CH3)3N
SO2NHCH3 +NaCL + H2O
NaOH
SO2N(CH3)2 +NaCL + H2O
NaOH
No reaction
42
((BIOCHEMISTRY))
Biochemistry is the study of life itself:
1. The interrelationships of the metabolism of carbohydrate, fats, and
proteins.
2. The need for enzymes, vitamins, and hormones in various body
processes.
3. The formation and excretion of waste products.
4. The properties and functions of the body’s fluids, including blood
and urine.
5. The cause and cure of certain pathologic conditions.
6. The study of human reproduction and its molecular basis as
evidenced by the study of DNA.
43
((Carbohydrates))
Carbohydrates (which contain the elements carbon, hydrogen, and
oxygen) are a class of organic compounds that includes sugars, starches, and
cellulose.
Carbohydrates were considered to be hydrates of carbon.
The formula for glucose, C6H12O6, was written as C6(H2O)6. Likewise sucrose,
C12H22O11, was written as C12(H2O)11.
Carbohydrates are defined as polyhydroxyaldehydes or polyhydroxyketones or
substances that yeild these compounds on hydrolysis.
Monosaccharides: Monossaccharides (mono-means one) are simple sugars. They cannot
be changed into simpler sugars upon hydrolysis (reaction with water).
Monosaccharidesd are called either aldoses or ketoses, depending
o
upon whether they contain an aldehyde (R-CHO) or a ketone (R-c-R)
group. Aldoses and ketoses are further classified according to the number
of carbon atoms they contain e.g. Aldopentose, ketohexose.
Geometric Isomers: are compounds with the same molecular
formula but different structural formulas owing to a restricted rotation
because of either adouble bond or a ring system.
H2
H2
CH3
H
H
H
C C
Cl
C C
Cl
Cis-1,2
dichloroethene
Cl
CH3
C
Cl
H
Trans – 1,2
H
C
C
C
C
C
H
H
H
CH3
Cis-1,2 dimethylcyclopropane
44
CH3
Trance – 1,2
Optical Isomers: are compounds with the same molecular formula but
with structures that are the mirror image of one another. Such isomers rotate the
plane of polarized light equally, but in opposite directions. The number of
optical isomers depends upon the number of asymmetric carbon atoms present
in a compound and may be calculated by using the formula 2n, where n is the
number of asymmetric carbons. e.g. glucose has four asymmetric carbons and
so has 24 or 16 optical isomers.
CHO
CHO
H
C
HO
OH
C
H
CH2OH
CH2OH
D- glyceraldehyde
L- glyceraldehyde
Glucose: (C6H12O6) is an aldohexases and may be represented
structurally as: H
6
1
H
2
HO
3
H
H
C = O
C
OH
C
H
4
C
OH
5
C
OH
H
H
OH
H
1
4
1
H
OH
H
H
H
4
HO
CH2OH
5
5
3
6
6
CH2OH
HO
OH
H
OH
3
2
2
H
OH
OH
ß-D-glucose
-D-glucose
CH2OH
D-glucose
Haworth projection
Fischer projection
1
CH2OH
2
C =O
HO
3
C
H
H
4
C
OH
H
5
C
OH
6
6
1
CH2OH
CH2OH
5
2
OH
H
H
OH
4
3
OH
H
CH2OH
-D-Fructose
D-Fructose
45
H
Reaction: 1. Oxidation: The reducing property is the basis of the test for sugar
in the uring and in blood (Benedict or Fehling’s solution contain
Cu(OH2))
A)
H
C= O
COOH
heat
H
C
OH
HO
C
H
Cupper (II)
Cupper(I)
H
C
OH
Hydroxide
H
C
OH
+2Cu(oh)2
H
C
OH
HO
C
H
oxide
H
C
OH
(red-orange ppt)
H
C
OH
Cu2O
(deep blue color)
+
CH2OH
CH2OH
D-glucose
D-gluconic acid
Tollen’s reagent:
glucose + AgOH
heat
gluconic acid + Ag + water
Tollen’s reagent
2.
Fermentation: ensymes
C6H12O6
glucose
Silver mirror
2C2H5OH + 2CO2
Ethyl alcohol
Oxygen- Carbon Dioxide Cycle in Nature:
Plants have the ability to pick up carbon dioxide from the air and water
from the ground to form carbohydrate, animals are unable to do this and must
rely on plants for their carbohydrate. Animals oxidize carbohydrates in their
bodies to yield carbon dioxide, water, and energy.
Plant photosynthesis
Energy + 6CO2 + 6H2O
C6H12O6 + 6O2
Animal metabolism
46
Disaccharides: Disaccharides (di- means two) are double sugar formed by the
combination of two monosaccharides. On hydrolysis they yield two simple
sugars.
Sucrose
Maltose
6
glucose + Fructose
glucose + glucose
6
CH2OH
5
H
5
H
H
1
4
HO
CH2OH
H
OH
3
H
+
OH
HO
3
H
2
H
OH
OH
ß-D-glucose
CH2OH
CH2OH
HO
H
OH
α -D-glucose
H
1
4
2
H
OH
H
H
H
H
H
OH
OH
H
+ H2O
OH
H
H
OH
H
O
H
OH
1,4- Linkage maltose
6CH2OH
H
CH2OH
H
H
H
OH
3
OH
HO
H
H
1
4
HO
H
OH
H
2
H
H
OH
1,2- Linkage
OH
α -D-glucose
O
+
6CH2OH
+ H 20
CH2OH
OH
2
5
H
OH
H
H
CH2OH
4
OH
OH
OH
H
3
H ß – fructose
OH
47
H
sucrose
CH2OH
The disaccharides are white, crystalline, sweet solid, optically active (are
rotate the plane of polarized light), soluble in water and they are too large to
pass through cell memberanes.
Reducing properties: In maltose, the aldehyde groups are at carbon 1 in each of the original
glucose molecules. Since the linkage is 1, 4, one free aldehyde group remains.
Therefor maltose acts as a reducing sugar.
In sucrose, the glucose part had the aldehyde at carbon 1 and the fructose
part had the ketone group at carbon 2. Since the linkage is 1,2, neighther group
is free. Therefor, sucrose is not a reducing sugar.
Polysaccharides: Polysaccharides are polymers of monosuccharides and have a high
molecular weight, insoluble in water, tasteless, and give negative tests for
reducing sugars.
Three common polysaccharides are: 1.
Starch: plants store their foods in the form of starch granules. Starch is
a mixture of the polysaccharides amelopectin and amylose. Amylopectin is a
branched polysaccharides present in starch to a large extent (80 to 85
percent). Amylose is a non branched polyaccharide present in starch to an
extent of 15 – 20 percent.
Starch gives a characteristic deep blue color with iodine when starch is
hydrolyzed, it forms: Starch
blue
erythrodextrine
red
(when iodine is added).
48
matlose
colorless
glucose
colorless
Cellulose: Wood, cotton, and paper are composed primarily of cellulose. It is the
supporting and structural substance of plants. Cellulose is a polysaccharide
composed of many glucose units. It is not effected by any of the enzymes
present in the human digestive system and so cannot be digested.
Cellulose dose not dissolves in water no in most ordinary solvents. It
gives no color test with iodine and give a negative test with copper (II)
hydroxide.
Glycogen: Glycogen is present in the body and it stored in the liver and the muscles,
where it serves as a reserve supply of carbohydrates.
Glycogen forms a colloidal dispersion in water and gives a red color with
iodine. It gives no test with copper (II) hydroxide. Glycogen is formed in the
body cells from molecules of glucose. This process is called glycogensis. When
glycogen is hydrolyzed into glucose , the process is called glycogenolysis.
glycogensis
Glucose
glycogen
glycogenalysis
49
Proteins
Proteins are high molecular weight compounds containing the
elements carbon , hydrogen , oxygen and nitrogen. Some proteins
also contain sulphur , phosphorus , iodine , iron , cupper and zinc .
Proteins are the chief constituents of all cells of the body and the
animals cannot synthesize protein from raw materials so they must
obtain their protein from plants or from other animals , which in turn
have obtained the proteins from plants .
Proteins serves to build new cells, to maintain existing cells, and to
replace old cells in the body ,protein is necessary for the formation of
the various enzymes and hormones in the body ,the oxidation of
protein yields 4 Kcal/g (Fat yield 9 Kcal/g and carbohydrat yeild 4
Kcal/g ).
Proteins are polymers built up from simple unites called amino
acids ,there are 20 known amino acids that can be produced by the
hydrolysis of protein .
Amino acids are organic acids having an amine group(-NH2) attached
to a chain containing an acid group on the alpha carbon that is , the
carbon atom next to the acid group .
COOH
NH2
α
C
L-amino acid
H
R
50
nitrogen – fixing
bacteria
Air nitrogen
denitrifying
bacteria
soluble nitrogen
compound in soil
Soil nitrates
synthesis
decay
animal wast
products
Plant protein
Animal protein
Nitrogen cycle in nature
Classification of amino acids:
A.A
Aliphatic A.A
Aromatic A.A
Hetrocyclic A.A
NH2
NH2
CH2
C
COOH
H C ═ C – CH2 – C–COOH
N
H phenylalanine
NH
H
histidine
CH
Neutral A.A
NH2
H─C─COOH
H
glyeine
Acidic A.A.
Basic A.A.
COOH
NH
NH3─C─H
H
NH2
H2N – C –- N – CH2 CH2 – C – COOH
CH2
arginine
COOH
aspartic acid
51
Amphoteric nature of amino acids :
Amino acids contain a – COOH group , which is acidic , and an
– NH2 group , which is basic . In solution , the carboxyl group can
donate a hydrogen ion to the a mino group , forming a dipolar ion ,
called zwitter ion .
R –CH – COOH
R – CH – COO‾
NH3+
NH2
amino acid
zwitter ion form of an amino acid .
Amino acids are amphoteric compounds that is, they can react
with either acids or bases . When amino acids are placed in a basic
solution, it forms a negatively charged ion that will be attracted
towards a positively charged electrode. In an acid solution, the amino
acid forms a positively charged ion that will be attracted towards a
negatively charged electrode .
H+
R –CH – COOH
NH3+
positively charged ion
R –CH – COO
NH3+
zwitter ion
( in acid solution )
−
OH‾
R –CH – COO−
NH2
negatively charged ion
( in basic solution )
Since amino acids are amphoteric, proteins, which are made up
of amino acids, are also amphoteric. This amphoteric nature of
proteins accounts for their ability to act as buffers in the blood, they
can react which either acids or bases to prevent an excess of either .
52
Synthesis of peptides :
C6H5 – CH2– OH + Cl – CO – Cl
benzyl alcohol
phosgene
HCl + C6 H5 – CH2 – O – CO – Cl
benzyloxycarbonyl chloride
H
C6 H5 – CH2 – O – CO – Cl + H – N – CH – COOH
CH3
L-analine
H
C6 H5 – CH2 – O – CO – N + CH – COOH + PCl5
CH3
Carbobenzoxy – L-alanine
H
H
C6 H5 – CH2 – O – CO – N + CH – COCl + H – N + CH – COOH
CH2
CH3
Carbobenzoxy – L-alanylchloride
C6H4OH
H
L-tyrosine
H
C6 H5 – CH2 – O – CO – N– CH – CO– – N + CH – COOH
CH3
H2
CH2
Carbobenzoxyl – L-alanyl – L – tyrosine
C6H4OH
H
C6 H5 – CH3 + CO2 + H2N – CH – CO – N – CH – COOH
CH3
CH2
C6H4OH
L-alanyl – L- tyrosine
53
Nucleic acid
Nucleic acids fall into two principle classes according to the
nature of the sugar they contain :
The deoxyribonucleic acids DNA and the ribose-containg
nucleic acids RNA, DNA is found in cell nuclei and in mitochondria,
RNA is found in the cytoplasm .
RNA and DNA are polymers composed of nitrogenous bases,
sugars and phosphoric acid, thus (bases–sugars–phosphate)n
Where n is a large number nucleic acid . The linkage between
the polymer units is the phosphate diester bond :
O
– sugar – O – P – O – sugar
OH
when the phosphate diester bond is hydrolyzed , they consist of
nitrogenous bases ,a sugar , and phosphate , this unit is called
nucleotide .
O
Base – sugar –
O–p–H
O
when the ester bond between the sugar and the phosphate group in a
nucleotide is hydrolyzed a nitrogenous base and sugar is obtained this
is called nucleotide .
54
base – sugar – OH
RNA
polymer
(base – ribose – phosphate) n
O
Nucleotide
base – ribose – O – P – OH
OH
Nucleotide Bases
base – ribose – OH
Purines :
adenine , guanine
Pyrimidines :
cytoine , uracil ( and others )
DNA
Polymer
( base – deoxyribose – phosphate ) n
O
Nucleotide
base – deoxyribose O – P – OH
Nucleotide Bases :
OH
base – deoxyribose – OH
Purines :
adenine , guanine .
Pyrimidines :
cytoine , thymine .
The sugars present in nucleic acids :
55
Both sugars are present in nucleic acids as the B – Furanoside
ring structure :
OHCH2 O
H
H
OH
H
H
H
OH
OH
OHCH2
H
H
OH
H
OH
OH
B – D – ribofuranoseB – D –
deoxyribofuranos
The bases found in nucleic acids are:
HO
OH
OH
C
C
N
CH
C
CH
HO
NH2
C
N
C CH3
C
CH
HO
N
Thymine
N
Uraci
N
C
N
CH
C
N
C
CH
OH
N
N
C
H2N C
C
C
HC
CH
N
Cytosine
NH2
C
N
N
N
CH
N
H
Guaninne
H
Adenine
56
The structure of nucleotide:
NH2
C
N
N
C
HC
C
N
CH
N
O
O
HO – P – O – CH2
OH
H
H
H
H
OH
OH
((Adenine nucleotide))
The bond between nucleotides is:
OH
O
base – sugar – OH
OH
HO – P – OH
HO – sugar – base
OH
OH
– H2O
O
OH
base – sugar – O – P – O – sugar – base
OH
57
Lipids
Lipids have the following general properties :
1. Insoluble in water .
2. Soluble in organic solvents such as alcohol, ether, acetone, and
carbon tetrachloride .
3. Contain carbon , hydrogen , and oxygen , sometimes contain
nitrogen and phosphorus .
4. Yield fatty acids on hydrolysis or combine with fatty acids to
form an ester .
5. Take part in plant and animal metabolism .
Fatty Acids :
Fatty acids are straight – chain organic acid , and its may be saturated
or unsaturated . The fatty acids that are found in natural fats usually
contain an even number of carbon atoms .
C17 H35 COOH
stearic acid, saturated, (plants and animals).
C17 H33 COOH
Oleic acid , unsaturated ( 1 double bond ) olive oil.
Linolic acid , unsaturated ( 2 double bond
C17 H31 COOH
)Linseed oil.
Oilc acid occurs in nature as the cis configuration, the trans forms is
called elaidic acid.
CH3
C H2
CH2
CH2
CH2
CH2
CH2
CH === CH
CH2
CH2
CH2
CH2
CH2
CH2
COOH
CH2
CH2
Oleic acid ( cis)
CH3
C H2
CH2
CH2
CH2
CH2
CH2
CH === CH
CH2
CH2
CH
CH2
elaidic acid ( trans ) .
58
CH2
CH2
CH2
COOH
CH2
Classification :
Lipids are divided into three main categories :
1.
Simple lipids :
Simple lipids are esters of fatty acids. The hydrolysis of a
simple lipid may be expressed as :
hydrolysis
Simple lipid + H2O
fatty acid + alcohol
If the hydrolysis of a simple lipid yields a fatty acid and glycerol ,
the simple lipid is called a fat .
If the hydrolysis of a simple lipid yields a fatty acid and high
molecular weight alcohol, the simple lipid is called a wax .
2. Compound lipids :
Compound lipids on hydrolysis yields a fatty acid , an alcohol ,
and some other type of compound . These compound lipid undergo
hydrolysis as follows :
hydrolysis
phosolipids + H2O
fatty acid + alcohol + phosphoric
acid + a nitrogen compound.
glycolipids + H2O
hydrolysis
fatty acid + a carbohydrate + sphingosine
( anitrogen –containing alcohol ) .
3. Derived Lipids :
Derived Lipids are compounds
derived from simple and
compound lipid on hydrolysis. Derived Lipids include such
substances as fatty acid , glycerol , other alcohols , and sterols ,
which are solid alcohols having a high molecular weight .
59
Fats and Oils :
Fats are esters formed by the combination of a fatty acid with
one particular alcohol , glycerol .
H
H
HO – C – H
C17H35 COO – C – H
3C17H35 COOH + HO – C – H
C17H35 COO – C – H
stearic acid
C17H35 COO – C – H
HO – C – H
H
H
glycerol
glyceryl tristearate ( a
fat ) Since stearic acid is a saturated fatty acid , the product is a fat .
As the degree of unsaturation of the fatty acid increases , the melting
point decreases . Fats with a melting point below room temperature
are called oils .
Iodine Number :
The iodine number of a fat or oil is the number of grams of
iodine that will react with the double bonds present in 100g of that fat
or oil . The higher the iodine number , the greater the degree of
unsaturation of the fat or oil . Fats have an iodine number below 70 ,
where as oils have iodine number above 70 .
Physical Properties :
Pure fats and oil are generally white or yellow solids and liquids
, respectively . they are also odorless and tasteless .
60
Fats and oils are insoluble in water but are souble in such organic as
benzene , acetone , and ether .
Fats and oils must be emulsified by bile in the body before they can
be digeted .
Chemical Reactions :
1- Hydrolysis :
When fats are treated with enzymes , acids , or bases , they
hydrolyze to form fatty acids and glycerol .
H
H
C15H31COO – C – H
C15H31COO – C – H + 3H2O
C15H31COO – C – H
HO – C – H
heat
3C15H31COOH HO – C – H
enzyme
palmitic acid HO – C – H
H
H
tripalmintin
glycerol .
2- Saponification:
Is the reaction of a fat with a strong base such as sodium
hydroxide to produce glycerol and the salt of a fatty acid .
H
H
C17H35COO – C – H
HO – C – H
C17H35COO – C – H + 3NaOH
3C17H35COONa + HO – C – H
sodium stearate
HO – C – H
a soap
H
glycerol
C17H35COO – C – H
H
tristearin
61
3- Hydrogenation :
Vegetable oils may be converted to fats by the addition of
hydrogen in the presence of a catalyst . This process is called
hydrogenation .
H
H
C17H33COO – C – H
C17H35COO – C – H
C17H33COO – C – H + 3H2
C17H35COO – C – H
C17H33COO – C – H
C17H35COO – C – H
H
triolein , an oil
(contains double bonds)
H
tristearin , a fat
(contains single bonds)
4- Acrolein Test :
The acrolein test , which is a test for the presence of glycerol , is
sometimes used as a test for fats and oils , since all fats and oils
contain glycerol .
When glycerol is heated to a high temperature , especially in the
presence of a dehydrating agent such as potassium bisulfate (KHSO4)
, a product called acrolein is produced .
H
H – C – OH
H – C – OH
H – C – OH
H
Heat
KHSO4
H–C=O
H–C
+ 2H2O
H–C
H
glycerol
acrolein
62
This substance is easily recognized by its strong pungent odor . When
fats or oils are heated to a high temperature or are burned , the
disagreeable odor is that of acrolein .
5- Rancidity :
Fats develop an unpleasent odor that taste when allowed to
stand at room temperature for a short period of time .
That is , they become rancid . Rancidity is due to two types of
reactions - hydrolysis and oxidation .
- Hydrolysis takes place between the fats and the water in the
presence of air microorganisms at room temperature . The products
of this hydrolysis are fatty acids and glycerol .
- Oxygen present in the air can oxidize some unsaturated parts of fats
and oils . If this oxidation reaction produces short chain acids or
aldehydes .
63
Vitamins
In addition to oxygen, water, protein, fats, carbohydrate and certain
inorganic salts, a number of organic compounds are also necessary
for the growth & health.
These compounds are known as vitamins, which defined as an
organic substance occurring in small amount in fresh native foods.
The vitamins have been classified into the: 1. Fat soluble vitamins (vitamin A, D, E and K).
2. Water soluble vitamins (vitamin C and B complex).
Vitamin A is found in fish liver oils, in butter, and in milk.
Vitamin A is an alcohol and occurs in two forms – vitamin A1 and
vitamin A2. The recommended daily adult dose of vitamin A is 5000
IV for adult male and 4000 IV for adult female.
A lack of vitamin A, produces keratinization in the membrances
of the eyes, digestive tract, respiratory tract, and genitourinary tract.
Vitamin A is stored in the liver .
CH3
CH3
1
2
3
4
CH3
CH3
CH = CH - C = CH- CH = CH - C = CH = CH2OH
CH3
Vitamin A1
Vitamin A2 contain an additional double bond between C3 and C4
B complex vitamins:
The B - complex represents a whole series of vitamins. Each
member of the B-complex vitamins has a different physiologic
64
activity. The B-complex, also called the vitamin B-family, contains
the following vitamins :
1. Vitamin B1 ----- thiamine
2. Vitamin B2 ----- riboflavin
3. Niacin
4. Pyridoxine
5. Pantothenic acid
6. Lipoic acid
7. Biotin
8. Folic acid
9. Inositol
10. p. Aminobenzoic acid
11. Cyanocobalamin – vitamin B12
Vitamin B1 (thiamine ) occurs naturally in yeast , milk , and
whole grains also may be made synthetically. A deficiency of this
vitamin causes a lack of appetite , failure of growth , and loss in
weight .
CH3
NH2
CH2 – CH2OH
CH2
N
+
N
Cl‾
S
N
CH3
N
Thiamin hydrochloride
65
Vitamin C {Ascorbic Acid}:
Fresh fruits and vegetables such as
oranges , lemons ,
grapefruit , …..e.x. are excellent sources of vitamin C . It can easily
be oxidized to dehydroascorbic acid . Both ascorbic acid and
dehydroascorbic acid are biologically active . A deficiency of vitamin
C produces a disease known as scurvy .
C═O
│
HO ─ C
║
HO ─ C
│
H─C
│
HO ─ C ─ H
│
CH2OH
C═O
│
O═C
│
O═C
│
H─C
│
HO ─ C ─H
│
CH2OH
O
Ascorbic acid
O
Dehydroascorbic acid
Vitamin D :Vitamin D is sometimes called “ the sunshine vitamin “ . The
richest sources of vitamin D are oils from such fish (sardines,
salmon).The vitamin D content of the body may be increased by
exposure of the skin to ultraviolet rays from the sun .
The D vitamins are a group of sterols. The two most important
vitamins in the D group are vitamin D2 (ergosterol ) and vitamin D3
(activated 7-dehydrocholesterol). Vitamin D1 originally called
vitamin D, was found to be a mixture of vitamins D2 and D3 .
The principal action of vitamin D is to increase the absorption
66
of calcium and phosphorus from the small intestines. It also functions
in the deposition of calcium phosphate in the bones and teeth , and is
necessary for normal growth and development .
Rickets, a disease primarily of infancy and childhood, was
previously thought to be due to a deficiency of vitamin D
CH3
CH3
CH3
│
│
CH─ CH ═ CH ─ CH─ CH
CH3 │
│
CH3
CH2
║
Vitamin D2
HO
CH3
CH3
│
CH─ CH2 ─ CH2 ─ CH2─ CH
CH3│
│
CH3
CH3
│
7-dehydrocholesterol
Ultraviolet light
HO
CH3
CH3
│
CH─ CH2 ─ CH2 ─ CH2─ CH
CH3 │
│
CH3
CH2
║
HO
Vitamin D3
67
Vitamin E :Vitamin E is found in milk, eggs, fish and corn oil. Vitamin E is
known to prevent sterility in animals. A deficiency of vitamin E in
male rats produce spermatozoa .
.There are several vitamins E .the most important is called αtocopherol; ß an δ-tocopherol are less active and 8-tocopherol is inactive
CH3 H2
HO
H3C
H2
O
CH3
CH2 ─ CH2 ─ CH ─ (CH2)3 ─ CH ─ (CH2)3 ─CH ─ CH3
│
│
│
CH3
CH3
CH3
CH3
α-tocopherol
Vitamin K :-
Vitamin K is found in spinach, liver, eggs and cheese It is
necessary for the production of prothrombin in the liver. When there
is a deficiency of vitamin K , there is a lack of prothrombin and thus a
prolonged clotting time for the blood .
There are two naturally occurring K vitamins, k1 and K2 .
Vitamin K1 differs from K2 only in the nature of the side chin.
K1 is produced in plants and K2 by intestinal bacteria .
O
║
║
O
CH3
CH3
CH3
CH3
CH3
│
│
│
│
CH2 ─ CH ═ C ─ (CH2)3 ─ CH ─ (CH2)3 ─ CH ─ (CH2)3 ─ CH ─ CH3
Vitamin K1
68
Enzymes are biological catalysts that increase the speed of a
chemical reaction but do not themselves change.
Enzymes are organic catalysts produced by living organism.
Each enzyme will affect only specific substances called substrates.
Enzymes are proteins and will undergo all the reactions of
protein. That is, enzymes may be coagulated by heat, alcohol, strong
acids, and alkaloidal reagents.
Properties of Enzymes Composition:
1. Effects of temperature:
The best temperature for enzyme function – the temperature at
which the rate of reaction involving an enzyme is the greatest – is
called the optimum temperature for that particular enzyme. At higher
temperatures, the enzyme will coagulated and be unable to function.
At temperatures below the optimum temperature, the rate of reaction
will be slower than the maximum rate. Many enzymes have an
optimum temperature near 40C°, or close to that of body temperature,
so that they function at maximum effecting in the body.
69
2. Effect of pH:
Each enzyme has a pH range at which it can best function. This
is called the optimum pH range for that particular enzyme. For
example, the optimum pH of pepsin, an enzyme found in the gastric
juice, is approximately 2, whereas the optimum pH of trypsin, an
enzyme found in the pancreatic juice, is near 8.2 , if the pH of a
substrate is too far from optimum pH required by the enzyme that
enzyme cannot function at all.
Activators and inhibitors:
Inorganic substances that tend to increase the activity of an
enzyme are called activators. For example, the magnesium ion, Mg2+,
is an inorganic activator for the enzyme phosphatase.
Substances that tend to decrease the activity of enzymes are
called inhibitors. Inhibitors may act by combining directly with the
enzyme and so effectively remove it from the substrate, or they may
react with the activator so that it, in turn, cannot active the enzyme.
Heat, changes in pH, alcohol, and alkaloidal reagents all can denature
protein. These are examples of nonspecific inhibitors, they affect all
enzymes in the same manner. Specific inhibitors affect one single
enzyme or group of enzymes. In this category are most poisonous
substances, such as cyanide, CN ‾, which inhibits activity of the
enzyme cytochrome oxidase.
70
Mode of Enzyme Activity:
How do enzymes act? Why are they so specific toward certain
substrates?
It is believed that enzyme activity occurs in two steps.
First, the active site of the enzyme combines with the substrate to
form an enzyme – substrate complex. This enzyme – substrate
complex then breaks up to form the products and the free enzyme that
can react again. According to this theory (the lock – and – key
method), the substrate must " fit " into the active site of the enzyme –
hence the specificity of that enzyme. (Fig.1). If some other substance
should fit into the active site of the enzyme, it could prevent that
enzyme from reacting with the substrate. Such a substance is called a
competitive inhibitor (Fig.2).
An example of a competitive inhibitor is sulfanilamide . It's structure
is similar to that of p – aminobenzoic acid , PABA , which certain
bacteria require for growth . The " sulfa " inhibits the growth of the
bacteria by competing for the active site on the enzyme .
71
Active site
A B
A B
Enzyme
Substrate
A
Enzyme – substrate complex
B
Products
Fig . 1 : Interaction of enzyme and substrate
Active site
Enzyme
competitive
inhibitor
Enzyme – inhibitor complex
Fig . 2 : Enzyme – inhibitor complex
72
Nomenclature :
Under the older system of naming enzymes the substrate was not
mentioned . Formerly enzymes were given names ending in – in (
Table 1 ) . The current system for naming enzyme utilizes the name
of the substrate or type of reaction involved , with the ending – ase
(Table 2) .
Table 1 : Enzymes named under the older system
Enzyme
Substrate
Rennin
Casein
Pepsin
Protein
Ptyalin
Carbohydrate
Table 2 : Enzymes and substrates
Proteases
Protein
Carbohydrases
Carbohydrates
Hydrolases
Hydrolysis reactions
73
Classification :
The commission on enzymes of the International Union of
Biochemistry has classified enzymes into six divisions .Each of these
divisions can be further subdivided into several classes :
1. Oxidoreductase: (enzymes that catalyze oxidation – reduction
reactions between two substrates) .
2. Transferases: (which catalyze the transfer of a functional group
between two substrates) .
3. Hydrolases: (which catalyze hydrolysis reactions) .
4. Lyases: (which catalyze the removal of groups from substrates by
means other than hydrolysis) .
5. Isomerases: which catalyze the interconversion of optical
,geometric ,or stractureal isomers .
6. Ligases: (which catalyze the coupling of two compounds with the
breaking of pyrophosphate bonds) .
74
CREATINE AND CRIATININE
Creatine, synthesized in the liver from three amino acid,
argnine, glycine, and methionine.
CH3 – N – CH2COOH ATP
ADP CH – N – CH COOH
3
2
C == NH
C == NH
NH2
Creatine
HNPO3
(Creatine)
Creatine phosphate
Excess H2O
H5PO4
CH3 – N – CH2
HN== C
C==O
NH
Creatinine
Creatine phosphate act as astronge from of energy, and
converted into a non – protein nitrogen wast producl creatinine which
removed from plasma in the urine.
75
UREA
Urea is the main product of protein metabolism in the body, and
this take place in the liver.
Krebs proposed a cyclic mechanism for urea synthesis
involving:
CO2
+
NH3
+
NH2
NH2
NH2
CO
C==NH
(CH2)3
NH +H2O HC – NH2
(CH2)3
(CH2)3
NH
(CH2)3
HC – NH2
HC – NH2
HC – NH2
COOH
COOH
COOH
Ornithione
citrulline
urginine
NH2
COOH
NH2
+
CO
NH2
Urea
ornithine
Two molecules of NH3 and one molecule of CO2 are converted to a
molecule of urea for each turn of the cycle.
76