chemistry book by supun ayeshmantha

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Content

Unit 1 –Atomic Structure 03

Unit 2 –Chemical Calculations 26

Unit 3 – State of matter 46

Unit 4 – Energetic 53

Unit 5 – S, P, D Blocks of Periodic table 62

Unit 6 – Organic Chemistry 91

Unit 7 – Hydrocarbons 103

Unit 8 – Alkyl Halides 114

Unit 9 – Oxygen containing organic compounds 117

Unit 10 –Nitrogen containing organic compounds 133


Unit 1 –Atomic Structure

Properties of Electrons, Neutrons and Protons Properties Electron Proton Neutron

Mass of particle 9.107 x 10-13 1.6725 x 10-27 1.6725 x 10-27

Relative mass 1/1840 1 1

Charge c -1.602 x 10-19 1.602 x 10-19 0

Relative Charge -1 +1 0

Types of Rays

Cathode rays Canal rays

Million Discovered by Oil drop experiment

Goldstyle used a perforated cathode and found this ray

They have a mass. Consist positive particles

Travel in straight lines Mass depend on residual gas.

Negatively charged Mass is equal to A.M. of gas.

Made up of particles. E/M ratio is different in different gases.

Generated from cathode.

E/M ratio is constant.

Radioactivity -Discovered by Bekeral. -Radioactivity is the spontaneous decompositional action of certain element into a stable condition.

-This produces three types of rays -Gamma -Beta -Alpha


Gold foil Experiment

-Gigan and Marsden under the guidance of Rutherford, directed a Alpha ray to gold foil

-Majority travelled through -1/8000 get deviated -1/20000 Travelled backward

Atom

-Rutherford measured deflected angle and calculated the radius of

nucleus as 10-13cm and 10-8cm as average atomic radius. -Rogan discovered, Atomic no = No of Protons = No of electrons -Chadwick directed alpha particles to Berilium and produced uncharged particles, after directed toward wax, it change into charged particles. He named them as Neutrons

Isotope -With same atomic no but different mass number. -Using Mass spectrometer, we can detect -No of isotopes. -Relative isotopic mass. -Relative abundance.

Spectrum -When E is energy, h is plank constant and f is the frequency then E =hf -There are two types of spectrums, -Continuous spectrum -Discontinuous spectrum 1-Absorption 2-Emmision


Continuous spectrum

-When white light send through Sodium vapour, a dark region can

be seen. This is due to absorption of energy

-This spectrum is known as Atomic absorption spectrum.

Emission spectrum

-When white light send through a excited sodium vapour, only one region can be seen.

-A combination of both the systems can form a complete spectrum.


Emission system of H2

Bohr Theory

-Every energy level has a definite energy. -When energy is supplied to electrons it gets excited and go to higher energy levels. -To stay stable, electrons release energy and come to lower energy levels. -When rotating in an level no absorption or release of energy.

Four Quantum numbers -Principle quantum no (n) No of main energy level -Azimuthal quantum no (l) Shape of the orbital n=1 S=0 0 n=2 S=0, P=1 0, 1 n=3 S=0, P=1, D=2 0, 1, 2


-Magnetic quantum no (m) In which orbital the electron exist in a certain sub energy level. S 0(l) 0 P 1(l) -1, 0, 1 D 2(l) -2, -1, 0, 1 , 2

F 3(l) -3, -2, -1, 0, 1 , 2, 3 -Spin quantum no (ms) The direction in which the orbital is

ms = ½ 0r - ⅟2


Arrangement of Electrons

-Hund’s rule Electrons will fill orbital by keeping their spins parallel. -Pauli’s exclusive principle Two electrons do not have similar quantum numbers. -Aufban’s Principle When arranging electrons, they go to the lowest energy level. The electrons arrange according to the following diagram.

Special Configurations

Half filled sub energy level is more stable than normal configuration.

Variation Ionization Energies -Ionization energy across the period increases, but Group (ii) and (iv) elements have higher IE. -Group (i) Minimum IE -Group (viii) Maximum IE - Group (v) > Group (iv) (Partially filled P orbital)


Electro negativity -The capacity or tendency of an atom to attract the shared pair of electrons is known as EN. -Variation En increases from left to right. En decreases down the group.


Chemistry of Bonds

Chemical bonds

H + H H2 Exothermic Reaction

H2 H + H Endothermic Reaction

Ionic bond

- Ionic bonding is the attraction of oppositely charged ions (cations and anions) in Large numbers to form a solid. Such a solid compound is called an

ionic solid. -When electronegativity difference is higher than 2.1 ,it shows ionic characters. - When electronegativity difference is equal to 2.1, then it is 50% ionic and 50% covalent. -In nature they form giant 3d lattice like NaCl, MgCl Physical properties

-High melting and boiling points. -Do not conduct electricity and heat.(Because the electrons are immobile) -Solid at room temperature.

Elements that can form ionic bonds

- (i) and (ii) groups form Anions.

- (vii) group forms Cations

Covalent Bond - A covalent bond is formed when two atoms share one or more pairs of electrons. -Covalent bonding occurs when the electro negativity difference, (EN), between Elements (atoms) is zero or relatively small.


- Ionic bonding cannot result from a reaction between two nonmetals, because their electro negativity difference is not great enough for electron transfer to take place. Instead, Reactions between two nonmetals result in covalent bonding.

Comparison between Ionic and Covalent Compounds

Sigma bond Pie bond

Axial overlapping Lateral overlapping

Symmetrical overlapping P-P overlapping

Overlap is greater Overlap is lesser

Bond is strong Bond is weaker

Electron density is high between nuclei

Less electron density

One atom relative to other can rotate

No movement


Dative bond

-In covalent bond the two atomic orbitals should have unpaired electrons -One should contain vacant orbital while other should have lone pair of electrons.

Polar Covalent bond

-If covalent molecule is formed between similar atoms electrons spread symmetrically. -If not electron doesn’t spread symmetrically. Due to that one atom get slight negative charge while other get slight positive charge. H+β - Cl-β

-The attractive forces between them are Static attraction Covalent bond force

-This type of bonds are more stronger than non polar bonds. -Physical properties They exist in simple molecules. Do not conduct electricity and heat. Melting and boiling points are very low.

Metallic bonds.

-The valency electrons of metallic elements release and form common stream of electrons and positive ions. -In metallic bonds, lattices are formed due to that. -As a result of delocalized electrons it conducts electricity. -The strength of a metallic bond depend on No of available electrons for bond. Charge density of Cation

Polarizability

-When an ideal ionic compound is considered, the anions and cations which are its components are regarded as existing in the form of regular solid spheres. -But depending on the nature of the cations and the anion which are the constituents of the ionic compound, the electron cloud of the anion gets attracted towards the electron cloud of the cations and as a result distortions occur in the electron cloud of the anion.


Factors which increase the polarizability 1) Cations : Smaller in size. Highly charged 2) Anion : Larger in size. Highly charged

Dot cross Structures The writing of Lewis formulas is an electron bookkeeping method that is useful as a first approximation to suggest bonding schemes. It is important to remember that Lewis dot formulas only show the

number of valence electrons, the number and kinds of bonds, and the order in which the atoms are connected. They are not intended to show the three-dimensional shapes of molecules and polyatomic ions.


Lewis Structures Chemical bonding usually involves only the outermost electrons of atoms, also called valence electrons. In Lewis dot representations, only the electrons in the outermost occupied s and p orbital’s are shown as dots. Paired and unpaired electrons are also indicated. All elements in a given group have the same outer-shell electron

configuration. It is somewhat arbitrary on which side of the atom symbol we write the electron dots. We do, however, represent an Electron pair as a pair of dots and an unpaired electron as a single dot


Octet Rule

Representative elements usually attain stable noble gas electron

configurations when they share electrons. In the water molecule eight electrons are in the outer shell of the O atom, and it has the neon electron configuration; two electrons are in the valence shell of each

H atom, and each has the helium electron configuration. Likewise, the C and O of CO2

In most of their compounds, the representative elements achieve noble gas configurations.

VALENCE SHELL ELECTRON PAIR REPULSION

(VSEPR) THEORY

Molecule (Example)

Total pairs

Bonded

pairs

Lone pairs

Shape Name

BeCl2 2 2 0 Linear

BeCl3 3 3 0 Trigonal plainer

O3 3 2 1 Angular

CF4 4 4 0 Tetrahedron

NH4 4 3 1 Trigonal

Pyramidal H2O 4 2 2 Angular

PH5 5 5 0 Trigonal by

pyramidal

SF4 5 4 1 See-saw

5 3 2 T-shape Xecl2 5 2 3 Linear

SF6 6 6 0 Octahedron

6 5 1 Square pyramid

XeCl4 6 4 2 Square Plainer


-There are three types of repulsions in-between atoms Lone-Lone pair repulsion Lone-Bond pair repulsion Bond-Bond pair repulsion

Hybridization

-The electron configuration of carbon is 1s2, 2s2, 2p2

- In the last shell of carbon there’s only 4 electrons and only 2 unpaired -Therefore carbon can form only 2 bonds.

-But carbon can form 4 bonds using all valency electrons. -Using hybridization we can explain this. SP3 Hybridization - Four sigma bonds are formed with 4 covalent bonds. - One S orbital get the energy of P orbital. - Tetrahedron in shape.


SP2 Hybridization

-One S orbital and 2 P orbitals are involved. -A sigma and a pie bond is formed.

SP1 Hybridization -One sigma and two pie bonds are formed.


Resonance Structures. -A molecule or polyatomic ion for which two or more Lewis formulas with the same -Arrangements of atoms can be drawn to describe the bonding is said to exhibit resonance.

- The three structures above are resonance structures of the carbonate ion. The relationship among them is indicated by the double-headed arrows. This symbol does not mean that the ion flips back and forth among these three structures. The true structure can be described as an average, or hybrid, of the three.

- The dashed lines include that some of the electrons shared between

C and O atoms are delocalized among all four atoms; that is, the four pairs of shared electrons are equally distributed among three C-O bonds.

-Example, Draw resonance structure of SO2


Formal Charge. -Formal charge is the hypothetical charge on an atom in a molecule or polyatomic ion; to find the formal charge, we count bonding electrons as though they were equally shared between the two bonded atoms. The concept of formal charges helps us to write

correct Lewis formulas in most cases. The most energetically favorable formula for a molecule is usually one in which the formal charge on each atom is zero or as near zero as possible. - The formal charge, abbreviated FC, on an atom in a Lewis formula is given by the relationship FORMAL CHARGE = (GROUP NO)-(NO OF BONDS – NO OF UNSHARED ELECTRONS)

Rules for applying formal charge. -Molecules FC = 0 -Poly Atomic ions FC = Charge it carries

Dipole Moments.

It is convenient to express bond polarities on a numerical scale. We indicate the polarity of a molecule by its dipole moment, which measures the separation of charge within the molecule. The dipole moment, µ, is defined as the product of the distance, d, separating

charges of equal magnitude and opposite sign, and the magnitude of the charge, q. A dipole moment is measured by placing a sample of the substance between two plates and applying a voltage. This causes a small shift in electron density of any molecule, so the applied voltage is diminished very slightly. Diatomic molecules that contain polar bonds, however, such as HF, HCl, and CO, tend to orient themselves in the electric field . This causes the measured voltage between the plates to decrease more markedly forthese substances, and we say that these molecules are polar.


Dipole – dipole Interaction.

-DDI causes between covalent molecules because of +ve and –ve attractions. -DDI effective only for short distances. -Approximately the DDI energy is 4 KJ/Mol -This is weaker than covalent and ionic bonds. -When temperature increases this become less important *Percentage of Ionic character = µ(Observed) x 100 /µ (Calculated)

Dispersion Interaction. -DI are weaker attractive forces only over extremely short distances because they vary as 1/d7. -Present in-between all types of molecules in condensed phase, but weaker for small molecules. -DI are the only forces in-between symmetrical non-polar

(noble gases) substances such as SO2 ,CO2 ,O2 and mono atomic molecules.

Dipole – Induced Interaction.

-A polar molecule may polarize a neutral molecule which lies in it’s vicinity and thus induces dipolarity molecule there. -In that molecule magnetic dipolarity occurs. -This induced dipole interact with first atom and thereby the 2 molecules are attracted each other.


-Without DI these cannot form liquids and solids. -DI increases when the size of the molecule increases.

Hydrogen Bond - Hydrogen bond is a special electrostatic interaction formed

by a hydrogen atom covalently bonded to an electro negative atom (x) , with an electronegative atom (Y) containing one or more lone pairs of electrons. - Hydrogen bonds become more stronger when X and Y are more electronegative atoms such as F, O, N and Cl. Strength of hydrogen bonds decreases as F>O>N>Cl. This is named as FONCl rule. In some special cases, when the positions of X and Y are taken by atoms other than the above ones, strong hydrogen bonds are formed - Importance of hydrogen bonding

Without HB water is a gas HB exist in many tissues, organs, skin and bones of animals. Play important role in determination of structure of protein. -There are two types of hydrogen bondings. Intermolecular hydrogen bond Intramolecular hydrogen bond Intermolecular hydrogen bonding. -When HB take place between two difference molecules of same compound such as in HF , H2O it’s called IHB -Homointermolecular hydrogen bonding -This type of H bonding is formed in-between same molecule.

-Heterointermolecular hydrogen bonding -This bonding take place in-between different compounds. Intermolecular hydrogen bonding. -If HB takes place in-between Hydrogen and the electronegative atom presence within the same molecule.


Lattices -A formation where the building units are attached to one another in an orderly pattern can be described as a lattice. - The presence of a formal pattern and the formation from a repetitive basic unit is a common feature of the lattices. - The substances with lattice arrangements can be classified according

to their building units as follows. Homoatomic lattices Heteroatomic lattices Non – polar molecular lattices

Polar molecular lattices Ionic lattice

- The bonding formed during the formation of the lattices are different, depending on the nature of the building unit of the lattice substance. - The nature of the bonding formed during the formation of the

lattice affects the physical properties of the lattice. - Diamond and graphite lattices which are formed from homogeneous atoms are examples for homoatomic lattices. - Silicon dioxide which is formed from heterogeneous atoms is an example for heteroatomic lattices. - Iodine crystals which are formed from nonpolar iodine molecules are examples for non- polarized molecular lattices. -Ice which is formed from polar molecules is an example for polar molecular lattice. -Sodium chloride which consists of sodium ions and chloride ions is an example for ionic lattices. - Homogeneous and heterogeneous atomic lattices are formed by atoms covalently bonding with one another. -Non polar molecular lattices are built by the bonding of non polar molecules to one another by attractive forces formed between

induced dipoles. -Polar molecular lattices are formed by the bonding together of polar molecules to one another by attractive forces developed between permanent dipoles. -Ionic lattices are formed by the bonding together of ions by strong electrostatic attractions developed between positively charged and negatively charged ions. - The substances composed of homogeneous and heterogeneous atomic lattices have a high hardness and also higher melting points/ boiling points because such lattices are formed by strong covalent bonds.


-There is no tendency to go into solution because the covalent bonds present in atomic lattices are very strong. - Electricity is not conducted through atomic lattices as they lack mobile electrons. -The substances with nonpolar molecular lattices consisting of molecules bonded to one another by weak Van der Waals forces have

a low hardness and their melting points/ boiling points are also low relative to substances formed by other lattices. -The substances formed by nonpolar molecular lattices dissolve in nonpolar solvents because they consist of nonpolar molecules and they do not conduct electricity as they do not possess mobile electrons. -The substances formed by polar molecular lattices consisting of molecules bonded by permanent dipole – permanent dipole attractive forces (or hydrogen bonds) have a high hardness and also higher melting points / boiling points compared to substances formed by

nonpolar molecular lattices. -The substances formed by polar molecular lattices show a high tendency to dissolve in polar solvents but do not conduct electricity due to the lack of mobile electrons. - The substances formed by ionic lattices bonded by strong electrostatic attractions show higher melting points/ boiling points and also a high hardness. -The substances with ionic lattices show a tendency to dissolve in polar solvents. -The substances consisting of ionic lattices do not conduct electricity in the solid state due to the lack of mobile electrons. -The substances with ionic lattices conduct electricity in the molten state or in solution because of the presence of mobile ions.


Atomic lattices The lattices formed by one type of atoms are known as homoatomic lattices. Diamond and graphite which are allotropic forms of carbon are examples for these. The lattices formed by the combination of atoms which are different from one another are called heteroatomic

lattices and silica (SiO2) is an example of this. Diamond - Diamond is formed by covalently bonding together of carbon atoms which are subjected to sp3 hybridization. This lattice is arranged so that each carbon atom is bonded by single covalent bonds to four other carbon atoms. Graphite -Graphite is formed by covalently bonding together of carbon atoms

which are subjected to sp2 hybridization. -The lattice is arranged so that each carbon atom is attached to three other carbon atoms by single covalent bonds. -Within the lattice structures, each carbon atom is surrounded by three other carbon atoms which are arranged so that they occupy the vertices of a planar triangle. - This is a two dimensional giant structure. -In graphite each carbons atom has a p– orbital which is not hybridized and also there is an unpaired electron in it. -The unhybrid p– orbital is situated perpendicular to the plane of the two dimensional lattice. -Therefore the p- orbitals containing these unpaired electrons help to form attractions between the two dimensional carbon layers.

Molecular Lattice -Some of the molecular lattices are formed by homo atomic molecules while some others are formed by heteroatomic molecules. -Considering the polarity of the molecule which forms the lattice, they can be classified as nonpolar molecular lattices and polar molecular lattices. -Iodine crystal is an example for a non polar molecular lattice. Iodine molecule is a large molecule.


-Therefore due to various reasons iodine molecules develop induced dipoles temporarily. -Iodine lattice is formed by attractions made between such induced dipole - induced dipole attractions. - Water molecule is polar and ice is a good example for polar molecular lattice. Due to the permanent dipole in -OH bond water

molecules are bonded by hydrogen bonds in ice.

Ionic Lattice

-Many ionic compounds exist naturally as solid substances with a crystalline structure. -This crystal lattice is formed by the packing of oppositely charged ions of the ionic compound in a definite pattern is space. -Electrostatic attractions exist between these oppositely charged ions. -Sodium chloride is a good example for an ionic lattice.


Unit 2 –Chemical Calculations

Significant Figures

-There is a difference between counting and measuring. Counting can be done with certainty while measuring is only stating the nearest value. The number of meaningful figures included in the numerical value of the statement of a certain physical quantity is known as the significant figures.

-Example : The length of a pen can be measured by different

measuring instruments and stated as follows. -Measurement by a ruler marked with centimeters 18 cm -Measurement by a meters ruler 18.1 cm

-Measurement by a vernier caliper 18.12 cm Therefore, the stated value after measurement by a measuring instrument is an approximate value and not a definite value. Here the first measurement is stated with 2 significant figures. The second measurement is stated with 3 significant figures and the third measurement with 4 significant figures. It is necessary to manipulate significant figures correctly in addition, subtraction, multiplication and division of numbers

Rules

-Nonzero digits are always significant. -Zero’s at beginning are not significant. -Zero’s between nonzero numbers are significant.

-Zero’s at the end of a number with decimals are significant

Relative Atomic Unit (AMU)

This defined as exactly the 1/12 of the mass of an atom of particular kind carbon atom.

AMU = Mass of atom x 12 / Mass of carbon


Percentage composition Examples: 158.04 g of KMnO4 contains 39.10 g of K, 59.94 g of Mn and 64.00 g of O.

Empirical formula The formula that shows the simplest whole number ratio between the number of atoms in agreement with the composition of a compound is its empirical formula

Example: Empirical formula of benzene is CH

Molecular formula The formula that shows the exact number of atoms in the molecule of a compound is its molecular formula. The molecular formula of water is H2O. The ratio between the empirical formula and the molecular formula of a compound is a whole number.


Composition of a compound

-Composition stated as the number of parts per million of parts

(ppm) = fraction x 106

-Composition stated as the number of parts per billion of parts

(ppb) = fraction x 10 9


Avogadro Constant (L)

-Avogadro constant is given as L= N/n, Where

N = number of particles n = amount of substance

Avogadro constant (L) = 6.022 x 1023 mol -1

Gas Constant (R) -Gas constant (R) is encountered during the study of ideal gases. -Consider a sample of gas. When pressure of the gas is P, volume of the gas is V amount of the gas is n mole and absolute temperature of the gas is T

PV = nRT R = 8.314 J K-1 mol -1

Faraday constant (F) -Faraday constant is defined as the molar charge of the proton.

F = eL

L – Avogadro constant e – Charge of the electron

Faraday constant (F) = 96485 C mol-1


Balancing Chemical Equation.

-There are three ways of balancing chemical equations.

Inspection method. Mathematical method. By using half reactions.

Inspection method Balancing an equation by balancing the number of atoms of each kind in the reactants and products is called inspection balancing. In this method, the elements in the reactants and products are separately balanced. Elements with the least number of atoms are first considered and molecules composed of atoms of same element.

Mathematical method Balancing equations when the reactants and the products are known

by comparing the coefficients is called the mathematical method.


Balancing equations using oxidation / reduction half reaction


Oxidation number -This represents the no of electrons lost or gain to change into compound from free state. -This gets positive when electrons are given. -This gets negative when electrons are gained. -In covalent bond it’s a imagination. Rule -Uncombined or free state elements oxidation number is zero. -F always -1 -Metals always positive. -Oxygen -2 , when in peroxides -1 -Hydrogen +1 in ionic hydrides -1

-Alkali metals +1 -Alkaline earth metals +2 -Halogen halides -1 -Sulphur in sulphides -2 -The algebraic sum of oxidation number is equals to the net charge. -Oxidation number may be a fraction. -Maximum oxidation number = Group no (Except O,F) -Minimum oxidation number = Group no -8 (Except metals)


Chemical Calculations -There are various ways in which the composition can be expressed. --They are

Mass fraction Volume fraction Mole fraction Mass/volume Moles /mass

-When the composition is expressed in terms of moles / volume it is called as concentration. -Composition of substances present in very small amounts is commonly expressed in terms of parts per millions (ppm) and parts

per billions (ppb). -ppm and ppb can be expressed as ppm = mass fraction x 106 and ppb = mass fraction x 109. ppm and ppb can be also be expressed as ppm = volume fraction x 106 and ppb = volume fraction x109. -Density of water is 1000 g dm-3. Density of a dilute aqueous solution can be considered as approximately equals to the density of water. Therefore, Mass of 1 dm3 of a solution = 1 kg = 1000 g = 1000000 mg -For such instances, mass/ volume ratio can also be expressed in ppm. As a mass fraction, 1 ppm means that 100000 mg of the mixture contains 1 mg of the particular substance for dissolve solutions. For dilute solutions,


CALCULATIONS BASED ON CHEMICAL EQUATIONS

- Let us again consider the combustion of methane in excess oxygen. The balanced chemical equation for that reaction is


-A chemical equation also indicates the relative

amounts of each reactant and productin a given chemical reaction. We showed earlier that formulas can represent moles of substances. Suppose Avogadro’s number of CH4 molecules, rather than just one CH4 molecule, undergo this reaction. Then the equation can be written


THE LIMITING REACTANT CONCEPT

-In the problems we have worked thus far, the presence of an excess of one reactant wasstated or implied. The calculations were based on the substance that was used up first,called the limiting reactant.


PERCENT YIELDS FROM CHEMICAL REACTIONS

- The theoretical yield from a chemical reaction is the yield calculated by assuming that the reaction goes to completion. In practice we often do not obtain as much product froma reaction mixture as is theoretically possible.

-The actual yield is the amount of a specified pure product actu-ally obtained from a given reaction. -The term percent yield is used to indicate how much of a desired product is obtainedfrom a reaction.


SEQUENTIAL REACTIONS

-Often more than one reaction is required to change starting materials into the desired product. This is true for many reactions that we carry out in the laboratory and for many industrial processes. These are called sequential reactions. The amount of desired product from each reaction is taken as the starting material for the next reaction.


CONCENTRATIONS OF SOLUTIONS

- Concentrations of solutions are expressed in terms of either the amount of solute present in a given mass or volume of solution, or the amount of solute dissolved in a given mass or volume of solvent.

- Percent by Mass Concentrations of solutions may be expressed in terms of percent by mass of solute,which gives the mass of solute per 100 mass units of solution. The gram is the usual mass unit.

Molarity -Molarity (M), or molar concentration, is a common unit for expressing the concentrations of solutions. Molarity is defined as the number of moles of solute per liter of solution:


DILUTION OF SOLUTIONS

-Recall that the definition of molarity is the number of moles of solute divided by the volume of the solution in liters:

USING SOLUTIONS IN CHEMICAL REACTIONS


Unit 3 – State of matter Behaviour of matter

Gases

Ideal gases

-The temperature, pressure, volume and the amount (moles) of substance of a gas are the factors that affect the behaviour of a gas. -The ideal gas equation can be described as a relationship of the above four variables regarding a gas.

Derivation of Boyle law from the ideal gas equation


Derivation of Charles law from the ideal gas equation

Derivation of Avogadro law from the ideal gas equation

Pressure

-Total pressure = PA + PB + PC +……..+ PN

PA = XA . PTotal

Where XA = Mole fraction


Kinetic molecular theory Assumptions of the molecular kinetic theory -The molecules of a gas are in a state of continuous random motion. - When molecules collide with one another and bounce off the total kinetic energy of the system remains the same.

-The pressure exerted by a gas is the result of collisions of the molecules on the walls of the container. -Molecular kinetic equation is

PV =⅟3mNC-2

Where m -Mass N -Moles

C-2 - Mean square velocity

Escape velocity -The minimum velocity at which any object should be projected from the surface of the Earth to enter into space completely escaping

from the gravitational attraction of the Earth is called its escape velocity. -When the mean velocity of a gas molecule exceeds its escape velocity it leaves the atmosphere of the Earth. -The mean velocity of light gases such as hydrogen and helium exceeds the escape velocity from the Earth and as a result they have left the atmosphere of the Earth.


Diffusion -The spreading of a certain type of molecules throughout space occupied by another type of molecules is called diffusion.

Example :- When a substance with a scent is kept inside a room diffusion takes place until the scent is distributed uniformly throughout the room.

-Rate of diffusion - solids < liquids < gases -It has been experimentally found that different gases diffuse with different rates. -The production of ammonium chloride by the diffusion of ammonia and hydrogen chloride molecules through air can be demonstrated by the following apparatus. -From this it is clear that the rate of diffusion of ammonia molecules with a low molecular mass is higher than that of hydrogen chloride molecules.

-Variation of the mean velocity of a gas with temperature is shown by the following Maxwell - Boltzmann curves.

HCl Ammonia (Conc.)


Dalton’s law of partial pressure -The contribution made by the constituent gases towards the total pressure of a mixture of gases is called their partial pressures.

-The pressure that a constituent gas of a mixture of gases would exert if it alone occupied the volume of the container of the mixture is called the partial pressure of that gas. -In a mixture of gases, the total pressure is equal to the sum of the partial pressures of each of the constituent gases. -If the partial pressures of individual gases in a mixture of gases A, B and C are PA, PB and PC respectively, total pressure of the mixture

PT = PA+ PB + Pc

Derivation of Dalton’s law of partial pressure from the ideal gas equation

In a mixture of gases A and B there are nA and nB moles of each gas respectively.

Total pressure = PA + PB + PC +……..+ PN


Compressibility factor -z is the compressibility factor. For ideal gases z = 1. But the fact that this value is not a constant for real gases is revealed by experimental data.

-The graph of the product PV against P for different gases at the temperature 273 K

-According to this graph it is clear that real gases approach ideal behaviour under conditions of low pressure. -The graph of PV/RT against P for a mole of hydrogen at various temperatures.

It is clear from the above graph that at high temperatures the real gases approach ideal behaviour.


Van der Waals equation

P = Pressure V = Volume n = amount (moles) of substance R = Universal gas constant T = Absolute temperature

a and b are constants (Van der Waals constants) for real gases Calculations using Van der Waals equation are not necessary

Calculation of molecular energy

-The transition kinetic energy of a particle is given as

E =½mNC-2

PV =⅟3mNC-2

E =½mNC-2

Therefore 2E = 3 PV & 2E = 3nRT


Unit 4 – Energetic

-The portion of the universe selected for the study is called the system

-All the rest other than the portion selected for the study is called the environment -The dividing line that separates the system and the environment is called the boundary.

-Systems where there is an exchange of energy and matter across the boundary are called open systems.

-Systems where only energy is exchanged across the boundary are called closed systems

-Systems where there is no exchange of both energy and matter across the boundary are called isolated systems.

Enthalpy -The quantity of heat supplied to a system or given out from a system

under the condition of constant pressure is called the enthalpy change (∆H). -This is a thermodynamic property and a function of state. -The enthalpy change (∆H ) associated with a reaction is given by the difference in enthalpy of the products and reactants.

∆H= HProduct - HReactant


-Enthalpy change associated with a reaction; If ∆H < 0 the reaction is exothermic If ∆H > 0 the reaction is endothermic

-Standard conditions ∆Hø

Temperature – 298k / 25oc Pressure - 1 x 105 Nm-2 / 1Pa / 1 atm

Standard enthalpy change.

Standard enthalpy of formation -∆HøF

-It is the enthalpy change that occurs when one mole of the compound is formed in the standard state from the constituent elements in the standard state. Example: – Standard enthalpy of formation of H2O (l) is = 286 kJ

mol-1.

Standard enthalpy of combustion -∆HøC

-It is the enthalpy change that occurs when one mole of an element or a compound in the standard state undergoes complete combustion in an excess of oxygen.

Standard enthalpy of bond dissociation -∆Hø

D

- It is the enthalpy change that occurs when a gaseous species in the standard state is converted into gaseous components by breaking a mole of bonds (this is stated with respect to a specified bond in a specified element or a compound).


Standard enthalpy of neutralization -∆HøNEU

-It is the enthalpy change that occurs when a mole of H+ ions in an aqueous solution reacts with a mole of OH - ions in an aqueous solution under the standard state to form a mole of H2O.

H+(aq) + OH-

(aq) H2O(l) -∆Hø = -57kJmol-1

Standard enthalpy of solution -∆Hø

SOL

- It is the enthalpy change that occurs when a mole of gaseous ions under the standard conditions changes into the solution form in the presence of an excess of the solvent.

Mn+ + Excess solvent Mn+ (solvent)

Standard enthalpy of hydration -∆HøHDY

-It is the enthalpy change that occurs when a mole of gaseous ions

under the standard conditions changes into the solution form in the presence of an excess of water.

Na+(g) + H2O (l) Na+

(aq)

Standard enthalpy of transition -∆HøTRA

-It is the enthalpy change that occurs when a substance under standard conditions gets changed from one phase to another phase of the same substance (Here the phase means especially an allotropic form. -The enthalpy changes related to changes in physical state which are considered as phase changes will be described separately.)

C (graphite) C (diamond)

Standard enthalpy of dissolution -∆HøDES

-It is the enthalpy change that occurs when a mole of a substance

under the standard conditions is dissolved in a solvent to form 1dm3 of solution.

NaCl(aq) + H2O(l) NaCl(aq) ∆HøDES= 1kJmol-1

Hess’s law -The enthalpy change that takes place in a chemical reaction takes a constant value through whatever the route the reaction was performed.


A+B ∆H 1 C+D

∆H 2 ∆H 3

E

∆H 1 = ∆H 2 + ∆H 3

Enthalpy diagrams

-The below diagram represent a enthalpy diagram

Example – Dram enthalpy diagram for HBr


Enthalpy of formation of ionic compounds.

Standard enthalpy of sublimation-∆Høsub

-It is the change in enthalpy that occurs when a mole of a solid element or a mole of a solid compound under the standard conditions is converted completely into the gaseous state.

Ca(s) Ca(g) ∆Høsub= 193 kJmol-1

Standard enthalpy of evaporation -∆Hø

vap -It is the change in enthalpy that takes place when a mole of a liquid compound or element under the standard conditions is converted into a mole of a gaseous compound/element. Br2(l) Br2(g) ∆Hø

vap= 30.91kJmol-1

Standard enthalpy of fusion -∆Hø

fus -It is the change in enthalpy that takes place when a mole of a solid

compound or element under the standard condition is converted into a mole of liquid compound/ element. Al(s) Al(l) ∆Hø

fus= 10.7kJmol-1

Standard enthalpy of atomization -∆Hø

atm -It is the change in enthalpy that takes place when a mole of an element under the standard conditions is converted into a mole of atom in the gaseous state. ½Cl2(g) Cl(g) ∆Hø

atm= 121kJmol-1

Standard enthalpy of first ionization -∆Hø

Ie1 -It is the change in enthalpy that takes place when a mole of unipositive ions are formed under standard conditions by removing an electron each that is most weakly bonded to the nucleus from a mole of gaseous atoms of an element.

Na(s) Na+(s) ∆Hø

Ie1= 496kJmol-1

Standard enthalpy of electron affinity (or electron gain) -∆Hø

ea - It is the change in enthalpy that takes place when a mole of uninegative ions are formed in the gaseous state under the standard conditions when electrons are given to a mole of atom in the gaseous state. (All the time second EA is positive) Cl(g) Cl-(g) ∆Hø

ea= -352kJmol-1


Standard lattice enthalpy of an ionic compound -∆Hø

L -It is the change in enthalpy that takes place when one mole of an ionic compound in the solid state under the standard conditions are formed from gaseous positive ions and negative ions.

Na+(g) Cl-(g) ∆Hø

L= -780kJmol-1

Average Standard bond enthalpy -Average bond enthalpies are approximate values. -When you want to assign a value to a standard enthalpy of dissolution of C-H bond in methane, the problem is different the energy required to break first C-H is not same as that, it require to remove H from CH3

- radical Example

CH4 CH3 . +H -∆Hø

=426kJmol-1

CH3 . CH2

.. +H -∆Hø=436kJmol-1

CH2 .. CH

... +H -∆Hø=451kJmol-1

CH ... C

.... +H -∆Hø=347kJmol-1

Average bond enthalpy = (426+436+451+347)/4 = 415kJmol-1

Bond S.A.B.E (347kJmol-1)

C-C +347

C=C +611

N-H +389

O-H +464

C-H +414

C-Br +292

C-O +360

For any reaction ∆Hø

R = [∆Hø of breaking bonds] – [∆Hø

of bond formation]


Born Haber Cycle -The born Haber cycle is a technique which occurs when an ionic

compound is formed. -The Born–Haber cycle involves the formation of an ionic compound from the reaction of a metal (often a Group I or Group II element) with a non-metal. Born–Haber cycles are used primarily as a means of calculating lattice energies (or more precisely enthalpies) which cannot otherwise be measured directly. - The lattice enthalpy is the enthalpy change involved in formation of the ionic compound from gaseous ions. -A Born–Haber cycle calculates the lattice enthalpy by comparing the standard enthalpy change of formation of the ionic compound (from

the elements) to the enthalpy required to make gaseous ions from the elements. This is an application of Hess's Law.

Example – Formation of lithium fluoride

The enthalpy of formation of lithium fluoride from its elements lithium and fluorine in their stable forms is modeled in five steps in the diagram ; -Atomization enthalpy of lithium -Ionization enthalpy of lithium -Atomization enthalpy of fluorine -Electron affinity of fluorine -Lattice enthalpy - The same calculation applies for any metal other than lithium or any non-metal other than fluorine.


- The sum of the energies for each step of the process must equal the enthalpy of formation of the metal and non-metal, ΔHf

-V is the enthalpy of sublimation for metal atoms (lithium)

-B is the bond energy (of F2). The coefficient 1/2 is used because the formation reaction is Li + 1/2 F → LiF. -IEM is the ionization energy of the metal atom -EAX is the electron affinity of non-metal atom X (fluorine) -UL is the lattice energy (defined as exothermic here) The net enthalpy of formation and the first four of the five energies can be determined experimentally, but the lattice energy cannot be measured directly. Instead, the lattice energy is calculated by subtracting the other four energies in the Born–Haber cycle from the

net enthalpy of formation. - The word cycle refers to the fact that one can also equate to zero the total enthalpy change for a cyclic process, starting and ending with LiF(s) in the example. This leads to

Enthalpy changes when Ionic compounds dissolve -When an ionic solid dissolve in a solvent , two types of enthalpy terms are involved. i) The ion must be separated - Ionic lattice energy is required. ii) The separate ions interact with molecules of solvent

(If solvent is polar , ions can be attracted)

∆Hø

Dissolution = ∆HøLattice dissolution + ∆Hø

solvation


Free energy And Entropy -A reaction which happens of it’s own accord without any external help is known as SPONTANEOUS REACTION. -The difference of enthalpy between product and reactants cannot be the only factor which decides whether the reaction is happened or

not. -Entropy of a system is a measure of the randomness of the system. -Entropy is a function of state and it depends only on the initial and final state if the system and is independent of the path of the change. -Spontaneous changes in an isolated system takes place with an increase in entropy. -As the entropy related to a certain system is a function of state, the change in entropy can be calculated by subtracting the initial value of entropy from the final value of entropy. -The entropy of perfect crystal at 0K is 0

-When degree of order decreases then entropy also decreases ∆S = ∆SProduct - ∆SReactants

When Pressure Entropy Volume Entropy Temperature Entropy -When disorder increases the entropy +ve CaO(s) + H2O(l) Ca(OH)2(s) ∆S –ve CaCO3(s) CaO2(s) + CO2(g)

-Both entropy and enthalpy change is important to decide weather a chemical reaction has occur or not.

G = H-ST

Where H-Enthalpy

S-Entropy T-Temperature in K G-Free energy -The sigh of G represents the nature of the reaction If G > 0 ; Not spontaneous G < 0 ; Spontaneous G = 0 ; Reaction under equilibrium


Unit 5 – S, P, D Blocks of

Periodic table

PHYSICAL PROPERTIES OF S AND P BLOCKS

Atomic radius -Atomic radius increases down the group

Covalent radius - When two atoms of the same element are covalently bonded, half the internuclear distance of these two atoms is called its Covalent radius. -The covalent atomic radius increases down a group and decreases from left to right along a period.

Covalent radius


Van der Waals radius -When two molecules are placed as close together as possible, half the distance between the two nuclei which are close to each other is called the Van der Waals radius. Van der Walls radius = d/2

Metallic radius

-Half the distance between two adjacent cation nuclei in the metallic lattice is the metallic radius. Metallic radius = d/2 -Nuclear charge and the shielding effect affect the atomic radius.

Valency and oxidation numbers -Formation of anions and cations depends on the number of electrons in the valency shell and ionization energy. -The elements of the group I, II and III, form cations while the elements of the groups V, VI and VII form anions. -The elements in the group IV do not form M4+ ions. The reason is the high aggregate of first, second, third and fourth ionization energies. -When the atomic number increases along a period the highest oxidation number increases. -The highest oxidation number that an element can have in a compound is equal to its number of valency electrons.


Ionization Energy - the minimum amount of energy required to remove the most loosely

bound electron from an isolated gaseous atom to form an ion with a 1+ charge.

Electron Affinity -The electron affinity (EA) of an element may be defined as the amount of energy absorbed when an electron is added to an isolated gaseous atom to form an ion with a -1 charge.


-Across a period from left to right the nuclear charge and atomic radius decreases and the ionization energy increases. Therefore, the tendency to form cations decreases and also the ability for reduction decreases across a period.

-Similarly the ability to form anions increases and also the ability for oxidation increases from left to right across a period

Electronegativity -The electronegativity (EN) of an element is a measure of the relative tendency of an atom to attract electrons to itself when it is chemically combined with another atom. -Elements with high electronegativities (nonmetals) often gain electrons to form anions. -Elements with low electronegativities (metals) often lose electrons to form cations.


S-Block Elements

-The elements which having electrons in last shell S orbital, is known as S-BLOCK element. -Their general electronic configuration is ns1 ,ns2 n=1 to 7

-Fr , Ra are radio active elements. -IA - alkaline metals, they react with water to form alkali. -IIA -alkaline earth metals, they oxides and react with water to form alkali and found in soil or water

-The total number of S block elements – 14 -Cs and Fr are liquid elements.

Physical properties

FLAME COLORS

Li Red Be Colorless

Na Yellow Mg Bri: white

K Lilac Ca Brick red

Rb Red Sr Crimson

Cs Blue Ba Apple green


Chemical properties Reactions with water -All the elements of the first group react with water liberating hydrogen and become hydroxides(Except Mg and Be).

Example: Na reacts rapidly with water liberating hydrogen.

-Group I metals are kept under non reactive medium to prevent them from being react with water vapour in the air. -When the Mg is reacted with warmed water, it react slowly.

- As the reactivity shown by Mg with water is lower compared to Na, it can be said that the metals of group II compared to metals of group I show a lower reactivity.

-Ca, Sr, and Ba react with water liberating hydrogen and forming the hydroxides.

-Be and Mg react with steam to form the oxides.

Reactions with air -There are several reactions with air of group I metals. 4Na(s)+O2(s) 2Na2O(s)

2Na(s)+O2(s) 2Na2O2(s)

-With CO2 in the air , Na2CO3 can form

Na2O(s) + CO2(g) Na2CO3(s)

-K, Rb and Cs react forming superoxides. K(s) + O2(g) KO2(s)

Rb(s) + O2(g) RbO2(s) Cs(s) + O2(g) CsO2(s)

Oxygen shows oxidation state O-2 in oxides Oxygen shows oxidation state O-1 in peroxides

Oxygen shows oxidation state O2-1 in superoxides

- When heated in air only Li of Group I reacts with nitrogen.

6Li(s) + N2(g) 2Li3N(s)


-When a clean piece of Mg ribbon and a small cut piece of Na are exposed to air Na tarnishes faster than Mg. Hence it is clear that the reactivity of Mg is lower than Na. -Accordingly it can be said that relative to metals of group I, the reactivity of group II metals with air is low

- Metals of the Group II when heated in air burn forming oxides and nitrides. 2Mg(s) + O2(g) 2MgO(s) 3Mg(s) + N2(g) Mg3N2(s) -For Be to react it should be heated to a very high temperature. Reactions with acids

-The metals of s- block elements can act as a reducing agent when reacting with acids. -When group I elements react with acids to form particular salt, they emits a large amount of heat and H2

-Reactions of group I elements with acids

-Reactions of group II elements with acids

P Block elements

-The elements which having electrons in last shell P orbital, is known as P-BLOCK element. -Their general electronic configuration is np2 p1-6 Where n= 2 to 6

-P orbital can accommodate a maximum of six electrons.


-The zero group contain Nobel gases -The total no of p – block elements in the periodic table is 30. -There are 9 gaseous elements (Ne,Ar,Kr,Xe,Rn,F2,Cl2,O2 and N2) -Gallium (Ga) and Bromine (Br) are liquids. -B to At are metalloids. Chemical properties

Reactions with water - When the elements of p - block are considered it can be observed that only the halogens react with water while reactivity decreases down the group. -Halogens dissolve in water and the reactivity with water decreases down the group. -Fluorine displaces oxygen in water.

2F2(g) + 2H2O(l) 4HF(aq) + O2(g)

-Chlorine reacts with water slowly because the activation energy is high.

Cl2(g) + 2H2O(l) HCl(aq) + HOCl(aq)

Oxidation no (O) (-1) (+1) - The halogens except F show a disproportionate reaction with water as shown above. Reactions with air - When the elements of the p - block are considered, Al when heated in air reacts forming the oxide and releasing a large quantity of heat.

4Al(s) + 3O2(g) 2Al2O3(s) -Al does not react with air at the room temperature because it is

covered by an oxide layer. - Carbon burns at high temperature forming CO2

C(s) +O2(g) CO2(g) -Si forms oxides when heated to a very high temperature. -N2 reacts with O2 only at very high temperatures to give NO.

N2(g) + O2(g) 2NO(g)


-White phosphorus reacts with air (O2). In a limited supply of oxygen P4(s) + 3O2(g) 2P2O3(s) In excess of oxygen P4(s) + 5O2(g) 2P2O5(s) -S burns in air to form SO2

S(s) + O2(g) SO2(g) Reactions with acids -Elements of p block react with acids showing a greater diversity.

-C, S and P react with hot concentrated H2SO4.

- C, S and P react with hot concentrated HNO3

The properties of compounds and their trends associated with s

and p block elements. -The solubility of a given pair of compounds can be compared in the following way


-Their solubility can be compared by comparing the change in Gibbs energy relevant to the above two occasions. Here, the change in Gibbs energy for MgSO4 is a negative value whereas for BaSO4 it is a positive value. Therefore, the solubility of MgSO4 which has a more negative change in Gibbs energy is relatively higher than

that of BaSO4. -When both values of Gibbs energy are negative values in a pair that is compared, the solubility of the substance with a larger numerical value is higher. -When both values of Gibbs energy are positive in a pair that in compared, the solubility of the substance with the smaller positive value in higher -When comparing the thermal stability of a given pair of compounds also, the above method is suitable.


-By comparing ∆Gø

1 and ∆Gø2 , it can be predicted that MgCO3 with

the smaller positive value would be subjected to thermal dissociation easily. - The dissociation temperature of MgCO3 is 540 0C whereas that of CaCO3 is about 900 0C.

-Almost all the salts belonging to group I of s block are soluble in water. -Information about the solubility of salts formed by group II metals belonging to the s- block is given in the following table.


- Acidic/basic/amphoteric nature of oxides, hydroxide and hydrides of elements of the 3rd period

-The acidic nature increases from left to right across the table Elements and compounds of s and p blocks

Properties of D block elements and compounds

PHOSPHOUROUS

- Allotropic forms of phosphorus

CO3- HCO3 NO2

- NO3- S-2 SO3

-2 SO4-2


.+.

- P4 is stored in water because it reacts with O2 in the air. But nitrogen exists in the atmosphere as a free gas because of the strength of N= N bond in it. - Oxy- acids of phosphorus

Hypo phosphorous acid -H3PO2

Ortho phosphorous acid -H3PO3

Phosphoric acid -H3PO4

OXYGEN

- Allotropic forms of oxygen :O = O: O -1/2:O: :O: -1/2

Oxygen Ozone

SULPHUR

-Allotropic forms of sulphur exist in two forms namely crystalline and amorphous. -Crystalline Sulphur All crystalline forms of sulphur consist of S molecules.


-Amorphous sulphur Examples are plastic sulphur and colloidal sulphur.

-Oxy - acids of sulphur

HYDROGEN PEROXIDE (H2O2)

-H2O2 as a reducing agent

-H2O2 as an oxidizing agent

SULPHUR DIOXIDE (SO2)

-It is a colorless gas with a higher density than air and with a pungent smell. -It is very soluble in water. -SO2 dissolves in water to form sulphuric(IV) acid. It is a weak acid.


SO2(g) + H2O(l) H2SO3(aq) -SO2 as an oxidizing agent

-SO2 as a reducing agent

-As a bleaching agent SO2(g) +2H2O(l) H2SO4(aq)+2H+

(aq) + 2e

X + 2H+ + 2e XH2 (Colored dye) ( Colorless compound)

HYDROGEN SULPHIDE (H2S) - It is a colorless gas slightly soluble in water. -It has the smell of rotten eggs. -An aqueous solution of H2S is somewhat acidic. - Evidence for the acidic nature (i) Reactions with sodium Excess

2H2S(g) + 2Na(s) 2NaHS(s) + H2(g) 2Na(s) + H2S(g) Na2S(s) + H2(g)

(ii) Reaction with sodium hydroxide

When the base is in excess 2NaOH(aq) + H2S(g) Na2S(s) + 2H2O(l)

When H2S is in excess


H2S(g) + NaOH(aq) NaHS(s) + H2O(l) -H2S reacts with many metal ions to give sulphides. This is used as a test for identification of metal ions.

Pb2+(aq) + H2S(g) PbS(s) + 2H+(aq)

Cu2+(aq) + H2S(g) CuS(s) + 2H+(aq)

-H2S as a reducing agent

2KMnO4(aq)+3H2SO4(aq)+5H2S(g)

K2SO4(aq)+5S(s)+2MnSO4(aq)+8H2O(l)

K2Cr2O7(aq)+4H2SO4(aq)+3H2S(g)

K2SO4(aq)+Cr2(SO4)3(aq)+3S(s)+7H2O(l)

-As3+ reacts with aqueous S2- ions to form a precipitate.

As3+(aq) + 3S2-

(aq) As2S3(s)

-H2S as an oxidizing agent

Cu(s) + H2S(g) CuS(s) + H2(g)

Distinguishing between SO2 and H2S

-When the gases are passed through aqueous H+/K2Cr2O7 , SO2 and

H2S turn the orange colour to green but due to the formation of S

with H2S the solution will not be clear.

-A filter paper moistened with aqueous Pb(CH3COO)2 turns

glistening black with H2S.

-When passed through aqueous H+/KMnO4 both gases turn the

purple color to colorless but with H2S the solution will not be clear

due to S formed.

-Petals of flowers are bleached by SO2 but not by H2S.


Halogens F2 - Pale yellow poisonous gas

Cl2 - Yellowish light green poisonous gas

Br2 - Reddish brown liguid

I2 - Shining black solid. Sublimes.

At - A radioactive element.

- Oxy acids of chlorine

-Hydrogen halides (HX)


Noble gases and their compounds -Boiling points are very low. -The boiling points increase with the increase in atomic number. -Polarizability appears in large atoms. -It was found by 1962 that Xe forms compounds with two

electronegative elements. Examples : XeF2, XeF4, XeF6, XeO3

Acidity of hydrogen halides in aqueous solution -Under dry conditions hydrogen halides are not acidic. However, their aqueous solutions are acidic. HCl(g) + H2O(l) H3O+

(aq) + Cl-(aq) - Hydrofluoric acid is a weak acid and all the other hydrogen halides are strong acids. The reason for this is the specially strong H - F bond.

Disproportionation in chlorine Cl2 (g) + dil. 2NaOH(aq) NaCl(aq) + NaOCl(aq) + H2O(l) (0) (-I) (+I) Cl2 (g) + conc. 6NaOH(aq) 5NaCl(aq) + NaClO3(aq) + 3H2O(l)


D-block Elements -D block elements contain partially filled or fully d sub orbital.

Ionization energy

Oxidation number - Oxidation numbers of elements from Sc to Zn (Common oxidation numbers are shown in bold letters)


Instances where d block elements and their compounds are used in

industries as catalysts.


Boiling and Melting points

Properties of compounds of d-block


-Using the solubility in acids and bases of the types of oxides stated above, they can be classified as acidic, basic or amphoteric. - The reaction of MnO4 in an acidic medium as an oxidizing agent with SO2

- The reaction of MnO4

- as and oxidizing agent in a basic medium with SO2

-Chromium


Properties of complex compounds of d – block - Complex ions formed by the elements Cr, Mn, Fe, Co and Cu with the ligands H2O, NH and Cl –

-The color of the complex varies depending on the central metal atom. Examples : [Cr(H2O)6]

3+ - Blue - violet [Fe(H2O)6]

3+ - Yellow brown -The color of the complex varies depending on the oxidation state of the central metal atom. Examples : [Fe(H2O)6]

2+ - Light green [Fe(H2O)6]

3+ - Yellow brown [Mn(H2O)6]

2+ - Light pink [Mn(H2O)6]

3+ - Violet

-The colour changes also when the ligands change. Examples : [Co(H2O)6]2+ Pink [Co(NH3)6]2+ Yellow brown

-The reaction of hydrochloric acid with aqueous Cu2+ ions.

Cu2+(aq) + 4HCl(aq) [Cu(Cl)4]

2-(aq) + 4H+(aq)

Blue Yellow -The reaction of hydrochloric acid with Co2+(aqueous)

Co2+(aq) + 4HCl(aq) [Co(Cl)4]

2-(aq) + 4H+

(aq) Pink Blue

-Oxy- anions of chromium and their conversion


CrO-24(aq) CrO-2

7(aq)

Yellow Orange

IUPAC nomenclature of complex compounds

(D block) -The complex compounds are considered simply under two categories.

-Cations are simple while the anions are complex. -Cations are complex while the anions are simple.

-Whatever the complex compound considered, common set of rules has to be followed in nomenclature.

As in the case of a simple inorganic compound, frist the cation

is named and then the anion afterwards. A space is kcpt between the name of the cation and the name of the anion.

The most important step is the identification of the complex

ion part of the compound. This can be positively or negatively charged. In the identified complex ion, the names of the groups or the ligands surrounding the central metal atom have to be found out. The name used depends on the charge of these ligands. 1) When there are neutral ligands specific names should be used for them. Examples : H2O - aqua

NH3 - ammine CO - carbonyl

NO - nitrosyl

2) When there are negative ligands the suffix - 'o' is added to their English name. Example : Cl- - chloro

CN - - cyano NO2

- - nitro OH - -hydroxo SCN- - thiocyanato H+ - hydrido O 2

- - oxo


3) When there are positive ligands the suffix" ium" is added to their English name. Example : +NH3 - NH2- Hydrazinium

When there are more than one ligand of the same type. in order to indicate the number of such ligands the name of the relevant number is used as a prefix before the name of the ligand. When there are 2, 3, 4, 5 and 6 ligands of the same type, the prefixes di-, tri,- terta- penta- and hexa- are used respectively. When the complex ion is positively charged or neutral, name of the metal is used and without leaving a space, the oxidation number should be shown by a capital Roman numeral within parantheses after the name.

Examples : Co3+ - cobalt(III) Fe2+ - iron(II) Cr6+ - chromium(VI)

Now the complex ion can be named. Here, the ligands are

named first and the metal next. There should be no spaces between the words in writing the name. Examples :

[Co(NH3)6]3+ - hexaamminecobalt(III) ion

[Fe(H2O)6]2+ - hexaaquairon(II) ion

[Cu(NH3)4]2+ -tetraamminecopper(II) ion

When several ligand types are joined to a complex ion in

naming the ligands they should be listed in the English alphabetical

order. Examples :

[Fe(CN)2 (NH3)4]+ - tetraamminedicyanoiron(III) ion

The complex ion part may be positively charged, negatively

charged or neutral. Depending on this the name also changes.

-When the complex ion part is positively charged or neutral, it is named using the name of the metal. Here also, it is necessary to indicate the oxidation number of the metal by a capital Roman numeral in parantheses.


Example : [Fe(CN)3 (NH3)4] This is neutral. Hence its name is tetraamminetricyanoiron. [Cu(H2O)6 ]

2+ This is positively charged. Its name is hexaaquacopper(II) ion

-When the complex ion part is negatively charged, the suffix 'ate' is added to the end of the name of the metal. Here also the oxidation number of the metal shoud be indicated by a capital Roman numeral inside parantheses. Examples :

[CoCl4]2- - tetrachlorocobaltate(II) ion [Co(CN)6]3- - hexacyanocobaltate(III) ion [CuCl4]2- - tetrachlorocuprate(II) ion

[Fe(CN)6]4- - hexacayanoferrate(II) ion [Fe(CN)6]3- - hexacayanoferrate(III) ion [Ag(CN)2]- - dicyanoargentate(I) ion [Cr(Br)6]3- - hexabromochromate(III) ion

The IUPAC name of any compound can be developed by following the rules studied so far systematically. Try to do this under two main groups. -It should be emphasized how a space is placed between the positively charged part and the negatively charged part and the name of the complex part is written as a single word without leaving spaces. Examples :

K3[Fe(CN)5 NO] -Potassium

pentacyanonitrosylferrate(II) Na2[ZnCl4] -Sodium tetrachlorozincate(II)

-When there is a complex cation and a simple anion there should be a space between the positively charged part and the negatively charged part. Examples : [Ag(NH3)2]Cl2 - Diamminesilver(I) chloride [Fe(OH)2(H2O)4]Br -Tetraaquahydroxoiron(III) bromide [CoCl(NH3)5](NO3)2 -Pentaamminechlorocobalt(III)nitrate(V)


Writing formula of a complex compound when its name is given. -As in the general method show the positively charged ion first and the negatively charged ion afterwards. -Always the complex ion part of the compound should be written within square brackets. -If there is a charge, it should be indicated outside the square bracket

at the upper end of the right side. The sign of the charge and its numerical value should be given. Example : [Fe(CN)5NO] -When the formula of the complex ion part is written first, the metal should be indicated and then the ligands. - In writing the ligands, the order of the English alphabet used during the nomenclature is not used here. Instead ligands are written according to their charges. That is in the order of negatively charged

ligands first, neutral ligands next and the positively charged ligands last. However, the various ligands present in a similar group should be arranged alphabetically. This is explained below.


Unit 6 – Organic Chemistry -Carbon has the ability to form a large number of compounds.

Because the Carbon -form single, double and triple bonds -having higher bonding energies. (Higher than Si–Same group)

Bond Bonding energy- kJmol-1

C-C 346

C=C 610

C=C 835

C-H 413

Si-Si 226

Si-H 318

-Carbon -Can form 4 covalent bonds. -Stable in room temperature. -Reaction of organic compounds with oxygen is highly Exothermic.

-Carbon compounds are thermodynamically unstable. But the values of activation energies associated with the reactions of organic compounds with oxygen are very high. Hence, according to chemical kinetics, organic compounds are stable and a large number of them occur naturally.

Functional Groups

-Compound which contain only C and H are known as

hydrocarbons. -On the basis of the structure, hydrocarbons are divided into two main groups


-The set of hydrocarbons consisting of open carbon chains only are named as aliphatic (acyclic) hydrocarbons.

-The aliphatic organic compounds are classified as alkanes, alkenes, and alkynes. -The cyclic organic compounds which are stabilized by forming a delocalized cloud of electrons are called aromatic compounds. -Benzene which is indicated by the molecular formula C6H6 is the simplest of aromatic hydrocarbon compounds. -In many organic compounds, when hetero atoms such as nitrogen and oxygen combine with the carbon chain, due to the difference in electro negativity between the carbon and the combined atoms, this group of atoms will impart the compound a characteristic reactivity. Such a group of atoms is called a functional group.

-The compounds are classified according to the functional group present in a molecule.


* In the IUPAC nomenclature, the halogen is not considered as a functional group

Naming aliphatic organic compounds -The International Union of Pure and Applied Chemistry -IUPAC, presented systematic method to name the organic compounds in 1949. -The name given to a compound according to the IUPAC nomenclature consists of several parts.


1. The suffix that is used to indicate the main functional group of the structure 2. Name of the chain that is used to identify the main carbon chain of the compound 3. The prefixes that are used to indicate the substituent groups

4. The numbers that are used to indicate the places at which the substituent groups, additional groups and the main functional groups are attached to the chain

-In the nomenclature of aliphatic compounds, by following the steps stated in the given order, the IUPAC name of a compound can easily be developed. Identifying the principal functional group Selecting the main chain

Selecting the root name for the principal chain Addition of the suffix for the remaining groups (additional

groups) to the name of the chain Addition of the suffix used to indicate the principal functional

group to the name of the chain Naming the substituent groups Adding the names of the substituent groups to the name of

the chain Numbering the carbon chain Showing the numbers that are used to indicate the positions

of the main functional group and the substituent groups in front of these groups.

The root names used for the compounds according to the number of carbon atoms and the name of corresponding hydrocarbon.


The series of functional groups arranged in the decreasing order of

their priority

Summary of IUPAC Rules for Naming Alkanes

1. Find the longest chain of C atoms. Choose the base name that describes the number of C atoms in this chain, with the ending -ane .


The longest chain may not be obvious at first if branches of different sizes are present. 2. Number the C atoms in this longest chain beginning at the end nearest the first branching. If necessary, go to the second branch closest to an end, and so on, until a difference is located. If there is branching at equal distances from

both ends of the longest chain, begin numbering at the end nearest the branch that is first in alphabetical order. 3. Assign the name and position number to each substituent. Arrange the substituent’s in alphabetical order. Hyphenated prefixes, such as tert- and sec-, are not

used in alphabetization of the substituents. 4. Use the appropriate prefix to group like substituent’s: di- = 2, tri- = 3, tetra- = 4, penta- = 5, and so on. Don’t consider these prefixes when alphabetizing attached groups. 5. Write the name as a single word. Use hyphens to separate numbers and letters (plus some hyphenated prefixes) and commas to separate numbers. Don’t leave any spaces.

Summary of IUPAC Rules for Naming Alkenes and Cycloalkenes

1. Locate the C atoms in the longest C chain that contains the double bond. Use the base name prefix with the ending -ene. 2. Number the C atoms of this chain sequentially beginning at the end nearer the double bond. Insert the number describing the position of the double bond (indicated by its first carbon location) before the base name. (This is necessary only for chains of four or


more C atoms, because only one position is possible for a double bond in a chain of two or three carbon atoms.) 3. In naming alkenes, the double bond takes positional precedence over substituents on the carbon chain. The double bond is assigned the lowest possible number.

4. To name compounds with possible geometric isomers, consider the two largest groups within the carbon chain that contains the double bond—these are indicated as part of the base name. The isomer in which the largest groups at each end of the CUC are located on opposite sides is called trans. If the largest groups are on the same side, the molecule is referred to as cis. Insert the prefix cis- or trans- just before the number of the double bond to indicate whether the largest groups are on the same or opposite sides, respectively, of the double bond.

5. For cycloalkenes, the double bond is assumed to be between C atoms 1 and 2, so no position number is needed to describe it.


Summary of IUPAC Rules for Naming Alkynes

Alkynes are named like the alkenes except for the following two

points. The suffix -yne is added to the characteristic root. Because the linear arrangement about the triple bond does

not lead to geometric isomerism, the prefixes cis- and trans- are not used.

Drawing compound according to IUPAC -Drawing the structural formula of a compound according to the

IUPAC nomenclature can be done by following the steps given below. Identifying the chain and drawing the chain according to that

name Numbering the chain Identifying the principal functional group and the remaining

groups according to the IUPAC name given and joining these groups to the correct places of the chain according to the number in front of these groups.

Placing hydrogen atoms in the chain structure so that each carbon atom has a valency of four.

Isomerism -Isomers are different compounds that have the same molecular formula; they have the same number and kinds of atoms arranged differently. -There are two major classes of isomers: structural (constitutional) isomers and stereoisomerism. For coordination compounds, each can be further subdivided as follows.


STRUCTURAL ISOMERISM

Chain isomerism -Chain isomerism occurs when the nature of the carbon chain changes for the same molecular formula in the same homologous series.

Examples:

Position isomerism -Though there is the same molecular formula, the same functional group and the same carbon chain when there is a change in the carbon atom to which the functional group is attached or a change in the location of the active position, then there occurs position isomerism. Examples:

Functional group isomerism -Functional group isomerism is the existence of structures with different functional groups for the same molecular formula. Examples:


STEREOISOMERS -Compounds that contain the same atoms and the same atom-to-atom bonding sequences, but that differ only in the spatial

arrangements of the atoms relative to the central atom, are stereoisomers. Complexes with only simple ligands can exist as stereoisomers only if they have coordination number 4 or greater. The most common coordination numbers among coordination complexes are 4 and 6, and so they will be used to illustrate stereoisomerism.

Geometric (cis–trans) Isomers - Geometric isomerism is one occasion where diastereo isomerism is seen. - In a C = C double bond due to the bond which exists in addition to the σ bond, these carbon atoms cannot freely rotate about the σ bond.

Therefore, although the carbon atoms are bonded in the same way, due to their three dimensional position, different configurations can exist. This is referred to as geometrical isomerism. -The different structures obtained here are called geometrical

isomers. - Geometrical isomers can be identified by the structural formula of the alkenes.

- If a ≠ b and p ≠ q, that compound exhibits geometrical isomerism. --In any of the above occasions if two identical groups are on the same side with reference to the axis of the double bond, that structure is


called the cis- isomer and if the groups are on opposite sides it is called the trans - isomer.

-When all the four groups a, b, p and q are different also it shows geometrical isomerism. But these isomers cannot be named as cis or trans. The nomenclature of these compounds is not relevant to the syllabus.

Enantiomer isomerism

-The isomers of which one is the mirror image of the other are known as enantiomers -A compound having a carbon atom which is joined to four different groups shows enatiomerism. -When plane - polarized light is passed through a solution containing only one enantiomer, the plane of polarization rotates. One enantiomer rotates the plane of polarization in one direction and the other enantiomer in the opposite direction. As the enantiomers rotate the plane of polarization, they are also known as optically active isomers.


Unit 7 – Hydrocarbons

Aliphatic Hydrocarbons.

Alkanes -This type contain C-C single bond. - If two consecutive members differ only by a CH2 unit, such a series of compounds is called a homologous series. -An alkane molecule is non polar or very weakly polar. The attractive forces between two non -polar molecules are the very weak Van der Waals forces. -The first few members of the series are gases at room temperature.

-Liquid and solid members are met on going down the series. -With the branching of the carbon chain Van der Waals forces become weak When alkane members with 5 carbon atoms are considered:

Alkenes - The general formula CnH2n+2 - The simplest alkenes contain one carbon–carbon double bond, C=C, per molecule. The general formula for noncyclic alkenes is CnH2n. The simplest alkene is C2H4, ethene, which is usually called by its common name, ethylene. - In alkenes also the physical properties are very much similar to those of alkanes. -The hybridization (sp2) and bonding at other double-bonded carbon atoms are similar. Both carbon atoms in C2H4 are located at the centers of trigonal planes. Rotation about C=C double bonds does not occur significantly at room temperature.


Alkynes - As the polarity of alkynes is also low, their physical properties, are very similar to those of the corresponding alkenes and alkynes. - The alkynes, or acetylenic hydrocarbons, contain one or more carbon–carbon triple bonds, -C=C-. The noncyclic alkynes with

one triple bond per molecule have the general formula CnH2n-2. The bonding in all alkynes is similar to that in acetylene.

Aromatic hydrocarbons

Benzene - Benzene is the simplest aromatic hydrocarbon. -Under the normal conditions benzene does not answer the tests for unsaturation. -Therefore, benzene cannot have a structure similar as that of a simple alkene or an alkyne. -Although the structure proposed for benzene by Kekule showed three double bonds for the molecule, benzene is not similar to the Kekule structure.

- Elemental analysis and determination of its molecular weight showed that the molecular formula for benzene is C6H6. The formula suggests that it is highly unsaturated. But its properties are quite different from those of alkenes and alkynes. - The bond length between two carbon atoms in benzene is the same and its value is 1.39 x 10-10 m. Further, the benzene molecule is completely planar. -But the length of a C=C double bond is 1.34x 10-10 m and the length of a C – C single bond is 1.54 x 10-10 m. -Therefore it is clear that the structure of benzene should be a hybrid of the resonance structures given below.


-The double – headed arrow, does not indicate that benzene changes between these two structures or that it is in equilibrium. It shows resonance. -The structure of benzene is very clearly explained by the molecular orbital theory

-All its C atoms have undergone sp2 hybridization. Here, as the electrons in the unhybridized p- orbitals exist as conjugate double bonds, they can overlap with the unhybridized p orbitals present on both sides.

-From this, there is the ability to form a circular electron cloud .There fore, due to the circular delocalization of the six 'p' electrons, the resonance structure of benzene is relatively more stable than the structure with three real triple bonds.

-The data for the standard enthalpy of hydrogenation also help to explain the stability of a benzene molecule. +H2 ∆Hø = -120 kJmol-1

C6H6 + H2 ∆Hø = -207 kJmol-1

-But if benzene possesses three double bonds, its standard enthalpy of hydrogenation should be 3 x (-120 kJ mol-1), that is -360 kJ mol-1. Hence, it is clear that benzene is more stable than its Kekule

structures by an amount equal to (360-208) = 152 kJmol-1


-Therefore, benzene doesn’t contain C=C 3 bonds. the structure is

drawn as below,

Chemical reactions of Alkanes, Alkenes and Alkynes

Reactions of Alkanes -The most important reaction in alkanes is combustion. -Inn alkanes there’s no special functional group. -All alkanes are covalent bonds -Alkanes doesn’t react with common lab reagents. (CN-,OH-) -Polarity is very low -As by-products they form CO2 and H2O

2C8H18 + 25O2 8CO2 + 9H2O ∆Hø = -11020kJmol-1

-The hydrocarbon gasoline (petrol) boils at 1500C The best fuel for best power is 2,2,4,Trimethylpentane ,This is known as iso octane and has assigned octane no 100.

Heptane -This fuel has low power and octane no is 0. -Octane number is calculated by the percentage of 2,2,4,Trimethylpentane contain in fuel. -Alkanes react with Cl2 , Br2 etc which undergo hemolytic fission.


-Homolysis of C-H bond is the first task of free radical generation.

During this process free radical containing a carbon is generated. -The probability for a Cl free radical to collide with CH4 molecule is high at the beginning and the before CH3Cl is formed. -Because the termination step of the chain can also take place in between the course of the reaction it is necessary to supply sunlight for the reaction to proceed continuously. Halogenations of alkanes -The density of π electrons which lie above and below the plane of the ethylene molecule is capable of attracting electrophilic reagents.


-When an electrophile gets attracted to the π electron cloud and bonds with a carbon, the other carbon will be attached only to three groups and hence another bond will be formed to complete the valency.

Reactions of Alkenes

-Alkenes react with hydrogen halides. Here, the hydrogen ions act like electrophiles and attack the double bond. During these electrophilic addition reactions, intermediate carbocations are formed. - Stability of carbocations is tertiary > secondary > primary. R R H R C+ R C+ R C+

R H H - When alkyl groups are attached to the positively charged C atom of the carbocation, the stability of the cation increases. The reason for this is the repulsion of electrons by the alkyl groups through the bonds towards the positively charged carbon atom to which they are attached. Here what happens is the stabilization as a result of the positive charge spreading completely throughout the ion (the alkyl groups are represented by R groups). -The are two types of alkyles i) A symmetrical ii) Symmetrical

Symmetrical Asymmetrical(Non alkyl)

- In the electrophilic addition reactions of a symmetric alkenes with hydrogen halides, two asymmetric carbocations can be formed after the bonding of the electrophile. -Out of these the more stable carbocation forms easily. -According to the stability of carbocations, the more stable carbocation is obtained when the electrophile gets attached to the carbon atom to which the highest numbers of hydrogen atoms are attached. -After studying the reactions of a large number of alkenes, this observation was generalized as Markownikoff’s rule.


-Of the hydrogen halides, only hydrogen bromide adds in the opposite way to this rule when there are peroxides in the reaction medium. The reason for this is that in the presence of peroxides the reaction of hydrogen bromide and the alkenes takes place via a free radical mechanism.

Mechanism of addition of bromine to alkenes -Due to the electron density present on the two sides of the plane of the alkene, an induced dipole forms in the bromine molecule. It is this induced dipole that acts as the electrophile.

Addition of sulphuric acid to alkenes and the hydrolysis of the product obtained.

-In the presence of Hg 2+ and dilute sulphuric acid , one molecule of water gets added on to alkynes.

-The immediate reorganization of the oxygen into the alkehyde is due to the high stability of C = O


Reaction of alkenes with cold alkaline KMnO4

Catalytic hydrogenation of alkenes -Under the normal conditions, hydrogen and alkenes do not react. But in the presence of finely divided Pt/Pd/Ni metal catalysts alkenes react with hydrogen to form the relevant alkanes.

Acidic nature of alkynes - In the alkynes, H-C=C-H and R-C=C-H the H attached to the C that forms the triple bond (terminal hydrogen) shows acidic

properties. -The acidic H in these alkynes can be displaced by metals. H-C=C-CH3

Na Na+-C =C-CH3 + H2

H-C=C-CH3 NaNH3 Na+-C =C-CH3 + NH3

H-C=C-CH3 NH

3 /CuCl

2 Cu-C=C-CH3 H-C=C-CH3

NH3

/AgNO3 Ag-C=C-CH3

Reaction types that Organic compounds can show -Substitution reactions An atom or a group replaces another. -Addition Reactions The molecule reacts to form another. -Elimination Reactions

One molecule reacts to form another one or more. -Rearrangement Reaction Rearranging , whenever the structure permits.

Types of reagents -In a covalent bond between A and B , If A is more electronegative , the distribution of electrons can be shown as A-β – B- β

-Due to that one side of the molecule become slightly polarized. -There are two types of reagents


1) Neucleophillic reagents -This type of reagents like positive charge

-They attack the centre of the positive charge Example- OH-,CN-

2)Electrophilic reagents -They attack the place where the electron density is high. Example- NO2

+, SO3+

Benzene -In benzene there’s a high electron density in both sides of the ring. -Due to that electrophilic reagents get attached on to it and form a carbo cation. -A electron deficiency occurs and H+ is eliminated to obtain cyclic delocalized electron clue.

NITRATION -Replacement of –H by –NO2 using NO2

+ (Nitronium) as an electrophile is known as nitration. -Nitronium ion is formed by using nitrating mixture of conc H2SO4 and conc HNO3 .

- Conc H2SO4 being a strong acid than HNO3 pretending to protonation of hydroxyl oxygen on HNO3

-Mechanism H2SO4 + H-O-NO2 HSO4

- + H O+ NO2

H NO2

+ NO2 + H+

H+ + HSO4- H2SO4

+ A+

A H

A + H+


CH3

FREDEL-CRAFTS ALKYLATION -During the reaction of benzene with alkyl halides, reaction requires a catalyst in form of Lewis acid. -Anhydrous AlCl3 is a strongest Lewis acid it pulls the halide in maximum strength.

+ R-X -Mechanism R-Cl + AlCl3 R-Cl+ -AlCl3

-

R-Cl+ + AlCl3 R+ +AlCl4

Consider R as CH3

CH3+-Cl- ++gg [AlCl4]

- HCl + AlCl3 +

HALOGENATION -When benzene react with a halogen carrier, a substitution of halogen in benzene ring take place. -Halogen molecule is polarized by the halogen carrier, the molecule function as a electrophile.

-Mechanism is similar to above diagram

OXIDATION

-Benzene doesn’t get oxidized by normal oxidizing agents like KMnO4/H+.

-But after a substitution it reacts.

Anhydrous AlCl3 R

CH3

H

+

+

CH3

H

Dry FeBr3/Febr2

Br

COOH

KMnO4/H+


Ni/1500c

-Tertiary alkyl groups doesn’t oxidize under normal condition in which primary and secondary alkyl groups get oxidized. -Only in higher reactions they get oxidize.

ELECTROPHILIC ADDITION RERACTIONS

-Although, alkenes undergo electrophilic reactions easily Benzene doesn’t show such reactions. -But in presence of a catalyst at about 150oC it react woth H2 gas to form Cyclohexane. + H2

DIRECTING ABILITY OF A SUBSTITUTE GROUP TO A MONO

SUBSTITUTENTED BENZENE

Orth,Para directing groups -If second substitution group get attached to Ortho or Para position

relative to first group, it’s a orthopara directing group. -Ortho , Para directing groups are –OH , -R , -NH2 , -OCH3 , Halogens

Meta directing groups -If substitutional group is a electron attracting group , it deactivate the ring , due to that second substitutional group take place in meta position - Meta directing groups are –NO2, -CHO , -COR, -COOH, -COOR


Unit 8 – Alkyl Halides

Alkyl Halides -Alkyl halides are named as primary, secondary or tertiary depending

on the nature of the carbon atom which carries the halogen atom.

- Alkyl halides are polar compounds. -Solubility of alkyl halides is water is very low. One reason for this is that they do not form hydrogen bonds with water.

-Due to the high eletronegativity of the halogen atom relative to the carbon atom, the C – X bond gets polarized. As a result, there is a deficiency of electrons in that carbon atom. Therefore, the nucleophiles attack this position.

Neucleophillic substitution reactions -Nuclephilic substitution reactions are characteristic of alkyl halides. Here, the carbon atom forms a new bond with the nuclephile and the halogen atom leaves as a halide ion.

- Any nucleophile can also act as a base. Therefore, there is an ability

to subject alkyl halides to an elimination reaction by removing an acidic hydrogen atom. -Although the hydrogen atoms attached to the carbon atom to which the halogen atom is attached are more acidic in nature, a stable compound is not formed by their elimination. Therefore, the hydrogen atoms with a slight acidity attached to the adjoining carbon atoms participate in elimination reactions. H H H H OH + H- C – C – X C = C + H2O +X H H H H


Chlorobenzene and vinyl chloride do not undergo nucleophilic substitution under the conditions which alkyl halides undergo these reactions. As reasons for this, mention the double bond nature of the C- X bond and the decrease in bond length due to the existence of the carbon atom on the sp2 hybridization in these compounds.

GRIGNARD REAGENT -Alkyl halides react with Mg in the medium of dry ether to form the Grignard reagent.

- Due to the polarization of bonds in RMgX, the carbon attached to magnesium acts as a strong nucleophile and a very strong base. -Therefore, to prevent the formation of alkanes by the donation of

protons by water a dry ether medium is used.

The strong nucleophilic features of the Grignard reagent can be shown by the following reactions.

PROPERTIES OF ALKYL HALIDES - To explain the nucleophilic substitution reaction of alkyl halides, the time interval between bond breaking and bond making steps can be considered. -When the breaking of the bond and the formation of the new bond take place simultaneously, the nucleophilic substitution reaction of the alkyl halide is considered as a one step reaction. -Accordingly, the one step reaction can be presented as follows:


- When the formation of the new bond takes place after the breaking of the bonds the nucleophlic substitution reaction of the alkyl halide is considered as a reaction that takes place by two steps. - Accordingly, the reaction that takes place by two steps can be presented as follows:

- The reaction that takes place by two steps goes through an intermediate carbocation. -On considering the stability of the carbocation formed, the tertiary alkyl halides which are able to from a more stable tertiary carbocation undergo nuclephilic substitution in two steps. -The primary alkyl halides undergo nucleophilic substitution

reactions in one step as they are unable to form a stable intermediate carbocation. -The procedure adopted by secondary alkyl halides will be determined by the nature of the reaction medium and the reaction conditions supplied.


Unit 9 – Oxygen containing

organic compounds

- Monohydric alcohols can be classified into three types as primary, secondary and tertiary

Physical properties

-In alcohols the –OH bond polarizes as R – 0 β- - H β+ . Hence, due to the inter molecular hydrogen bonds formed between alcohol molecules, their boiling points have higher values compared to the alkenes and ether with comparable relative molecular masses. -The boiling point increases in going down the alcohol series.

-The above diagram shows how the intermolecular hydrogen bonds

exist in ethanol. -Alcohols are soluble in water. The solubility of alcohols in water which is a polar solvent is facilitated by the – OH group. The non polar alkyl group in the alcohol molecule is a hindrances to the solubility in water. - In going down the homologous series of alcohols the size of the nonpolar alkyl group gradually increases compared to the –OH groups. Accordingly the solubility in water gradually decreases.


REACTIONS INVOLVING CLEAVAGE OF THE O-H BOND Reaction with sodium

- Alcohols behave as weak acids and react with sodium liberating hydrogen and forming sodium alkoxides. The alkoxide ion is a strong nucleophile and also a strong base. Reaction with carboxylic acids (Acylation of alcohols)

Where G represent reversibility - Alcohols react with carboxylic acids to form esters. For this esterification reaction, concentrated H2SO4 acid acts as a catalyst. - Nucleophilic substitution reactions that take place by the cleavage of C-O bond in alcohols -Alcohols react with PCl3 or PCl5 to give alkyl chlorides.

- Reaction with alkyl halides


Alcohols react with HBr to give corresponding alkyl bromides. Here the Br- ion from HBr is the nucleophilic reagent.

Reaction with anhydrous ZnCl2 and conc. HCl (Lucas test) -Here, ZnCl2 acts as a catalyst. Due to the alkyl halide formed as the product, a turbidity appears in the reaction medium. In relation to the time taken for turbidity to appear, the rates of reaction of primary, secondary and tertiary alcohols with the Lucas reagent can be compared. -This test is restricted to primary, secondary and tertiary alcohols which are soluble in water.

-The reaction rate of alcohols with the Lucas reagent is a follows:

-Under the supplied conditions the above reaction takes place in two steps. Here, the tertiary alcohol forms a more stable intermediate and therefore the tertiary alcohol in the presence of the Lucas reagent forms a turbidity in a very short time.

Alcohols undergo an elimination reaction with conc. H2SO4 (Dehydration of alcohols)

-The reaction in which a molecule of water is eliminated from an

alcohol is the dehydration of alcohols. Here, an alkene is formed as the product of the reaction.

Oxidation of alcohols


The product of oxidation depends on primary, secondary or tertiary nature of the alcohol.

Primary alcohols

Secondary alcohol

-Normally the tertiary alcohols do not undergo oxidation in the presence of condition that oxidize primary and secondary alcohols. - It is because in tertiary alcohols due to the lack of hydrogen on the

carbon attached to the oxygen, it is the carbon - carbon bond that has to be broken. -Oxidation of alcohols is carried out with H+/KMnO4 or H+/K2Cr2O7 or H+/CrO3. -In the presence of the above reagents the aldehydes undergo further oxidation very easily. Therefore, it is not easy to obtain aldehydes using these reagents. When there is a need to obtain the aldehyde, PYRIDENIUM CHLOROCHROMATE [C5H5NH.CrO3Cl] can be used.

-When primary alcohols are oxidized in the presence of pyridenium chlorochromate the corresponding aldehyde is obtained as the product. Here it does not get further oxidized to carboxylic acid.

Phenol -In aromatic compounds, the compounds in which an -OH group is joined to a carbon atom of the benzene ring instead of a hydrogen atom are called phenols. -The phenols are more acidic relative to the alcohols. It means that the equilibrium point of alcohols shown above is more shifted towards the right. -The reason for this is that the stability of phenoxide ion relative to phenol is greater than the stability of the alkoxide ion relative to the alcohol.


-The phenoxide ion is more stable because its negative charge gets delocalized by resonance. In alkoxide ion there is no such charge dispersion. -The higher acidity of phenols is confirmed by the following examples too.

- Although an alcohol reacts with sodium it does not react with NaOH. But phenol reacts with sodium as well as with NaOH 2C2H5OH + 2Na 2C2H5O

-+Na + H2 C2H5OH + 2NaOH No reaction 2C6H5OH + 2Na 2C6H5O

-+Na + H2 C6H5OH + 2NaOH 2C6H5O

-+Na + H2O -Because the carbon atom in phenol exists in the sp2 hybridization

state and due to the double bond nature developed as a result of

resonance in the bond that oxygen forms with the π electron cloud

of the benzene ring, the carbon- oxygen bond in phenol is shorter than the carbon- oxygen bond in an alcohol. -Therefore, unlike alcohols phenols do not undergo nucleophilic substitution reactions.

Effect of the -OH group on the reactivity of the benzene ring in

phenol -Due to the delocalization of the lone pairs of electrons which were on the oxygen atom of phenol in the benzene ring, the ring has become very reactive towards electrophilic reagents. -The O-H group of phenol is ortho, para directing.

-When the electrophilic substitution reactions of phenol are compared with the corresponding reactions of benzene along with the relevant conditions, it is clear that the benzene ring of phenol, had become more reactive towards electrophiles. -Consider the following example

(i) Reacts immediately with bromine water to give a white precipitate of 2,4,6 –tribromophenol.


(ii) For nitration of phenol even dilute HNO3 is sufficient.

Aldehydes and Ketones -Aldehydes and ketones contain the carbonyl group. -In aldehydes, at least one H atom is bonded to the carbonyl group. -Ketones have two carbon atoms bonded to a carbonyl group. Models of formaldehyde (the simplest aldehyde) and acetone (the simplest ketone) are shown below

-The simplest ketone is called acetone. Other simple, commonly

encountered ketones are usually called by their common names. -These are derived by naming the alkyl or arylgroups attached to the carbonyl group.


-Aldehyde can oxidize easily relative to ketones, it means aldehydes are more reactive. -In carbonyl compounds polarization occurs, but in the ketone molecule two alkyle groups are repelling and reduces the charge.

-But in aldehydes one alkyle group and hydrogen atom repel lesser than above, therefore it is more reactive. Mechanism of the addition of HCN to aldehydes and ketones. -This reaction is carried out by adding a dilute mineral acid into a mixture of the carbonyl compound and an aqueous solution of sodium cyanide. Here the CN- ion acts as the nucleophile.

Mechanism of the reaction with Grignard reagent (RMgX)


Mechanism of the reaction with Brady’s reagent (2,4-DNP)

-Brady’s reagent is 2,4,-DNP (Dinitrophenylhydrozone) -Nucleophilic addition

-Dehydration The above intermediate product undergoes dehydration as soon as it is formed to give the final product, which is 2,4–dinitrophenylhydrozone.

-This reaction is used as a test to distinguish between aldehydes and ketones.

Reduction

(i) Reduction by LiAlH4 or NaBH4 -Here, the aldehydes and ketones get reduced to alcohols.


(iii) Reduction by Zn(Hg) conc. HCl (Clemenson reduction) -Here, the aldehydes and ketones can be reduced to the corresponding hydrocarbons.

OXIDATION OF ALDEHYDES

Oxidation by Tollen’s reagent -Tollen’s reagent, [Ag(NH3)2]

+ is prepared by adding dilute ammonium hydroxide to the precipitate of silver oxide formed by the addition of a few drops of dilute sodium hydroxide to an aqueous solution of silver nitrate.

-Oxidation by Tollen’s reagent or the silver mirror test is used to distinguish between aldehydes and ketones.

Oxidation by Fehling solution -A solution of basic cupric tartarate is known as Fehling solution. -This is a dark blue aqueous solution. -When a few drops of an aldehyde are added to this reagent and heated, the blue colour of the solution gradually disappears and a brick red precipitate of cuprous oxide is formed.


-Aldehydes and ketones can be distinguished from each other by

Fehling solution too.

Oxidizing by oxidizing agents H+/KMnO4 and H+/K2Cr2O7

-Aldehydes get oxidized to carboxylic acids by oxidizing agents such as acidified potassium dichromate of acidified or alkaline potassium permanganate.

In the presence of aldehydes the pink color of H+ /KMnO4 solution become colorless. -The orange color of H+/K2Cr2O7 solution turns green. By using these reagents also aldehydes and ketones can be distinguished from each other.

Self condensation reactions -Aldehydes and ketones undergo condensation with presence of NaOH

-Due to the polarization of the C = O group in aldehydes and ketones, the electrons get pulled towards it and as a result the C – H bonds in the carbon atoms adjacent to it get weakened and these hydrogen in the presence of alkalis can get eliminated forming carbanions.

-The above carbanion attacks the carbon atom of the carbonyl group of an unionized aldehyde molecule as a nucleophile.


Hence aldehydes and ketones with hydrogen undergo self-condensation reactions. Example : Reaction of acetaldehyde in the presence of aqueous NaOH

-The addition products obtained above undergo dehydration easily. Examples :

Carboxylic acids -The functional group of the carboxylic acids is the carboxyl group. In the carboxyl group there is a carbonyl group and a hydroxyl group attached to that carbon atom.


-Carboxyl group is a polar functional group Due to the polarity of C - O and O – H groups it forms intermolecular hydrogen bonds. -Relative to the aldehydes and ketones with equal relative molecular masses, the boiling points of carboxylic acids are higher.

-The ability to form dimer structures where carboxylic acid molecules are attached by hydrogen bonds as pairs is also a reason for the high boiling points of carboxylic acids. -Existence of carboxylic acids as dimer structures attached by hydrogen bonds is shown below.

REACTIONS OF CARBOXYLIC ACIDS -Reactions involving the cleavage of O – H bond

A Comparison of the reactions of alcohols, phenols and carboxylic acids with sodium, sodium hydroxide and sodium bicarbonate is given in the table below


-The above compounds react with Na2CO3 in the same way that they react with NaHCO3. -The acidic strengths of alcohols, phenols and carboxylic acids vary as follows.

Alcohols < Phenols< Carboxylic acids

-The carboxylic acids attain the following equilibrium in an aqueous solution

-The equilibrium point of the above equilibrium is situated more shifted towards the right side relative to the corresponding equilibrium attained by the phenols. -The reason for this is that the stability of the carboxylate ion relative

to the carboxylic acid is greater than the stability of the phenoxide ion relative to phenol. -The carboxylate ion is a resonance hybrid of the following structures.

-The carboxylate ion is more stable because its resonance structures are more stable than the resonance structures of the phenate ion. -This stability is due to the existence of the negative charge on the electronegative oxygen atoms in the carboxylate ion

Reactions involving cleavage of the C – O bond -With PCl3/ PCl5


-With alcohols

-Reduction of carboxylic acids with LiAlH4

Carboxylic acids give alcohols when reduced with LiAlH4 which is a very strong reducing agent but are not reduced by NaBH4

Reaction with acid derivatives Reactions of acid chlorides Reaction with sodium hydroxide

-Acid chlorides react with NaOH to form the corresponding carboxylic acid. -It exists in the reaction medium as the carboxylate ion.

-Mechanism of the reaction

Reaction with water -Acid chlorides react with water to form the corresponding carboxylic acid.


Reaction with primary amines

Reaction with alcohols

-Acid chlorides react with alcohols to form esters

Reaction with phenol

Reaction with NH3 -Acid chlorides react with ammonia to form amides

Reactions of esters

i) Esters undergo hydrolysis with dilute acids to form the corresponding carboxylic acid and the alcohol.


ii) Esters when reacted with aqueous NaOH form the sodium salt of the corresponding carboxylic acid and the alcohol.

Reactions of amides With NaOH

When amides are warmed with an aqueous solution of NaOH, NH3 is liberated and the sodium salt of the corresponding carboxylic acid is formed


Unit 10 –Nitrogen containing

organic compounds -There are three types of nitrogen containing organic compounds.

Amides. Amines. Amino acids. -And also there are organic compounds like dizoniumchloride, DNA and napthol containing nitrogen.

Amines -Amines can be defined as compounds where alkyl or aryl groups are

attached to nitrogen instead of hydrogen atoms in ammonia. -Amines are classified as primary, secondary and tertiary. -Unlike the alkyl halides and alcohols, the amines are classified according to the number of alkyl or aryl groups attached to the nitrogen atom. -The compounds in which an alkyl or an aryl group is attached instead of one of the three atoms of hydrogen in ammonia are called primary amines. -The compounds in which two alkyl or aryl groups are attached instead of two atoms of hydrogen in ammonia are called secondary ammines and the compounds in which three alkyl or aryl groups are attached instead of the three atoms of hydrogen are called tertiary amines.

- The compounds in which at least one aryl group is attached to the nitrogen atom are called aryl amines.


- Amines act as nucleophile due to the lone pair of electrons on the nitrogen atom. -The following are some of the reactions of primary amines with various reagents acting as nucleophiles.

With acid chlorides

With aldehydes and ketones

(The above reaction corresponds to the reaction of aldehydes and ketones with the Brady’s reagent.) With alkyl halides

-The secondary amine thus formed reacts with alkyl halide to form a tertiary amine and a quaternary ammonium salt.

Accordingly, a mixture of products is obtained in the above reaction. With NaNO2/HCl (Nitrous acid)

As alkyl diazonium chloride is unstable it easily turns into the alcohol.


Basicity of amine with other organic compounds. -Dilute mineral acids convert amines into their salts. These salts react with aqueous hydroxides to regenerate the amine. -Hence it is clear that although amines are more basic than water less basic than hydroxide ions.

- Amines are more basic than alcohols

-As nitrogen is less electronegative than oxygen it has a higher

tendency to donate lone pair of electrons. -Hence the stability of the alkyl ammonium ion relative to the amine is stronger than the stability of the alkyl oxonium ion relative to the alcohol. -The reason for this is that an atom with low electronegativity can bear a positive charge more easily. -Aliphatic primary amines are more basic than aniline. -The reason for the low basicity of aniline is because the lone pair of electrons on the nitrogen of aniline is not easily available to a proton due to it being delocalized with the aromatic ring by resonance. -Amides are less basic than amines. -It is because the pair of electrons on the nitrogen of the amide group is delocalized with the carbonyl group by resonance.

Basicity OH- (Hydroxyl ion) > R-NH2 > H2O

-Due to the high electronegativity of oxygen, amides are less basic than even the aromatic amines.


Reactions of Dizonium salt. - Aniline reacts with nitrous acid to give phenol.

- Aromatic diazonium salt is more stable than aliphatic diazonium salt. -Therefore, when this reaction is carried out at low temperatures, the conversion of the aromatic diazonium salt to the phenol can be

prevented.

REPLACEMENT REACTIONS OF THE DIAZONIUM GROUP


Reactions in which the diazonium salt reacts as an electrophile

- Diazonium salt gives an orange colored compound with phenol and a red colored compound with β- naphthol

chemistry book by supun ayeshmantha

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