PREFACE

KVS is a pace setter organization in the field of school education & always strives to provide quality education to the students.Preparation & printing of study-material is one of the most important aspects of quality education.

I feel great pleasure in presenting this study-material for class XII chemistry subject-“CHEM-BOOSTER”

CHEM-BOOSTER is basically designed to equip the students with ideal guidance to improve their grades. It provides “THE EASY WAY TO LEARN CHEMISTRY”

The unique features of this study-material are- Adequate number of important terms, definitions & formulae. Organic chemistry has been covered under sections like-nomenclature, named-

reactions, distinguishing–tests & Q-Answers which will be very useful to the students.

Q-answers are provided for every chapter, this will help students to learn & write the answers correctly in board exams.

Efforts have been made for thorough coverage of syllabus and make this material error free.

I hope this material will help students to get through Board exam with flying colours and teachers will find this book as good assignment material for the students.

I am very thankful to the Deputy Commissioner(Ahmedabad region) – Shri.P.Devakumar and Assistant Commissioners(Ahmedabad region) – Shri.P.Madan, Shri.Y.P.Singh, for their motivation and encouragement.

I appreciate the valuable contribution of following team members – 1. Dr. Gunjan Gaur PGT Chem. K.V. ONGC Baroda2. Mrs. Sabiha Shaikh PGT Chem. K.V. No.1, Surat3. Mr. R. R. Singh PGT Chem. K.V. EME Baroda4. Mrs. Samtan Negi PGT Chem. K.V. AFS Baroda

My Best wishes to all the users of this material.

Rajni TanejaPRINCIPAL

KV AFS BARODA

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INDEX

PHYSICAL CHEMISTRY1) THE SOLID STATE 42) SOLUTIONS 53) ELECTROCHEMISTRY 54) CHEMICAL KINETICS 5

INORGANIC CHEMISTRY1) GENERAL PRINCIPLES & PROCESSES OF ISOLATION OF ELEMENTS 32) THE p-BLOCK ELEMENTS 83) THE d-BLOCK AND f-BLOCK ELEMENTS 54) COORDINATION COMPOUNDS 3

ORGANIC CHEMISTRY1) HALOALKANES AND HALOARENES 42) ALCOHOLS,PHENOLS AND ETHERS 43) ALDEHYDES, KETONES AND CARBOXYLIC ACID 64) ORGANIC COMPOUNDS CONTAINING NITROGEN 4

GENERAL CHEMISTRY1) SURFACE CHEMISTRY 42) BIOMOLECULES 43) POLYMERS 34) CHEMISTRY IN EVERYDAY LIFE 3

-----------------------X---------------------------------X-----------------------------------X------------- PHYSICAL CHEMISTRY MATERIAL PREPARED BY – MRS. SAMTAN NEGI INORGANIC CHEMISTRY MATERIAL PREPARED BY – DR. GUNJAN GAUR ORGANIC CHEMISTRY MATERIAL PREPARED BY – MRS.SABIHA SHAIKH GENERAL CHEMISTRY – MR. R.R.SINGH

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PHYSICAL CHEMISTRY

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SOLID STATE

Q1. Crystalline solids are anisotropic in nature. What does this statement mean?Ans. Crystalline solids are anisotropic in nature means physical properties like

electrical conductivity, refractive index. etc., are show different values when measured along different directions in the same crystal.

Q2. How do metallic and ionic substances differ in conducting electricity?Ans. Metallic substances conduct electricity through electrons while ionic

substances conduct electricity in molten state or in solution through ions.Q3. What type of interactions hold the molecules together in polar molecular

solid?Ans. Dipole – dipole attractions.Q4. What is the total number of atoms per unit cell in a primitive cubic cell body

cetred cubic cell, facecentred cubic cell?Ans. Total number of atoms per unit cell in primitive cubic cell is 1

Total number of atoms per unit cell in body cetred cubic cell is 2Total number of atoms per unit cell in face centred cubic cell is 4

Q5. What is meant by the term 'coordination number'?What is the coordination number of an atom in:

(a) primitive cubic unit cell(b) body-centred cubic unit cell.

(c) face centred cubic unit cell (cubic close-packed structure)Ans. The number of nearest neighbours of any constituent particle present in the

crystallattice is called its coordination number.The coordination number of an atoms in a

(a) primitive cubic unit cell is 6(b) in a body-centred cubic structure is 8

(c) face centred cubic unit cell is 12Q6. Niobium crystallises in body-centred cubic structure.If density is 8.55gcm-3,

calculate atomic radius of niobium using its atomic mass 93 u.

Ans. It is given that the density of niobium, d = 8.55 g cm−3Atomic mass, M = 93 gmol−1As the lattice is bcc type, the number of atoms per unit cell, z = 2We also know that, NA = 6.022 × 1023 mol−1

= 3.612 × 10−23 cm3So, a = 3.306 × 10−8 cmFor body-centred cubic unit cell:

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=1.432 × 10−8 cm= 14.32 × 10−9 cm= 14.32 nm

Q7.

Ans.

Q8.

Ans.

Q9. Silver crystallises in fcc lattice. If edge length of the cell is 4.07 × 10−8 cm and density is10.5 g cm−3, calculate the atomic mass of silver.It is given that the edge length, a = 4.077 × 10−8 cmDensity, d = 10.5 g cm−3As the lattice is fcc type, the number of atoms per unit cell, z = 4We also know that, NA = 6.022 × 1023 mol−1Using the relation:

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= 107.13 gmol−1Therefore, atomic mass of silver = 107.13 u

Q10. Iron has a bcc unit cell with a cell edge length of 286.65 pm. The density of iron is 7.874 g cm-3 .To calculate Avogadro’s number.(At. Mass of iron is 56 g mol-1)

Ans. Given: Z = 2A = 286.65 pmD = 7.874 g cm-3

M = 56 g mol-1NA =?d = ZM / (a)3 X NA7.875 = 2 X 56 / (286.65)3 X 10-30 X NANA = 2 X 56 / (286.65)3 X 10-30 X 7.874

NA = 6.054 X 1023

Q11. A cubic solid is made of two elements X and Y. Atoms of Y are at the corners of the cubeand X at the body-centre. What is the formula of the compound?What are thecoordination numbers ofX and Y?It is given that the atoms of Y are present at the corners of the cube.

Therefore, number of atoms of Y in one unit cell.It is also given that the atoms of X are present at the body-centre.Therefore, number of atoms of P in one unit cell = 1This means that the ratio of the number of X atoms to the number of Y atoms, X:Y =1:1Hence, the formula of the compound is XY.The coordination number of both X and Y is 8.

Q13(i)(ii) (iii) (iv)(v)

Explain the following terms with suitable examples:Schottky defectFrenkel defectInterstitialsF-centresFerromagnetism

(i) Schottky defect: Schottky defect is basically a vacancy defect shown by ionicsolids. In this defect, an equal number of cations and anions are missing to maintainelectrical neutrality. It decreases the density of a substance. Ionic substances containing similarsizedcations and anions show this type of defect. For example: NaCl, KCl, CsCl, AgBr,etc.

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(ii) Frenkel defect: Ionic solids containing large differences in the sizes of ions showthis type of defect. When the smaller ion (usually cation) is dislocated from its normal site to an interstitial site, Frenkel defect is created. It creates a vacancy defect as well asan interstitial defect. Frenkel defect is also known as dislocation defect. Ionic solids suchasAgCl, AgBr, AgI, and ZnS show this type of defect.

(iii) Interstitials: Interstitial defect is shown by non-ionic solids. This type of defect is created when some constituent particles (atoms or molecules) occupy an interstitial siteof the crystal. The density of a substance increases because of this defect.

(iv) F-centres: When the anionic sites of a crystal are occupied by unpaired electrons,the ionic sites are called F-centres. These unpaired electrons impart colour to thecrystals. For example, when crystals of NaCl are heated in an atmosphere of sodiumvapour, the sodium atoms are deposited on the surface of the crystal. The Cl ions diffusefrom the crystal to its surface and combine with Na atoms, forming NaCl. During thisprocess, the Na atoms on the surface of the crystal lose electrons. These releasedelectrons diffuse into the crystal and occupy the vacant anionic sites, creating F-centres.

Ferromagnetism: The substances that are strongly attracted by a magnetic fieldare called ferromagnetic substances. Ferromagnetismarises due to spontaneous alignment of magnetic moments in the same direction.Ferromagnetic substances can be permanently magnetised even in the absence of a magnetic field. Some examples of ferromagneticsubstances are iron, cobalt, nickel, gadolinium, and CrO2.

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Q14. Define Doping? Classify each of the following as being either a p-type or an n-type semiconductor:(i) Ge doped with In (ii) B doped with Si.

Ans. Doping: It is a process by which impurity is introduced in semiconductors to enhance its conductivity. (i) Ge (a group 14 element) is doped with In (a group 13 element). Therefore, a hole will be created and the semiconductor generated will be a p-type semiconductor.(ii) B (a group 13 element) is doped with Si (a group 14 element). So, there will be an extra electron and the semiconductor generated will be an n-type semiconductor.

Q15. Why Frenkel defect not found in pure alkali metal halides?Ans. This is because the alkali metal cations have large size which cannot fit into

the interstitial sites.Q16. What changes occur When AgCl is doped with CdCl2?Ans. Cationic vacancy is generated or metal excess defect.

SOLUTIONS

Q1. Differentiate between molality and molarity of solution. What is the effect of change intemperature of a solution on its molality and molarity?

Ans. Molality: is the number of moles of solute per thousand grams of solvent where as Molarity is the number of moles of solute dissolved in one litre of solution.Molality is independent of temperature whereas Molarity is functionoftemperature because volume depends on temperature and the

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mass does notor Molarity decreases with increase of temperature.

Q2. Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20%(mass/mass) aqueous KI is 1.202 g mL-1.

Ans. (a) Molar mass of KI = 39 + 127 = 166 g mol−120% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g ofsolution.That is,20 g of KI is present in (100 − 20) g of water = 80 g of water

Therefore, molality of the solution= 1.506 m

= 1.506 m= 1.51 m(b) It is given that the density of the solution = 1.202 g mL−1

Volume of 100 g solution

= 83.19 mL= 83.19 × 10−3 LTherefore, molarity of the solution

= 1.45 M

(c) Moles of KI

Moles of water

Therefore, mole fraction of KI

= 0.0263

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Q3. Concentrated nitric acid used in laboratory is 68 % nitric acid by mass in aqueous solution.What should be the molarity of such sample of acid if the density of the solution 1.504 g/ ml?

Ans. Given: 68% HNO3 by mass means mass of HNO3= 68 g; d =1.504 g/ ml Molar mass of HNO3 = 53 g/molMolarity ofssolution = % X d X 10 / Molar mass of HNO3 = 68X 1.504 X 10 / 63 = 1022.723 / 63 = 16.23 M

Q4. a) State Henry’s law and mention its two important applications.b) Which of the following has higher boiling point and why? 0.1M NaCl or ) 0.1M Glucose

Ans. a) Henry’s law: The solubility of gas in a liquidis directly proportional to the pressure of the gas.

Applications: 1. Solubility of CO2 is increased at high pressure. 2. Mixture of He and O2 are used by deep sea divers because Less soluble than nitrogen.

b) 0.1 M NaCl, because it dissociates in solution and furnishes greater number of particles per unit volume while glucose does not dissociate.

Q5. CCl4 and water are immiscible where as ethanol and water are miscible in all proportions. Correlate this behavior with molecular structure of these compounds.

Ans. CCl4 is non-polar compound, where as water is polar compound. CCl4 can neither form H- bond with water molecules nor can it break H- bonds between water molecules, therefore , it is insoluble in water.Ethanol is polar compound and can form H- bond with water, which is a polar solvent, therefore it is miscible with water in all proportions.

Q6. State the Raoult's law in its general form in reference to solutions. How does Raoult's law become a special case of Henry’s law?

Ans. Raoult's law states that for a solution of volatile liquids, the partial vapourpressure of each component in the solution is directly proprtional to its molefraction in its solution. PA = P0A× XAPB = P0B × XBHenry’s law:The partial pressure of the volatile component or gas is directly proportional to its mole fraction in solution. P = KH × X Only the proportionality constant KH differs from P0A . Thus, Raoult’s law becomes a special case of Henry’s law in which KH becomes equal to P0A.

Q7. Two liquids A and B boil at 1450C and 1900C respectively. Which of them has higher vapour pressure at 800c?

Ans. Liquid A with lower boiling point is more volatile and therefore will have higher vapour pressure.

Q8. Differentiate between Ideal solution, Non- ideal solution, negative and positive deviations.

Ans. Ideal solution Non- ideal solution Negative deviations

positive deviations

i) Those solutions which follows Raoult’s law under all conditions of temperature

Those solutions which does not obey Raoult’s law for all concentration.

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and pressure.PA = P0A× XAPB = P0B × XB PA = P0A× XA

PB = P0B × XB

PA> P0A× XAPB> P0B × XB

PA<P0A× XAPB< P0B × XB

ii) Hmix = 0 Hmix = 0 Hmix> 0 i.e., +Ve Hmix< 0 i.e., -Veiii) Vmix = 0 Vmix = 0 Vmix> 0 i.e., +Ve Vmix<0 i.e., -Vevi) A-A = B-B = A-B (intermolecular forces between solvent- solvent and solut-solute and solvent – solute are same)

A-A = B-B = A-B A-B > A-A and B-B

A-B < A-A and B-B

v)Examples: methanol and ethanol, benzene and toluene

Acetone and benzene, CCl4 and C6H6, HNO3 and water

Acetone and benzene, CCl4 and C6H6

HNO3 and water, chloroform and acetone

Q9. Non-ideal solutions exhibit either positive or negative deviations from Raoult’s law. What are these deviations and why are they caused? Explain with one example foreach type.

Ans. When the vapour pressure of a non-ideal solution is either higher or lower than that predicted by Raoult’s law, the solution exhibits deviations.These deviations are caused because of unequal intermolecular attractive forcesbetween solute-solvent molecules and solute-solute or solvent-solvent molecules. Positive deviation eg: mixture of ethanol and acetone, carbon-disulphide and acetoneNegative deviation eg: Chloroform and acetone, nitric acid and water.

Q10. Two liquids A and B on mixing produce a warm solution. Which type of deviation from Raoult’s law does it show?

Ans. Warming up of the solution means that the processof mixing is exothermic, i.e Hmix = - Ve. This implies that the solution shows a negative deviation.

Q11. Define the following terms:(i) Azeotropess (ii) Colligative properties (iii) Van’t Hoff Factor (iv)

Isotonic solutionAns. (i) Azeotrope : the are binary mixtures having same composition in

liquid and vapour phase and boil ay constant temperature.(ii) Colligative Properties : Those properties which depend on the

number of solute particles but not upon their nature are called Colligative Properties.

(iii) Van’t Hoff Factor(i) : It may be defined as the ratio of normal molecular mass to the observed molecular mass of the solute.

(iv) Isotonic solution : Two solutions having same osmotic pressure at a given temperature are called Isotonic solution.

Q12. Define the terms, ‘osmosis’ and ‘osmotic pressure’. What is the advantage of using osmotic pressure as compared to other colligative properties for the determination of molar masses of solutes in solutions?

Ans. The flow of solvent from solution of low concentration to higher concentration 11

through semipermeable membrane is called osmosis.The hydrostatic pressure that has to be applied on the solution side to prevent the entry of the solvent into the solution through the semipermeable membrane is called the Osmotic Pressure.Advantage: i) Unlike other colligative properties, osmotic pressure is used toDetermine the Molar mass of macromolecules/polymers likeprotein.ii) Molarity of the solution is used instead of molality. iii) Compared to other colligative properties, its magnitude islargeeven for very dilute solutions.

Q13. a) Define Reverse Osmosis and state the condition resulting in reverse osmosis.b) Mention a large scale use of phenomenon called reverse osmosis?

Ans. (a) Pressure larger than osmotic pressure is applied to the solution side,i.e, P>π (osmotic pressure).

b) Desalination of sea waterQ14. What mass of NaCl (molar mass = 58.5 g mol–1) must be dissolved in 65 g of

waterto lower the freezing point by 7.500 C ? The freezing point depression constant, Kf for water is 1.86 K kg mol–1. Assume van’t Hoff factor for NaCl is 1.87.

Ans. Given: Tf = 7.50C : Weight water = 65 g:Kf = 1.86 × K kg mol-1 : van’t Hoff factor for NaCl is 1.87Weknow that: Tf = i Kf m 7.5 = 1.87× 1.86 × w × 1000/ 58.5 × 65 W = 8.2 g

Q15. Derive An equation to express that relative lowering of vapour pressure for a solution is equal to the mole fraction of the solute in it when the solvent alone is volatile.

Ans. For a solution of volatile liquids Raoult’s law, is given as P = PA + PBIf solute (componentB) is nonvolatile then P = PA = P0A XA = P0A(1 – XB) (XA + XB = 1 ) P = P0A - P0AXB P0AXB = P0A – PP0A – P / P0A = XBThus, relative lowering of vapour pressure is equal to the mole fraction of non-volatile solute.

Q16. (i) 1.00 g of a non-electrolyte solute dissolved in 50 g of benzene lowered the freezing point by 0.40 K. find the molar mass the solute. (Kf for benzene is 5.12 K kg mol-1)

(ii) Why is freezing point depression of 0.1M sodium chloride solution nearly twice that of 0.1 M glucose solution?

Ans. (i) Given WB = 1.00g; WA = 50g; Kf = 5.12 K kg mol-1 ; Tf = 0.40KTf = Kf × WB × 1000 / MB × WA (in grams)MB = Kf × WB × 1000 / WA × TfMB = 5.12 × 1 × 1000 / 0.40 × 50MB = 256g mol-1

(ii) NaCl being an electrolyte, dissociate almost completely to give Na+ and Cl- ions in solutions where as glucose being non-electrolyte, not

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does dissociate. Therefore, the number of particles in 0.1 M NaCl solution is nearly double( i= 2) than that in 0.1 M glucose solution. Freezing point depression, is a colligative property, therefore freezing point depression of 0.1M sodium chloride solution nearly twice that of 0.1 M glucose solution.

Q17. A solution prepared by dissolving 1.25 g of oil of winter green (methyl salicylate) in 99.0 g of benzene has a boiling point of 80.31 °C. Determine the molar mass of this compound. (B.P. of pure benzene = 80.10 °C and Kb for benzene = 2.53 °C kg mol–1)

Ans. Tb = (80.31 – 80.10)0C = 0.210C or 0.21 KΔTb = Kb m0.21 = 2.53 x 1.25g x 1000 / 99 x MM =152 g mol -1Where M is molar mass of the solute

Q18. What would be the value of Vant’s Hoff factor for a dilute solution of K2SO4 and K4[Fe(CN)6] in water?

Ans. In dilute solution , For K2SO4 2K+ + SO42

Vant’s Hoff factor(i)= Number of particles after dissociation / Number of particles before Dissociation= 3 / 1I = 3For K4[Fe(CN)6 4K+ + Fe(CN)64-

i = 5Q19. A solution prepared by dissolving 8.95mg of a gene fragment in 35.0mL. of

water has an osmotic pressure of 0.335 torr at 25°C. Assuming that the gene fragment is a non-electrolyte, calculate its molar mass.

Ans. osmotic pressure () = CRT = n2RT / V = w2 RT / V M2 M2 = w2 R T / VM2 = 8.95x10–3 x 0.0821 x 298 x 760 x 1000 / 0.335 x 35M2 = 14193.3 g mol–1 or 1.42x104g mol-1

Q20. The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase.

Ans. It is given that:P0A = 450 mm of HgP0B = 700 mm of HgP Total= 600 mm of HgFrom Raoult’s law, we havePA = P0A× XAPB = P0B × XB

Therefore, total pressure, PTotal = PA + PB

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Therefore, XB = 1 - XA = 1 – 0.4 XB = 0.6

Q21. The depression in freezing point of water observed for the same amount of acetic acid, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain briefly.

Ans. Among H, Cl, and F, H is least electronegative while F is most electronegative. Then, F can withdraw electrons towards itself more than Cl and H. Thus, trifluoroacetic acid can easily lose H+ ions i.e., trifluoroacetic acid ionizes to the largest extent. Now, the more ions produced, the greater is the depression of the freezing point. Hence, the depression in the freezing point increases in the order:Acetic acid < trichloroacetic acid < trifluoroacetic acid.

Q22. Determine the amount of CaCl2 (i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27°C.

Ans. We know that, () = iCRT = in2RT / V = iw2 RT / V M2 w2 = V M2 / i RGiven: = 0.75 atm; V = 2.5 litre ; i = 2.47 ; T = 273 + 27 = 300KHere,R = 0.0821 L atm K-1m0l-1M2 = 1 40 + 2 × 35.5= 111g mol-1w2 = 0.75 × 2.5 × 111 / 2.47 × 0.0821 = 3.42 gHence, the required amount of CaCl2 is 3.42 g.

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ELECTROCHEMISTRY

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Q1. Define conductivity and molar conductivity for the solution of an electrolyte. Discuss their variation with concentration.

Ans. Conductivity of a solution is defined as the conductance of a solution of 1 cm in length and area of cross-section 1 sq. cm. The inverse of resistivity is called conductivity or specific conductance. It is represented by the symbol κ. If ρ is resistivity, then we can write:

Units is S m-1.Conductivity always decreases with a decrease in concentration, both for weak and strong electrolytes. This is because the number of ions per unit volume that carry the current in a solution decreases with a decrease in concentration.Molar conductivity of a solution at a given concentration is the conductance of volume V of a solution containing 1 mole of the electrolyte kept between two electrodes with the area of cross-section A and distance of unit length.

/\m =K× 1000 / CUnits is S cm2 mol-1.

Molar conductivity increases with a decrease in concentration. This is because the total volume V of the solution containing one mole of the electrolyte increases on dilution.The variation of /\m with√ c for strong and weak electrolytes is shown in the following plot:

Q2. Express the relation among the cell constant, the resistance of the solution in the cell and the conductivity of the solution. How is the conductivity of a solution related to its molar conductivity?

Ans. k = 1/R (l/A)Where k is conductivity, R is resistance and l/A is cell constant/\m = k/CWhere /\m is molar conductivity and C is concentration of the solution.

Q3. The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm−1. Calculate its molar conductivity.

Ans. Given,κ = 0.0248 S cm−1

c = 0.20 M

Molar conductivity,

= 124 Scm2mol−1

Q4. The resistance of a conductivity cell containing 0.001M KCl solution at 298 K is 1500 M. What is the cell constant if conductivity of 0.001M KCl solution at 298 K is 0.146 × 10−3 S cm−1?

CHEMICAL KINETICS

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Q1. Define 'order of a reaction'. Identify the order of a reaction if the units of its rate constant are:(i) L–1 mol s–1

(ii) L mol–1 s–1

Ans.

Order of a reaction: The sum of powers of the concentration terms of the reactants in the rate law expression is called the order of that chemical reaction.rate = k[A]P [B]q

Order of reaction = P+qi) zero orderii) second order

Q2. Distinguish between 'rate law' and 'rate constant' of a reaction.

Ans.

Rate Law is the expression in which reaction rate is given in terms of molar concentration of reactants with each term raised to some power which may or may not be same as the stoichiometric coefficient of the reacting species in a balanced chemical equation, whereas the rate constant is defined as the rate of reaction when the concentration of the reactant(s) is unity.

Q3. A reaction is of first order in reactant A and of second order in reactant B. How is the rate of this reaction affected when (i) the concentration of B alone is increased tothree times (ii) the concentrations of A as well as B are doubled?

Ans.

Rate = k[A][B]2(i) When the concentration of B is increased to 3 times, then rate would beRate = k[A][3B]2

= 9k[A][B]2= 9 times the initial Rate1 i.e. rate is increased 9 times(ii) When the concentration of A as well as B are doubled, then rate would beRate = k[2A][2B]2= 8k[A][B]2= 8 times the initial Rate i.e. rate is increased 8 times

Q4. (a) Explain the following terms:(i) Rate of a reaction(ii) Activation Energy

(iii) Molecularity of a reaction(b) The rate of a reaction increases four times when the temperature changes from 300 K to 320 K. Calculate the energy of activation of the reaction, assuming that it does not change with temperature. (R = 8.314 J K–1 mol–1)

Ans.

(a)(i) Rate of a reaction- Rate of Change of concentration of reactant or product with time is called rate of reaction(ii) Activation Energy – Minimum energy which the reacting molecules should acquire so that they react to give product is called activation energy.

(iii) Molecularity – Number of molecules taking part in rate determining step of a reaction is called molecularity.

(b) log k2/ k1 = Ea × [T2- T1] / 2.303 R T1 × T2

log 4 = Ea × 320- 300 / 2.303 x 8.314 × 300 x 320

0.6020 = Ea×20 / 2.303 x 8.314 x 96 x 103

Ea = 55336.7 J mol-1 Ea = 55.33 kJ mol-1

Q5. The rate constant for a reaction of zero order in A is 0.0030 mol L –1 s–1. How long will it take for the initial concentration of A to fall from 0.10 M to 0.075 M ?

Ans.

[R]t = - kt + [R]00.075M = - (0.0030 mol L-1 s-1)t + 0.10M

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INORGANICCHEMISTRY

UNIT 6 GENERAL PRINCIPLES AND PROCESS OF ISOLATION OF ELEMENTS

Q.1.Discuss following a. Ore b. Hydraulic washing c. Electromagnetic separation A.1. a. The mineral from which the metal is conveniently and economically extracted is called ore. Eg: bauxite, heametite. b. . The process by which lighter earthly particles are feed from the heavier ore particles by washing with water.

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c. Electromagnetic separation This method of concentration is employed when either the ore or the impurities associated with it are magnetic in nature.eg haemetite,pyrolusite. Q.2. Discuss a. Froth floatation.b. Leaching.A.2.a.Froth floatation. This method is based upon the fact that the surface of sulphide ores is preferentially wetted by oils while that of gangue is preferentially wetted by water. Depressant are used to prevent certain types of particles from forming the froth with the air bubbles. For example, sodium cyanide is used as depressant. 4NaCN + ZnS -- Na2[Zn(CN)4] + Na2Sb. Leaching. This is a chemical method of concentration and is useful in the case when the ore is soluble in some suitable solvent. In this method the powered ore is treated with certain reagents in which the ore is soluble but the impurities are not soluble. The impurities left undissolved are removed by filtration. Al2O3.2H2O + 2NaOH + 2H2O -- 2Na[Al(OH)4] + 4H2O 2Na[Al(OH)4] + CO2 Al2O3.XH2O + 2NaHCO3 Al2O3.XH2O Al2O3 + XH2O

Q.3. Discuss calcinations and roasting.A.3.Calcination. It is a process of heating the ore strongly either in a limited supply of air or in the absence of air.

For eg: ZnCO3 ZnO + CO2 Roasting. It is a process of heating the ore strongly in the presence of excess of air at a temperature below the melting point of the metal. 2ZnS + 3O2 2ZnO + 2SO2

Fe2O3 + 2Al ------ 2Fe + Al2O3 Cu2S +2 Cu2O------ 6Cu + SO2

Q.4. What do you understand by a. Auto reduction b. Hydrometallurgy c.ElectrometallurgyA.4 a.Auto reduction. Reduction can be carried out by the auto reduction in which

the anions associated with the help in the reduction. This is used for the reduction of sulphide ore of Pb, Hg and Cu. In this case, no reducing agent is required. The metal is obtained either by simple roasting or by the reduction of its partly oxidized form. HgS + O2 ------ Hg + SO2 PbS + 3O2 ----- 2Pb + 2SO2

b. Hydrometallurgy. The process of extraction of metals by dissolving the ore in a suitable chemical reagent and the precipitation of the metal by more electrop- ositive metal is called hydrometallurgy. Ag2S + 4NaCN -- 2Na[Ag(CN)2] + Na2S 2Na[Ag(CN)2] + Zn – 2Ag + Na2[Zn(CN)4] Na2[Zn(CN)4] + 4NaOH –- Na2ZnO2 + 4NaCN c. Electrometallurgy. The process of extraction of metals by electrolysis process is called electrometallurgy.

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At cathode Na+ + 1e- -- Na At anode 2Cl- ----------- Cl2 + 2e-

Q.5. Discuss a. Electrorefining b. Zone refining c. Vapour phase refining d. Van arkel method.A.5. a.Electrorefining. This is the most general method for the refining of metals and based upon the phenomenon of electrolysis. In this method electricity is passed through the electrolytic

solution.copper is refined by this method. At cathode Mn+ + ne- → M At anode M → Mn+ + ne-

b.Zone refining. This method is used for the metals which are required in very high purity e.g Ge.This method is based upon the principle that the impurities are more soluble in the melt than in the solid state of the metal.

c. Vapour phase refining.(Mond’s process) This method on the fact that certain metal are converted to their volatile compounds while the impurities are not affected during compound formation. The two requirements are :

(a)The metal should form a volatile compound (b) The volatile compound should be easily decopopsed.

Ni + 4CO → Ni(CO)4 → Ni + 4CO IMPURE PURE

d. Van arkel method. This method is similar to the above method and used for getting the ultra pure metal. Ti + 2I2 → TiI4 → Ti + 2I2

Q.6. Which is the cheapest and most abundant reducing agent which is used in

the extraction of metal?A.6. Carbon in form of coke.Q.7. Why is zinc and not copper used for the recovery of metallic silver from

its cyanide complex [Ag(CN)2]?A.7. Zinc is more electropositive than silver and therefore, zinc displace silver from its

solution. 2[Ag(CN)2] + Zn → [Zn(CN)4]2- + 2AgOn the other hand copper is less electropositive than silver and therefore, cannot displace silver from its solution.

Q.8. Why does CaO react with with SiO2 to form a slag?A.8. CaO is basic oxide whereas SiO is acidic oxide. Their reaction is acid-base reaction as.

CaO + SiO2 → CaSiO3. SlagQ.9. Why is the formation of sulphate in calcinations sometimes advantageous?A.9. Sulphates are usually water soluble and the gangue remains insoluble. Therefore,

the desired metals is leached away as soluble sulphate from insoluble gangue.Q.10. How does NaCN act as a depressant in preventing ZnS from forming the froth?A.10. NaCN forms a layer of zinc complex, Na2 [Zn(CN)4] on the surface of ZnS and

thereby prevents it from the formation of the forth.Q.11. At a site, low srade copper ores are available and zinc and iron

scraps are also available. Which of the two scraps would be more suitable for reducing the leached copper ore and why?

22

A.11. Since zinc lies above iron in electrochemical series, it is more reactive than iron. As a result, if zinc scraps are used, the reduction will be fast. However, zinc is a costlier metal than iron. Therefore, it is advisable and advantageous to use iron scraps.

Q.12. Lime stone is used in the manufacture of pig iron from haematite. Why?A.12. Haematite is an ore of iron contain silica (SiO2) as the main impurity. The

purpose of limestone is to remove SiO2 as calcium silicate (CaSiO3) slag. CaCO3 → CaO + O2 Limestone CaO +SiO2 → CaSiO3 Basic flux Q.12. What is the composition of copper matte?A.12. Copper matte contains cuprous sulphide (Cu2S) and iron sulphide (FeS).

Q.13. What is cupellation?A.13. Cupellation is a method used for refining of those metals which contains

impurities of other metals which form volatile oxides. For example, removal of last traces of lead from silver.

Q.14. How is the granular zinc obtain?A.14. Granular zinc is obtained by pouring molten zinc in cold water.Q.15. What is Kroll process?A.15. The production of titanium metal at 900K from TiCl4 by reduction with Mg in

argon atmosphere. TiCl4 (g) + 2Mg → Ti (s) + 2MgCl2 (l)Q.16. Galena (PbS) and cinnabar (HgS) on roasting often give their

respective metals but zinc blende (zinc) does not. Explain?A.16. On roasting sulphides are partly converted to their respective oxides. Since

the oxides of lead and mercury are unstable, these bring about the reduction if their respective sulphides to the corresponding metals.

PbS + 2PbO → 3Pb + SO2 Unstable HgS + 2HgO → 3Hg + SO2 Unstable

However, zinc oxide is stable and it does not reduce ZnS to Zn.

Q.17. What is the stabilizer in froth floatation process? Give examples.A.17. Stabilizer in froth floatation process stabilizes the forth. For example, aniline, cresol, etc.Q.18. Name three forms of iron. How these three forms do differs?A.18. Three forms of iron are : cast iron, wrought iron and steel. These differ in their carbon content.Q.19. What is the actual reducing agent of haematite in the furnace?A.19. Carbon monoxide, COQ.20. Why do not metals occure as nitrates in the nature?A.20. Nitrates of almost all metals are soluble in water.Q.21. Why copper matte is put in silica lined converter?A.21. Matte has Cu2S and FeS .FeS is oxidized to FeO which combines with silica to form FeSiO3.Q.22. Why alumina can not be reduced by carbon?A.22. At high temperature alumina reacts with carbon to form aluminium carbide. 2Al2O3 + 6C → Al4C3 + 3CO2Q.23. Name the following.

(a) Any two metals which never occur in native state. (b) Any two metals which never occur in uncombined state.

23

A.23. (a) sodium, calcium (b) gold, platinumQ.24. What is the main difference between red & white bauxite ores?A.24 Red bauxite contains iron oxides as the main impurity while white bauxite

contains silica as the main impurity.

*****

24

UNIT- 7P – BLOCK ELEMENTS

GROUP 15Q. 1. N does not form NCl5.Why?A. 1. Due to absence of d-orbital, it cannot extend its octet.Q. 2. NCl3 is easily hydrolysed.Why?A. 2. Due to presence of vacant of orbital in Cl atom, it can form co-ordinate bond with H2O.Q. 3. N2 is highly inert. Why?A. 3. Due to high bond- dissociation energy.Q. 4. Why is white P kept under water?A. 4. Ignition temperature of white P is low (303K); on exposure to air, it spontaneously

catches fire forming P4O10.Q. 5. H3PO3 act as a reducing agent but H3PO4 does not .Why?A. 5. H3PO3 has one P−H bond, therefore it acts as reducing agent but H3PO4 does not have any P−H bond.Q.6. First element of each group differs in its properties from other elements of the group .Why?A.6. Because of: - 1). Small size 2). High ionisation energy 3). Absence of d-orbital.Q.7. HNH angle is higher then HPH , HAsH & HSbH angle.Why?A.7. N is highly electropositive it attracts bonded pair of electron more toward itself,

therefore it minimum between lp-bp electros bond. Angle is high.Q.8. White P is more reactive than red P. A.8. White P has P4 Tetrahedral Structure and under stain. Red P has polymeric structure

and becomes giant molecule.Q.9. Bismuth is strong oxidising agent in penta valent state?A.9. +5 oxidation state is less stable than +3 due to inert pair effect.Q.10. NO becomes brown when released in air. A.10.No reacts with O2 to form Brown gas NO2

2NO + O2 2NO2Q.11. Why does R3P=O exist but R3N=O does not (R= alkyl group)?A.11. On the other hand P due to the presence of d orbitals forms pπ – dπ multiple bonds

and hence can expands its covalency beyond four. Therefore P forms R3P=O in which covalency of P is five.

Q.12. PCl5 is ionic in solid state.A.12.It exists as [PCl4]+ [PCl6]-

Q.13. NH3 acts as a ligand.A.13.In NH3, Nitrogen has a lone pair of electron.

Q.14. H3PO2 acts as monobasic acids.25

A.14.Because it has only one replaceble H-atom

Q.15. NH3 has higher proton affinity than PH3.A.15.Due to smaller size, high electron density and early availability of electron on N in

NH3.

Q.16. Discuss preparation of HNO3 by Ostwald Process.A.16. Ostwald process Pt/Rh 4NH3 + 5O2 → 4NO + 6H2O condition : 500 K/9 bar 2NO + O2↔2NO2 3NO2 + H2O → 2HNO3 +NO

Q.17. What is the maximum covalence of nitrogen?A.17. 4Q.18. PH3 has lower boiling point than NH3. Why?A18. Due to presence of hydrogen bonds in NH3.

GROUP 16Q.19. Mention three areas in which sulphuric acid plays an important role.A.19.1.In the manufacturing of fertilizers 2. It is used in storage batteries

3.It is used in petroleum refining.

Q. 20. Sulphur disappears when boiled with an aqueous alkaline solution of Na2SO3. Why? A.20. Na2SO3 + S → Na2S2O3.Q.21. SF6 is known & not SCl6.A.21.Because of larger size of Cl atom, 6Cl atoms cannot be accommodated around S.Q.22. H2S is less acidic then H2Te.Why?A.22. Because H −Te bond is weaker then H −S bond.Q.23. Sulphur in vapour state exhibits paramagnetic behavior?A.23.In vapour state S exist in S2 form like O2 has two unpaired electrons in antibonding π¿ orbital.Q.24. Thermal stability of H2O is more than H2S.A.24.Due to H – bonding in H2O.

Q.25. SF6 is well known but SCl6 is not known.A.25.Due to small size of S, six large Cl atoms can not be accommodated around S atom.Q.26. Write preparation of sulphuric acid through contact process.A.26. Contact process S + O2 → SO2 V2O5 2 SO2+ O2 → 2SO3 key reaction ∆rH = -196.6 kJmol-1 SO3 + H2SO4 →H2S2O7 (oleum) H2S2O7 + H2O → 2H2SO4

GROUP 1726

Q.27

A.27

Why is F2O referred to as a fluoride but Cl2O is an oxide?Because F is more electronegative than O and O is more electronegative than Cl.

Q .28. Bleaching by Cl2 is permanent & SO2 is temporary.A.28. Cl2 bleaching is by oxidation, SO2 bleaching is by reduction, hence the product can be re-oxidised.Q.29. F2 is the strongest oxidizing agent. Why?A.29. It is due to low enthalpy of dissociation of F2 & high enthalpy of hydration of F¯ion.Q.30. HI cannot be prepared by heating KI with concentrated H2SO4.Why?A.30. HI is strong reducing agent & reduces H2SO4 to SO2 & gets oxidised to I2.Q.31. ICl is more reactive then I2?A.31. I−Cl bond is weaker then I−I bond , therefore ICl is more reactive.

Q.32. Si F6−2

is known but Si cl6−2

is not known,A.32.Due to smaller size of F, steric repulsion is less in Si F6

−2

Q. 33. The order of increasing oxidising ability is HIO4<HClO4<HBrO4.Why?A. 33. Because standard electrode potential of XO4¯/XO3¯ is highest in HBrO4.Q. 34. HClO4 is most acidic amongst HClO, HClO3 & HCO2.Why?A. 34. Because it’s conjugate base is most stable due to resonance

HClO4 ↔ H+ + ClO4¯ Q.35. Bond dissociation energy of F2 is less than that of Cl2.A.35.Due to smaller size, larger electron – electron repulsion.Q.36. HF is a weaker acid than Hcl in water.A.36.Due H – bonding in HF, H+ ions are not formed easily.Q.37. Addition of Cl2 to KI solution gives a brown colour but excess of Cl2

turns it colourless.A.37.Due to the following reactions.

KI + Cl2 Kcl + I2 (Brown)in excess Cl2 + I2 + H2O HCl + HIO3 Colourless

Q.38.

A.38.

The halogens are coloured.Why?This is Because halogens absorb radiations in the visible region. This results in the

excitation of valence electrons to a higher energy region. Since the amount of energy required for excitation differs for each halogen, each halogen displays a different colour

GROUP 18

Q. 39. Only Xe among noble gases forms compounds. Why?A. 39. Because of its low ionization energy.

27

Q. 40. Boiling point of noble gases increase with the increase in atomic number. Why?A. 40. Because magnitude of van-der-waalۥs forces increases with increase in atomic size.Q. 41. Ne is generally used in waning signal illumination. Why?A. 41. Because it is visible through mist & fog from long distances.Q.42. Noble gases are Chemically inert.A.42.Reasons (i) Stable configuration

(ii) High I.E.(iii) Very low electron affinity.

Q.43. Why is helium used in diving apparatus?A.43. Because of its very low solubility in blood.

Structures Of Compounds

1) NH3

2) White Phosphorus

3) Red Phosphorus

4) PCl3

5) PCl5

28

6)O3

7) S6 8) S8

10) HOClO9) SO2

12)HOClO2

11)HOCl

13)HOClO3

14)BrF3

29

15)XeF2

16) XeF4

17) XeOF4

18) XeF6

19) XeO3

20) H4P2O7

30

21)H3PO3 22) H3PO2

23) H3PO4

24)H2SO3

25)H2SO4

31

26) H2S2O8

27) N2O4

28) N2O

29)NO

30) N2O5

31) NO2

32) H2S2O7

33)ICl4

32

34)IBr2 -

35)BrO-3

33

UNIT-8 THE d & f BLOCK ELEMENTS

Q.1. Transition metals and their compounds are good catalysts.A.1. (1) They form reaction intermediates which follow the path of low activation

energy.(2) They provide suitable large surface area for adsorption of reactant.

Q.2. Enthalpy of atomisation of transition elements is very high.A.2. Because of large number of unpaired electrons in their atoms. They have

stronger bonding between the atoms.Q.3. Explain why transition metals are paramagnetic?A.3. The ions of transition metals generally contain one or more unaired electrons hence the compounds of transition metal are paramagnetic i.e., they are weakly attracted by magnetic field. The paramagnetic character is directly related to the value of magnetic moment, m which intern depends upon the number of unpaired electrons.

Q. 4. Why the melting points of transition elements are high ?A.4. The melting points of transition elements are high due to the presence of strong intermetallic bonds and covalent bonds.Q. 5. Why Zn, Cd and Hg are not regarded as transition elements ?A.5. Because they have the completely filled d-subshell with outer electronic configuration (n –1)d10 ns2.Q.6 . Why the compounds of transitions elements are coloured ?A.6. The colour of compounds of transition elements depend upon the unpaired

electrons present in d-orbitals fo transition element. If d-orbitals are completely vacant or completely filled the compounds will be colourless, but if any unpaired electron is present in d-orbitals, the compound d transition. Thewill be coloured due to d → unpaired electron is excited from one energy level to another energy level with in the same d-sub-shell. For this purpose, the energy is absorbed from visible region of radiation. The complementary part of the absorbed light i.e., reflected light will decide the colour of the compound.

Q.7. Why the transition elements act as catalyst ? Give two examples.A.7. (a) Transition metal show variable oxidation states, therefore, they can from

intermediate products of difficult reactant molecules.

34

(b) Transition elements are capable to form interstitial compounds due to which they can absorb and activates the reacting molecules.Example :(a) V2O5 is used for the oxidation of SO2 to SO3 in contact process of H2SO4.(b) Ni is used as a catalyst in the hydrogenation of alkenes and alkynes

Q.8. Why transition elements form(a) Interstitial compounds and(b) Alloys

A.8. (a) Interstitial compounds : Transition elements form large number of intestitial compounds. In these compounds small size atoms like hydrogen, carbon, nitrogen , nitrogen, boron etc. occpy the empty space of metal lattice. The small entraped atom in the interstics form the bonds with metals due to which malleability and ductility of the metals decrease, whereas tensile strength increases.(b) Alloys : Tranisition element forms alloys with each other because they have alomost similar sizes. Due to similar sizes atoms of one metal in the crystal lattice can easily take up the position of the atom of transition elements. Alloys are more resistant to corrosion than the constituent elements, and usually harder with higher melting point.

Q. 9. Why is K2Cr2O7 generally preferred over Na2Cr2O7 in volumetric analysis although both are oxidising agents ?

A.9. Because Na2Cr2O7 is hygroscopic, hence it is difficult to prepare its standard solution for volumetric analysis, but because of nonhygroscopic nature of K2Cr2O7 its standard solution can be prepared.

Q. 10. Why do Zr and Hf exhibit similar properties ?A.10. Because of the lanthanoid contraction Hf has similar size to Zr, therefore

both Zr and Hf exhibit similar properties.Q.11. Give two uses of lanthanoid compounds.A.11. (i) Misch metal is pyrophoric and used in gas lightness, tracer bullets and shell.

(ii) Oxides of neodymium and praseodymium are used for making colour glasses.Q. 12. All scandium salts are white. Why ?

35

A.12. Because they have no electron in d-orbital, thus no d - d transition is possible. Due to this reason all salts of Sc3+ are white.

Q.13. The first ionisations energies of the 5d-transition elements are higher then those of 3d and 4d Transition elements.

A.13.Due to lanthanoid contraction.Q. 14. Transition metals are known to form complexes.Discuss. A.14.Because of small size and presence of vacant ‘d’ orbitals.Q.15. Name a transition metal which does not exhibit variation in oxidation state in its compounds.A.15.Zinc in its compounds shows an oxidation state of + 2 only.Q.16. Name a transition metal which exhibit maximum nuber of oxidation states in its compounds.A.16. MnQ. 17. Why do the d-block elements exhibit a larger number of oxidation states than f-block elements ?A.17. Because the energy of ns-electron and (n – 1) d-electrons are nearly same, therefore, ns electrons as well as (n – 1) d-electrons can take part in bond formation in transition elements. In f-block elements last electron goes to the f-orbitals of second order outer most shell, thus the difference between the energy of ns-electron and (n – 2) f-electrons increases. Due to this reasons all the (n – 2) f-electrons cannot take point in bond formation.Q. 18. Explain why the first ionisation energies of the elements of the first transition service do not very much with increasing atomic numbers.A.18. With the increasing atomic number, d-electrons add one by one in (n – 1) shell or penultimate shell. The screening effect of these d- electrons shield the outer s-electrons from inward nuclear pull. The effect of the increase in nuclear charge with the increase in atomic number is opposed by the shielding effect of d-electrons . thus due to these counter effect there is a vary little variations in the values of ionization energies of first transition series.Q.19. Discuss extraction of potassium dichromate from chromite ore.A.19. Preparation:-

36

Q.20. Draw structure of chromate ion and dichromate ion.A.20.

Q.21. What is the effect of pH on the colour of the solution of potassium dichromate?A.21. A lower pH, the colour of the solution is orange due to the solution is orange due to the presence of dichromate ions (Cr2O72–). But in alkaline PH, the colour of the solution changes to yellow due to the conversion of dichromate ions to chromate ions.

Q.22. Discuss oxidizing action of potassium dichromate in acidic medium.

A.22. Q.23. Discuss extraction of potassium permanganate from pyrolusite ore.A.23. Potassium permanganate KMnO4

PREPARATION :-

37

Q.24. In what way the electronic configuration of transition elements are different from those of representative elements ?A.24. In representative elements last electron goes either s-subshell or p-subshell of last orbit, while in transition elements last two orbits are incomplete. The outer electronic configuration of representative element is either ns1–2 or ns2 np1–6, while in transition element outer electronic configuration is (n – 1) d1–10 ns1–2.Q. 25. What is lanthanoid contraction ? Discuss its cause and consequences.A.25. The steady decrease in atomic and as we move from La to Lu, is known as lanthanide contraction.

Cause : In lanthanoids, last electron enters the 4f-sub-shell i.e., second last shell. The shape of f-orbitals is very much diffused, because of this reason the mutual shielding of 4f-electrons is very little. The nuclear charge increases at each step due to the increase in atomic number, while the mutual shielding effect of 4f-electrons is comparatively negligible. This causes a decrease in size of the 4f sub-shell with the increase in atomic number.

Consequences : Following are the main consequence of lanthanoid contraction.(a) Basic strength of oxides and hydroxides of lanthanoids : The basic strength of

oxidesand hydroxides of lanthanides decrease with the increase in atomic number the size of lanthanum ion (Ln3+) decreases due to lanthanide contraction. The decrease in size of Ln3+,increases the covalent character i.e., decreases the ionic character between Ln3+

and OH–

ions, consequently the basic strength of oxide and hydroxide will decrease.

38

(b) Similar sizes of second and third transition series elements : Normally the atomic radii

increases with the increase in atomic number in the same sub-group, but after lanthanides

the atomic radii of the elements of second and third transition series are almost similar.

The expected increase in size from second to third transition series is cancelled by the decrease

in size due to lanthanide contraction. Due to the similar atomic of radii of second and third

transition series elements they resemble each other very closely.(c) Separation of lanthanoids : The properties of lanthanides are very similar,

therefore, it isdifficult to separate them. However due to lanthanoid contraction, decrease in

size oflanthanoids make the separation possible by ion exchange methods.

Q.26. Write differences between chemistry of Lanthanoids and Actinoids A.26. Differences between Lanthanoids and Actinoids

Lanthanoids Actinoids1 Electrons are successively

added in 4f orbitals.1 Electrons are successively added

in 5f orbitals.2 The chemistry of

lanthanoide elements is fairly similar

because of the large energy difference between4f and 3d subshells.

2 There is a considerable difference in the chemistry

of actinide elements. This is due to very small difference in 5f and

6d energy levels.

3 4f electrons have greater shielding effect.

3 5f electrons have poor shielding effect.

4 4f orbitals have higher binding energies.

4 Binding energies of 5f orbitals are lower.

5 Lanthanoids show lesser number of oxidation states.

5 Actinoids show a variety of oxidation states.

6 They do not have much tendency for complexes

6 They have much tendency for complexes formation.

39

formation.7 Except promethium, all the

lanthanoids are non radioactive.

7 All the actinoids are radioactive.

8 Lanthanids do not form oxocations.

8 Actinoids form oxocations.

9 Compounds of lanthanoids are less basic.

9 Compounds of actinoids are more basic.

10. Electronic configuration is 6s2 5d0-1 4f1-14

10. Electronic configuration is 7s2 6d0-1 5f1-14

Q.27. For the first row of the transition metals the E° value are:Metal V Cr Mn Fe Co Ni Cu E o [M+2 /M] – 1.18 – 0.91 – 1.18 – 0.44 – 0.28 –2 0.25 + 0.34 volts (V)Give suitable explanation for the irregular trend in these values

A.27. The irregular trend in the (M+2 /M) E values for the first row transition metals is due to the irregular variation in the ionisation and sublimation energies across the series.Q.28. Of the ions Co2+, Sc3+ and Cr3+ which ones will give coloured aqueous solutions and how will each of them respond to a magnetic field and why?

(Atomic numbers: Co = 27, Sc = 21, Cr = 24)A.28. The electronic configurations of the given ions are,

Co2+ : 1s2 2s2 2p6 3s2 3p6 3d7 No. of unpaired electrons = 3Sc3+ : 1s2 2s2 2p6 3s2 3p6 3d0 No. of unpaired electrons = 0Cr3+ : 1s2 2s2 2p6 3s2 3p6 3d3 No. of unpaired electrons = 3Co2+ and Cr3+ ions will give coloured aqueous solutions. Co2+ and Cr3+ are paramagnetic, and Sc3+ is diamagnetic. Therefore, Co2+ and Cr3+ ions will get attracted to the magnetic field, whereas Sc3+ ion will be repelled by the magnetic field.

Q.29. (a) Assign reason for each of the following:(i) Ce3+ can be easily oxidised to Ce4+.

(ii) E° for Mn3+/Mn2+ couple is more positive than for Fe3+/Fe2+.

40

(b) Mention two uses of potassium permanganate in the laboratory.

(Atomic number: Mn = 25, Fe = 26, Ce = 58) A.29. (i) Ce3+ has only one electron in its 4f orbitals. Due to extra stability of completely empty orbitals belonging to an energy level as compared to having only one electron in it, Ce3+ tends to lose its only electron from 4f orbital and get oxidised to Ce4+.

(ii) Mn3+ has a d 4 configurtion, so it has greater tendency to accept one electron to acquire d 5 configuration. On the other hand, Fe3+ has a d5 configuration which is more stable than the d 6 configuration of Fe2+. As a result, reduction of Fe3+ to Fe2+ is not favoured. Since, E° values reflect the reduction tendency, therefore, E° value for Mn3+/Mn2+ couple is more positive than Fe3+/Fe2+.

(b) The uses of potassium permanganate in the laboratory are(i) As an oxidising agent,(ii) In volumetric estimation of reducing agents such as Fe2+ salts, oxalic

acid etc.

****

UNIT 9CO-ORDINATION COMPOUNDS

IMPORTANT QUESTIONS

Q. 1. Define coordination compound.A.1 Coordination compounds are those addition compounds which retain their

identity in dissolved state also like solid state. In coordination compounds the atoms or group of atoms have been attached to central metal atom or ion beyond the number possible according to electrovalent or covalent bonding.

Q. 2. What do you mean by ambidentate ligands ?A. 2. Any ligand which has two or more donor atoms but only one donor atom is

attached to the metal ion at a time, during complex formation, is known as ambidentate ligand.

41

Q. 3. What is the relation between the molar conductivity of the solution of the complex compound and total number of ions?

A. 3. The total number of ions given by a complex can be predicted with the help of molar conductivity of the solution of complex compound. Molar conductivity of the solution of complex compound depends upon number of ions by a complex in solution.

Q. 4. What do you mean by organometallic compounds ?A.4 . Any compound which contains at least one metal-carbon bond is called

organometalliccompound.

Q.5. How many isomers are there for octahedral complex [CoCl2(en) (NH3)2]+.

A. 5. To geometrical isomers i.e., Cis and tans. cis isomer has two optical isomers i.e., d-cis isomer and l-cis isomer.

Q. 6. Write the IUPAC name for the linkage isomer of [Co (NH3)5. O.NO]Cl2.

A.6. The linkage isomer will be [Co(NH3)5 NO2]Cl2 and its IUPAC name will be Pentamminitrito-N-cobalt (III) chloride.Q. 7. Write the IUPAC name of the ionisation isomer of [Co

(NH3)5SO4] Br.A.7. The ionisation isomer will be [Co (NH3)5 Br]SO4 and its IUPAC name will be

Pentamminebromidocobalt (III) sulphate.Q.8. How is ammonia molecule a good ligand ?A.8. Nitrogen of ammonia has one lone pair of electrons. Because of the small

size of nitrogen, the tendency to donate this electron pair is also high, thus NH3 is a good ligand.

Q. 9. What do you mean by inner orbital complex and outer orbital complexes ? Give examples of each.A. 9. If the d-orbitals used in the hybridization are of lower shell than the s-and

p-orbitals, the (n–1)d2sp3 type of complexes are formed and are called inner orbital or low spin complexes. For example in K4[Fe (CN)6], 3d, 4s and 4p orbitals take part in hybridisation, therefore, it is inner orbital complex. If the d-orbitals used in the hybridization are of the same principal energy level as that of s-and p-orbitals, the complexes are of nsp3nd2 type and are called outer orbital complexes or high spincomplexes. For example in [FeF6]3–, 4s, 4p and 4d orbitals take part in hybridization, therefore, it is outer orbital complex.

Q.10. It is true that a cyclic complex is usually more stable than an open one. Substantiate your answer with an example.

A. 10. Cyclic complexes i.e., chelates are more stable than open complexes. This is because of reduced strain due to the formation of 5 or 6 membered ring including metal ion. Moreover in cyclic complex, the ligand is attached with two or more bonds with metal ion, hence more bonds have to break. Due to this reason cyclic complexes are more stable. The copper tetraammine complex is less stable than the copper ethylenediamine

42

complex although in both the cases nitrogen atom is the donor. Many cyclic complexes like Ni-dmg, Chlorophyll, Haemoglobin are very stable towards dissociation, while the noncyclic complexes of these metal ions are less stable anddissociate easily.

Q. 11. Mention the factors on which stability of the complexes depends.

A. 11. Following factors affect the stability of the complexes.(i) Basic nature of the ligand : Greater the basic nature of the ligand greater will be the stability. The copper complex with CN– is more stable than the complex with NH3 (less basic).(ii) Charge on central metal ion : Greater the charge on central metal ion greater will be the stability. [Fe (CN)6]3– is more stable than [Fe (CN)6]4– due to the higher charge on Fe.(iii) Number of ring structures in complex : If a ring is formed during complexion, it provides extra stability. That is why chelates are more stable.

Q. 12. Define cis and trans isomerism in complexes.A. 12. C is and trans isomerism i.e., geometrical isomerism arises due to the

different arrangement of ligands around the central metal ion. In cis isomer two identical ligand molecules are adjacent to each other i.e., on same side while in trans isomer two identical ligand molecules are diametrically opposite to each other. These two isomers differ in physical and chemical properties of each other and can be separated by some chemical and physical methods. This type of isomerism is not found in tetrahedral complexes but is common in square planner and octahedral complexes.

Q. 13. What is the importance of coordination compounds in industry and therapy?

A. 13. In Industry(i) Photography : In photography, the excess of silver salts has to be remove to fix the image on the negative and to prevent their further reduction. For this purpose the negative is immersed in a bath containing sodium thiosulphate solution, where excess of silver forms the soluble complex and is washed away.(ii) In electroplating : Cyanocomplexes of copper, silver, gold etc, are used for electroplating of these metals. Since the metal complexes release the metal ions slowly, therefore, a thin and uniform coating of the metal can be deposited on desired object with the help of electroplating. In chemotherapyMany coordination compounds are successfully used as chemotheropeutic

drugs.(i) Cis-platinum is used in the treatement of cancer.(ii) Metal complexes of Ni, Fe, Co with some nitrogen and sulphur containing ligands are very good antituberculosis agents.

Q. 14. Explain geometrical isomerism with reference to square planar complexes giving one example. How is that tetrahedral

43

complexes with simple ligand donot exhibit geometrical isomerism?

A. 14. Geometrical isomerism arises due to the different geometrical arrangement of at least two different ligands in square planner complexes. It is shown by MA2X2, MABX2, MA2XY and MABXY types identical ligands occupy adjacent position,cis-isomer is formed and when they occupy opposite positions trans-isomer is formed. In tetrahedral complexes, this isomerism is not found because the relative position of the ligands with respect to each other will be the same.

Q.15. Explain how [Pt (NH3)2 Cl2] and [Pt (NH3)6]Cl4 will differ in their hybridisation states of Pt in these complexes.

A.15 . Electrolytic conductance of the complexes depends upon the number of ions given by the complexes in solution. We know that any ligand present in the coordinate sphere will not ionize when it is dissolved in water. Only these groups will form ions which are present in ionic sphere of the complex, hence; [Pt (NH3) 2Cl2] aq has no ions [Pt (NH3)]6Cl4 aq has [Pt (NH3 )6 ] +4 (aq) + 4Cl - (aq) ( 5 ions)threfore, the electrolytic conductivity of [Pt (NH3)6]Cl4 will be higher than [Pt (NH3)2 Cl2]. Hybridizationof Pt in [Pt (NH3)2 Cl2] is dsp2 and in [Pt (NH3 )6 Cl4 is d2 sp3.

Q. 16. Giving a suitable example describe the importance of the formation of complex compounds in :(i) the estimation of hardness of water.(ii) the extraction of a particular metal from its natural source.

A. 16. (i) Hard water contains magnesium and calcium ion which form stable complexes with EDTA.Because the values of stability constants of Mg and Ca complexes of EDTA are different, therefore, the selective estimation of these ions is also possible. The hardness of water can be detected by titrating it against EDTA solution.(ii) Silver can be extracted from its ores by treatment with solution cyanide solution. Silve forms a soluble complex. Ag + (aq) + 2NaCN (aq) Na[Ag (CN2)] aq + Na+ (aq)

Q. 17. Describe each one of the following :(a) Magentic behaviour of Ni (Co)4

A. 17. (a) In Ni (Co)4 the oxidation state of Ni is zero. CO ligands create strong ligand field , thus the unpaired electrons of nickel get paired. Because there is no unpaired electron in the complex, so it would be diagmagnetic in nature.

Q.18. Indicate the types of isomerism in the following complexes. (i) [Co(NH3)5NO2](NO3)2 (ii) [Co(en)3 ]Cl3 A.18. (i) Ionisation and Linkage (ii) Optical

Q.19. Define the terms with example (i) Homoleptic complexes (ii) Ambidentate ligands

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A.19. (i) The complexes in which only one kind of ligands are bonded with central metal atom.

(ii) Monodentate ligands which can co-ordinate with central metal atom through more than one site.

-NO2 and –ONO, -CN and –NC Q.20. Write IUPAC name of the following complexes-

COMPLEX IUPAC NAME[Co(NH3)4Cl(NO2)Cl Tetraamminechloridonitrito-N-Cobalt(III)chloride

Pt(NH3)2Cl(NH2CH3)]Cl Diamminechlorido(methylamine)Platinum(II)chlorideK [Ag(CN)2] Potassium dicyanidoargentate(I)

[Ag(NH3)2]Cl Diamminesilver(I) chloride[CrCl2(H2O)4]NO3 Tetraaquadichloridochromium(III) nitrate

[PtCl2(NH3)2] Diamminedichloridoplatinum(II)[PtClNO2(NH3)4]SO4 Tetraamminechloridonitrito-N-platinum(IV) sulphate

[Cr(H2O)6]Cl3 Hexaaquachromium(III) chloride

[Cu(NH3)4]SO4 Tetramminecopper(II) sulphate[Ag(CN)2]– Dicyanidoargentate(I)

[AIH4]– Tetrahydridoaluminium(III)[CoCl2(en)2]2SO4 Dichloridobis(ethylenediamine)cobalt(III) sulphate

Na3[Co(NO)6] Sodium hexanitrosylcobalate(III)[CoCl(NO2)(en)2]+ Chloridobis(ethylenediamine)nitrito-N-

cobalt(III)Na2[SiF6] Sodium hexafluoridosilicate(IV)

[Co(NO2)3(NH3)3] Triamminetrinitrito-N-cobalt(III)[Co(NH3)6]Cl3 Hexaamminecobalt(III) chloride

[PtCl2(NH3)4]Br2 Tetraamminedichloridoplatinum(IV) bromideK2[HgI4] Potassium tetraiodomercurate(II)

Na[Au(CN)2] Sodium dicyanidoaurate(I)K4[Fe(CN)6] Potassium hexacyanidoferrate(II)

K2[PtF6] Potassium hexafluoridoplatinate(IV)K[Ag(CN)2] Potassium dicyanidoargenate(I)

K[PtCl5(NH3)] Potassium amminepentachloridoplatinate(IV)K3[Fe(CN)6] Potassium hexacyanidoferrate(II)K4[Fe(CN)6] Potassium hexacyanidoferrate(III)K4[Mo(CN)8] Potassium octacyanidomolybedate(IV)

NH4[Cr(NCS)4(NH3)2] Ammonium diamminetetrathiocyanato-N-chromate(III)

K3[Fe(C2O4)3] Potassium trioxalatoferrate(III)Na2[Ni(edta)] Sodium ethylenediaminetetraacetonickelate(II)

[Cu(H2O)2(NH3)4]SO4 Tetraamminediaquacopper(II) sulphate[Ni(CO)4] Tetracarbonylnickel(0)

[Pt(NH3)4][PtCl4] Tetraammineplatinum(II)

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tetrachloridoplatinate(II)[CoCl2(NH3)4]3[Cr(CN)6] Tetraamminedichloridocobalt(III)

hexacyanidochromate(III)

Q.21. Draw figure to show splitting of d –orbitals in an a.octahedral crystal field b. in tetrahedral fieldA.21.

a.

b.

Q.22. Write the co-ordination number and d-orbital occupation of central metal ion in the following complexes.2 (i) [Mn(H2O)6]SO4 (ii) K3[Co(C2O4)3] Discuss the importance of co-ordination complexes in the following fields. (i) Medicinal field (ii) Metallurgical fieldA.22. (i) CN=6, t2g3 eg2 (ii) CN = 6 , t2g6 eg0 Medicinal- cis platin [PtCl2(NH3)2] ,

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Metallurgy- Ag2S + NaCN → Na[Ag(CN)2] + Na2S Na[Ag(CN)2] + Zn → Na2[Zn(CN)4] + Ag

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ORGANIC CHEMISTRY

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IMPORTANT ORGANIC COMPOUNDS

COMPOUND FORMULA COMMON MAME IUPAC NAMEHALO COMPOUNDSR-X

CH3 -ClCH3-CH2-ClCH3-CH2-CH2-Cl(CH3)2 -CH-ClCH3-CH2- CH2 -CH2-Cl(CH3)2 -CH- CH2ClCH3-CH2- CH(Cl) –CH3(CH3)3 –C-ClC6H5 –Cl

C6H5 - CH2-Cl

C6H11Cl

Methyl chlorideEthyl chloriden-Propyl chlorideIsopropyl chloriden-Butyl chlorideIsobutyl chloridesec-Butyl chlorideter-Butyl chloridePhenyl chloride

Benzyl chloride

Cyclohexyl chloride

ChloromethaneChloroethane1-Chloropropane2-Chloropropane1-Chlorobutane1-Chloro-2-methylpropane2-Chlorobutane2-Chloro-2-methylpropaneChlorobenzene

Benzyl chloride

ALCOHOLSR-OH

CH3 -OHCH3-CH2-OHCH3-CH2-CH2-OH(CH3)2 -CH-OH

Methyl alcoholEthyl alcoholn-Propyl alcoholIsopropyl alcohol

MethanolEthanolPropan-1-olPropan-2-ol

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CH3-CH2- CH2 -CH2-OH(CH3)2 -CH- CH2OHCH3-CH2- CH(OH) –CH3(CH3)3 –C-OHC6H5 -OHC6H5 - CH2-OHCH2 = CHOH

n-Butyl alcoholIsobutyl alcoholsec-Butyl alcoholtert-Butyl alcoholPhenolBenzyl alcoholVinyl alcohol

Butan-1-ol2-Methyl propan-1-olButan-2-ol2-Methylpropan-2-olPhenolBenzyl alcoholEthen- 1-ol

ETHERSR-O-R’

CH3 –O - CH3CH3-CH2-O – CH2-CH3C6H5 - O- CH3

Dimethyl etherDiethyl etherAnisole

MethoxymethaneEthoxy ethaneMethoxybenzene

ALDEHYDESR- CHO

HCHOCH3 -CHOCH3-CH2-CHOCH3-CH2-CH2-CHOC6H5 -CHO

FormaldehydeAcetaldehydePropionaldehydeButyraldehydeBenzaldehyde

MethanalEthanalPropanalButanalBenzaldehyde

KETONESR-CO - R’

CH3 –CO - CH3C6H5 –CO - CH3C6H5 –CO - C6H5

AcetoneMethyl phenyl ketoneDiphenyl ketone

PropanoneAcetophenoneBenzophenone

CARBOXYLIC ACIDSRCOOH

HCOOHCH3 -COOHCH3-CH2-COOHC6H5 –COOH

Formic acidAcetic acidPropionic acidBenzoic acid

Methanoic acidEthanoic acidPropanoic acidBenzoic acid

ACID DERIVATIVESACID CHLORIDES

AMIDES

ESTERS

CH3 –COClC6H5 –COClCH3 –CONH2C6H5 -CONH2CH3 -COO CH2-CH3

Acetyl chlorideBenzoyl chlorideAcetamideBenzamideEthyl acetate

Ethanoyl chlorideBenzoyl chlorideEthanamideBenzamideEthyl ethanoate

AMINESRNH2

Primary aminesCH3 - NH2CH3-CH2- NH2CH3-CH2-CH2- NH2C6H5 - NH2Secondary amines(CH3)2 -NHTertiary amines(CH3)3 -N

Methyl amineEthylaminen-PropylamineAniline

Dimethyl amine

Trimethyl amine

MethanamineEthanaminePropan-1-amineBenzenamine/Aniline

N-Methylmethanamine

N,N-

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Dimethylmethanamine

NITRILESRCN

CH3 - C NCH3-CH2-CNC6H5 -CN

Methyl cyanideEthyl cyanidePhenyl cyanide

Ethane nitrilePropane nitrileBenzene nitrile

NITRO COMPOUNDSRNO2

CH3-CH2- NO2C6H5 -NO2

NitroethaneNitrobenzene

ISONITRILESRNC

CH3 - NCCH3-CH2-NCC6H5 -NC

Methyl isocyanideEthyl isocyanidePhenyl isocyanide

Methyl carbylaminesEhtyl carbylaminesPhenyl carbylamine

NAME REACTIONS1. Sandmeyer’s reaction Benzene diazonium salt with cuprous chloride or cuprous bromide results in the replacement of the diazonium group by –Cl or –Br.

2. Wurtz reaction Alkyl halides react with sodium in dry ether to give hydrocarbons containing double the number of carbon atoms present in the halide.

2 RX+ 2 Na R-R +2 NaX

3. Wurtz-Fittig reactionA mixture of an alkyl halide and aryl halide gives an alkylarene when treated with

50

sodium in dry ether.

4. Fittig reactionAryl halides when treated with sodium in dry ether give diphenyl.

5. Finkelstein reactionR-X + NaI → R-I + NaX X=Cl, Br

6. Swarts reactionH3C-Br +AgF → H3C-F + AgBr

7. Kolbe’s reaction When phenol reacts with sodium hydroxide, phenoxide ion generated undergoes

electrophilic substitution with carbon dioxide and forms salicylic acid.

8. Reimer-Tiemann reactionOn treating phenol with chloroform in the presence of sodium hydroxide, a –CHO group is introduced at ortho position of benzene ring.

9. Williamson synthesis An alkyl halide is treated with sodium alkoxide to form ether. CH3Br + C2H5ONa CH3 – O - C2H5 10. Rosenmund reduction.

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Acyl chloride (acid chloride) is hydrogenated over catalyst, palladium on barium sulphate to give aldehyde..

 11. Clemmensen reduction Aldehydes and ketones are reduced to alkanes on treatment with zinc- amalgam and concentrated hydrochloric acid. CH3 –CHO CH3 - CH312. Wolff-Kishner reduction Aldehydes and ketones are reduced to alkanes on reaction with hydrazine followed by heating with potassium hydroxide in ethylene glycol. CH3 –CO - CH3 CH3 –CH2 - CH3

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13. Aldol condensation  Aldehydes and ketones having at least one α-hydrogen undergo a reaction in the presence of dilute alkali form β-hydroxy aldehydes (aldol) or β-hydroxy ketones (ketol), respectively..

14.  Hell-Volhard-Zelinsky reaction Carboxylic acids having an α-hydrogen are halogenated at the α-position on treatment with chlorine or bromine in the presence of small amount of red phosphorus to give α-halocarboxylic acids.

15. Hoffmann bromamide degradation reaction Primary amines can be prepared by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide.The amine so formed contains one carbon less than that present in the amide.

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Important reactions

i) Chloroform is slowly oxidised by air in the presence of light to an extremely poisonous gas, carbonyl chloride, also known as phosgene. It is therefore stored in closed dark coloured bottles completely filled so that air is kept out.

ii) Phenol is manufactured from the hydrocarbon, cumene (isopropylbenzene)

iii) Carbylamine reactionAliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanides or carbylamines which are foul smelling substances. Secondary and tertiary amines do not show this reaction.

iv) Coupling reaction Benzene diazonium chloride reacts with phenol or aniline to form azo dyes.

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DISTINCTION TESTS

S NO

NAME OF THE COMPOUND

CHEMICAL REAGENT

OBSERVATION PAIRS OF COMPOUNDS THAT CAN BE DISTINGUISHED

1 R - X Aqueous NaOH & AgNO3 solution

PrecipitateAgCl – WhiteAgBr – Pale yellowAgI – Dark yellow

1) Ethyl chloride & chlorobenzene Ethyl chloride gives white ppt Chlorobenzene does not give ppt.

2) Cyclohexylbromide & bromobenzene Cyclohexylbromiide gives pale yellow ppt Bromobenzene does not give the test.

3) Benzyl chloride & chlorobenzene Benzyl chloride gives white ppt Chlorobenzene does not give ppt.

4) Vinyl iodide & Allyl iodide Vinyl iodide gives pale yellow ppt. Allyl iodide does not give the test.

2 R - OH Lucas test – conc HCl & anh ZnCl2

Iodoform test – NaOH &I2Note: When both the alcohols given are primary / sec apply this test

10 – Does not react.20 – Forms turbidity after few min30 - Forms turbidity immediately

Yellow ppt of iodoform (CHI3)

To distinguish primary , sec & ter alcoholsPrimary – Methanol, Ethanol, n – alkyl alcohols / alkan – 1-ol, benzyl alcoholSecondary – Isopropyl alcohol, alkan – 2 – olTertiary – ter- buytl alcohol (2- methyl – propan – 2- ol)

Alcohols with CH3 – CH – group give this test OH1) Ethyl alcohol & methyl alcohol (both are 10) Ethyl alcohol gives yellow ppt Methyl alcohol does not give this test.

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2) Pentan – 2- ol & Pentan – 3- ol (both are 20) Pentan – 2- ol gives yellow ppt Pentan – 3- ol does not give this test.

3 Phenol Neutral ferric chloride

Violet colour 1) Phenol & ethyl alcohol Phenol gives violet colour Ethyl alcohol does not give this test

4 Aldehydes Tollens test – ammoniacal AgNO3 , warm

Iodoform test – NaOH &I2Note: When both the compounds given are aldehydes/ ketones apply this test

Silver mirror

Yellow ppt of iodoform (CHI3)

Aldehydes give positive test.1. Acetaldehyde (Propanal) & Acetone (Propanone) Acetaldehyde (Propanal) gives Ag mirror Acetone (Propanone) does not give this test.

Ald. & ket. with CH3 – C – group give this test O

1) Acetaldehyde & benzaldehyde Acetaldehyde gives yellow ppt. Benzaldehyde does not give this test.1) Acetaldehyde & formaldehyde Acetaldehyde gives yellow ppt. Formaldehyde does not give this test.3) 2 – Pentanone & 3 – Pentanone 2 – Pentanone gives yellow ppt. 3 – Pentanone does not give this test.4) Acetophenone & benzophenone Acetophenone gives yellow ppt. Benzophenone does not give this test.

5. Carboxylic acids

Sodium bicarbonate test

Effervescence due to evolution of CO2 gas

All carboxylic acids give this test

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Formic acid ( Methanoic acid)

Tollens test Effervescence due to evolution of CO2 gas

HCOOH is the only carboxylic acid that gives this testHCOOH and CH3COOHHCOOH gives this test . CH3COOH does not give this test

6 Amines

Primary amines

Aniline

Hinsberg test –Benzene sulphonyl chloride + NaOH

Carbylamine test – Alcoholic KOH + CHCl3

Azodye test –NaNO2 + HCl + phenol ( 00 to 5 0 C)

10 –Soluble20 – Insoluble30 – does not react with Hinsberg reagent

Very unpleasant smelling isocyanide gas evolves

Reddish orange dye

To distinguish primary , sec & ter aminesAlkyl amine – 10

Dialkylamine -20 (N- alkyl alkanamine)Trialkylamine( N,N – dialkyl alkanamine - 30

To distinguish primary amines from other amines1)Ethylamine & DiethylamineEthylamine gives this test.Diethylamine does not give this test2) Aniline & DimethylamineAniline gives this test.Diethylamine does not give this test

Aniline & BenzylamineAniline gives this testBenzylamine does not give this test.

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CHAPTER 10HALOALKANES AND HALOARENES

VSA (one mark each)

1.What are ambident nucleophiles ? Give two eg. Nucleophiles that possess two nucleophilic centres. Eg CN- & NO2 -

2. What are enantiomers ? Stereoisomers related to each other as non-superimposable mirror images.

3. What is chirality ? The property of objects that are non-superimposable on their mirror images.

4. What is called racemisation ?The process of conversion of enantiomer into a racemic mixture (mixture of d & l in equal proportions )

5. Why is chloroform stored in closed dark colour bottle ? In the presence light & air it forms poisonous gas , phosgene. 2CHCl3 +O2 2COCl2 + 2HCl

6. Alkly halides though polar are immiscible with water. Why ? Because they do not form intermolecular H bond with water.

7. Chlorobenzene(haloarene) is less reactive than ethyl chloride (haloalkane) towards nucleophilic substitution reaction. Why ?

In chlorobenzene C – Cl bond acquires partial double bond character due to resonance. So it is difficult to break the bond.

8. Among isomeric dihalobenzenes, the para-isomers have high melting compared to their ortho- and meta-isomers. Why? It is due to symmetry of para-isomers that fits in crystal lattice better as compared to ortho- and meta-isomers.

9. Arrange the following set of compounds in order of increasing boiling points. 1-Chloropropane, Isopropyl chloride, 1-Chlorobutane. Isopropyl chloride < 1-Chloropropane < 1-ChlorobutaneNOTE : SN1 : 10 < 20 < 30 Tertiary cabocation is more stable SN2 : 30 < 20 < 10 Steric hindrance is less in primary

58

10. Which one in the following pairs would undergo SN2 reaction faster?

Because iodine is a better leaving group.

11. Which one in the following pairs would undergo SN1 reaction faster?

12. An alkyl halide having molecular formula C4H9Cl is optically active .What is its name? Ans.= 2-chlorobutane

13. P-Dichlorobenzene has higher melting point than o and m-isomer.Why?Ans=The P-isomer being more symmetrical fits closely in the crystal lattice.

SA1 (two marks each)

1. Explane why haloarenes are less reactive towards nucleophilic substation reactions than Haloalkanes.

(a)Due to resonance in benzene ring there is partial double bond in C-X bond which is difficult to break.

(b)It is difficult for electron rich nucleophiles to approach electron rich benzene ring in haloarenes.

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2. How do the products differ with ethyl bromide react separately with ;

(a) Alc. KCN and Alc. AgCN.(b)Aq. KOH and Aic. KOH.

Ans. (a) Alc.KCN C2 H5 CN+ KBr

C 2H5Br Alc.AgCN C2 H5 NC+ AgBr

(b)

Aq.KOH C2H5Br + KBr C2H5 Br

Alc.KOH CH2=CH2 + KBr +H2O

3. Write the structure of major organic product; C2H5ONa(a) (CH3 ) 3 CH-CH(Br)CH 2CH 3 Ethanol /Heat

(b) CH3CH2Cl + SbF3 HeatAns. (a) CH3-C=CH-CH2-CH3 CH3

(b) CH3CH2F

4. Out of C6H5CH2Cl and C6H5 CHClC6H5 which is more easily hydrolysed by aqueous KOH and why

Ans. C6H5CHClC6H5 will get hydrolysed easily because carbo cation formed Will be stabilized by resonance effect of two phenyl groups.

5. (a) Give the IUPAC name of the following ;

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H3C CH3

H

H CH3 Br (b) Write Wurtz fittig reaction .

Ans. (a) 4-Bromo-3-methyl –pent-2-ene.

(b) R + 2NaX 2Na + X-R

Dry ether

 6. Arrange the following compounds in order of reactivity towards SN2 displacement: 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane 2-Bromo-2-methylbutane (ter) < 2-Bromopentane(sec) < 1-Bromopentane (pri) Because of less steric hindrance in 1-Bromopentane7. a) How alkyl halides become darker on standing in the presence of sunlight b) Which is better nucleophile Br‾ or I‾ and why.

ANS:- (a) Alkyl halides are unstable and liberate iodine . (b) I‾ because of large size and low electronegativity .

X

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SA –II (three marks ) 1.(a) Write the IUPAC names of organic compounds ‘A’,’B’,’C’ and in the following:-

Br2 Alc. KOH H2O/H2SO4

CH3 ─CH─CH2─CH2 ‘B’ ‘C’ ‘D’

Br

b) Which isomer of C4H9Br has the lowest boiling point?

Ans. a) A=But-2-ene

B=2,3-Dibromobutane

C=But-2-yne

D=Butan-2-one

b) CH3

׀ CH3 - C– CH3 has lowest boiling point.

׀ Br

3. Convert the following:- (a) ethanol to propanenitrile (b) propene to 1-propanol (c) 1-Bromopropane to 2-Bromopropane

Ans a) PCl5 KCN CH3CH2OH CH3CH2Cl CH3CH2CN

HBr/peroxide KOH b)CH3CH=CH2 CH3CH2CH2Br CH3CH2CH2OH

alc. KOH HBr c) CH3CH2CH2Br CH3CH=CH2 CH3CHBrCH3

‘A’Alc. KOH

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4) An organic compound “A” having molecular formula C4H8 on treatment with dil H2SO4 gives “B” which on treatment with conc. HCl and anhydrous ZnCl2 gives “C” and on treatment with sodium ethoxide gives back “A”. Identify A, B,C.

Ans.A=But-2-ene

B=Butan-2-ol C=2-Chlorobutane

5. Write the Structure of major products:-

(a)

Br

CH2Br

CH(CH3)2

+ HNO3 Conc. H2SO4

+ Br2

NaOH

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64

CHAPTER 11ALCOHOLS, PHENOLS AND ETHERS

VSA (one mark each)

Topic:- Reasoning QuestionsVSA (one mark each)

1. Phenol is acidic in nature.Ans. Phenol is acidic in nature because:

a) phenol , due to resonance, the positive charge rests on oxygen making the sharedpair of electrons more towards oxygen and hydrogen as H+

b) The carbon attached to OH is SP2 hybridize and is more electronegative, this decreases the electron density on oxygen, increasing the polarity of O-H bond and ionization of phenol.c) The phenoxide ion formed by loss of H+ is more resonance stabilized than phenol itself.

2. Phenol has a smaller dipole moment than methanol.Ans. In phenol due to electron rich benzene ring the C-O bond is less polar whereas in

methanol the C-O bond is highly polar. Therefore the dipole moment of methanol is higher than phenol.

3. o- nitrophenol has lower boiling point ( is more volatile ) than p – nitrophenol.Ans. P- nitrophenol has intermolecular hydrogen bonding which increases the boiling

point while in o- nitro phenol due to presence of intra molecular hydrogen bonding, there is a decrease in boiling point and increase in volatility.

4. Methanol is miscible with water while iodomethane is not.

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Ans.

5. Alcohols have higher boiling points than isomeric ethers.

Ans. Alcohols can form intermolecular hydrogen bonds due to their high polarity whereas, ether cannot. Therefore alcohols have higher boiling points than isomeric ethers.

6. Ethers are soluble in water alkanes are not.Ans. Ethers can form H- bonding with water molecule whereas alkenes cannot.

Therefore ethers are soluble in water and alkanes are not.

7. The order of acidic strength in alcohols is R CH2OH > R2 CHOH > R3 COHAns. In alcohols, the acidic strength is due to polar nature of O-H bond. An electron

releasing group e.g., alkyl groups, increases electron density on oxygen tending to decrease the polarity of O-H bond. This decreases the acid strength. Therefore the order of acid strength is

Methanol can form intermolecular hydrogen

bonding with water but there is no hydrogen

bonding in iodomethane and water. Therefore

methanol in miscible in water.

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8. During preparation of ester from alcohol and acid, water has to be removed as soon as it is formed.

Ans.

9. Ethers cannot be prepared by dehydration of secondary or tertiary alcohols.Ans. For secondary and tertiary alcohols, elimination competes over substitution and

alkenes are formed on acidic dehydration as the reaction follows Sn1 mechanism. Therefore the acidic dehydration of secondary or tertiary alcohols does not give ethers.

10. Reaction of anisole with HI gives methyl iodide and phenol.Ans.

11. tert butyl alcohol has lower b.pt than n-butyl alcohol. Why ?Ans. In tert butyl alcohol there is more branching and so less van der Waals forces due

to less surface area.

12. Alcohols are more soluble in water than hydrocarbons of comparable molecular mass. Why ?

Ans. Because alcohols form hydrogen bond with water.

13. o-nitrophenol is more acidic than o-methoxyphenol. Why ?

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Ans. Because of electron withdrawing nitro group in o-nitrophenol there is effective delocalisation negative charge.

14. Alcohols are weaker acids than water. Why?Ans. Water is a better donor than alcohols. RO - + H2O ROH + OH –

15. Phenol is more acidic than alcohols. Why?Ans. Phenoxide is more stabilised by resonance than alkoxide ion.

16. What is denaturation of alcohols ?Ans. Commercial alcohol is made unfit for drinking by adding copper sulphate/

pyridine.

17. Arrange the following compounds in increasing order of their acid strength:Propan-1-ol, 2,4,6-trinitrophenol, 3-nitrophenol, phenol, 4-methylphenol

Ans. Propan-1-ol < 4-methylphenol < phenol < 3-nitrophenol < 2,4, 6-trinitrophenol.

18. Write the names of reagents and equations for the preparation of the following ether by Williamson’s synthesis: 2-Methoxy-2-methylpropane

Ans. Reagents : Ethyl bromide & sodium tert butoxide C2H5Br + (CH3)3CONa (CH3)3CO C2H5 + NaBr

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ALCOHOLS, PHENOLS AND ETHERSTopic:- Conversions

SA-I (two marks each)

1. Ethene to 1,2 -ethanediolAns.

2. Phenol to SalicyldehydeAns.

3. Butanol to Butanoic acid

4. Ethanol to propanone

5. Phenol to salicylic acidAns.

6. Methanol to Ethanol

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7. Ethanol to propanol

8. Phenol to Benzyl AlcoholAns.

9. Ethanal to propan -2- olAns.

10. l – propanol to 2 – bromo propaneAns.

ALCOHOLS, PHENOLS AND ETHERSTopic:- Identification Question

SA-II (three marks each)

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1. An organic compound ‘ A ‘ having molecular formula C3 H6 on treatment with aq. H2SO4 give ‘B’ which on treatment with Lucas reagent gives ‘C’. The compound ‘C’ on treatment with ethanolic KOH gives back ‘ A’ .Identify A, B , C .

Ans.

2. An organic compound A (C6H6O) gives a characteristic colour with aq. FeCl3 solution. (A) On reacting with CO2 and NaOH at 400k under pressure gives (B) which on acidification gives a compound (C) .The compound (C) reacts with acetyl chloride to give (D) which is a popular pain killer. Deduce the structure of A,B,C & D.

Ans.

3. An organic compound (X) when dissolved in ether and treated with magnesium metal forms a compound Y. The compound, Y, on treatment with acetaldehyde and the product on acid hydrolysis gives isopropyl alcohol. Identify the compound X. What is the general name of the compounds of the type Y.

Ans. The compound X is CH3Br and Y is CH3MgBr The compounds of the type ‘Y’ are called Grignard reagent.

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4. A compound ‘A’ with molecular formula C4H10O on oxidation forms compound ‘B’ gives positive iodoform test and on reaction with CH3MgBr followed by hydrolysis gives (c). Identify A, B & C.

Ans. The compound ‘B’ is obtained by oxidation of C4H10O and gives positive iodoform test and also reacts with CH3MgBr , it must be methyl Ketone , it must be methyl ketone having four carbon atoms i.e, CH3COCH2CH3 .This can be obtained by oxidation of 2 – butanol i.e , CH3 CH CH2 CH3 Therefore ,

thereactions are:

5. An aromatic compound (A) having molecular formula C6H6O on treatment with CHCl3 and KOH gives a mixture two isomers ‘B’ and ‘C’ both of ‘B’ & ‘C’ give same product ‘D’ when distilled with Zn dust. Oxidation of ‘D’ gives ‘E’ of formula C7H6O2. The sodium salt of ‘E’ on heating with soda lime gives ‘F’ which may also be obtained by distilling ‘A’ with zinc dust. Identify compounds ‘A’ to ‘F’ giving sequence of reactions.

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6. Compound ‘A’ of molecular formula C5H11Br gives a compound ‘B’ of molecular formula C5H12O when treated with aq. NaOH. On oxidation the compound yields a mixture of acetic acid & propionic acid. Deduce the structure of A, B & C.

Ans.

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The reactions are:

Ans.

****************************************************************************************

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CHAPTER 12ALDEHYDES, KETONES AND CARBOXYLIC ACIDS

VSA (one mark each)

1. The b.pts of aldehydes & ketones are higher than ethers of comparable molecular masses. Why ?

Ans. Due to stronger dipole – dipole interactions in aldehydes & ketones.

2. Acetaldehyde (ethanal) and acetone (propanone) are miscible with water. Why?Ans. Because they form hydrogen bond with water.

3. Arrange the following set of compounds in order of increasing boiling points. CH3CH2CH2CHO, CH3CH2CH2CH2OH, CH3CH2 OCH2 CH3 ,CH3CH2CH2 CH3Ans. CH3 ,CH3CH2CH2 CH3 < CH3CH2 OCH2 CH3 < CH3CH2CH2CHO < CH3CH2CH2CH2OH

4. Ketones are less reactive than aldehydes towards nuclephilic addition reaction. Why?

Ans. Two electron releasing alkyl groups in ketones reduce the electrophilicity of carbonyl. Also due to steric hindrance nucleophilic attack is not easier.

5. Benzaldehyde is less reactive than propanal towards nucleophilic addition reaction. Why ?

Ans. Resonance in benzaldehyde reduces the electrophilicity of carbonyl carbon.

6. (i) Cyclohexanone forms cyanohydrin in good yield but 2,2,6- trimethylcyvlohexanone does not.

Ans. The three methyl groups reduce the elecrophilicity of carbonyl and also offer steric hindrance to the nucleophilic attack(ii) In semicarbazide ,the NH2 group directly attached to the -C = O group is not involved in the formation of semicarbazones.Give reason.The lone pair of electrons of the – NH2 group directly attached to –C = O group is involved in resonance. Hence it does not act nucleophile

7. Arrange the following compounds in increasing order of their reactivity towards nuclephilic addition reaction.

Methanal, propanone, ethanolAns. Propanone < Ethanal < Methanal

8. Acetaldehyde undergoes aldol condensation. Why ?Ans. Because it contains alpha hydrogen.

9. Formaldehyde & benzaldehyde do not undergo aldol condensation. Why ?Ans. Because they do not contain alpha hydrogen.10. Formaldehyde & benzaldehyde show Cannizaro reaction. Why ?Ans. Because they do not contain alpha hydrogen.

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11. What is formalin ? Mention its uses.Ans. 40% formaldehyde is formalin. It is used to preserve biological specimen and to

prepare bakelite.

12. Carboxylic acids are higher boiling liquids than alcohols. Why ?Ans. Carboxylic acids form more extensive hydrogen bond than alcohols and exisi as

dimer.

13. Lower members of carboxylic acids are miscible with water. Why ? Ans. Because they form hydrogen bond with water.

14. Carboxylic acids are more acidic than phenol. Why ? OR Ka of carboxylic acid is more than that of phenol.Why ? OR pKa of carboxylic acid (acetic acid) is more than that of phenol. Why?Ans. Carboxylate ion is more stabilised than phenoxide ion due to resonance with

negative charge on more electronegative oxygen atom.

15. Carboxylic acids do not show the reactions of carbonyl group. Why ?Ans. Due to resonance the double bond character is reduced in carboxylic acids. 16. What is decarboxylation reaction ?Ans. Carboxylic acids loose CO2 to form hydrocarbons when their sodium salts are

heated with sodalime. sodalime CH3COONa CH4 + CO2 + Na2CO3

17. During the preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst, the water or the ester should be removed as soon as it is formed. Why ?

Ans. The preparation of esters from a carboxylic acid and an alcohol in the presence of an acid catalyst is a reversible reaction. So, the water or the ester should be removed as soon as it is formed to shift the equilibrium towards the ester formation (forward) according to Le Chatelier’s principle.

18. Exercise 12.12 page no 378(i) Di-tert-butyl ketone < Methyl tert-butyl ketone < Acetone < Acetaldehyde (reactivity towards HCN)(ii) (CH3)2CHCOOH < CH3CH2CH2COOH < CH3CH(Br)CH2COOH < CH3CH2CH(Br)COOH(acid strength)(iii) 4-Methoxybenzoic acid < Benzoic acid < 4-Nitrobenzoic acid < 3,4-Dinitrobenzoic acid (acd strength)

19. Why aldehydes and ketones have lower boiling point than alcohols.Ans. Alcohol molecules are associated with intermolecular H-bonding whereas they are

not.

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20. Compare the acid strength of the following acids-HCOOH , CH3COOH , C6H5COOH

Ans. HCOOH > C6H5COOH > CH3COOH

21. Why chloroacetic acid stronger than acetic acid.Ans- Cl is electron withdrawing group thus chloroacetate ion is more stable than

acetate ion.

22. Which of the aldehyde undergo Cannizaro reaction.Ans. Aldehydes which do not have alpha hydrogen.e.g. HCHO.

SA-I (two marks each)

1. Although phenoxide ion has more number of resonating structures than carboxylate ion, carboxylic acid is a stronger acid than phenol.

Ans- In carbooxlate ion –ve charge is delocalized over two oxygen atoms which are higly electronegative whereas in phenoxide ion –ve charge is delocalized over only one oxygen atom. Carboxylate ion is more stable phenoxide ion that is why carboxylic acids are more acidic than phenols.

2. State(i) Why benzaldehyde does not undergo aldol condensation.(ii) How an acid amide may be converted to the parent acid.Ans (i) Because of absence of ɑ - Hydrogen O O H+

(ii) CH3 – C –NH2 + H2O CH3 – C- OH + NH3

3. Complete the following

(i)

(ii)Ans (i)

+ H2O

COOH

SO3/H2SO4

COOH

Conc HNO3

Conc H2SO4

COOH

SO3/H2SO4

COOH

SO377

(ii)

+ H2O

4.. What are nucleophilic addition reactions- How would you account –Aldehydes are more reactive than ketones towards nucleophiles?Ans. Reactions which are intiated by nucleophiles are called nucleophilic addition reactions. O OH CH3-C-H + HCN CH3-CH -CN (acetaldehyde

Cynohydrin)Aldehydes are more reactive because they are more polar than ketones due to lesser no. of electron releasing alkyl groups, also steric hinderance in aldehydes is less than in ketones.

SA-II (three marks each)1. Write the major products of the following reactions:(i) Nitration of benzaldehyde (ii) Sulphonation of benzaldehydeAns (i) Nitration Nitration of benzaldehyde with a mixture of conc. H2SO4 and HNO3 forms 3-nitro- benzaldehyde and 3, 5-dinitrobenzaldehyde.

Benzaldehyde 3-nitro- benzaldehyde 3, 5-dinitrobenzaldehyde(ii) Sulphonation. Sulphonation of benzaldehyde with conc. H2SO4 forms 3- benzaldehyde sulphonic acid and then 3- benzaldehyde-1, 5-disulphonic acid.

Conc HNO3

Conc H2SO4

COOH

NO2

CH

273 – 283 K

HNO3 + H2SO4

CHHNO3 + H2SO4

CH

NO2 NO2 NO2

CH CH

CH

H2SO4

SO3HH3OS

COOH

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Benzaldehyde 3- benzaldehyde sulphonic 3- benzaldehyde-1, acid 5-disulphonic acid

1. What happens when-a. Acetophenone is treated with bromine in the presence of anhdy AlCl3.b. Acetophenone is nitrated.

Ans-

Br2/anhyd AlCl3

a.Br

b.

3-nitroacetophenone 3,5-dinitroacetophenone

3. Complete the following reactions- Oa. R-C-H Zn/HCl,4[H] -------

Ob. CH3-C-NH2 HNO2 ------

NO2

H2SO4

SO3H

COCH3COCH3

COCH3COCH3COCH3

HNO3/H2SO4

NO2

NO2

HNO3/H2SO4

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Ans- O

a. . R-C-H Zn/HCl,4[H] R-CH3 + H2O

O

b. . CH3-C-NH2 HNO2 CH3COOH + N2 + H2O

4. Carry out the following conversion—a. Acetylene to Acetaldehydeb. Toluene to Benzaldehyde

Ans—a. CHΞCH + H2O H2SO4/HgSO4 CH2=CHOH CH3CHO

b. [O] CrO2Cl2 + H2O

CH3CH

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CHAPTER 13ORGANIC COMPUNDS CONTAINING NITROGEN

VSA (one mark each)

1. Ammonolysis is not preferred for preparation for primary amines.Why ?Ans. Because it gives mixture of amines.

2. Alkyl amines have higher b.pts than alkanes of comparable molecular mass. Why?

Ans. Because alkyl amines form intermolecular hydrogen bond.

3. Alcohols have higher b.pts than alkyl amines of comparable molecular mass. Why?

Ans. Because in alcohols the strength of H – bond is more than that in amines since O is more electronegative than N.

4. Arrange the following set of compounds in order of increasing boiling points. n-Butylamine(10), Diethylamine(20),Ethyl dimethylamine(30) [isomeric amines]Ans. Ethyl dimethylamine(30) < Diethylamine(20) < n-Butylamine(10) Reason - 30 amine – No H bond

In primary amine 2 hydrogen atoms are available for H- bond. In sec amine only one H atom available for H- bond.

5. (i) Ethylamine (aliphatic amines) is a stronger base than aniline (aromatic amine). Why ?

Ans. In ethyl amine the electron releasing alkyl group increases the electron density on nitrogen. In aniline the electrons on N is involved in resonance.

(ii) pKb of aniline is more than that of methyl amine. Why?Ans. In methyl amine the electron releasing alkyl group increases the electron density

on nitrogen. In aniline the electrons on N is involved in resonance.

6. Amines are less acidic than alcohols . Why ?Ans. RO - is more stable than RNH –

7. Why cannot primary amines be prepared by Gabriel phthalimide ?Ans. Aryl halides do not undergo nucleophilic substitution with the anion formed by

phthalimide.

8. Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide. Why ?

Ans. Methylamine is more basic than water. So it forms OH- with water. The OH- ions react with ferric chloride and gives ppt.

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9. Aniline does not undergo Friedel- Crafts reaction. Why ?Ans. In Friedel- Crafts reaction AlCl3 which is used a catalyst is a Lewis acid. It reacts

with aniline and forms salt.

10. Nitration of aniline gives a substantial amount of m-nitroaniline besides ortho and para derivatives. Why ?

Ans. During nitration, in presence of strong acidic medium aniline is protonated to form the anilinium ion which is meta directing.

11. Diazonium salts of aromatic amines are more stable than those of aliphatic amines. Why ?

Ans. Arenediazonium ion is stabilised by resonance.

12. Arrange the following: (13.4 page no 400)(i) (C2H5)2NH > (C2H5 )3 N > C2H5NH2 (Basic strength in aqueous medium)(ii) (CH3)2NH > CH3NH2 > (CH3)3 N > NH3(Basic strength in aqueous medium)(iii) C6H5NH2 > (C2H5 )3 N > (C2H5)2NH > C2H5NH2 (Basic strength in gaseous phase)(iv) p-nitroaniline < Aniline < p-toluidine (Basic strength)(v) Diethylamine(20) < n-Butylamine(10) < n- butylalcohol (boiling point)

NOTE Hydrocarbons < Ethers < Ald & Ketones < Amines < Alcohols < Carboxylic acid

(boiling pts)

Electron withdrawing group increases the acidic strength Higher Ka, lower pKa higher is the acidic strength

Electron releasing group increases the basic strength Higher Kb, lower pKb higher is the basic strength

13. Arrange the following In decreasing order of basic strength:C6H5NH2,C6H5(CH3)2,(C2H5)2NH and CH3NH2

Ans- (C2H5)3N > CH3NH2 > C6H5NH2 > C6H5N(CH3)2 .

14. Why do amines react as nucleophiles ?Ans. It is because they have lone pair of electrons .

15. Give reason why secondary amines are more basic than primary amines ?Ans. In secondary amines , there are two alkyl groups which increase electron density on N more than “N” present in primary amine in which there is one alkyl group.

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TWO MARKS QUESTIONS : -

Q1. Write chemical equations for the following conversations :(a). CH3-CH2-Cl- into CH3- CH2 –NH2(b). C6H5-CH2-Cl into C6H5-CH2-CH2-NH2

A1. (a) CH3-CH2-Cl ethanolic NACN CH3-CH2-C Ξ N reduction CH3-CH2-CH2-NH2(b). C6H5-CH2-Cl ethanolic NACN C6H5-CH2-CΞ N H2/Ni C6H5-CH2-CH2-NH2

Q2. Give reasons :(a). Electrophilic substitution in case of aromatic amines takes place more

readily than benzene.(b). Nitro compounds have higher boilings points than hydrocarbons having

almost same molecular mass.Ans. (a) It is because –NH2 group is electron releasing, i.e. activating and increase

electron density at o-and-p-positions . Electrophilic substitution reactions takes place faster.

(b).Nitro compounds are more polar than hydrocarbons , therefore, have more van der Waals” forces of attraction than hydrocarbons, hence higher boiling point.

5.How will you convert Aniline to Benzonitrile?Ans-

NaNO2+HCl KCN 0 – 5 0C CuCN Aniline Benzonitrile

CΞNN2+Cl-NH2

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THREE MARK QUESTION-1. a. Explain why is so that methyl bromide reacts with KCN to give mostly methyl

isocyanides?b. Why nitro alkanes have higher boiling points than the corresponding hydrocarbons?

c. Before nitration, aniline is converted to Acetanilide?Ans. a. KCN is an ionic compounds it generates CN- which replace Br-.AgCN is covalent

comp,it forms bond through N forming isocyanides.b. Nitro alkanes are more polar than the hydrocarbons.c. Aniline get oxidized with HNO3,therefore it is converted in to acetanilide

before nitration.

2. Account for the following :-(i) Pkb of aniline is morethan that of methylamine.(ii) Ethylamine is soluble in water whreas aniline is not.(iii)Why has aniline a weak basic nature than aliphatic amines.

Ans- (i) aniline has an electron withdrawing phenyl group so it is the weaker base than ammonia. Methyl group in methylamine is electron donating group so it is stronger base than ammonia.

(ii) ethyl group in ethylamine is comparatively a small group and causes no hindrance in formation of hydrogen bonding and hence it is soluble in water.(iii) Due to resonance in aniline.

4. Arrange the following(i) in decreasing order of the pKb value:

C2H5NH2, C6H5NHCH3,(C2H5)2NH AND CH3NH=(ii) in decreasing order of basic strength: C6H5NH2, C6H5N(CH3)2,(C2H5)2NH and CH3NH2

(iii) in increasing order of basic strength: (a) aniline,p-nitroaniline and p-toylidine(b) C6H5NH2,C6H5NHCH3,C6H5CH2NH2

Ans) (i) C6H5NH2>C6H5NHCH3>C2H5NH2>(C2H5)2NH (ii) (C2H5)2NH>CH3NH2>C6H5NH2>C6H5N(CH3)2

(iii)-(a) p-nitroaniline<aniline<p-toulidine(b)C6H5NH2<C6H5NCH3<C6H5CH2NH2

5. (a) Give possible explanation for each of the following:-(i) the presence of a base is needed in the ammonolysis of alkyl halides.(ii)aromatic primary amines cannot be prepared by Gabriel phtaliminde synthesis(b)Write the IUPAC name of CH3-N-C-CH3 - C2H5

Ans- (a)(i) To remove the HX produced during the reaction. (ii)Aryl halides do not undergo nucleophilic substitution with POTASSIUM

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PHTHALIMIDE (b) N-Ethyl-N-methylethanamide.

85

GENERAL CHEMISTRY

86

UNIT 5 SURFACE CHEMISTRYSurface Chemistry – The branch of chemistry which deals with the study of nature of surfaces and the changes occurring on the surfaces is called surface chemistry.The phenomenon of attracting and retaining the molecules of a substance on the surface of a liquid or a solid resulting into higher concentration of the molecules on the surface is called adsorbate and the substance on which it is adsorbed is called adsorbent. The reverse process i.e. removal of the adsorbed substance from the surface is called desorbtion (which can be brought about by heating or reducing the pressure). The adsorption of gases on the surface of metals is called occlusion.DIFFERENCE BETWEEN ADSORPTION AND ABSORPTION

ADSORPTION ABSORPTIONIt is a surface phenomenon. It is a bulk property.Conc. of Adsorbate is more on the surface than in the bulk.

The adsorbate/substance is uniformly distributed throughout.

It is rapid in beginning and its rate slowly decreases until equilibrium is attained.

It occurs at a uniform rate.

E.g. Silica Gel. Aluminum Gel, Colloidal etc.

e.g. CaCl2 etc.

Types of Adsorption – Depending upon the nature of forces between particles of Adsorbate and Adsorbent, Adsorption is of two types:

1. Physical Adsorption or Vander Waal’s Adsorption or Physiorption :

When a gas is held on the surface of solid by Vander Waal’s forces without resulting in the formation of any chemical bond between Adsorbate and Adsorbent called Physical Adsorption.2. Chemical or Chemisorption (Langmuir Adsorption) :

When a gas is held on to the surface of the solid by forces similar to those of a chemical bond (covalent or ionic), the type of Adsorption is called chemical Adsorption.DIFFERENCE BETWEEN PHYSICAL AND CHEMICAL ADSORPTION

PHYSICAL CHEMICALDefinition: Definition:Reversible process. Irreversible process.Rate of Adsorption decreases with increase in temperature.

Rate of Adsorption initially increases then decreases by increase in temperature.

It is not specific in nature. It is highly specific in nature.No compounds formed on its surface. Surface compounds are formed.Forms multimolecular layer. Forms unimolecular layer.Enthalpy of Adsorption is the order of 20-40 KJ mol-1.

Enthalpy of Adsorption is the order of 200-400 KJ mol-1.

Increase with increase surface area It also increases with increase

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of adsorbent. surface area of adsorbent.

Adsorption of Gases on Solids:-The extent of adsorption of a gas on a solid surface depends upon the following factors:-

i. Nature of Gas : - The easily liquefiable gases are adsorbed more than the permanent gases (H2, N2, O2, He, Ne). Liquifiable Gases – NH3, SO2, SO3, HCL etc.Liquefaction of gas TC (critical temperature).

The minimum temperature above which a gas cannot be liquefied even by applying very high pressure is called critical temperature (TC).

ii. Nature of Adsorbent : - The extent of adsorption also depends upon the nature of adsorbent. E.g. Activated charcoal or animal charcoal can be adsorbed those gases easily which are easy to liquefy.

Permanent gases can be adsorbed on the surface of transition metal.iii. Effect of Pressure:-

Isotherm – A graph between the amount of gas adsorbed per gram of the adsorbent (x/m) and the equilibrium pressure of the adsorbate at constant temperature is called Adsorption.

At value of Ps, of Equilibrium pressure, x/m reaches its maximum value & then remain constant even pressure is increased. It is saturation state

and pressure called saturation pressure.

x = amount of as adsorbatem = mass of adsorbentP = pressure, P0 = independent pressure.

a. At low pressure x/m P = Kp.b. At high pressure x/m P0 = Kp0 = K.c. At intermediate x/m = Kp1/n 1/n = 0 to 1 (0.1 to 0.5) probably

log(x/m) = log(K) + 1/n(log(P))log(x/m) = 1/n(log(P)) + log(K)

y = mn + c

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A graph between x/m VS log(P), a straight line with slope equal to 1/n and ordinate intercept (at y-axis) equal to log(K).

PHYSICAL

CHEMICAL

Catalysis – A substance which accelerates the rate of

89

a chemical reaction without undergoing any change in its chemical composition or mass during the reaction called catalyst.

The Phenomenon of increasing the rate of a reaction with the help of a catalyst is known as catalysis.2KClO3 2KCl + 3O2

By adding small amount of MnO2, decomposition becomes faster and occurs at lower temperature.2KClO3 2KCl + 3O2

Types of Catalysis:-A. Homogeneous Catalysis – When a catalyst mixes homogenous with the reactants

& this form a single phase, catalyst called homogenous & this type of catalysis called Homogenous Catalysis.

I. SO2 oxidised to SO3 in presence of NO(g).2SO2(g) + O2(g) 2SO3(g)

II. Hydrolysing Ester

CH3COOC2H5(l) + H2O(l) CH3COOH + C2H5OH

III. Hydrolysis of pure sucrose.

C12H22O11 + H2O C6H1206 + C6H12O6 { Glucose + Fructose }=Soln

IV. Preparation of Di-ethyl ether in the presence of conc. H2SO4.

2C2H5OH C2H5-O-C2H5 + H2OAll reactants and catalyst are in same phase.

B. Heterogeneous Catalysis – When a catalyst exist in different phase then that of reactants called heterogeneous catalyst and process called heterogeneous catalysis.

2SO2(g) + O2(g) 2SO3 [Contact process of H2SO4]

In such type of catalysis, the catalyst is generally solid while reactant are gases and reaction from surface of solid catalyst. Hence, it is also called surface catalysis.e.g. N2(g) + 3H2(g) 2NH3

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Oxidation of NH3

4NH3(g) + 5O2(g) 4NO + 6H2OHydrogenation of Vegetable Oil

Oil + H2 Vegetable Ghee

Enzyme Catalysis:-Enzyme are the biological catalyst when the rate of bio-chemical reaction.Enzymes are proteins. There are certain substances which may increase the reactivity of enzyme; they are called activator or co-enzyme.These co-enzymes are generally metal ions like zinc, magnesium, sodium, potassium, etc.

Inversion of consumptionC12H12O11 + H2O C6H12O6(aq) + C6H12O6(aq)

C6H12O6 2C2H5OH(aq) + 2CO2(g)2(C6H10O5)n + nH2O nC12H22O11(aq)

C12H22O11 + H2O 2C6H12O6

(NH2)2CO + H2O 2NH3 + CO2

Milk Curd

Proteins Amino acidsCharacteristics:-a. Enzymes have very high efficiency(106)b. Very small amount of enzyme is reactive.c. Very specific in nature.d. Enzymes are active at optimum temperature and PH.

Temperature at which enzyme activity is maximum called optimum temp [37 0C] around PH =7.

e. The activity of most enzyme catalyst is closely regulated.

Mechanism of enzyme Activity:-The reactant molecule (substrate) binds itself to the active site on the surface of enzyme. The molecules of reactants fit into these cavities (active site) like key fit in a specific lock. This active binding results in the formation of enzyme substrate complex (activated complex).

I. E + S ES

II. ES EP

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III. EP E + P

Product

1. Binding of enzyme with substrate to form enzyme substrate.

E + S ES2. Enzyme substrate converted into enzyme product.

E + S ES3. Product gets released from surface of enzyme.

EP E + P

Applications of enzymes:-Enzymes are used in the Fermentation of Carbohydrates

Sucrose Glucose + SucroseGlucose Ethyl Alcohol + CO2

DIFFERENCE BETWEEN A TRUE SOL N , COLLOIDAL AND SUSPENSION TRUE SOLUTION COLLOIDAL SUSPENSION

Particle size <1nm( <10 A0 )

Between 1nm – 1000nm(10-9 – 10-6)

>100nm. i.e. >10-6

Particles are invisible Scattering of light by particles is observed under ultra-microscope(Tyndale effect)

Visible to naked or under a microscope.May or may not show Tyndale effect.

Particles do not settle down

Do not settle down Settle down on standing

Appear clear and transparent

Translucent Opaque

Do not show Brownian movement

Show Brownian movement

Show Brownian movement

Homogeneous Appear homogeneous but heterogeneous

Heterogeneous

Brownian movement: - Movement of particle in zigzag form when they collide with solvent particles.Tyndale effect: - When a beam of light passes through colloidal solution then scattering of light takes place, it is called Tyndale effect.Classification of colloidal: -

Collodial System

Dispersion

Dispersal Medium

True Solution

Solute

Solution

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Dispersed phase means the substance means the substance distributed in the

dispersion medium in the form of colloidal particles. Dispersion medium in which the substance is dispersed in the colloidal particles. The colloidal system thus obtained is sometimes called a colloidal dispersion or

disperse phase. Depending upon the nature of interaction between the particles of dispersed

phase and dispersion medium, Colloids are of two types:1. Lyophilic colloids2. Lyophobic Solution

Lyophilic colloids:- (liquid loving) – Substances like gum, gelatin, starch, rubber etc. which when mixed with suitable liquids as the dispersion medium directly form colloidal solution called lyophilic and solutions formed called lyophilic solutions of water take as dispersion medium, solution called hydrophilic solution.Lyophobic solution:- (liquids hating) – In this type of solutions, dispersion phase has very little affinity for the dispersion medium. These solutions are less stable than hydrophilic solutions.E.g. Solutions of metals and their insoluble compounds like sulphides and oxides.If water is taken as dispersion medium, solution is called hydrophilic.

DIFFERENCE BETWEEN LYOPHOBIC & HYOPHILIC SOLUTIONLYOPHOBIC HYOPHILIC

They are irreversible solutions Reversible solutionsLess stable and get coagulated, by heating or by agitating

Quite stable, are not easily coagulated by electrolysis

Not much hydrated Highly hydratedCannot prepared directly, prepared by special method only

Prepared easily by directly mixing with the liquid dispersion medium

Viscosity is almost same as that of the medium

Viscosity is higher than that of medium

Surface tension is nearly same as Surface tension is usually lower than

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that of dispersion medium that of the dispersion mediumSolution generally formed by metals, their sulphides etc.(inorganic substances)

Usually formed by organic substances like starch, gum, proteins etc.

They carry either +ve or-ve charge Carry no charge

CLASSIFICATION – DEPENDING UPON MOLECULAR SIZETypes of colloids:-a. Multi molecular colloids :-When on dispersion of a substance in the dispersion

medium, a large number of atoms or smaller molecules of the substance(< 1mm) aggregate together to form species having size in colloidal range, the species thus formed are called multimolecular colloids. E.g. a gold solution consists of particles of various sizes which are clusters of various gold atoms.

A solution of sulphur consists of colloidal particles which are aggregates of S8 molecules.

b. Macro molecular colloids :- These are the substances which on dissolution form solution in which the dispersed phase have the size of particles of colloidal range. Such substances are called macromolecular colloids. E.g. colloidal solution of starch, protein, cellulose, carbohydrates etc. These solutions are quite stable & versatile with True solution.Manmade: Polythene, Nylon, synthetic rubber, etc.

c. Associated colloids :- These are substances which when dissolved in the medium behave as normal electrolyte at low concentration but behave as colloid at higher concentration.

Soap solution at low concentration act as electrolyte but at higher concentration is colloid.

The aggregated particles thus formed are micelles. The formation of micelles takes place only above a particular temp called Kraft Temp (Tk) and above a particular concentration called critical micelle concentration (CMC). Kraft Temperature:-The minimum temperature at which micelle formation takes place called Kraft Temperature.CMC (Critical Micelle Concentration):- The concentration at which micelle concentration takes place.

Mechanism of micelle formation:- Micelle are generally formed by amphiphilic molecules which have both lyophilic as well as lyophobic ends.[e.g. Soap solution]. Such molecules are thrown as surface active molecules or surfactant molecules.

Soap sodium stearate C11H35COO-Na+

C17H35 - long hydrocarbon part is hydrophobic tail.COO- is hydrophilic head.

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Concentration below CMC [ 3 x 10-3 mol L-1] act as good electrolyte & ionize as C17H35COO- & Na+ ions.As concentration increases CMC, hydrophobic part starts receding away from the solvent and are made to approach each other & COO- part interact to water molecule. This ultimately leads to the formation of cluster a large having dimension of colloidal particles. In each such cluster a large number of stearate gaps are clumped together in a spherical manner, so that their hydrocarbon parts interact with one another but COO- part remain projected in water Na+. Cleansing action of soap:-Suppose any oil patch is sticking of cloth. When soap dissolves in water, it dissociated into two parts. The stearate ions array around oily patch and hydrophilic parts project outside the grease/oily patch/droplet.As hydrophilic part is polar can interact with water molecules, present around oily droplet as a result oily droplet pulled away from surface of cloth into water to form ionic micelle which is then washed away with the excess of water.Bredig’s Method [Electro-disintegration]This method is mainly used for the preparation of metal solution. The metal whose solution to be prepared is made as two electrodes immersed in dispersion medium. Dispersion medium kept in dispersion medium. Dispersion medium kept in ice cold bath. An electric spark is introduced between the electrodes. The metal vapourises and condensed immediately to form colloidal solution. The colloidal solution is unstable therefore a small amount of stabilizer is added.Peptization Method:-A process of converting a freshly prepared ppt. into colloidal form by shaking it with dispersion medium in the presence of small amount of electrolyte.The electrolyte used for this purpose is called peptizing agent.Fe(OH)3 + FeCl3 Fe(OH)3Fe3+ (colloidal solution)Fe(OH)3 + FeCl3 Colloidal particles of Fe(OH)3 Cause of Peptization:- As electrolyte is added to a freshly precipitated substance, the particles of ppt. preferentially adsorb one particular type of ions of electrolyte (ions common with the precipitated substance). As a result they get dispersed due to repulsion. It gives particles colloidal size.

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Electrophoresis:- The phenomenon/process of moment of the colloidal particles towards the opposite electrode by passing electric current is called Electrophoresis.Emulsions:- Emulsions are kind of colloidal solutions in which both dispersion phase and dispersion medium are liquids. Any two immiscible liquids form an emulsion.If two immiscible liquids do not mix well the emulsion is unstable and it require a substance for stability called Emulsification.

Types of Emulsification:-i. Oil in water O/W: In this type of emulsion oil acts as

dispersed phase and water act as dispersion medium. E.g. milk is emulsion of soluble fat in water.

ii. Water in Oil W/O: In the case water acts as dispersed phase and oil act as dispersion medium.

Identification of Emulsion:-I. Dye Test :- To the emulsion some oil soluble dye added, if background

becomes coloured, the emulsion is O/W.II. Dilution Test :- If emulsion can be diluted with H2O, it shows that H2O is

dispersion medium & emulsion is O/W. If water form a separate layer emulsion is W/O.

Applications Cleansing action of soap Digestion of fats, (fat emulsify by alkaline solution) Metallurgical process(both floatation process) Disinfectants. Building roads. Emulsion asphalt & H2O.

De-Emulsification:- It is a process of decomposing an emulsion back into its constituent’s liquids. It can be done by different methods like centrifugation, filtration, boiling, freezing and by addition of some chemicals.Gels:- Gel is a colloidal system in which a liquid is dispersed in a solid. The process of gel formation called gelation. Gels are of two categories.

a. Elastic Gels :- Gels which have the property of elasticity e.g. gelatin, agar-agar, starch.

b. Non-Elastic Gel :- Gels which do not have property of elasticity e.g. silica gel.

Application of Colloids:- Purification of drinking water :- The water obtained from natural sources often

contains suspended impurities. Alum is added to such water to coagulate the suspended impurities and make water for drinking purpose.

Rubber Industry :- Latex is a colloidal solution of rubber particles which are –vely charged. Rubber is obtained by coagulation of latex.

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Industrial products :- Paints, inks synthetic plastics, rubber, graphite lubricants, cement etc. are all colloidal solutions.

Important Questions1. Why substances like Pt. & Pd. are used to carry electrolysis of aqueous

solution?

Ans: They absorb H2 gas on the electrolysis of aqueous solution.2. Why Physiorption decrease with increase temperature?

Ans:3. Why powdered

substances are more effective adsorbed than crystalline forms?

Ans: Due to more surface area. Surface area adsorption.4. Why hydrolysis of ester slow at beginning & than become fast?

Ans: RCOOR’ + H2O RCOOH + R’OH Acid produced in the reaction acts as a catalyst

(auto-catalyst). In the case of homogenous auto-catalyst , rate increases with passage of time. Rate=K[CH3COOC2H5][CH3COOH].5. What is the role of desorption in the process of catalysis?

Ans: In heterogeneous catalysis, the products remain adsorbed on the surface of catalyst.

Desorption makes the surface of the solid catalyst free for fresh adsorption of the reactants on the surface.6. What modification can you suggest in Hardy-Schulz law?

Ans: According to this law, greater is valency of ion, stronger is its power to coagulate colloidal solution.-Here not only valency of ion but concentration of ion should be considered.

7. Why adsorption always exothermic?

Ans: In adsorption surface energy of adsorbent decreases which appears as heat. Therefore, adsorption is always an exothermic process.When a gas adsorbed on the surface of a solid, its energy decreases. ∆S=-ve, i.e. ∆G=T∆S, for spontaneous process ∆G=-ve.Here if ∆S=-ve. -T∆S is +ve. Therefore if ∆H=-ve, then ∆G=-ve.Hence adsorption is always exothermic.8. Explain what is observed:

i. When a beam of light is passed through a colloidal solution.

Ans: It appears bright at night angle from the incident beam. It is called Tyndall effect, it takes place due to scattering of light by colloidal particles.

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ii. An electrolyte NaCl is added to ferric oxide solution.

Ans: The =ve charged particles of Fe(OH)3 get coagulated by the oppositely charged Cl- ion of NaCl.

iii. Electric current is passed through a colloidal solut5ion.

Ans: By passing electric current colloidal solution then charged colloidal particles more towards oppositely charged electrodes, this phenomenon called Electrophoresis.

BiomoleculesBiomolecules may be defined as complex lifeless chemical substances which form the basis of life i.e. they not only build up living systems (creatures) but are also responsible for their growth, maintenance and their ability to reproduce. E.g. carbohydrates, fat, protein, nucleic acid, starch, glycogen etc.

Carbohydrates/Saccharides: General formula is Cx (H2O) y Polyhydroxy aldehydes and ketones are called carbohydrates.

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Mutarotation: Spontaneous change in specific rotation of an optically active compound. D-glucose D-mannose D-fructoseAnomers: Pairs of isomers having different configuration at C-1 e.g. α- and β- glucose

Inversion of SugarSucrose on hydrolysis gives dextro-rotatory glucose and laevo-rotatory fructose. The laevo rotation of fructose is more than dextro rotation of glucose; hence the resulting solution becomes laevo-rotatory. Since, the hydrolysis of Sucrose is accompanied by a change in the sign of optical rotation from dextro rotatory to laevo rotatory, the overall process called inversion of Sugar and equimolar mixture of D-(+) glucose and D-(-) fructose called invert sugar. Protein:- High molecular mass complex biopolymers of amino acids.

Hydrolysis hydrolysis Protein Polypeptides α – aminoacidsStructure of 𝛂 –amino acids:

Monosaccharide(Reducing & Non hydrolysable e.g. Glucose & Fructose

e.g. Sucrose Hydrolyzed into D-glucose & D-Fructose

Non-Reducing

Lactose Hydrolyzed into D-Glucose & D-Galactose

Maltose Hydrolyzed into 2 moles of D-Glucose

Reducing

Polysaccharidee.g. Starch & Cellulose (Non-Reducing), Insoluble in water and do not have sweet taste

Oligosaccharide

Non-SugarSugar (soluble in water & sweet in taste)

CARBOHYDRATES

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(R- Different side chains for different amino acids)Zwitter ion – Dipolar ion carrying positive as well as negative charge C CAmino acidsEssential amino acids: Amino acids which cannot be synthesised in the body and must therefore be obtained through diet are known as essential amino acids.Example: - Lysine, Valine, leucine etc.

Non-essential amino acids: amino acids which can be synthesised in the body and are not required in our diet are known as non-essential amino acids.Example: - Glycine, Alanine, Proline etc.

Isoelectric Point: pH at which all amino acid molecules exist as neutral ion and do not migrate towards any electrode.pH< isoelectric point – Positive ion predominates which moves towards cathodepH > isoelectric point – Negative ion predominates and it moves towards anode.

Peptide bond: Amide bond (-CONH-) which combine two amino acid molecules. Structure of protein:

a) Primary structure determine the sequence of amino acidsb) Secondary structure :𝛂- helix: H-bonding exist between >NH and >C=O groups of different

polypeptidesChain gives right handed helical structure.β – Structure: H – bonding exist between >NH and >C=O groups of different polypeptide chains which may be parallel or antiparallel to each other.

c) Tertiary structure: Overall folding of polypeptide chains gives two molecular shapes i.e. fibrous or globular.Functions of proteins: 1.As enzymes catalyze biochemical reactions 2. As hormones regulate metabolic process 3. As antibodies protect the bodies against toxic substances Nucleic Acids: Polynucleotides which are of two types 1) DNA 2) RNANucleotide: Nucleoside + PhosphateNucleoside: Base + SugarSugar : Deoxyribose & Ribose

NH3+

COO-

CH3

R

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Denaturation of proteins: Disruption of the native conformation of protein due to changes in pH, temp and presence of salts. It disrupts higher structure of protein without affecting the primary structureNative Protein: Energetically most stable form of protein with definite configuration and biological activity.Structure of DNA:

1. Double helical structure made up of two polynucleotide strands which are antiparallel to each other.

2. Both stands are held together by H-bonds. A combine with T (A=T) and C with G (C=G)

Genetic code: Hereditary information in an organism is encoded in form of special sequence of bases on polynucleotide chain. It is coma less & universal. It is degenerate.Codon: Combination of 3 nitrogenous bases which code a particular amino acid sequence in protein synthesis.Replication: Duplication of DNATranscription: Process of Synthesis of RNA from DNATranslation: Synthesis of protein by RNAMutation: Sudden chemical changes in DNA molecule which lead to synthesis of protein with altered amino acids sequence.

Distinguish BetweenDNA RNA

1 contains deoxyribose sugar. Contains ribose sugar

2 Contains A,T,G,C as bases. Contains A,U,G,C as bases

3 Has double helix structure Has single helix structure

4 Can replicate Can not replicate

Globular Proteins Fibrous ProteinsThey consists of polypeptide chains, partly of helical structure folded around itself in 3-dimensions so as to give a spherical shape.

They have long and || polypeptide chains half together in linear thread like form (fibers) which is held by intermolecular H-bond & disulphide bonds.

Soluble in water. Insoluble.

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Sensitive to small change in temp & pH. e.g. Elbumines in egg, insulin.

Stable to moderate change in temp & pH.

Enzymes Co-Enzymes1 They are biological catalyst require to catalyze

biochemical reactionsThey are non protein part which increases the activity of enzymes.

2. All enzymes are globular proteins e.g. Lactase, Invertase,

Coenzymes are non proteinouse.g. Metal ions like Zn2+, Mg2+,Mn2+ etc

Starch Cellulose1. It is a white amorphous powder It is a colourless amorphous powder2. It is a mixture of two polysaccharides ,

amylase(straight chain) & amylopectin (branched chain)

It is a straight chain polysaccharide

3. Its solution in water gives blue colour with iodine solution.

Its solution in water does not give blue colour with iodine solution.

4. It has D-Glucose unit joined together by α-glycosidic linkages

It has D-Glucose unit joined by β-glycosidic linkages.

Amylose Amylopectin1. Straight chain polymer madeup of α – D

glucose units.Branched chain polymer of α – D glucose units

2. Gives blue colour with I2 solution Does not give blue colour with I2 solution

3. Having 100 – 300 D-glucose units Having only 25 – 30 D-glucose units

Important reaction of Glucose

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Evidence against open chain structure:- i. Aldehyde group present but glucose does not react with NaHSO3 & NH3.ii. Glucose does not give the Schiff’s Test & 2,4-DNP test for aldehyde.iii. Glucose does not react with Grignard reagents.iv. Glucose penta-acetate does not react hydroxyl amine, which shows that

aldehyde group is absent in glucose penta-acid.v. Glucose exist in two sterioisomeric forms ( & ).vi. An aqueous solution of glucose shows Mutarotation i.e. specific rotation

gradually decreases from +1100 to 52.50 in case of α-glucose and increase from 19.70 to 52.50 in case of β-glucose.

All these observations indicate that free aldehydic group is not present in glucose.

(II) The penta-acetate of glucose does not react with hydroxylamine. It indicate that free aldehyde group is not present in glucose.Avitaminosis: Condition of vitamin deficiency.Hypervitaminosis: Excess intake of vitamins.Fat soluble vitamin: Vitamin A, D, E, & KWater soluble vitamins: Vitamin B & CHypoglycemic Factor: Insulin is also known as hypoglycemic factor & decrease glucose concentration in bloodSaturated Fatty acids: e.g. Palmitic acid & stearic acidUnsaturate Fatty acid: e.g. Oliec acid, linolenic acid

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S No.

Name of vitamin Source Deficiency disease

1 Vitamin A(bright eye vitamin)

Fish oil particularly shark liver oil, liver

Xerophthalmia i.e cornea hardening of eye

2 Vitamin B1 (thiamin)

Yeast, milk, green vegetables etc

Beri-beri (a disease of nervous system)

3 Vitamin B2(riboflavin)

Yeast, vegetables, milk, egg white liver and kidney

Dark red tongue (glossitis), dermatitis cheilosis (fissuring at corners of mouth and lips)

4 Vitamin B6 (pyridoxine)

Cereal, grams, molasses, yeast, egg yolk and meat

Severe dermatitis, convulsions

5 Vitamin H ( Biotin) Yeast, liver, kidney and milk Dermatitis, loss of hair and paralysis

6 Vitamin B12 Liver of ox, sheep, pig, fish etc Pernicious anaemia7 Vitamin C Citrus fruits, green vegetables Scurvy

8 Vitamin E Wheat germ oil, cotton seed oil Sterility9 Vitamin K Cereals, leafy vegetable Haemorrhagic conditions10 Coenzyme Q10 Chloroplasts of green plants Low order of immunity of body

against many diseases

Some Important QuetsionsQ1.How are vitamins classified? Name the vitamin responsible for coagulation of blood.Ans: - vitamins are classified into two groups depending upon their solubility in water or fat.1. Water soluble vitamins: - These include vitamin B-complex and vitamin C. 2. Fat soluble vitamins: - These include vitamins A,D,E and K. They are stored in liver and adipose. Vitamin K is responsible for coagulation of blood.

Q2.What are nucleic acids? Mention their two important functions.Ans: - Nucleic acids are biomolecules which are found in the nuclei of all living cells in form of nucleoproteins or chromosomes.Nucleic acids are of two types; deoxyribonucleic acid (DNA) and ribonucleic acid (RNA).The two important functions of nucleic acids are as follows:1. DNA is responsible for the transmission of hereditary effects from one generation to another. This due to unique property of replication, during cell division when two identical DNA strands are transferred to the daughter cells.2.DNA and RNA are responsible for synthesis of all proteins needed for the growth and maintenance of our body.

Q3.What is meant by reducing "reducing sugars "?

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Ans:- Carbohydrates which reduce Tollen's reagent or Fehling solution are called reducing sugars. This is due to the presence of free aldehydic group e.g.,glucose. Q4. What is invert sugar?Ans :- Invert sugar is a equimolar mixture of (+) glucose (+52.50) and (-) fructose (-92.40).It is obtained by the hydrolysis of sucrose .Since sucrose is dextrorotatory , but after hydrolysis the mixture is laevorotatory, the mixture is called invert sugar.Q5. Describe denaturation of protein.Ans :- Denaturation :- When the natural conformation of a protein is disturbed in which it does not change in primary structure but loses some or all of its secondary, tertiary or quaternary structure is called denaturation of protein. As a result of denaturation, the bioactivity of protein is lost. Protein denaturation is caused by change of temperature, change of pH, addition of soluble salts like ammonium sulphate, magnesium sulphate, addition of solvents like water, alcohol etc.

Q6. Describe Peptide linkage or Peptide bond.Ans :- Peptide bond :- Proteins are condensation polymers of α-amino acids which are the same or different α-amino acids are connected by peptide linkage. Chemically, a peptide bond is an amide linkage formed between -COOH group of one α-amino acid -NH3 group of other α-amino acid by loss of a water molecule.Q7. What is the difference in structures of α-D(+) glucose and β-D(+) glucose?Ans: These are Anomers and differ in the configuration of H-atom and –OH group about C1 atom.Q8. What is difference between a nucleotide and nucleoside?Ans: A nucleoside contains a pentose sugar and base (purine or pyrimidine) while in nucleotide a phosphoric acid component is also present.Q9. Name two RNA molecules found in the cells of organism.Ans: Messenger RNA (m-RNA) and Ribosomal RNA (r-RNA).Q10. Give one use of enzyme streptokinase in medicines.Ans: This enzyme can dissolve blood clots. It is useful medicine for checking heart attacks due to blood clotting.

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POLYMERSVery short answer type questions:

1. Define the term homopolymerisation giving an example.Ans. Polymerisation process involving only single monomeric species is known as homopolymerisation. For example formation of polyethene from ethane.2. Define the term Polymerisation.Ans. It is a process of formation of a high molecular mass polymer from one or more monomers by linking together a large number of repeating structural units through covalent bonds.3. Write the name and structure of the monomer of polymer: PVC.Ans. The monomer of PVC is vinyl chloride.

CH2=CH-Cl4. Write the monomer of polyethene.Ans. Ethene (CH2=CH2)5. What does the designation 6,6 means in the name of Nylon-6,6?Ans. 6,6 means that both the monomers are of 6 carbon each.6. Give an example of elastomers.

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Ans. Natural rubber of Buna-S.7. What is the primary feature for a molecule to be a monomer in a condensation

polymerisation reaction?Ans. Monomer should have more than one functional groups.8. Give the chemical name of Teflon.Ans. Polytetrafluoroethylene.9. What are the monomer units of Bakelite?Ans. Phenol and Formaldehyde.10. What does SBR stands for?Ans. Styrene Butadiene Rubber.11. Give an example of step growth polymer.Ans. Terylene or Dacron.12. Name a polymer used to make cups for hot drinks.Ans. Polystyrene is used to make disposable cups for hot drinks as it does not become soft at a temperature near boiling point of water.13. Write the name and structure of monomer of natural rubber.Ans. Isoprene or 2-Methylbuta-1,3-diene.14. What are the monomer units of terylene?Ans. Ethylene glycol and terepthalic acid.15. What does PMMA stands for?Ans. Polymethylmetaacrylate.

SHORT ANSWER TYPE QUESTIONS:

1. Write the structure of the monomer of each of the following polymers:(i) Nylon-6(ii) Teflon(iii) Neoprene

2. Differentiate between thermoplastic and thermosetting polymers. Give one example of each.Ans.Thermoplastic polymer Thermosetting polymer

1. They are linear polymer without cross-linking.

1. They have three dimensional network of covalent bonds with cross links.

2. Upon heating these polymers become soft and becomes hard again on cooling.

2.Upon heating they retain their strength and does not become soft.

3. These plastics can be recycled. They can be melted and reused.

3.These plastics cannot be recycled.

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4.Example: PVC, Nylon, Polystrene.

4.Example: Phenol formaldehyde resin, urea formaldehyde resin.

3. Explain each of the following giving suitable example of each:(i) Elastomers(ii) Condensation polymers(iii)Addition polymersAns. (i) Elastomers: Polymers which are rubber like with elastic properties are

called Elastomers.(ii) Condensation polymers: The polymer formed by the repeated condensation reaction between two same or different bifunctional monomeric units with the elimination of simple molecules like water, ammonia etc., are called as condensation polymers.(iii)Addition polymers: The polymers formed by the repeated addition of monomer molecules possessing double or triple bond are called addition polymers.

4. Mention two important uses of each of the following polymers.(i) Bakellite(ii) Nylon-6,6(iii) PVC

Ans. (i) Bakellite: For making electrical switches, sockets, plugs etc. (ii) Nylon-6,6: For making bristles for brushes, ropes, in textile industries. (iii)PVC For making insulation for electrical wires, for making foot wear, vinyl flooring, buckets, toys etc.

5. Name the sub-groups into which the polymers are classified on the basis of magnitude of intermolecular forces.

Ans. On the basis of magnitude of intermolecular forces polymers can be classified as: a. Elastomers b. Fibers c. Thermoplastic polymers d. thermosetting polymers.6. Write the name and structures of the monomers of the following polymers:

Buna-S, Dacron and Neoprene Ans. Polymer MonomerBuna-S CH2=CH-CH=CH2 (Butadiene), CH2=CH-C6H5 (Styrene)Dacron HOCH2-CH2OH ( Ethylene Glycol), HOOC-C6H4-COOH

(Terepthalic acid)Neoprene CH2=C(Cl)-CH-CH3 (Chloroprene)

7. What are biodegradable polymers? Give two examples.Ans. Polymers which can be easily degraded by micro-organisms present in nature are called biodegradable polymers. Common examples are poly-β-hydroxybutyrate-co-β-hydroxyvalerate (PHBV), Nylon-2-nylon-6.8. Distinguish between ‘Chain growth polymerisation’ and ‘Step growth

polymerisation’ and give one example of each.Ans.Chain growth polymerisation Step growth polymerisationOnly addition reaction adds Any two molecular species present

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repeating unit one at a time. can react.Monomers are unsaturated compounds like alkenes, alkadienes etc.

Monomers are generally bifunctional organic compounds.

This type of polymerisation occurs without the loss of any small molecule.

Loss of simple molecule like water or ammonia takes place.

Example: polymerisation of ethene to form polyethene.

Example: formation of nylon-6,6 from hexamethylene diamine and adipic acid.

9. What is the structural difference between LDPE and HDPE?Ans. LDPE (Low density polyethylene) has a branched structure hence does not packs well. It is a transparent polymer with moderate tensile strength and toughness. It is widely used as packing materials thin plastic bags, flexible pipes, squeeze bottles etc.HDPE (High density polyethylene) has a linear structure and hence packs well. It is a translucent polymer with greater toughness, higher tensile strength than LDPE. It is used in the manufacture of Buckets, mugs, toys, housewares etc.

10. Give the differences between natural rubber and vulcanised rubber.OR

How does vulcanisation change the character of the natural rubber?Ans.

Natural rubber Vulcanised rubberIt becomes soft at temperature above 335K and brittle at temperature below 283K

It has a much wider useful temperature range.

It shows high water absorption capacity.

It absorbs water to a much lesser extent.

It is highly soluble in non-polar solvents.

It is less soluble in non-polar solvents.

It is readily attacked by oxidising agents.

It is not readily attacked by oxidising agents.

11. Why are the numbers 6,6 and 6 put in the names of nylon-6,6 and nylon-6?Ans. Nylon-6,6 is obtained from hexamethylene diamine and adipic acid. As each of these monomers contains six carbon each. Nylon-6 is obtained by polymerisation of caprolactum of its hydrolysis product, aminocaproic acid. As there is only one monomer with 6 carbon atoms, the nylon so obtained is known as nylon-6.12. Distinguish between homopolymer and copolymer. Give example of each.Ans. Homopolymer: A polymer obtained by polymerisation of only one type of monomer is called as homopolymer. For example Nylon-6 is obtained from caprolactum only.

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Copolymer: A polymer obtained by polymerisation of two types of monomer is called as copolymer. For example, Nylon-6,6 is obtained by copolymerisation of hexamethylene diamine and adipic acid.

13. How is Dacron obtained from its monomers? Give one use of Dacron.Ans. Dacron is obtained by condensation polymerisation of ethylene glycol and terepthalic acid with the elimination of water molecules. The reaction is carried out at 420-440K in the presence of a catalyst consisting of mixture of zinc acetate and antimony trioxide.nHOCH2CH2OH + nHOOC-C6H4-COOH 4⃗20−440K ,Zn(OCOCH 3)2 , Sb2O3 [-O-CH2-CH2-O-CO-C6H4-CO-]

(Dacron)14. What are elastomers? Explain giving an example?Ans. These polymers have weakest intermolecular forces of attractions. These are highly elastic. These can be stretched due to weak intermolecular forces. They regain their original shape when stress is removed due to cross links. For example: Buna S.

15. How do you explain the term functionality of a monomer?Ans. For a substance to act as a monomer, it must have at least two reactive sites or bonding sites. The number of reactive sites in a monomer is referred to as its functionality. In a olefin, the double bond can be considered as a site for two free valencies. When a double bond is broken two single bonds become available for bonding. Other bifunctional monomers are aminocaproic acid, pthalic acid, ethylene glycol etc.16. What are natural and synthetic polymers? Give examples of each type.Ans. Natural Polymers: Polymers which are found in natural living organisms, i.e., animals and plants are called natural polymers. For example: proteins, cellulose, nucleic acids, resins etc.Synthetic polymers: Man made polymers are called as synthetic polymers. For example: Plastics, PVC, Nylon-6,6 etc.17. Describe the preparation of Buna-S frim its monomers.Ans. Buna-S is a synthetic polymer obtained by polymerisation of butadiene and styrene. Sodium is used as a polymerising agent.nCH2=CH-CH=CH2 + nC6H5CH=CH2 N⃗a, heat [-CH2-CH=CH-CH(C6H5)-CH2-]n18. Explain the role of initiator in the free radical polymerisation of an alkene.Ans. In free radical polymerisation of alkenes an initiator like benzoyl peroxide, acetyl peroxide etc are used. The initiator undergoes homolytic fission in the presence of heat or light to produce free radicals.

C6H5COO-OOCC6H5 H⃗omolyticfission 2C6H5COO.

(benzoyl free radical)

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As this redical reacts with another molecule of ethene, another bigger size free radical is formed and chain propagates to form polyethene.

19.Explain the following terms with suitable examples:(i) Polyamides(ii) Polyesters(iii) Macromolecules

Ans. (i) Polyamides: Polyamides are polymers with amide linkage in their chains. For example silk, wool, nylon-6, nylon-6,6.

(ii) Polyesters: Polymers with ester linkages in their chains are called as polyesters. For example terylene, glyptal, etc.

Macromolecules: Molecules of large size are generally called as macromolecules. Most of these are polymers formed by small repeating structural units.

20.Write the uses of the following polymers:(i) Melamine(ii) Teflon(iii) PANAns. Melamine: it is used in the manufacture of unbreakable crockery.

Teflon: it is used in making oil seals, gaskets and for non-stick coated utensils.PAN ( Polyacrylonitrile): it is used as a substitute for wool and for making commercial fibres.

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CHEMISTRY IN EVERYDAY LIFE

VERY SHORT ANSWER TYPE QUESTIONS:1. What is meant by narrow spectrum antibiotics?Ans. Antibiotics which are mainly effective against gram positive of gram negative bacteria are called as narrow spectrum antibiotics. For example Penicillin G is a narrow spectrum antibiotic.2. What is meant by broad spectrum antibiotics?Ans. These are medicines which are effective against several different types of micro-organisms. For example tetracycline, chloramphenicol etc.3. What do you understand by the term ‘Chemotherapy’?Ans. Use of chemicals for therapeutic effect is called as Chemotherapy.4. Name a drug used in case of mental depression?Ans. Equanil, barbituric acid derivatives such as seconal, neuronal are used as antidepressents.5. Give an example of Narcotic which is used as analgesic.Ans. Morphine is a narcotic which is used as an analgesic i.e. to reduce pain.6. What is the role of Bithional in toilet soaps?Ans. Bithional acts as antiseptic when added to soap. Thus, it increases the antiseptic properties of the soap.7. Define antipyretic. Give two examples.Ans. These are the chemicals which are used to bring down the body temperature during high temperature. For example: Asprin, Paracetamol.8. What is tincture of iodine? For what purpose it is used?Ans. Alcohol water solution containing 2-3% iodine is known as tincture of iodine. It is used as antiseptic.9. What precautions will you take before administering penicillin to a patient?

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Ans. The patient should be tested for sensitivity to penicillin before it is administered.10. Give an example of non-ionic detergent.Ans. Lauryl alcohol ethoxylate.11. Name the chemical responsible for antiseptic properties of Dettol.Ans. Chloroxylenol and terpineol in a suitable solvent.12. To which class of drugs cimetidine and ranitidine belongs?Ans. Anti-histamines.13. Name a food preservative which is most commonly used by food producers.Ans. The most common food preservative used by food producers is sodium benzoate.14. Describe antiseptics giving suitable examples.Ans. Antideptics are the chemical substances which prevent the growth of micro-organism and may even kill them. Boric acid, hydrogen peroxide, bithional, is common antiseptics.15. How are transparent soaps manufactured?Ans. Transparent soaps are manufactured by dissolving soap in ethanol and evaporating the excess solvent.16. What are artificial sweetening agents? Give two examples.Ans. These are the chemical substances which are sweet in taste but do not add to calorie intake. Some common examples are aspartame saccharin, alitame etc.17. Why the use of aspartame is is limited to cold food and drinks?Ans. This is because aspartame decomposes at baking and cooking temperatures. Hence aspartame can be used only with cold drinks and foods.18. Name the additive added to shaving soap to prevent rapid drying.Ans. Glycerol.19. Name an analgesic which is also antipyretic.Ans. Asprin.20. Name an artificial sweetener which appear and taste like sugar and is

stable at cooking temperature.Ans. Sucrolose.

SHORT ANSWER TYPE QUESTIONS:1. What are biodegradable and non-biodegradable detergents? Give one example

of each. Ans. Detergents which are easily degraded by micro-organisms present in the rivers, ponds etc., are called as biodegradable polymers. For example sodium-dodecylbenzene sulphonate.

Non-biodegradable detergents are not easily degraded by mico-organisms. For example sodium-1,1-dimethyldecylbenzene sulphonate.

2. State the reason for each of the following:a. Soap do not work in hard water.b. Synthetic detergents are better than soaps.

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Ans. a. ordinary soaps are sodium or potassium salts of higher fatty acids like stearic acid, palmitic acid. Hard water contains calcium and magnesium ions. These ions form insoluble scum with soap and useless as cleansing agent.

b. soaps do not work in hard water but synthetic detergents are effective in cleaning even in hard water.3. Explain the following giving one example of each type:

a. Antacids b. Disinfectants c. Enzymes Ans. (i) Antacids: The chemicals which control the release of acid in gastric juices or neutralises the excess acid in gastric juices are called antacids.(ii) Disinfectants: The chemical substances which are used to kill microorganisms but they cannot be applied on living tissues are called as disinfectants. For example 1% solution of phenol in water is a disinfectant.(iii) Enzymes: Proteins which perform the role of biological catalysts and catalyse biological reactions are called enzymes. For example, an enzyme maltase catalyses the hydrolysis of maltose to glucose.4. Explain the following giving one example of each type:

a. Tranquilizers b. Food preservativesAns. a. Tranquilizers: These are chemicals used for the treatment of stress and mild or even severe mental diseases. For example, veronal, serotonin, derivatives of barbituric acid etc.b. Food preservatives: Chemicals which are used to protect food items against micro-organisms are called food preservatives. For example, sodium benzoate, sodium metabisulphite etc.5. Mention one use of the each of the following:a. Ranitidine b. Paracetamol c. Tincture of iodineAns. Ranitidine: Antihistamine /Antacid

Paracetamol: Analgesic, antipyreticTincture of iodine: Antiseptic

6. Describe the following giving one example:a. Antifertility drugs b. Antioxidants

Ans. Antifertility drugs: These drugs are used to prevent unwanted pregnancies. For example Norethindrone

Antioxidants: Chemicals which prevent the oxidation of food stuff during storage etc., are called as antioxidants. Vitamin C and E are natural oxidants. BHA ( Butylated Hydroxy Anisole) is a synthetic antioxidant.7. what type of drug is Chloramphenicol, Equanil and asprin?Ans. Chloramphenicol is an antibiotic, Equanil is a tranquilizer and asprin is an analgesic.

8. Describe and illustrate with an example- a detergent.Ans. Chemical substances which help in removing dirt by emulsifying oil of grease are called as detergents. Detergents help in lowering the surface tension of the water. Sodium salt of p-dodecylbenzene sulphonate is the most common example of synthetic detergent. 9. How is antibiotic different from antiseptics? Give one example of each.

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Ans. Antibiotic can be taken orally or injected into the body while an antiseptic can only be applied to the skin, outside the body. Erythromycin is an antibiotic while chloroxylenol is an antiseptic.10. Sleeping pills are recommended by doctors to the patients suffering form sleeplessness, but it is not advisable to take its doses without consultation with the doctor. Why?Ans. Not only the sleeping pills, in fact most of the drugs taken in the doses higher than recommended may cause harmful effect and act as poison. Therefore, a doctor should always be consulted before taking medicine.11. How antiseptics differ from disinfectants. Give one example of each.Ans.

Antiseptic Disinfectant1. Antiseptic can kill or prevent the growth of micro-organism.

1. Disinfectants can kill of prevent the growth of micro-organism.

2. Antiseptics do not harm the living tissues. Therefore they can be applied on the skin.

2. Disinfectants are toxic to the living tissues. Therefore they cannot be applied on the living tissues.

3. Antiseptics are used for dressing of wounds, cuts and in the treatment of skin diseases.

3. Disinfectants are used for disinfecting floors, toilets, drains, instruments etc.

4. Examples: Furacine, boric acid, tincture of iodine.

4. Examples: 1% aqueous solution of phenol.0.4ppm chlorine.

12. Write the IUPAC name of Asprin, also write its usage. Why it not be taken on an empty stomach?Ans. IUPAC name of Asprin is 2-Acetoxybenzoic acid. In the stomach, it gets hydrolysed to give salicylic acid , which on empty stomach can damage the lining of the stomach.13. Why are Cimitidine and ranitidine better antacid than sodium hydrogen carbonate?Ans. Excessive intake of sodium hydrogen carbonate can make stomach alkaline and trigger the production of even more acid. Also these treatments controls only the symptoms of hyperacidity and not the cause. Cimetidine, ranitidine are better antacids because they control the cause of hyperacidity. It is observed that hormone Histamine stimulates the production of pepsin and hydrochloric acid in the stomach. Drugs like cimetidine and ranitidine prevent the interaction of histamine with the receptors present in the stomach wall. This results in production of lesser amount of acids and controlling hyperacidity.14. What are analgesic medicines? How they are classified when they are commonly recommended for use?Ans. The chemical substance which are used to relieve pain are called analgesics. These are of two types:(i) Non-narcotic (ii) NarcoticNon-narcotic drugs are effective in relieving skeletal pain preventing heart attack, viral inflammation etc. narcotic drugs are recommended for the post-operative pains, cardiac pain, terminal cancer etc.15. Describe the following with an example:(i) Edible colours (ii) cationic detergents

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Ans: Edible colours: These are synthetic or natural dyes used to impart attractive colours to food, drugs etc. common examples are carotene, diazonium salt of sulphanilic acid.

Cationic detergents: cationic detergents are quaternary ammonium salts of amines with acetates/ chlorides/ bromides as anions. For example cetyltrimethyl ammonium bromide.16. Explain why diabetic patients are advised to take artificial sweetners instead of natural sugar?Ans. Artificial sweeteners like Aspartame, alitame are either not metabolised by the body or do not produce carbohydrates like glucose when metabolised. Thus these artificial sweeteners help in controlling the blood glucose level.17. Account for the following:(i) Asprin drug helps in preventing the heart attack.(ii) Detergents are non-biodegradable while soaps are biodegradable.Ans. (i) Asprin is found to help in prevention of heart attack because of its anti-blood clotting action. The blood clots if formed, are mainly responsible for heart attack.

(ii) Common detergents which contain branched chain hydrocarbon part in their molecules are non-biodegradable. This is because micro-organisms present in the waste water cannot metabolise branched chain hydrocarbons. On the other hand soaps contain linear chain hydrocarbons, which can be biodegraded more easily and water pollution is prevented.

18. In chemotherapy, what are target molecules?Ans. Drugs usually interact with biological macromolecules such as carbohydrates, proteins, lipids, nucleic acids. These are called target molecules. Drugs are designed to interact with specific targets so that these have the least chance of affecting other targets. This minimises the side effects and localises the action of drug.19. What are antihistamines? Explain how they act on human body?Ans. Antihistamines are drugs which interfere with the natural action of hormone histamine by competing with histamine for binding sites of receptors where histamines exert its effect. They are also called as anti-allergic, drugs used to treat allergy. Allergic reactions like skin rashes are caused due to liberation of histamine in the body that is why these drugs are also called as anti-histamines.20. Can sulphanilamide be classified as an anti-biotic? Give suitable reasons.Ans. Yes, according to the modern definition of antibiotic, an antibiotic is a substance produced wholly or partially by chemical synthesis, which in low concentration inhibits the growth or destroy micro-organisms by intervening in their metabolic processes. Sulphanilamide fits in the definition as it is chemically synthesized and destroys micro-organisms.

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HOTSCHAPTER 10

HALOALKANES AND HALOARENES

1. Draw the structure of most stable carbocation of the formula C5H11+ .CH3

Ans: CH3—CH2—C+--CH3.

2.(a)Convert Nitrobenzene to m-Bromoiodobenzene.(b)Draw the structure of the product of the Bromination of following compounds in presence of FeBr3.

(b)Because –NO2 deactivates the ring but OCH3 activates.

3(a) Identify ‘A’ and ‘B’

i) NaNH2 Na/NH3

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CH3-C≡CH ‘A’ ‘B’ ii) CH3I

(b) Distinguish between CHCl3 and CHI3 Ans (a) A=But-2-yne B= But-2-ene

(b) Dissolve them in 50% alcohol solution.Iodoform will give yellow ppt with AgNo3 because C-I bond is weaker than C-Cl bond whereas CHCl3 does not react.

4. Write the major products:- F

(a) NO2

CH3׀

(b) CH3-C-CH2Br C2H5OH׀CH3

(c) COOC6H5

Br2 FeBr3

Ans (a) OCH3

NO2

CH3

CH3ONa

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׀ (b) CH3 - C - CH2 – CH3

׀ OC2H5

(c) COOC6H5

Br

5. Two isomeric compounds ‘A’ and ‘B’ have same formula C11H13OCl. Both are unsaturated give the same compound ‘C’ on catalytic hydrogenation and produce 4-chloro-3-exthoxy benzoic acid on vigorous oxidation. ’A’ exits in two geometrical isomers ‘D’ and ‘E’ but B does not. Identify A to E. Which has higher melting point out of D and E

Ans CH = CH – CH3

OC2H5 Cl

CH2 - CH = CH2

B = OC2H5 Cl

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C= CH2CH2CH3

OC2H5 Cl

D= H H

Cl C = C

CH3 OC2H5

(Cis)

E= H CH3

Cl C = C

H OC2H5

( trans)E has higher Melting point due to symmetry.

6. Convert the following:- (a) ethanol to propanenitrile (b) propene to 1-propanol (c) 1-Bromopropane to 2-Bromopropane

Ans a) PCl5 KCN CH3CH2OH CH3CH2Cl CH3CH2CN

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HBr/peroxide KOH

b)CH3CH=CH2 CH3CH2CH2Br CH3CH2CH2OH

alc KOH HBr c) CH3CH2CH2Br CH3CH=CH2 CH3CHBrCH3

7) An organic compound “A” having molecular formula C4H8 on treatment with dil H2SO4 gives “B” which on treatment with conc. HCl and anhydrous ZnCl2 gives “C” and on treatment with sodium ethoxide gives back “A”.Identify A, B,C.

Ans.A=But-2-ene

B=Butan-2-ol C=2-Chlorobutane

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CHAPTER 11ALCOHOLS, PHENOLS AND ETHERS

1. An organic compound ‘ A ‘ having molecular formula C3 H6 on treatment with aq. H2SO4 give ‘B’ which on treatment with Lucas reagent gives ‘C’. The compound ‘C’ on treatment with ethanolic KOH gives back ‘ A’ .Identify A, B , C .

Ans.

2. An organic compound A (C6H6O) gives a characteristic colour with aq. FeCl3 solution. (A) On reacting with CO2 and NaOH at 400k under pressure gives (B) which on acidification gives a compound (C) .The compound (C) reacts with acetyl chloride to give (D) which is a popular pain killer. Deduce the structure of A,B,C & D.

Ans.

3. An organic compound (X) when dissolved in ether and treated with magnesium metal forms a compound Y. The compound, Y, on treatment with acetaldehyde and the product on acid hydrolysis gives isopropyl alcohol. Identify the compound X. What is the general name of the compounds of the type Y.

Ans. The compound X is CH3Br and Y is CH3MgBr The compounds of the type ‘Y’ are called Grignard reagent.

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4. A compound ‘A’ with molecular formula C4H10O on oxidation forms compound ‘B’ gives positive iodoform test and on reaction with CH3MgBr followed by hydrolysis gives (c). Identify A, B & C.

Ans. The compound ‘B’ is obtained by oxidation of C4H10O and gives positive iodoform test and also reacts with CH3MgBr , it must be methyl Ketone , it must be methyl ketone having four carbon atoms i.e, CH3COCH2CH3 .This can be obtained by oxidation of 2 – butanol i.e , CH3 CH CH2 CH3 Therefore ,

thereactions are:

5. An aromatic compound (A) having molecular formula C6H6O on treatment with CHCl3 and KOH gives a mixture two isomers ‘B’ and ‘C’ both of ‘B’ & ‘C’ give same product ‘D’ when distilled with Zn dust. Oxidation of ‘D’ gives ‘E’ of formula C7H6O2. The sodium salt of ‘E’ on heating with soda lime gives ‘F’ which may also be obtained by distilling ‘A’ with zinc dust. Identify compounds ‘A’ to ‘F’ giving sequence of reactions.

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6. Compound ‘A’ of molecular formula C5H11Br gives a compound ‘B’ of molecular formula C5H12O when treated with aq. NaOH. On oxidation the compound yields a mixture of acetic acid & propionic acid. Deduce the structure of A, B & C.

Ans.

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The reactions are:

Ans.

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CHAPTER 12ALDEHYDES, KETONES AND CARBOXYLIC ACIDS

1. How are formalin and trioxane related to methanal?Ans- Formalin is 40% aq solution of methanal whereas trioxane is trimer of methanal.

2. How is acetone obtained from ethanol?Ans-

CH3CH2OH CH3CHO CH3CHCH3 CH3CCH3

3. Carboxylic acids do not give characterstics reactions of carbonyl group,explain why?

Ans In carboxylic acids,carbonyl group is involved in resonance O O-

R-C- O- H R-C=O+-H Therefore it is not a free group. But no resonance is possible in aldehydes and ketones.

4. Name on reagent used to convert phenol into salicyladehyde?Ans. CHCl3/NaOH at 343 K.

5. Why does benzoic acid not undergo Friedel Crafts reaction?Ans. Due to the deactivation of the benzene ring by electron withdrawing effect of the

-COOH group.

i) CH3MgBr

ii) H2O/H+

Cu

575 K

Cu

575 K OH O

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CHAPTER 13ORGANIC COMPUNDS CONTAINING NITROGEN

1. Why methylamine is a stronger base than trimethylamine?Ans- Due to steric strain.Addition of protonincreases the crowing on the nitrogen

which increases the steric strain. Trimethylamine have the more strain,consequently lesser tendency of protonation,so basicity is reduced.

2. Why do amines dissolve in mineral acids?Ans- The nitrogen atom in amines contains a lone pair of electron which it can donate.

Thus they accept a proton from mineral acid to form a salt which is water soluble.R-N:H2 + H+Cl- [ R-NH3]+Cl- Water soluble salt.

3. Electrophilic substation in case of aromatic amines takes place more readily than benzene?

Ans- Amino group( -NH2) in aniline is electron pushing group.It activates the benzene ring due to +I effect, and it stabilized by resonance .Thus aromatic amines undergo electrophilic substation at ortho and para position more readily than benzene.

4. How would convert methylamine into ethylamine?Ans- HI CH3NH2 + HNO2 CH3OH CH3I+KCN CH3CN methanol iodomethane Ethanenitrile

CH3CH2NH2Ethanamine

5. Amines are basic substances while amides are neutral?Ans- Due to presence of lone pair of electronson nitrogen of amines,they are basic in nature where as nitrogen of amides acquires positive charge due to resonance with carbonyl group which makes it neutral.

LiAlH4

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