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CBSE Class 12 Pre Board 2021 : Chemistry - with ANSWERS / MARKING SCHEME (Kendriya Vidyalaya (KV), Kanpur Cantt)

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KENDRIYA VIDYALAYA CLASS 12(CHEMISTRY) Pre- board 1 (2020-2021) M.M. :70 TIME: 3 Hours GENERAL INSTRUCTIONS: read the following instructions carefully. (a) There are 33 questions in this question paper. All questions are compulsory (b) Section A: Question numbers 1 to 2 are case- based questions having four MCQs or Reason Assertion type based on the given passage each carrying 1 mark. (c)Section A: Question 3 to 16 are MCQs and Reason type questions carrying 1 mark each. (d) Section B: Question 17 to 25 are short answer questions and carry 2 marks each. (e) Section C: Question number 26 to 30 are short answer questions and carry 3 marks each. (f) Section D: Question numbers 31 to 33 are long answer questions and carry 5 marks. (g) There is no overall choice. However, internal choices have been provided. (h) Use of calculators and log tables is not permitted. SECTION A (OBJECTIVE TYPE) 1. Read the passage given below and answer the following questions: Ethers are inert and stable compounds. It can be symmetrical and unsymmetrical .Both symmetrical and unsymmetrical ether can be prepared by the Williomson synthesis reaction. R-X + RONa ROR + NaX Ethers can also be prepared by dehydration of alcohol. But we can get only symmetrical ether by this method.The formation of ether follow SN2 mechanism. Alkoxy group (- OR) is ortho, para directing and activates the aromatic ring towards electrophillic substitution . The following questions are multiple choice question. Choose the most appropriate answer: 1(i). What is the main product for the reaction? (CH3)3CBr + Na-OCH3 ------ (a) (CH3)2C=CH2 (b) (CH3)3C-OCH3 (c) (CH3)3C=CH2 (d) (CH3)3C-ONa 1. (ii). Write the product of Friedal Craft Alkylation reaction of anisole? (a) 3-Methoxy- toluene (b) 2- Methoxy benzene and 4- Methoxy benzene (c) 2- Methoxy toluene and 4- Methoxy toluene (d) 3- Methoxy toluene 1. (iii) . What is the IUPAC name of the compound? C6H5O(CH2)6-CH3 (a) 1- Heptoxybenzene (b) Benzeneheptanoxy (c) 1- Phenoxyheptane (d) 1- Methylhexanebenzene 1(iv).p. What is the product of reaction (CH3)3C-OC2H5 + HI ------ (a) (CH3)3C-OH + C2H5I (b) (CH3)3C-H + C2H5OI (c) (CH3)3C-I + C2H5OH (d) (CH3)2CH + CH3-C2H5-I OR 1.(iv). q. What is the product of the reaction:CH3-CH2-CH2-O-CH3 + HI ---- (a) CH3-CH2-CH2-OH + CH3I (b) CH3-CH2-CH2-I + CH3OH (c) CH3-CH2-I + CH3-O-CH3 (d) CH3-I + CH3-CH2-O-CH3 2. Read the passage given below and answer the following questions: Colloidal solution contain the particles of size between 1- 1000 nm. Colloidal solutios are of two types. One is lyophillic sol and other is lyophobic sol. Lyophilic sol is more stable as compare to lyophobic sol. During the preparation of colloidal sol, some electrolyte we have to add. But during purification of these colloidal solution , we have to remove the extra amount of electrolytes. Charges are present on the colloidal particles which is responsible for the stability of these colloidal sol. But if we add the oppositely charged electrolyte, then coagulation will occur and the colloidal sol will become unstable. In these questions (Q.No. 2.(i)-(iv), a statement of assertion followed by a statement of reason is given. choose the correct answer out of the following choices. (a) Assertion and reason both are correct and reason is correct explanation of assertion. (b) Both assertion and reason are correct statements but reason is not correct explanation of assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. 2. (i) Assertion: An ordinary filter paper impregnated with colloidian solution stops the flow of colloidal particles. Reason: Pore size of the filter paper becomes more than the size of colloidal particles. 2. (ii).p. Assertion: Colloidal solutions show colligative properties. Reason: Colloidal particles are larger in size. OR 2.(ii).q. Assertion: Colloidal solutions do not show Brownian motion. Reason: Brownian motion is responsible for stability of sols. 2. (iii) Assertion : Lyophillic sols are more stable than lyophobic sols. Reason : Lyophillic sol contains charge on particles whereas no charge is present on the particles of lyophobic sols. 2. (iv)Assertion : Coagulation power of Al3+ is more than Na+. Reason : Greater the valency of the flocculating ion added, greater is its power to cause precipitation ( Hardy Schulze rule). Following questions (Q. No. 3- 11) are multiple choice questions carrying 1 mark each: 3. If limiting molar conductivity of Ca2+ and Cl are 119.0 and 76.3 S cm2 mol-1, then the value of limiting molar conductivity of CaCl2 will be (a) 195.3 S cm2 mol-1 (b) 271.6 S cm2 mol-1 (c) 43.3 S cm2 mol-1 (d) 314.3 S cm2 mol-1. 4.p. On oxidation with a mild oxidising agent like Br2/H20, the glucose is oxidized to (a) saccharic acid (b) glucaric acid (c) gluconic acid (d) valeric acid OR 4.q. The melting points of amino acids are higher than the corresponding halo-acids because (a) amino acids exist as zwitter ions resulting in strong dipole-dipole attraction (b) amino acids are optically active (c) due to higher molecular mass of -NH2 group molecular mass of amino acids is higher (d) they interact with water more than halo-acids and have salt like structure 5. Osmotic pressure of a solution is 0.0821 atm at a temperature of 300 K. The concentration in moles/litre will be (a) 0.33 (b) 0.666 (c) 0.3 10 (d) 3 -2 6.p. Which one of the following characteristics of the transition metals is associated with higher catalytic activity? (a) High enthalpy of atomisation (b) Paramagnetic behaviour (c) Colour of hydrate ions (d) Variable oxidation states OR 6.q.The correct order of E0 M2+/M values with negative sign for the four successive elements Cr, Mn, Fe and Co is (a) Fe > Mn > Cr > Co (b) Cr > Mn > Fe > Co (c) Mn > Cr > Fe > Co (d) Cr > Fe > Mn > Co 7.p. What is the end product in the following sequence of reactions? (a) Aniline (b) Phenol (c) Benzene (d) Benzenediazxonium chloride OR 7.q. Among the following: I. CH3NH2 II. (CH3)2NH III. (CH3)3N IV. C6H5NH2 Which will give the positive carbylamine test? (a) I and II (b) I and IV (c) II and IV (d) II and III 8.p. The correct IUPAC name of the coordination compound K3[Fe(CN)5NO] is (a) Potassium pentacyanonitrosylferrate (II) (b) Potassium pentacyanonitroferrate (II) (c) Potassium nitritopentacyanoferrate (IV) (d) Potassium nitritepentacynanoiron (II) OR 8.q. Correct formulae of tetraaminechloronitroplatinum (IV) sulphate can be written as (a) [Pt(NH3)4 (ONO) Cl]SO4 (b) [Pt(NH3)4Cl2NO2]2 (c) [Pt(NH3)4 (NO2) Cl]SO4 (d) [PtCl(ONO)NH3(SO4)] 9. Which of the following has magnetic moment value of 5.9 BM? (a) Fe2+ (b) Fe3+ (c) Ni2+ (d) Cu2+ 10. Identify X and Y in the follow following sequence (a) X = KCN, Y = LiAlH4 (b) X = KCN, Y = H3O+ (c) X = CH3Cl, Y = AlCl3, HCl (d) X = CH3NH2, Y = HNO2 11. Examine the given defective crystal (a)) It is Frenkel defect and density will increase (b) It is Schottky chottky defect and density will decrease (c)) It is Frenkel defect and density will decrease (d) It is Schottky defect and density will increase In the following questions(Q. No. 12- 16) a statement of assertion followed by a staterment of reason is given. Choose the correct answer out of the following choices. Note : In the following questions a statement of assertion followed by a statement of reason is given. Choose the correct answer out of the following choices. (a) Assertion and reason both are correct and reason is correct explanation of assertion. (b) Both assertion and reason are correct statements but reason is not correct explanation of assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. (e) Both assertion and reason are wrong statements. 12. Assertion(A) : glucose and glucose are anomers of each other. Reason(R) : and glucose differ at C-2 position. 13. Assertion (A): Both rhombic and monoclinic sulphur exist as S8 but oxygen exists as O2. Reason (R): Oxygen forms p -p multiple bond due to small size and small bond length but p -p bonding is not possible in sulphur. 14.p. Assertion : When methyl alcohol is added to water, boiling point of water increases. Reason : When a volatile solute is added to a volatile solvent elevation in boiling point is observed. OR 14. q. Assertion : When NaCl is added to water a depression in freezing point is observed. Reason : The lowering of vapour pressure of a solution causes depression in the freezing point. a 15. Assertion (A): Ortho and para isomers of nitro phenol can not be separated by steam distillation. Reason (R): Both ortho and para isomers of nitro phenol have the similar boiling point. 16. Assertion (A): Aldehydes and ketones, both react with Tollen s reagent to form silver mirror. Reason (R): Both, aldehydes and ketones contain a carbonyl group. SECTION B Tht following questions, Q.No. 17-25 are short answer type and carry 2 marks each. 17. Write the structural formula of A, B, C and D in the following sequence of reaction: OR 17. Chlorobenzene is extremely less reactive towards a nucleophilic substitution reaction. Give two reasons for the same. 18. 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) . 19. Compare the following complexes with respect to structural shapes of units, magnetic behaviour and hybrid orbitals involved in units : [Co(NH3)6]+3, [Cr(NH3)6]3+, [Ni(CO)4] (At. nos. : Co = 27, Cr = 24, Ni = 28) OR 19. Explain the following terms giving a suitable example in each case : (i) Ambident ligand (ii) Crystal field splitting in an octahedral field 20. 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? OR Half-life for a first order reaction 693 s. Calculate the time required for 90% completion of this reaction. 21. (a) For a reaction A + B P, the rate law is given by, r = k[A]1/2 [B]2. What is the order of this reaction? (b) A first order reaction is found to have a rate constant k = 5.5 10-14 s-1. Find the half life of the reaction. 22. Explain the mechanism of acid catalysed hydration of an alkene to form corresponding alcohol. OR Explain the mechanism of following reaction. 23. What happens when: (i) SO2 gas is passed through an aqueous solution of Fe 3+ salt? (ii) XeF4 reacts with SbF5? 24. How will you carry out the following conversions : (i) 2-Bromopropane to 1-bromopropane (ii) Benzene to p-chloronitrobenzene 25. An element with density 11.2 g cm-3 forms a f.c.c. lattice with edge length of 4 10-8 cm. Calculate the atomic mass of the element. (Given : NA = 6.022 1023 mol-1) SECTION C The following questions, Q.No. 26-30 are short answer type II and carry 3 marks each. 26. Explain the following observations : (i) Zn2+ salts are colourless. (ii) Copper has exceptionally positive E0 M2+/M value. (iii) The metallic radii of the third (5d) series of transition elements are virtually the same as those of the corresponding members of the second series. OR 26. (i) Which metal in the first transition series (3d series) exhibits +1 oxidation state most frequency and why? (ii) Which of the following cations are coloured in aqueous solutions and why? SC3+, V3+, Ti4+, Mn2+. (Atomic no. Sc = 21, V = 23, Ti = 22, Mn = 25) 27. Account for the following : (i) Primary amines (R-NH NH2) have higher boiling point than tertiary amines (R3N). (ii) Aniline does not undergo Friedel Crafts reaction. (iii) (CH3)2NH is more basic than (CH3)3N in an aqueous solution. OR 27. (i) Give reasons : (a) Aniline is a weaker base than cyclohexyl amine. (b) It is difficult to prepare pure amines by ammonolysis of alkyl halides. (ii) Arrange the following in increasing order of basic strength. Aniline, p-Nitroaniline Nitroaniline and pp-Toluidine 28. (i) Why does presence of excess of lithium makes LiCl crystals pink? (ii) Analysis shows that nickel oxide has the formula Ni0.98 O1.00. What fractions of nickel exist as Ni2+ and Ni3+ ions? 29. Define the following as related to proteins : (i) Peptide linkage (ii) Primary structure (iii) Denaturation 30. Complete the following chemical reac reaction equations . (a) (b) Draw the structures of the following compounds (i) BrF3 (ii) XeF4 SECTION D The following questions, Q.No. 31-33 are long answer type and carry 5 marks each. 31. Account for the following: (i) Bond angle is NH4+ is higher than that in NH3. (ii) ICl is more reactive than I2. (iii) Bond dissociation energy of F2 is less than that of Cl2. (iv) H2S is more acidic than H2O. (v) SF6 is kinetically inert. OR 31. Give reasons for the following: (i) BiCl3 is more stable than BiCl5. (ii) Bleaching of flowers by Cl2 is permanent while that of SO2 is temporary. (iii) Noble gases have very low boiling points. And out of the noble gases, which noble gas has the lowest temperature. (iv) Sulphur in vapour state exhibits paramagnetism. (v) Complete the following reaction:Cl2 + F2 (excess) 32. (i) The cell in which the following reaction occurs: 2Fe3+ (aq) + 2I (aq) 2Fe2+ (aq) + I2(s) has Eo Cell = 0.236 V at 298 K. Calculate the standard Gibbs energy of the cell reaction. (Given: 1F = 96,500 C mol-1) (ii) Calculate the potential of hydrogen electrode in contact with a solution with pH equal to 10. (iii) There are two electrolytes A and B. The electrolyte A show small increase in molar conductivity with dilution and electrolyte B show large increase in molar conductivity with dilution. So find out the weak and strong electrolyte out of A and B. OR 32. (i) A voltaic cell is set up at 25 C with the following half cells : Ag+ (0.001 M) | Ag and Cu2+ (0.10 M) | Cu What would be the voltage of this cell? (E0cell = 0.46 V) [ Given log 10 =1] -4 -5 -1 (ii) Conductivity of 2.5 10 M methanoic acid is 5.25 10 S cm . Calculate its molar conductivity, degree of dissociation and dissociation constant. Given : 0(H+) = 349.5 Scm2 mol-1 and 0(HCOO ) = 50.5 Scm2 mol-1. 33. (i) An organic compound A which has characteristic odour, on treatment with NaOH forms two compounds B and C . Compound B has the molecular formula C7H8O which on oxidation with CrO3 gives back compound A . Compound C is the sodium salt of the acid. C when heated with soda lime yields an aromatic hydrocarbon D . Deduce the structures of A , B , C and D . (ii) Give reason : Electrophilic substitution in Benzoic acid takes place at meta position. OR 33. Write the reactions involved in the following reactions: (i) Clemmensen reduction (ii) Cannizzaro reaction (iii) Hell-Volhard Zelinsky reaction (iv) Wolff-Kishner reduction (v) Etard reaction CLASS 12( CHEMISTRY) MARKING SCHEME FOR PREBOARD 1(2020-2021) 1.(i) A 1.(ii) C 1.(iii) C 1.(iv) C OR A 2.(i) C 2.(ii) B OR D 2.(iii) C 2.(iv) D 3. B 4. C OR A 5. C 6. D OR C 7. A OR B 8. A OR C 9. B 10. A 11. B 12. C 13. A 14. E OR A 15. E 16. D 17. CH3-CH(Cl)CH3 CH3-CH2-CH2Br (B) CH3-CH2-CH2-Mg-I > CH3-CH=CH2 (A) > > CH3-CH2-CH2I (C) > (D) (1/2 mark for each compound) OR The reasons are: (i) Due to resonance/diagrammatic representation, C Cl bond acquires a partial double bond character. As a result, the C Cl bond in chlorobenzene is shorter and hence stronger. Thus, cleavage of C Cl bond in benzene becomes difficult which makes it less reactive towards nucleophilic substition. (ii) Due to repulsion between nucleophile and electron rich arenes. (1+1) 18. . Given : w2 = 1.25 g, w1 = 99 g Tb = 80.31 80.10 C = 0.21 C Kb = 2.53 C kg mol-1 According to the formula : Tb= Kb X m 1/2 1/2 Tb= Kb X M2 = Substituting these values in the formula, we get M2 = . . . =3162.5/ 20.79 = 152 g mol-1 [Co(NH3)6]+3 Octahedral shape, d2sp3 hybridisation, diamagnetic Formation of [Co(NH2)6]+3 oxidation state of Co is +3. 19. (i) 1 (2) Or 19. (i) Ambidentate ligand : The monodentate ligands with more than one coordinating atoms is known as ambidentate ligand. Monodentate ligands have only one atom capable of binding to a central metal atom or ion. For example, the nitrate ion NO2 can bind to the central metal atom/ion at either the nitrogen atom or one of the oxygen atoms. Example : SCN thiocyanate, NCS isothiocyanate 1 (ii) Crystal field splitting: It is the splitting of the degenerate energy levels due to the presence of ligands. When ligand approaches a transition metal ion, the degenerate d-orbitals split into two sets, one with lower energy and the other with higher energy. This is known as crystal field splitting and the difference between the lower energy set and higher energy set is known as crystal field splitting energy (CFSE) 1 20. For a zero order reaction, 1 1 OR (1mark for calculating K and 1 mark for calculating t) 21. (a) According to the formula : r = k[A]1/2 [B]2 (1mark for each part) 22. Acid catalysed hydration : Alkenes react with water in the presence of acid as catalyst to form alcohols Mechanism : It involves three steps : (i) Protonation of alkene to form carbocation by electrophilic attack of H3O+ (2) OR (2) 23. (i) In this sulphur dioxide acts as a reducing agent and reduces Fe3+ to Fe2+. 2Fe3+ + SO2 + 2H2O 2Fe2+ + SO42- + 4H+ (ii) XeF4 + SbF5 [XeF3]+ [SbF6] 1 1 24. Answer: (1+1) 25. Given : p = 11.2 g cm-3, a = 4 10-8 cm For fee lattice, Z = 4 Using formula, (1/2 mark for value of Z, for formula and 1 mark for calculation) 26. (i) Zn2+ salts are colourless due to absence of unpaired electrons in its ground state and ionic state i.e. Zn 2+ = [Ar] 3d104s04p0 ( 1 mark) (ii) The E0M2+/M for any metal is related to the sum of the enthalpy changes taking place in the following steps : M(g) + aH M(g) ( aH = enthalpy of atomization) M(g) + iH M2+(g) ( iH = ionization of atomization) M2+(g) + aq M2+(g) + hydH ( hydH = hydration atomization) Copper has high enthalpy of atomization (i.e. energy absorbed and low enthalpy of hydration (i.e. energy released). Hence E 0M2+/M for copper is positive. The high energy required to transform Cu(s) to Cu 2+(aq) is not balanced by its hydration enthalpy. (1 mark) (iii) Because of very small energy gap between 5f, 6d and 7s subshells all their electrons can take part in bonding and shows variable oxidation states. ( 1 mark) OR (i) Copper exhibits + 1 oxidation state more frequently i.e., Cu+1 because of its electronic configuration 3d 104s1. It can easily lose 4s1 electron to give stable 3d10 configuration. ( 1 mark) (ii) Sc3+ = 4S0 3d0 = no unpaired electron V3+ = 3d2 4S0 = 2 unpaired electron Ti4+ = 3d0 4s0 = no unpaired electron Mn2+ = 3d5 4s0 = 5 unpaired electron Thus V3+ and Mn2+ are coloured in their aqueous solution due to presence of unpaired electron. ( 2 mark) 27. (i) Due to presence of two H-atoms on N-atom of primary amines, they undergo extensive intermolecular H-bonding while tertiary amines due to the absence of a H-atom on the N-atom, do not undergo Hbonding. As a result, primary amines have higher boiling points than 3 amines. (ii) Aniline being a Lewis base reacts with Lewis acid AlCl 3 to form a salt. As a result, N of aniline acquires positive charge and hence it acts as a strong deactivating group for electrophilic substitution reaction. Consequently, aniline does not undergo Freidel Craft reaction. (iii) Due to more steric hindrance in (CH3)3N it is less basic than (CH3)2NH. (1+1+1) OR Answer: (i) (a) In aniline, the lone pair of electrons on the N-atom is delocalised over the benzene ring. As a result, the electron density on the nitrogen decreases. But in cyclohexylamine, the lone pair of electrons on N-atom is readily available due to absence of -electrons. Hence aniline is weaker base than cyclohexylamine. (b) Because the primary amine formed by ammonolysis itself acts as a nucleophile and produces further 2 and 3 alkyl amine. (ii) p-Nitroaniline < Aniline < p-Toluidine (1+1+1) 28. (i) This is due to metal excess defect due to anionic vacancies in which the anionic sites are occupied by unpaired electrons (F-centres). (1) (ii) 98 Ni-atoms are associated with 100 O atoms. Out of 98, Ni-atoms, suppose Ni present as Ni2+ = x Then Ni present as Ni3+ = 98 x Total charge on x Ni2+ and (98 x) Ni3+ should be equal to charge on 100 O2- ions. Hence, x 2 + (98 x) 3 = 100 2 or 2x + 294 3x = 200 or x = 94 Fraction of Ni present as Ni2+ = (94/98) 100 = 96% Fraction of Ni present as Ni3+ = (4/98) 100 = 4% (2) 29. (i) Peptide linkage : A peptide linkage is an amide linkage formed between COOH group of one a-amino acid and NH2 group of the other a-amino acid by loss of a molecule of water. The-CO-NH-bond formed is called petide linkage. (ii) Primary structure : Proteins may have one or more polypeptide chains. Each polypeptide in a protein has amino acids linked with each other in a specific sequence and it is this sequence of amino acids that is called the primary structure of that protein. (iii) Denaturation : Due to coagulation of globular protein under the influence of change in temperature, change in pH etc., the native shape of the protein is destroyed and biological activity is lost and the formed protein is called denaturated proteins and the phenomenon is denaturation. (1+1+1) 30. (a) (1) (b) (2) 31. (i) Because in NH4+ ion there is no lone pair of electrons which is present in NH3 due to which lone pair pair-bond bond pair repulsion occurs and bond angle decreases from 109 28 109 28 to 107.3 . (ii) Because I-Cl Cl bond is weaker than II-1 1 bond as a result of which ICl breaks easily to form halogen atoms which readily bring about the reaction, hence nce more reactive. (iii) Due to smaller size of F than Cl as a result of which electronelectron electron repulsions between the lone pairs of electrons are very large than that of Cl, hence bond dissociation energy of F 2 is less than that of Cl2. (iv) Since the size of sulphur is more than oxygen, S-H bond length increases and hence bond dissociation energy of S-H is less than O-H. Therefore S-H easily loses H+ and thus is more acidic than H 2O. (v) Because SF6 is showing steric hindrance due to 6 (six) fluorine atoms which make it unable to react further with any other atom. (1+1+1+1+1) OR (i) BiCl3 is more stable than BiCl5 due to inert pair effect because as we move down the group, the stability of +3 oxidation state increases and of +5 decreases. (ii) Chlorine bleaches the material by oxidation hence it is permanent while SO2 bleaches the material by reduction and as the material is exposed to air, it gets oxidised and the colour is restored, hence it is temporary. (iii) Due to presence of weak Van der waal forces of attraction, noble gases have very low boiling point. Helium has the lowest boiling point (4.2K). (iv) Because sulphur in vapour state has two unpaired electrons in the antibonding * orbitals like O2. (v) Cl2 + 3F2 (excess) 2ClF3 32. (i) 2Fe3+ (aq) + 2I (aq) 2Fe2+ (aq) + I2 (s) For the given reaction, n = 2, E = 0.236 V (1+1+1+1+1) Using formula G = -nF E cell = -2 96,500 C mol-1 0.236 V G =-45.55 kJ mol-1 (2) (ii) For hydrogen electrode, ctrode, H+ + e 1/2H2 Applying Nernst equation, (2) (iii)) A is a strong electrolyte and B is a weak electrolyte. (1) OR (i)) The reaction takes place in cell as . = 0.46 x5 = 0.46 0.1475 = 0.3125 (2.5) (ii) Concentration is 2.5 x 10-4 M K = 5.25 10-5 Scm-1. cm=K Dissociation constant = C 2/(1- ) = 2.5 X 10-4X (0.525)2/(1-0.525) = 1.45 X 10 -4 mol L-1 (2.5) 33. (i) (A) gives characteristic odour which on treatment with NaOH and forms two compounds B and C. (4) (ii) The benzene ring of benzoic acid undergoes electrophilic substitution reaction such as nitration, sulphonation etc. Since the COOH group in benzene is an electron withdrawing group, therefore it is meta directing group. (1) OR (i) Clemmensen reduction. The carbonyl group of aldehyde and ketones is reduced to CH2 group on treatment with zinc amalgam and concentrated hydrochloric acid. (ii) Cannizzaro reaction. Aldehydes, which do not have an -hydrogen atom undergo self oxidation and reduction on treatment with cone, alkali and produce alcohol and carboxylic acid salt. (iii) RCH2COOH () , ( ) > RCH(X)COOH (X= Cl, Br) (iv) Wolff-Kishner reduction (v) Etard reaction (1+1+1+1+1)

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