CET 20 - 21 CHEMISTRY DETAILED SOLUTIONS.pdf - CREATIVE P.U COLLEGE

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CREATIVE LEARNING CLASSES, KARKALA K-CET DETAILED SOLUTIONS - 2021 DEPARTMENT OF CHEMISTRY SET – C3 1. For the reaction A(g) + B(g) C(g) + D(g); ∆H = - Q KJ The equilibrium constant cannot be disturbed by (A) Addition of A (B) Addition of D (C) Increasing of pressure (D) Increasing of temperature Ans : C Since number of moles of gaseous products and reactants remain same, change in presence cannot disturb the equilibrium. 2. An organic compound 'X' on treatment with PCC in dichloromethane gives the compound Y. Compound ‘Y' reacts with I2 and alkali to form yellow precipitate of triiodomethane. The compound X is (A) CH3CHO (B) CH3COCH3 (C) CH3CH2OH (D) CH3COOH Ans : C PCC I2 / Alkali CH 3 - CH 2 - OH CH 3CHO CHI 3 (X) (Y) Yellow ppt 3. A compound ‘A’ (C7H8O) is insoluble in NaHCO3 solution but dissolve in NaOH and gives a characteristic colour with neutral FeCl3 solution. When treated with Bromine water compound ‘A’ forms the compound B with the formula C7H5OBr3. ‘A’ is OH OH OH OH A) B) C) D) CH3 CH3 CH3 Ans : B Compound A (molecular Formula C7H8O) is insoluble in water and dil. NaHCO3, but dissolves in dil. NaOH and gives colour with aq. FeCl3. Hence it is phenol. On treatment with Br2 it readily gives precipitate of C7H5OBr3. Therefore, compound is m – cresol. HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 4. In set of reactions, Identify D CH3COOH  SOCl  A  2Benzene B  HCN  C  H O D 2 Anh AlCl3 COOH CN CH3 OH CH3 A) OH B) OH COOH CH3 CN CH3 C) OH D) Ans : D CN COOH COCH 3 H3C OH H3C OH SOCl 2 C6 H6 HCN H2 O CH 3COOH CH 3COCl (A) Anh. AlCl 3 (B) (C) (D) 5. Ka values for acids H2SO3, HNO2, CH3COOH and HCN are respectively 1.3 x 10-2, 4 x 10-4, 1.8 x 10-5 and 4 x 10-10, which of the above acids produces stronger conjugate base in aqueous solution? (A) H2SO3 (B) HNO2 (C) CH3COOH (D) HCN Ans : D As Ka decreases the acidic character also decreases and its conjugate base become more stronger. 6. HgSO 4/ H2SO4 PCC Acetaldehyde (i) SnCl 2 / HCl + (ii) H3O A, B and C respectively are (A) ethanol, ethane nitrile and ethyne (B) ethane nitrile, ethanol and ethyne (C) ethyne, ethanol and ethane nitrile (D) ethyne, ethane nitrile and ethanol HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA Ans : C HgSO 4/H2SO4 HC CH PCC CH 3 - CH 2 - OH CH 3CHO SnCl 2 CH 3CN + H3O 7. The reagent which can do the conversion CH3COOH  CH3 – CH2 – OH is (A) LiAlH4 / ether (B) H2, Pt (C) NaBH4 (D) Na and C2H5OH Ans : A LAH/Ether CH3COOH CH3CH2OH 3 (i) CH MgBr 2 6 C 2 4  B  Conc H SO (i) B H 8. CH3CHO   A  (ii) H3O+ Δ (ii) H2O, OH- A and C are (A) Identical (B) Position isomers (C) Functional isomers (D) Optical isomers Ans : B OH B 2H6 CH 3MgBr Conc. H 2SO4 CH 3CHO H3C H CH 2 = CH - CH 3 - CH 3CH 2CH 2OH + H2O, OH H3 O CH3 (A) (B) (C) Hence (A) and (C) are positional isomers. 9. Which of the following is not true for oxidation? (A) addition of oxygen (B) addition of electronegative element (C) removal of hydrogen (D) removal of electronegative element Ans : D Removal of electromagnetic element is reduction. 10. Which is the most suitable reagent for the following conversion? O O || || CH3 - CH = CH - CH2 - C - CH3   CH3 - CH = CH - CH 2 - C- OH (A) Tollen's reagent (B) Benzoyl peroxide (C) I2 and NaOH solution with subsequent acidification (D) Sn and NaOH solution Ans : C + I2 / NaOH / H CH3 - CH = CH - CH2 - CO - CH3 CH3 - CH = CH - CH2 - COOH HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA alc.NH 3  A  3 B 2CH Cl 11. C6H5CH2Cl  The product B is (A) N, N- Dimethyl phenyl methanamine (B) N, N- Dimethyl benzenamine (C) N- Benzyl – N – methyl methanamine (D) phenyl - N, N - dimethyl methanamine Ans : A Alc NH 3 2CH 3Cl C6H5CH 2Cl C6H5CH 2NH 2 C6H5 - CH 2 - N (CH 3)2 12. The method by which aniline cannot be prepared is (A) Nitration of benzene followed by reduction with Sn and con HCI (B) Degradation of benzamide with bromine in alkaline solution (C) Reduction of nitrobenzene with H2 / Pd is ethanol (D) Potassium salt of phthalimide treated with chlorobenzene followed by the hydrolysis with aqueous NaOH solution Ans : D Aniline cannot be prepared by this method because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide. 13. Permanent hardness cannot be removed by (A) Using washing soda (B) Calgon's method (C) Clark's method (D) lon exchange method Ans : C Clark’s method is used to remove temporary hardness. 14. A hydrocarbon A(C4H8) on reaction with HCI gives a compound B(C4H9Cl) which on reaction with 1 mol of NH3 gives compound C(C4H10N). On reacting with NaNO2 and HCl followed by treatment with water, compound C yields an optically active compound D. The D is CH 2 -CH3 CH 2 -CH3 | | (A) CH3  C H (B) CH3 - C -H | | Cl OH CH 2 -CH3 CH 2 -CH3 | | (C) CH3 - C -H (D) CH3 - C -H | | NH 2 H Ans : B HCl 1mol NH3 CH3 - CH = CH - CH3 CH3 - CH(Cl) - CH2 - CH3 CH3 - CH(NH2) - CH2 - CH3 (A) (B) (C) NaNO 2 / HCl CH3 - CH(OH) CH2 - CH3 H2 O (D) Optically Active HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 15. RNA and DNA are chiral molecules, their chirality is due to the presence of (A) D-Sugar component (B) L-Sugar component (C) Chiral bases (D) Chiral phosphate ester unit Ans : A Deoxy ribose and ribose sugars are D-chiral sugars in DNA and RNA. 16. The property of the alkaline earth metals that increases with their atomic number is (A) Ionisation enthalpy (B) Electronegativity (C) Solubility of their hydroxide in water (D) Solubility of their sulphate in water Ans : C Due to decrease in lattice energy of the hydroxide salt and size of cation increases down the group. 17. Primary structure in a nucleic acid chain contains bases as G A T G C……The chain which is complementary to this chain is (A) G G T G A….. (B) T G A A G….. (C) C T A C G….. (D) T T T A G….. Ans : C G A T G C      C T A C G 18. In the detection of II group acid radical, the salt containing chloride is treated with concentrated Sulphuric acid, the colourless gas is liberated. The name of the gas is (A) Hydrogen chloride gas (B) Chlorine gas (C) Sulphur dioxide gas (D) Hydrogen gas Ans : A In II group acid radical Cl2 gas liberated reacts with conc. H2SO4 give HCl gas which is colourless. 19. The number of six membered and five membered rings in Buckminster Fullerene respectively Is (A) 20,12 (B) 12, 20 (C) 14, 18 (D) 14, 11 Ans: A In Buck minster fullerene there are 26 membered and 12 5-membered rings. 20. In Chrysoberyl, a compound containing Beryllium, Aluminium and oxygen, oxide ions form cubic close packed structure. Aluminium ions occupy 1/4th of tetrahedral voids and Beryllium ions occupy 1/4th of octahedral voids. The formula of the compound is (A) BeAlO4 (B) BeAl2O4 (C) Be2AlO2 (D) BeAlO2 Ans : B Be is at octahedral void. Therefore Be 1 4 Al is at tetrahedral voids HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA ∴ Al 1 ×2 =Al 1 4 2 Oxide is at CCP   Be 1 Al 1 O1   4  4 2  = Be Al2 O4 21. The correct statement regarding defects in solids is (A) Frenkel defect is a vacancy defect (B) Schottky defect is a dislocation defect (C) Trapping of an electron in the lattices leads to the formation of F-centre (D) Schottky defect has no effect on density. Ans : C Trapping of an electron in the anionic vacancy leads to the formation of F – center 22. A metal crystallises in BCC lattice with unit cell edge length of 300 pm and density 6.15 gcm-3. The molar mass of the metal is (A) 50 gmol-1 (B) 60 gmol-1 (C) 40 gmol-1 (D) 70 gmol-1 Ans: A  × a 3× NA M= z 6.15  (3 108 )3  6.022 1023  2 = 49.99gmol  50 gmol-1 -1 23. Henry's law constant for the solubility of N2 gas in water at 298 K is 1.0 x 105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure is (A) 4.0 x 10-4 (B) 4.0 x 10-5 (C) 5.0 x 10-4 (D) 4.0 x 10-6. Ans : A PN = Ptotal ×  N 2 2  5  0.8 = 4 atm W.K.T, P = KH  n N2 P = KH n H2O n N2 4 =1×105 × 10 5 nN2  4 10 10 = 4×10-4 mol 24. A pure compound contains 2.4 g of C, 1.2 x 1023 atoms of H, 0.2 moles of oxygen atoms. Its empirical formula is (A) C2HO (B) C2H2O2 (C) CH2O (D) CHO HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA Ans : D 2.4 C= = 0.2 mol 12 O = 0.2mol 1.2×1023 H = = 0.2 mol 6.022×1023 ∴ By taking simple ratio of numbers of moles of elements C H O 0.2 0.2 0.2 0.2 0.2 0.2 C1H1O1 = CHO 25. Choose the correct statement (A) KH value is same for a gas in any solvent (B) Higher the KH value more the solubility of gas (C) KH value increases on increasing the temperature of the solution (D) Easily liquefiable gases usually has lesser KH values Ans : C Henry’s constant is temperature dependent. 26. The KH value (K bar) of Argon (l), Carbondioxide (II) formuldehyde (III) and methane (IV) are respectively 40.3, 1.67, 1.83 × 10-5 and 0.413 at 298 K. The increasing order of solubility of gas in liquid is (A) I < II < IV < III (B) III < IV < II < I (C) I < III < II < IV (D) I < IV < II < III Ans : A The relationship between KH and solubility is According to Henry’s law P =K χ B H B  For dilute solution n A >>> n B  n PB =K H . B nA PB .n A Solubility = n B = KH 1 i.e solubility  KH ∴ (i) Argon KH = 40.3 (ii) carbon dioxide = 1.67 (iii) Formaldehyde = 1.83 x 10-5 (iv) Methane = 0.413 ∴ III > IV > II > I HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 27. The vapour pressure of pure liquids A and B are 450 and 700 mm of Hg at 350 K respectively. If the total vapour pressure of the mixture is 600 mm of Hg, the composition of the mixture in the solution is (A) xA = 0.4, xB = 0.6 (B) xA = 0.6, xB = 0.4 (C) xA= 0.3, xB = 0.7 (D) xA = 0.7, xB = 0.3 Ans : C Given PA0  450 MM of Hg PB0  700 MM of Hg PTotal  600 MM of Hg According to Raoult’s law, PTotal = (PA0 - PB0 ) χ A + PB0 600  (450  700)χ A  700 250χ A = 100 100 A   0.4 250 χ B =1 - χ A = 0.6 PA = PAo .χ A = 450 × 0.4 = 180 mm PB = PBo .χ B = 700 × 0.6 = 420 mm 180 Mole fraction of liquid A =  0.3 180  420 ∴ Mole fraction of liquid B = 1 – 0.3 = 0.7 28. Consider the following electrodes P  Zn2 (0.0001 M) / Zn Q  Zn2 (0.1 M) / Zn R  Zn2 (0.01 M) / Zn S  Zn2 (0.001 M) / Zn Eo Zn / Zn 2  0.76 V Electrode potentials of the above electrodes in volts are in the order (A) P  S  R  Q (B) S  R  Q  P (C) Q  R  S  P (D) P  Q  R  S Ans : C For concentration cells electrode potential ∝ concentration of solution 29. The number of angular and radial nodes in 3p orbital respectively are (A) 3,1 (B) 1,1 (C) 2,1 (D) 2,3 Ans : B Number of angular nodes for 3p = 1 Radial nodes = 0 – l – 1 =3–1–1 =1 ∴ 1, 1 HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 30. The resistance of 0.01 m KCI solution at 298 K is 1500  . If the conductivity of 0.01m KCl solution at 298 K is 0.146 103 S cm1. The cell constant of the conductivity cell in cm1 is (A) 0.219 (B) 0.291 (C) 0.301 (D) 0.194 Ans: A G = K. R = 0.146 x 10-3 x 1500 = 219 x 10-3 = 0.219 cm-1 31. H2(g)  2AgCI(s) 2Ag(s)  2HCI(aq) E0cell at 250C for the cell is 0.22 V. The equilibrium constant at 25o C is (A) 2.8 107 (B) 5.2 108 (C) 2.8 105 (D) 5.2 104 Ans : A H 2(g) + 2AgCl 2Ag(s) + 2HCl(aq) For the above reaction n = 2 nFE o log K C = 2.303 RT 2 × 96500 × 0.22 = 2.303 × 8.314 × 298 log K C = 7.44 KC = Antilog 7.44 = 2.8 x 107 32. For a reaction A+2B  Products, when concentration of B alone is increased half life remains the same. If concentration of A alone is doubled, rate remains the same. The unit of rate constant for the reaction is (A) S1 (B) L mol1S1 (C) mol L1S1 (D) atm1 Ans : A R = k[A]0 [B]1 R mol L-1×S-1 -1 k= 1= =S [B] mol L-1 33. The third ionisation enthalpy is highest in (A) Alkali metals (B) Alkaline earth metals (C) Chalcogens (D) Pnictogens Ans : B Alkaline earth metals : GEC (Noble gas) ns2 On removing 2 electrons it attains stability. 34. If the rate constant for a first order reaction is k, the time (t) required for the completion of 99% of the reaction is given by 4.606 2.303 0.693 6.909 (A) t  (B) t  (C) t  (D) t  k k k k HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA Ans : A 2.303 100 t= log k 1 4.606 t= k 35. The rate of a gaseous reaction is given by the expression k [A][B]2 . If the volume of vessel is reduced to one half of the initial volume, the reaction rate as compared to original rate is 1 1 (A) (B) (C) 8 (D) 16 16 8 Ans : C R2 = K [2A] [2B]2 R1 = K [A] [B]2 on comparing we get R2 = 8R1 36. The correct IUPAC name of …………. (A) 4-Ethyl-1-Fluoro-2-nitrobenzene (B) 1-Ethyl-4-Fluoro-3-nitrobenzene (C) 3-Ethyl-6-Fluoronitrobenzene (D) 5-Ethyl-2-Fluoronitrobenzene Ans : D NO 2 F CH3 5 - ethyl – 2 – fluoronitrobenzene 37. Higher order (>3) reactions are rare due to (A) Shifting of equilibrium towards reactants due to elastic collisions (B) Loss of active species on collision (C) Low probability of simultaneous collision of all reacting species (D) Increase in entropy as more molecules are involved Ans : C 38. Arrange benzene-hexane and ethyne in decreasing order of their acidic behavior (A) Benzene > n-hexane > ethyne (B) n-hexane > Benzene > ethyne (C) ethyne > n-hexane > Benzene (D) ethyne > Benzene > n-hexane Ans : D More the % of S-charcter of C more will be acidic nature. ∴ Ethyne > Benzene > n – hexane sp sp2 sp3 39. A colloidal solution is subjected to an electric field than colloidal particles more towards anode. The amount of electrolytes of BaCl2, AlCl3 and NaCl required to coagulate the given colloid is in the order (A) NaCl > BaCl2 > AlCl3 (B) BaCl2 > AlCl3 > NaCl (C) AlCl3 = NaCl = BaCl2 (D) AlCl3 > BaCl2 > NaCl HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA Ans : A Greater the valence of the flocculating ion added, greater is its power to cause coagulation. And coagulatory power follows order AlCl3 > BaCl2 > NaCl Higher the coagulating power smarter the quantity is need. ∴ NaCl > BaCl2 > AlCl3 40. Which of the following is an incorrect statement? (A) Hydrogen bonding is stronger than dispersion forces (B) Sigma bonds are stronger than π – bonds (C) Ionic bonding is non-directional (D) σ- electrons are referred to as mobile electrons Ans : D 41. Zeta potential is (A) Potential required to bring about coagulation of a colloidal sol. (B) Potential required to give the particle a speed of 1 cm S-1 (C) Potential difference between fixed charged layer and the diffused layer having opposite charges (D) Potential energy of the colloidal particles. Ans : C 42. Which of the following compound on heating gives N2O? (A) Pb(NO3)2 (B)NH4NO3 (3) NH4NO2 (D) NaNO3 Ans : B NH4 NO3  Δ  N2O + 2H2O 43. Which of the following property is true for the given sequence NH 3> PH3> AsH3> SbH3> BiH3? (A) Reducing property (B) Thermal stability (C) Bond angle (D) Acidic character Ans : B & C As size of central atom increases thermal stability decreases down the group. Bond angle decreases down the group due to decrease in bond pair bond pair repulsion. 44. The correct order of boiling point in the following compound is (A) HF > H2O > NH3 (B) H2O > HF > NH3 (C) NH3> H2O > HF (D) NH3> HF > H2O Ans : B H2O > HF > NH3 Move the electronegativity more will be boiling point and more the number of H-bonds per molecule more will be boiling point. H2O has surrounding 4 H2O molecules but HF has two molecules. 45. XeF6 on partial hydrolysis gives a compound X, which has square pyramidal geometry ‘X’ is (A) XeO3 (B) XeO4 (C) XeOF4 (D) XeO2F2 HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA Ans : C XeF6  H2 O  XeOF4 46. A colorless, neutral, paramagnetic oxide of Nitrogen ‘P’ on oxidation gives reddish brown gas Q, Q on cooling gives colorless gas R, R on reaction with P gives blue solid S. Identify P, Q, R, S respectively. (A) N2O NO NO2 N2O5 (B) N2O NO2 N2O4 N2O3 (C) NO NO2 N2O4 N2O3 (D) NO NO N2O4 N2O5 Ans : C Oxidation Cooling NO NO  NO   N O N O 2 2 4 2 3 47. Which of the following does not represent property stated against it? (A) CO+2 < Fe+2 < Mn+2 - Ionic size (B) Ti < V < Mn – Number of oxidation states (C) Cr+2 < Mn+2 < Fe+2 – Paramagnetic behavior (D) Sc > Cr > Fe – Density Ans : D 48. Which one of the following is correct for all elements from Sc to Cu? (A) The lowest oxidation state shown by them is +2 (B) 4s orbital is completely filled in the ground state. (C) 3d orbital is not completely filled in the ground state (D) The ions in +2 oxidation states are paramagnetic Ans : D Ions in +2 oxidation state have unpaired electrons hence they are paramagnetic. 49. When the absolute temperature of ideal gas is doubled and pressure is halved, the volume of gas (A) Will be half of original volume (B) Will be 4 times the original volume (C) Will be 2 times the original volume (D) Will be 1/4th times the original volume Ans : B P1V1 P2 V2 = T1 T2 P 2T V2 =V1× 1 × 1 = 4V1 0.5P1 T1 V2 =4 V1 50. Which of the following pairs has both the ions colored in aqueous solution? (Atomic numbers of Sc = 21, Ti = 22, Ni = 28, Cu = 29, Mn = 25) (A) Sc3+, Mn2+ (B) Ni2+, Ti4+ (C) Ti3+, Cu+ (D) Mn2+, Ti3+ Ans : D Mn2+ and Ti+3 contains unpaired electrons. 3d5 3d1. These are coloured. HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 51. For the crystal field splitting in octahedral complexes, (A) The energy of the eg orbitals will decrease by (3/5) ∆o and that of the t2g will increase by (2/5) ∆o (B) The energy of the eg orbitals will increase by (3/5) ∆o and that of the t2g will decrease by (2/5) ∆o (C) The energy of the eg orbitals will increase by (3/5) ∆o and that of the t2g will increase by (2/5) ∆o (D) The energy of the eg orbitals will decrease by (3/5) ∆o and that of the t2g will decrease by (2/5) ∆o Ans : B 52. Peroxide effect is observed with the addition of HBr but not with the addition of HI to unsymmetrical alkene because (A) H-I bond is stronger that H-Br is not cleaved by the free radical (B) H-I bond is weaker than H-Br bond so that Iodine free radicals combine to form Iodine molecules (C) Bond strength of HI and HBr are same but free radicals are formed in HBr (D) All of the these Ans : B H – Br undergoes homolysis to form free radical. But Iodine free radicals have greater tendency to combine. • • • R - O - O - R  homolysis  2R - O H - I  Peroxide H+I • • • R - O + H - Br   R - O - H + Br I + I   I2 53. The IUPAC name of Co  NH3 5  CO3  Cl is (A) Pentaamminecarbonatocobalt (III)Chloride (B) Carbonatopentamminecobalt (III) Chloride (C) Pentaamminecarbonatocobaltate (III) Chloride (D) Pentaammine cobalt (III) Carbonate Chloride Ans : A Pentaammine carbonato cobalt (III) chloride 54. Homoieptic complexes among the following are   A  K 3  Al  C2O4 3  ,  B  CoCl2  en 2   C  K 2  Zn  OH 4  (A) A only (B) (A) and (B) only (C) (A) and (C) only (D) (C) only Ans : C K3[Al(C2O4)3] and K2[Zn(OH)4] In these complexes central metal surrounded by one type of ligands. HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 55. The correct order for wavelengths of light observed in the complex ions CoCl  NH3 5  , Co  NH3 6  and Co  CN 6  2+ 3+ 3- is (A) CoCl  NH3 5  > Co  NH3 6  > Co  CN 6  2+ 3+ 3- (B) Co  NH3 6  > Co  CN 6  > CoCl  NH3 5  3+ 3- 2+ (C) Co  CN 6  > CoCl  NH3 5  > Co  CN 6  3- 2+ 3- (D) Co  NH3 6  > CoCl  NH3 5  > Co  CN 6  3+ 3- Ans : A [CoCl(NH3)5]2+ > [Co(NH3)6]3+ > [Co(CN)6]3- In [CoCl(NH3)5] less CFS and [Co(CN)6]3- shows maximum CFS. ∴ Wavelength absorbed by [Co(CN)6]3- is less and [CoCl(NH3)5]+2 is more. 56. CH3 Br2 A UV light O 2N The compound A (major product) is Br CH3 CH3 A) B) O 2N O 2N CH3 CH3 Br C) D) O 2N Br O 2N Ans : B Br CH3 CH3 Br2 / UV light O 2N O 2N 57. Bond enthalpies of A2, B2 and AB are in the ratio 2 : 1 : 2. If bond enthalpy of formation of AB is -100 KJ mol-1. The bond enthalpy of B2 is (A) 100KJ mol-1 (B) 50KJ mol-1 (C) 200 KJ mol-1 (D) 150KJ mol-1 Ans : C A2 + B2 ⟶ 2AB Let B.E of A2 = 2x, B2 = x, AB = 2x ∆H = (A2 + B2) – 2AB = 2x + x – 2(2x) -100 x 2 = 3x – 4x -200 = -x x = 200 kJ mol-1 HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA 58. The order of reactivity of the compounds C6H5CH2Br, C6H5CH(C6H5)Br, C6H5CH(CH3)Br and C6H5C(CH3)(C6H5)Br in SN 2 reaction is CH3 H H H A) H5C6 Br < H5C6 Br < H5C6 Br < H5C6 Br C6H5 C6H5 CH3 H H H H CH3 B) H5C6 Br < H5C6 Br < H5C6 Br < H5C6 Br H CH3 C6H5 C6H5 H H H CH3 C) H5C6 Br < H5C6 Br < H5C6 Br < H5C6 Br CH3 H C6H5 C6H5 H H H CH3 < < H5C6 Br < D) H5C6 Br H5C6 Br H5C6 Br C6H5 H CH3 C6H5 Ans : A For SN 2 reduction reactivity order of alkyl halides is 1° > 2° > 3° HBr 59. The major product of the following reactions is CH2 = CH - CH2 - OH  product Excess (A) CH3 – CHBr - CH2Br (B) CH2 = CH - CH2Br (C) CH3 – CHBr - CH2 - OH (D) CH3 – CHOH - CH2OH Ans : A HBr (Excess) CH2 = CH - CH2 - OH CH3 - CH(Br) - CH2 - Br 60. CH3 CH3 CH3 H3C C - O - OH + H + O2 A+B H2O The product ‘A’ gives white precipitate when treated with bromine water. The product ‘B’ is treated with Barium hydroxide to give the product C. The compound C is heated strongly form product D. The product D is (A) 4-Methylpent-3-en-2-one (B) But -2-enal (C) 3-Methylpent -3-en-2-one (D) 2-Methylbut-2-enal Ans : A HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA OH + CH 3COCH 3 (A) (B) OH OH Br Br + 3Br2 + 3HBr (A) Br OH Ba(OH) 2 heat CH 3COCH 3 H3C CH 2COCH 3 H3C CH - CO - CH 3 (B) CH3 CH3 (C) (D) The Product (D) is 4 –methyl pent – 3 – en – 2 – one Department of Chemistry  Mr. Adarsha M. K  Mr. Vishwanath Math  Mr. Yogisha Vasishta  Mr. Nagaraja H CREATIVE EDUCATION FOUNDATION MOODBIDRI (R), KARKALA Website : www.creativeedu.in Phone No. : +91 9019844492 HKS P. U COLLEGE, HASSAN CREATIVE P. U COLLEGE, KARKALA