Chemistry - 12 Sample Paper 1

SAMPLE Paper 1 [Easy Concept] Time Allowed : 3 hrs.] [Max. Marks : 70 General Instructions: (a) There are 33 questions in this question paper. All questions are compulsory. (b) Section A: Q. No. 1 to 2 are case-based questions having four MCQs or Reason Assertion type based on given passage each carrying 1 mark. (c) Section A: Question 3 to 16 are MCQs and Reason Assertion type questions carrying 1 mark each (d) Section B: Q. No. 17 to 25 are short answer questions and carry 2 marks each. (e) Section C: Q. No. 26 to 30 are short answer questions and carry 3 marks each. (f) Section D: Q. No. 31 to 33 are long answer questions carrying 5 marks each. (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: [1 × 4 = 4] The evaporation of salt in water will be difficult due to interaction between salt ions and water molecules, therefore boiling temperature increases. Salt ions attract water molecules more strongly preventing water molecules from gathering, thus causes freezing of water. According to equation NaCl + H2O → NaOH + HCl, bonds occurring between OH– and Na+/H+ and Cl– in water cause freezing point depression and boiling point elevation. Salt ions prevent water molecules from being in an order by entering between them. As a result of this, freezing temperature decreases. In the case of boiling, as some of the heat given is spent for removing water molecules which hydrate salt ions, boiling temperature increases. (Reference: This journal is @ The Royal society of chemistry 2009, chem.educ. Res Prac. 2009, 10 273-280, 275.) The following questions are multiple choice questions. Choose the most appropriate answer: (i) Molal elevation constant Kb for water is 0.52 K/m. 0.2 molal solution of glucose will boil at (a) 100.52° C (b) 100.052° C (c) 101.04° C (d) 100.104° C (ii) Which of the following is/are colligative property? (a) Elevation in boiling point (b) Depression in freezing point (c) Osmotic pressure (d) All of these (iii) 45 g of ethylene glycol (C2H6O2) is mixed with 600 g of water. The depression in freezing point is equal to [Kf = 1.86 K/m, Molar mass of C2H6O2 = 62 g/mol] (a) 1.2 K (b) 2.2 K (c) 4.4 K (d) 0.6 K 50 Together with®  EAD Chemistry—12

(iv) Which of the following has highest boiling point? (a) 0.1 m urea solution in water (b) 0.2 m solution of sucrose in water (c) 0.3 m solution of fructose in water (d) 0.5 m solution of glucose in water Or Which of the following has highest freezing point? (a) 0.1 m urea (b) 0.2 m sucrose (c) 0.3 m fructose (d) 0.5 m glucose 2. Read the passage given below and answer the following questions: [1 × 4 = 4] Metal catalysed oxidation of organic substrate is gaining importance as viable alternative to metal promoted stoichiometric oxidations. We have recently demonstrated that cobalt(II) Schiff's base complexes act as effective catalyst in oxidising olefinic substrates in presence of O2 and an aldehyde. These catalysis are however only effective during oxidation of double bond or allylic oxidation. Co(II) porphyrin complex is extremely versatile in oxidising wide range of organic substrates by employing a combination of 2-methyl propanal and dioxygen. (Copyright @ 1996, Published by Elsevier Science Ltd printed in Great Britain, Ajay K.Mandal, Vibha Khanna, Javed Iqbal, Department of chemistry, IIT Kanpur, 208016, India)[P. 3769] In these 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 statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. (i) Assertion: Acetone on reaction with I2 and NaOH gives CHI3 and CH3COONa. Reason: Acetone is difficult to oxidise as compared to acetaldehyde. (ii) Assertion: CH3—CH==CH—CHO gets oxidised to CH3—CH==CH—COOH by Tollens’ reagent. Reason: Tollens’ reagent is mild oxidising agent, does not affect olefinic double bond. (iii) Assertion: PCC is used to convert allyl alcohol to propenal. Reason: Allyl alcohol is CH2==CHOH. (iv) Assertion: p-fluoro toluene oxidised to p-fluoro benzaldehyde by CrO3 in presence of acetic anhydride, followed by hydrolysis. Reason: (i) CrO2Cl2/CS2, (ii) H2O/H+ can also be used to convert toluene to benzaldehyde. Or Assertion: Isobutyl alcohol on oxidation with Cu/573 K gives 2-methyl propanal. Reason: Cu/573 K is strong oxidising agent: Following questions (No. 3–11) are multiple choice questions carrying 1 mark each: 3. Which of the following will have highest Λ° (limiting molar conductivity)? (a) HCl (b) CH3COOH (c) NaCl (d) CH3COONa Sample Papers 51

4. Which parts of amino acid molecules are linked through hydrogen bonds in the secondary structure of proteins? (a) NH2 group (b) —COOH group (c) —C— and —NH—groups (d) None of the above O Or Which of the following is not reducing sugar? (a) Glucose (b) Sucrose (c) Lactose (d) Maltose 5. 98% mHo2Sl–O1)4 by mass has density 1.84 g cm–3. The molarity of solution is (Molar mass of H2SO4 is 98 g (a) 18.4 M (b) 36.8 M (c) 9.2 M (d) 4.6 M 6. Which of the following is considered as transition metal?  (a) Zn (b) Cd (c) Hg (d) Sc Or Which of the following is most paramagnetic? (a) Cr3+ (b) Mn2+ (c) Fe2+ (d) Ni2+ [Atomic numbers Cr = 24, Mn = 25, Fe = 26, Ni = 28] 7. CH3CH2C≡≡N  partialOhHyd–rolysis→ ‘A’ NaOH+Br2→ ‘B’ ‘A’ and ‘B’ respectively are OO (a) CH3CH2—C—NH2, CH3CH2CH2NH2 (b) CH3—CH2—C—NH2, CH3CH2NH2 OO (c) CH3—CH2—C—OH, CH3CH2OH (d) CH3—C—NH2, CH3NH2 Or Aniline reacts with Br2(aq) to give (a) o-Bromoaniline (b) p-bromoaniline (c) m-bromoaniline (d) 2,4,6-tribromoaniline 8. EDTA is a (a) monodentate ligand (b) bidentate ligand (c) ambidentate ligand (d) hexadentate ligand Or The formula of potassium tri- oxalato chromate (III) is (a) K2[Cr(C2O4)3] (b) K[Cr(C2O4)3] (c) K3[Cr(C2O4)3] (d) K4[Cr(C2O4)3] 52 Together with®  EAD Chemistry—12

9. The atomic radius of Ag is closest to (a) Au (b) Ni (c) Hg (d) Cu 10. Which of the following will undergo SN1 mechanism at faster rate? (a) CH3Cl (b) CH3—CH—CH3 CH3 Cl (c) CH3—C—CH3 (d) CH3—CH—CH2—CH3 Cl Cl 11. Which of the following is molecular solid? (a) Graphite (b) Silicon carbide (c) Iodine (d) Silicon In the following questions (Q. No. 12 – 16), 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 statements and reason is correct explanation for assertion. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (c) Assertion is correct statement but reason is wrong statement. (d) Assertion is wrong statement but reason is correct statement. 12. Assertion: Two strands in double strand helix structure of DNA are complementary to each other. Reason: Hydrogen bonds are formed between specific pair of bases. 13. Assertion: The acidic strength of hydrides of group 16 varies in the order: H2O < H2S < H2Se < H2Te Reason: The bond dissociation enthalpy of hydrides of group 16 decreases in the order. H2O > H2S > H2Se > H2Te 14. Assertion: CH3CH2OH is more acidic than CH3OH. Reason: CH3O– is more stable than C2H5O. 15. Assertion: p-nitro chlorobenzene undergo nucleophilic substitution reaction faster than chlorobenzene. Reason: —NO2 group is electron releasing group. 16. Assertion: Aquatic species are more comfortable in warm water. Reason: Different gases have different KH at the same temperature. Or Assertion: Water and HCl form maximum boiling azeotropes. Reason: Azeotropes are constant boiling mixtures which distill out unchanged in their composition. SECTION B The following questions, Q. No 17–25 are short answer type and carry 2 marks each. 17. With the help of resonating structures, explain the effect of presence of nitro group at p-position in chlorobenzene. Sample Papers 53

Or Carry out the following conversions in not more than 2 steps. (i) Aniline to bromobenzene (ii) 1-bromopropane to 2-bromopropane. 18. Explain the following: (i) Animal hide is soaked in tannin. (ii) Adsorption of moisture by leather is physisorption. (iii) Sugar is decolourised by heating sugar solution with charcoal. 19. (a) Write down the IUPAC name of the following complex: [Cr(NH3)4(H2O)2]Cl3 (b) Write the formula for the following complex: Potassiumhexacyanidoferrate (II). Or Discuss the bonding in the coordination entity [Cr(H2O)2(C2O4)2] on the basis of valence bond theory. Also comment on the geometry and spin of the given entity. (Atomic number of Cr = 24) 20. A flask contains a mixture of compounds 'A' and 'B'. Both the compounds decompose by first order kinetics and t½ of 'A' and 'B' are 300 and 180 seconds respectively. If concentration of 'A' and 'B' are equal initially, calculate the time that required for concentration of A be 4 times B. Or 2A + B → A2B + C Determine the rate law expression for the reaction from the following data: (A) (B) rate I 0.1 0.1 6 × 10–3 mol L–1 s–1 II 0.1 0.2 2.4 × 10–2 mol L–1 s–1 III 0.2 0.1 1.2 × 10–2 mol L–1 s–1 21. For a first order reaction the rate constant for decomposition of N2O5 is 6 × 10–4 s–1. Calculate its half-life. 22. Convert (i) 2-Propanol to 1-Propanol (ii) Ethanol to diethyl ether. 23. Calculate the number of lone pairs on the central atom in XeF6 molecule and predict the shape on the basis of VSEPR theory. 24. If the edge length of an FCC crystal lattice is 400 pm, calculate the radius of atom. 25. The following haloalkanes are hydrolysed in presence of aq. KOH: (i) 2-chlorobutane (ii) 3-chloro-3-methylhexane Which of the above is most likely to give (a) an inverted product (b) a racemic mixture? 54 Together with®  EAD Chemistry—12

SECTION C Q.No 26–30 are Short Answer Type II carrying 3 marks each. 26. (a) Why is Cr3+ more stable than Cr2+? (b) Why is Cu2+ more stable than Cu+? (c) Why do transition metals form complex compounds? Or (a) Which lanthanoid shows +4 oxidation state and why? (b) Why is ionisation enthalpy of 5d series more than 3d and 4d? (c) What is most common oxidation state shown by lanthanoid? 27. Complete the following: CH3CH2NH2 HNO2 → A SOCl2 → B KCN→ C LiAlH4 → D CHCl3/KO→H E Or (a) Convert benzene to aniline. (b) Distinguish between C2H5NH2 and (C2H5)2NH by suitable chemical test. (c) Arrange CH3NH2, NH3, (CH3)2NH, (CH3)3N in increasing order of basicity. 28. When heated above 916° C, iron changes from BCC crystalline form to FCC without the change in radius of atom. Calculate the ratio of density of the crystal before and after heating. 29. (a) Represent glycine in zwitter ion form. (b) Make two tripeptides by using glycine H2N­—CH2—COOH, alanine CH3—CH—COOH and aspartic acid HOOC—CH2—CH—­ NH2. NH2 COOH 30. (i) Oxygen does not form OF6, why? (ii) Why is F2 gas, whereas I2 solid? (iii) Why is HI stronger acid than HCl? SECTION D Q.No 31 to 33 are long answer type carrying 5 marks each. 31. (a) Calculate the emf of the following cell reaction at 298 K: Mg(s) + Cu2+ (0.0001 M) → Mg2+ (0.001 M) + Cu(s) The standard Ec°ell = 2.71 V. [log 10 = 1] (b) What happens when (i) H2SO4 is electrolysed. Write the reaction at cathode as well as anode. (ii) Ag NO3 is electrolysed using silver electrodes. What will be reactions taking place at cathode and anode? Or (a) The specific conductance of a 0.01 M solution of acetic acid at 298 K is 1.65 × 10–4 ohm–1 cm–1. The molar conductivity) for H+ ion and CH3COO– ion molar conductance at infinite dilution (limiting mol–1 respectively. Calculate its molar conductivity are 349.1 ohm–1 cm2 mol–1 and 40.9 ohm–1 cm2 and degree of dissociation. Sample Papers 55

(b) Solutions of CH3COOH and KCl are diluted. Λ°m of KCl increases to smaller extent while that of CH3COOH increases to larger extent comparatively. Which of the two is a strong electrolyte? Justify your answer. 32. (i) Answer the following questions. (a) Write the balanced chemical equation for the reaction of Cu with conc HNO3. (b) Draw the shape of IF5. (ii) A gas 'X' is used in discharge tube and fluorescent bulb for advertisement display purpose. It does not form any compound. Its boiling point is 27.1 K. Identify the gas and why is its boiling point is low. Gives one use of bulbs filled with this gas. Or Answer the following questions. (a) Arrange the following in increasing order of thermal stability. HCl, HF, HBr, HI (b) Explain the ring test for NO3–. (c) A pungent smelling gas 'X' reacts with HCl to form white fumes of 'Y'. 'Y' on reaction with slaked lime produces gas 'X'. Aqueous solution of 'X' gives deep blue colouration with CuSO4 solution of Z. Identify 'X', 'Y' and 'Z'. 33. (a) Draw the structure of the following derivatives? (i) 2, 4-DNP of benzaldehyde (ii) Propanonoxime (b) Complete the following reactions. CH2—CH3 (i) KMnO4 KOH, D (ii) CH3COCl Anhy. AlCl3 (c) Carboxylic acids are stronger acids than phenol, Justify. Or (a) An organic compound (A) with molecular formula C4H8O2 was hydrolysed with dil. H2SO4 to give a carboxylic acid (B) and an alcohol (C). 'C' on dehydration gives ethene and 'C' on oxidation gives 'B' back. Identify A, B and C and write equation for the reaction involved. (b) How will you convert ethanal to the following compounds: (i) Ethanol (ii) Ethane 56 Together with®  EAD Chemistry—12

Answers 1. (i) (d) DTb = Kb × m = 0.52 × 0.2 = 0.104 Boiling point of solution = 100 + 0.104 = 100.104° C (ii) (d) All of there are colligative properties as they depend upon number of particles of solute and not on nature of solute. (iii) (b) DTf = Kf × WB × 1000 = 1.86 × 45 × 1000 = 2.2 K. MB WA 62 600 (iv) (d) Q DTb ∝ m \\ 0.5 m Glucose solution will have highest boiling point. Or (a) Q DTb ∝ m, DTf is lowest in 0.1 m urea. \\ Freezing point is highest. 2. (i) (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. (ii) (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. (iii) (c) Assertion is correct statement but reason is wrong statement. (iv) (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. Or (c) Assertion is correct statement but reason is wrong statement. 3. (a) HCl because mobility of H+ is higher, therefore HCl has highest Λ°. 4. (c) H-bond is formed between C O and NH group. Or (b) Sucrose is not reducing sugar because it does not have free aldehyde group. 5. (a) M = Percentage by mass × d × 10 Molar mass = 98 ×1.84 × 10 = 18.4 M. 98 6. (d) Sc, because it has incompletely filled d-orbital. Or (b) Mn2+ has 5 unpaired electrons. O 7. (b) CH3—CH2C≡≡N  OH–→ CH3—CH2—C—NH2  NaOH + Br2→ CH3—CH2—NH2 ‘A’ ‘B’ Or NH2 NH2 Br Br (d) + 3Br2 (aq) → Br Sample Papers 57

8. (d) EDTA is hexadentate ligand because it has 6 donor atoms. Or (c) K3[Cr(C2O4)3] is correct formula. 9. (a) The atomic radius of Ag is closest to Au due to lanthanoid contraction, atomic size of 4d and 5d series close to each other. 10. (c) Tertiary halides form stable carbocation, so react faster. 11. (c) Iodine is molecular solid because it has weak intermolecular forces of attraction. 12. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. 13. (a) Assertion and reason both are correct statements and reason is correct explanation for assertion. 14. (d) Assertion is wrong statement but reason is correct statement. [Hint: Methanol is more acidic than C2H5OH since CH3O– is more stable than C2H5O–.] 15. (c) Assertion is correct statement but reason is wrong statement. 16. (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. Or (b) Assertion and reason both are correct statements but reason is not correct explanation for assertion. Cl Cl OH Cl OH Cl OH Cl OH 17. + HO– Slow step ≡ → N N N N N OO OO OO OO OO OH Fast + Cl– step → N OO —NO2 group stabilises intermediate carbanion, increases rate of nucleophilic substitution reaction NH2 N+2Cl– Or Br (i) NaN0O–25°+ CHCl → Cu/HBr → Aniline Bromobenzene (ii) CH3CH2CH2Br KOH(alc) CH3—CH==CH2 HBr CH3—CH—CH3 → → 1-bromopropane Br 2-bromopropane 58 Together with®  EAD Chemistry—12

18. (i) Animal hide soaked in tannin results in hardening of leather due to mutual coagulation. (ii) It is because it forms multimolecular layer with low enthalpy of adsorption. (iii) Animal charcoal adsorbs colour from sugar to get decolourised. 19. (a) Tetraamminediaquachromium(III) chloride. (b) K4[Fe(CN)6] Or [Cr(H2O)2(C2O4)2]– Cr(24) : [Ar] 4s13d5 Cr3+(24) : [Ar] 4s0 3d 3 It has d 2sp3 hybridisation, octahedral shape and paramagnetic in nature. 20. t½ of A = 300 seconds. After 900 seconds, 3 half lives, 13 = 1 th of A will be left. c2m 8 t½ of B = 180 seconds. After 900 seconds, 5 half lives, 15 = 1 th of B will be left. c2m 32 Conc. of A : Conc. B 1 : 1 8 32 4 : 1 Or Rate = k[A]x [B]y 6.0 × 10–3 = k50.1?x 50.1?y ⇒ 1 = 1 ⇒ 2y = 22 ⇒ y = 2 2.4 × 10–2 k50.1?x 50.2?y 4 2y Also, 6.0 × 10 –3 = k50.1?x 50.1?y ⇒ 1 = 1 ⇒ 2x = 21 ⇒ x = 1 1.2 × 10 –2 k50.2?x 50.1?y 2 2x Rate law = k[A] [B]2 = k[A] [B]2 21. Given, k = 6 × 10–4 s–1 t1/2 = 2.303 log [R] 0 = 2.303 log  2 = 2.303× 0.3010 = 0.693 ×104 = 6930 = 1155 seconds. k [R] 0/2 6 # 10−4 6 ×10−4 6 6 22. (i) CH3—CH—CH3 Conc.H2SO4→ CH3—CH==CH2 (ii)(Hi)2BO22H/O6 H–→ CH3CH2CH2OH D OH (ii) 2C2H5OH Con4c1.3HK2SO4 → C2H5OC2H5 + H2O Sample Papers 59

23. There are 6 bonded pairs and one lone pair of electrons. F F F Xe F FF It has distorted octahedral shape due to presence of lone pair of electrons. 24. In FCC, 4r = 2 a 4r = 2 × 400 = 1.414 × 400 4r = 565.6 ⇒ r = 565.6 = 141.4 pm. 4 25. (i) 2-chlorobutane will give inverted product due to back side attack of OH– following SN2 mechanism. (ii) 3-chloro-3-methylhexane will give racemic mixture due to formation of carbocation with equal probability of attack by OH– from both sides leading to racemic mixture by SN1 mechanism. 26. (a) Cr3+(t32g) has half-filled t2g orbitals having lower energy than Cr2+(3d4). (b) Cu2+ has higher hydration enthalpy which can overcome second ionisation enthalpy of Cu+ ion. (c) It is due to small cation, higher charge and availability of d-orbitals of suitable energy. Or (a) Ce shows +4 oxidation state as it has stable electronic configuration (4f 0). (b) It is due to lanthanoid contraction, the effective nuclear charge increased, therefore, ionisation enthalpy of 5d series is higher. (c) +3 is common oxidation state shown by lanthanoids. 27. CH3CH2NH2 HNO2 CH3CH2OH SOCl2 CH3CH2Cl KCN CH3CH2C≡N   Ethanamine ‘A’ ‘B’ ‘C’ Ethanol 1-Chloroethane Propane nitrile LiAlH4 CH3CH2CH2NH2CHCl3/KOH CH3CH2CH2NC ‘D’ ‘E’ Propyl isocyanide Propan-1-amine Or NO2 NH2 (a) + conc. HNO3 con3c3.3HK2SO4→ Sn/HCl → Benzene Aniline (b) Add CHCl3 and KOH. C2H5NH2 will give offensive smelling compound, ethyl isocyanide, whereas (C2H5)2NH will not react with CHCl3 and KOH. (c) NH3 < (CH3)3N < CH3NH2 < (CH3)2NH is increasing order of basic character. Q pKb  4.7 4.22 3.38 3.29. H igher the pKb lesser will be basic character. 60 Together with®  EAD Chemistry—12

28. d1 = Z×M = 2 × 56 for BCC a3× NA e 4r 3 NA 3 o× d2 = Z×M = (2 4 × 56 for FCC a3× NA 2 r)3 × NA \\ 29. (a) H3N—CH2—COO– d1 = 0.918. d2 O OO (b) —HNCH2C­—NH—CH—C—NH—CH—CH2—C— CH3 COOH I OO O —HN—CH—C—­ NH—CH2—C—NH—CH—CH2—C— CH3 COOH II 30. (i) Oxygen does not have d-orbitals, therefore, OF6 does not exist. (ii) F2 is smaller in size, less surface area, therefore, weak van der Waals’ forces of attraction between molecules therefore, gas as compared to larger size I2 having more van der Waals’ forces of attraction, therefore, solid. (iii) H—I has lower bond dissociation enthalpy than HCl due to longer bond length as I is bigger than Cl. 31. (a) Ecell = Ec°ell – 0.0591 log [Mg2+] = 2.71 V – 0.0591 log [10–3] 2 [Cu2+] 2 [10–4] = 2.71 V – 0.0295 × log 10 = 2.71 V – 0.0295 = 2.68 V  [Q log 10 = 1] (b) (i) H2SO4(aq)  → 2H+ + SO42– At cathode: 2H+ + 2e–  → H2(g) At anode: 2H2O  → 4H+ + O2 + 4e– At cathode H2 will be liberated whereas at anode O2 will be formed. (ii) At anode: Ag  → Ag+ + e– At cathode: Ag+ + e–  → Ag(s) Or (a) L° = l°H+ + l°CH3COO– = 349.1 + 40.9 = 390 ohm–1 cm2 mol–1 Lm = 1000 κ = 1000 ×1.65×10–4 = 16.5 ohm–1 cm2 mol–1 M 0.01 Sample Papers 61

a = Λm = 16.5 = 0.0423  ⇒  a = 0.0423 × 100% Λ° 390 ⇒ a = 4.23% (b) KCl is strong electrolyte. Limiting molar conductivity increases only to small extent for a strong electrolyte as on dilution because number of ions do not increase much, only mobility of ions increases. CH3COOH is weak electrolyte, its Λm° increases to large extent because degree of dissociation increases, therefore, number of ions in total volume of solution increases along with mobility of ions increases. 32. (i) (a)C u + 4HNO3(Conc) → Cu(NO3)2 + 2NO2 + 2H2O F FF (b) I FF (ii) Gax 'X' is Neon. Its boiling point is low due to weak van der Waals’ forces of attraction. Neon bulbs are used in botanical gardens and in greenhouses. Or (a) HI < HBr < HCl < HF is increasing order of thermal stability. (b) To the salt solution, add freshly prepared FeSO4 solution. Add conc. H2SO4 very slowly and carefully along sides of test tube. A brownish black ring formed confirms the presence of NO–3. (c) 'X' is NH3, Y is NH4Cl and Z is [Cu(NH3)4]SO4. NH3 + HCl → NH4Cl 'X' 'Y' (White fumes) Ca(OH)2 + 2NH4Cl D CaCl2 + 2NH3 + 2H2O → 'X' CuSO4 + 4NH4OH → [Cu(NH3)4]SO4 + 4H2O 'Z' Deep blue colouration 33. (a) (i) C6H5CH==N—NH NO2 CH3 NO2 (ii) CH3—C==NOH 62 Together with®  EAD Chemistry—12

CH2CH3 COOH KKOMHnO, D4→ (b) (i) O O + CH3—C­—Cl C­—CH3 (ii) AlCl3→ (c) It is because carboxylate ions are more stable than phenoxide ions due to delocalisation of charge on two oxygen atoms. Or CH3COOC2H5 + H2O H2SO4→ CH3COOH + C2H5OH 'A' 'B' 'C' C2H5OH ConcH2SO4→ CH2==CH2 +H2O D 'C' C2H5OH KHM2SnOO44→ CH3COOH O 'C' 'B' (b) (i) CH3—C—H LiAlH4→ CH3CH2OH O Zn(Hg) (ii) CH3—C—H Conc HCl → CH3—CH3 Sample Papers 63