Chemistry-1.AITS-PART TEST-1-MAILING-Questions

CHEMISTRY Time: Two Hours Maximum Marks : 100 Note: i) This paper has SEVEN W questions. ii) Separate answers are to be given on the separate pages. iii) Attempt all questions. iv) Use of logarithmic table is PERMITTED v) Use of calculator is NOT PERMITTED Useful Data: Gas Constant R = 8.314 J mol−1 K−1 = 0.0821 lit atm mol−1 K−1 = 2 Cal mol−1 Avogadro's Number Na = 6.023 × 1023 \Planck’s constant h = 6.625 × 10−34 J sec. 1 Faraday = 96500 Coulomb 1 calorie = 4.2 Joule Atomic No: Ca = 20, C = 6, O = 8, K = 19, Cl = 17, N = 7, S = 16, Na = 11, Cu = 29, Co =27, Mn = 25. Atomic Masses: Ag =108, Mn = 55, Cr = 52, Ca = 40, C = 12, O = 16, K = 39, Cl = 35.5, N = 14, S = 32, Na = 23, H = 1, I = 127, Mg = 24, Ne = 20, Br = 80, I = 127, As = 75, Cu = 63.54 Name : ☐☐☐☐☐☐☐☐☐☐☐☐☐☐☐☐☐ Enrol No. : ☐☐☐☐☐☐☐☐☐☐☐☐☐ 1. a) Which one of the two compounds have high pKa value? Give reason. [4] b) Identify the missing products in the following reactions. In cases where more than one product is possible give only the major product. [3 + 3] i) ii) c) Aldol condensation as the key step [6] 2. a) Treatment of diethyladipate C2H5O2C(CH2)4CO2C2H5, with sodium ethoxide in absolute ethanol followed by neutralization with aqueous acid yields a compound (A) having the formula C8H12O3. Acid hydrolysis of (A) followed by strong heating produces a compound (B) having M.F. C5H8O, which responds to D.N.P. reagent and on reaction with Amalgamated Zinc and HCl gives a compound (C) having M.F. C5H10. Identify the compounds (A), (B), (C) and give mechanism of the formation of (A) from diethyladipate. [8] b) Write down the structures of (A) and (B) from the following datas: [6] 3. a) Complete the following: i) ii) [4] b) Compound C9H11OBr (A) gives iodoform test. (A) on treatment with aqueous KOH followed by acidification gives compound C9H2O2 (B). (B) on reaction with HIO4 gives benzaldehyde and compound (C). Identify compounds (A), (B) and (C). Reactions are not needed. [6] c) Deduce the structure of a compound C10H14, that is hydrogenated with three equivalents of H2/Pd to give 1-isopropyl-4-methyl cyclohexane, and, on reductive ozonolysis, gives the following products: [6] 4. a) An alkane (A) with the M.F. C6H14 reacts with chlorine in the presence of u.v. light to yield three isomeric monochloro derivatives B, C and D. Out of these only C and D undergoes dehydro halogenation with sodium ethoxide in ethanol to produce an alkene. Moreover C and D yields the same alkene E (C6H12). Hydrogenation of E produces A. Treating E with HCl produces a compound F that is an isomer of B, C and D. Treating F with Zn and acetic acid gives a compound G, that is isomeric with A. Propose structures for A to G. [8] b) A neutral organic compound (A) having molecular formula C3H6O2 gives positive Tollen’s test. (A) reacts with HCN to yield an optically active compound (B). (B) on hydrolysis yields a monobasic acid (C) having molecular formula C4H8O4. (C) is an optically active compound. Identify A, B and C with logical reasoning. [6] 5. Give reasons for the following statements a) Explain why neopentyl chloride, (CH3)3CCH2Cl, a 1° RCl, does not participate in typical SN2 reactions b) Account for the following PhCH2CH(CH3)2 + Br2 PhCHBrCH(CH3)2 with little or no PhCH2CBr(CH3)2 formed. [3 + 3] 6. a) An organic compound A(C6H12O) forms an oxime but does not reduce Tollen’s reagent. A on reduction with sodium amalgam forms an alcohol B which on dehydration forms chiefly a single alkene C. The ozonolysis of C produces D and E. The compound D reduces Tollen’s reagent but does not answer iodoform test. The compound E does not reduce Tollen’s reagent but answer iodoform test. What are the structures of the above compound? Explain the reactions. [8] b) An optically active compound M, C3H7O2N, forms a hydrochloride, but dissolves in water to give a neutral solution. On heating with sodalime, M yields N, C2H7N; both M and N react with nitrous acid, the former yielding a compound P, C3H6O3, which on heating is converted to Q, C6H8O4. Account for the above reactions and suggest how M may be synthesized. [6] 7. a) Explain the following i) is less basic than CH3 – CH2 – ii) In SN1 reactions, racemic mixture is obtained iii) iv) CH3 – COCl is more reactive than CH3 –CO – O – CO – CH3 towards nucleophilic substitution. v) Explain the reaction vi) C6H5 is a weak o.p.– directing group. [12] b) Compound (A) having the empirical formula C5H4O2, gives the Cannizaro reaction to form (B) & (C). Compound (B) has a equivalent weight of 112.08. Acetonic participates with (A) in a condensation reaction with a dilute base to form (D) which has the formula C8H8O2. Oxidation of (C) gives (B). When (B) is heated at 200-225°C, (E) which has the formula C4H4O is formed. Hydrogen under pressure in the presence of Ni reduces (E) to (F). Vigorous treatment of (F) with HBr gives (G) which is a dibromoalkane. Treatment of (G) with KCN followed by hydrolysis and acidification gives adipic acid (1,6-hexanedioic acid). Give structures of (A) to (G). [8] 6