Practical Organic Chemistry-02-Solved Problems

SECTION - I SUBJECTIVE TYPE PROBLEMS Problem 1 : In a Victor Meyer’s determination, the following observations have been made : Mass of compound = 0.17 g; Volume of air collected = 34.2 mL; Temperature = 15°C; Atmospheric pressure = 750 mm; Vapour pressure of water at 15°C = 13 mm. Calculate the vapour density and molecular mass of the compound. Solution : Given V1  34.2 mL V2  ?  P1  (750 13)  737 mm P2  760 mm NTP conditions T  (15  273)  288 K T  273 K  1 2  By gas equation, V2  7371 34.2  273 288 760 = 31.4376 mL W  0.17  60.08 V2  0.00009 31.4376  0.00009 Mol.mass= 2 × Vapour density = 2 × 60.08 = 120.16. Problem 2: An organic substance (0.2115 g) on complete combustion gave 0.4655 g of carbon dioxide and 0.2533 g of water. Determine the percentage composition of the compound. Solution : Mass of CO2 formed = 0.4655 g Mass of H2O formed = 0.2533 g Mass of the organic substance taken = 0.2115 g Percentage of carbon = 12  mass of CO2 100  12  0.4655 100  60.03 44 mass of compound 44 0.2115 Percentage of hydrogen 2  mass of H2O 100  2  0.2533 100  13.30 18 mass of compound 18 0.2115 Percentage of oxygen = 100 – (60.03 + 13.30) = 26.67 Composition of the substance : C = 60.03%; H = 13.30%; O = 26.67% Problem 3: 0.2 g of an organic compound was analysed by Kjeldahl’s method. Ammonia evolved was absorbed in 60 mL N/5 H2SO4. Unused acid required 40 mL of N/10 NaOH for complete neutralisation. Find the percentage of nitrogen in the compound. Solution : Mass of oganic compound = 0.2 g Unused acid rquired = 40 mL N NaOH 10 40mL N NaOH  40 mL N H SO  20mL N H SO 10 10 2 4 5 2 4 Acid used for absorption of ammonia  (60  20)mL N H SO  40mL N H SO 5 2 4 1.4  1  40 5 2 4 Percentage of nitrogen = 1.4  N1  V  5  1.4  40  56 W 0.2 0.2  5 Problem 4 : 0.156 g of an organic compound on heating with fuming HNO3 and AgNO3 gives 0.235 g of AgI. Calculate the percentage of iodine in the compound. Solution : Mass of organic compound, W = 0.156 g Mass of AgI, W1 = 0.235 g AgI  235g I 127 g Percentage of iodine = 127  W1 100  127  0.235 100  81.41 235 W 235 0.156 Problem 5 : 0.38 g of a silver salt of a dibasic acid on ignition gave 0.27 g of silver. Calculate the molecular mass of the acid. Solution : Mass of silver salt = 0.38 g Mass of silver = 0.27 g Eq. mass of silver salt  Mass of silver salt Eq. mass of silver Mass of silver Eq. mass of silver salt = 0.38 108 0.27 or E  107  0.38 108 0.27 or E   0.38 108 107  45 (E = Eq. mass of acid)  0.27  Mol. mass of the acid = Eq. mass × basicity = 45 × 2 = 90 Problem 6 : An organic compound contains 40% C. 6.66% H and rest oxygen. Its vapour density is 30. Calculate its empirical and molecular formulas. Solution : Calculation of empirical formula : Element Percentage At. Mass Relative Number of atoms Simplest Ratio Carbon 40.0 12 40  3.33 12 3.33  1 3.33 Hydrogen 6.66 1 6.66  6.66 1 6.66  2 3.33 Oxygen 53.34 16 53.34  3.33 3.33  1 (by difference) 16 3.33 Empirical formula = CH2O Empirical formula mass = (12 +2 + 16) = 30 Mol. mass = 2 × V.D. = 2 × 30 = 60 n  Mol. mass  60  2 Emp. mass 30 Molecular formula = 2 × (Empirical formula) = 2 × (CH2O) = C2H4O2 Problem 7 : A dibasic organic acid gave the following results : C = 34.62%. H = 3.84%. 0.1075 g of this acid consumes 20 mL of 0.1 N NaOH for complete neutralisation. Find out the molecular formula of the acid. Solution : Calculation of empirical formula : Element Percentage At. Mass Relative Number of atoms Simplest ratio of atoms Carbon 34.62 12 34.62  2.88 12 2.88  1 3  3 2.88 Hydrogen 3.84 1 3.84  3.84 3.84  1.33 3  4 1 2.88 Oxygen 61.54 (by difference) 16 61.54  3.84 16 3.84  1.33 3  4 2.88 Empirical formula of the acid = C3H4O4. Empirical formula mass = (3 × 12) + (4 × 1) + (4 × 16) = 104 Calculation of molecular mass : 20 mL 0.1 N NaOH  0.1075 g acid 20 × 0.1 mL 1 N NaOH  0.1075 g acid So 1000 mL 1 N NaOH = 0.1075 1000g acid  53.75 g acid 20  0.1 Eq. mass of the acid = 53.75 Mol. mass of the acid = Eq. mass × basicity = 53.75 × 2 = 107.50 n  Mol. mass  107.50  1 Emp. mass 104.0 Molecular formula = C3H4O4 Problem 8 : 2.65 g of a diacidic base was dissolved in 500 mL of water. 20 mL of this solution N required 12 mL of 6 HCl solution. Calculate the equivalent mass and molecular mass of the base. Solution : 12mL N HCl solution  20 mL of base solution 6 or 2mL 1 N HCl solution  20 mL of base solution or 2  500 mL 1 N HCl solution  500 mL of base solution 20 or 50 mL 1 N HCl solution  2.65 g of base 1000 mL 1 N HCl solution  2.65 1000  53 g base 50 Eq. mass of the base = 53 Mol. mass of the base = Eq. mass × acidity = 53 × 2 = 106 Problem 9 : During the test for nitrogen in Lassaigne’s filtrate, sometimes red colouration is obtained when ferric chloride is added. Solution : When nitrogen and sulphur both are present in an organic compound, sodium thiocyanate is formed which gives red colouration with FeCl3. Problem 10 : What is the role of copper sulphate and potassium sulphate in Kjeldahl’s process for the estimation of nitrogen in an organic compound? Solution : Copper sulphate acts as a catalyst while potassium sulphate raises the boiling point of H2SO4. SECTION - II SINGLE CHOICE PROBLEMS Problem 1 : In the Lassaigne’s test for the detection of nitrogen in an organic compound, the blue or green colour is due to the formation of (a) K3[Fe(CN)6] (b) K4[Fe(CN)6] (c) Fe(CN)6 (d) Fe4[Fe(CN)6]3 Solution : Ferric ferrocyanide, Fe4[Fe(CN)6]3. And. (d Problem 2 : When an organic compound containing phosphorus is fused with fusion mixture, it gives (a) Na2HPO4 (b) Na3PO4 (c) NaH2PO4 (d) Na3PO3 Solution : Fusion of organic compound with fusion mixture (Na2CO3 + K2CO3) converts P into Na3PO4. Ans. (b) Problem 3 : In organic compounds, phosphorus is estimated as (a) Mg2P2O7 (b) Mg(NH4)PO4 (c) Mg3(PO4)2 (d) H3PO4 Solution : Magnesium pyrophosphate, Mg2P2O7. Ans. (a) Problem 4 : Kjeldahl’s method cannot be used for the estimation of nitrogen in (a) Pyridine (b) Nitro compounds (c) Azo compounds (d) All the three abvoe Solution : Kjeldahl’s method cannot be used for the estimation of N in nitro and azo compounds and compounds containing N in the ring (i.e., pyridine) since these compounds are not completely converted into (NH4)2SO4 during digestion. Ans. (d) Problem 5 : 0.28 g of a nitrogenous compound was kjeldahlised to produce 0.17 g of NH3. The percentage of nitrogen in the organic compound is (a) 5 (b) 30 (c) 50 (d) 80 Solution : % of N  14  0.17 100  50. 17 0.28 Ans. (c) Problem 6 : The simplest formula of a compound containing 50% of element X (at. wt. 10) and 50% of element Y (at. wt. 20) is (a) XY (b) XY2 (c) X2Y (d) X2Y3 Solution : X : Y  50 : 50  2 :1 or E.F.  X Y. 10 20 2 Ans. (c) Problem 7 : The presence of carbon in an organic compound is detected by heating it with (a) sodium metal to convert it into NaCN (b) CaO to convert it into CO which burns with a blue flame (c) CuO to convert it into CO2 which turns lime water milky (d) Cu wire to give a bluish-green flame. Solution : Heating with CuO converts C into CO2. Ans. (c) Problem 8 : The empirical formula of a compound is CH2O and its molecular weight is 60. The compound is (a) CH3CHO (b) CH3CH2OH (c) CH3OH (d) CH3COOH Solution :  M.F.  E.F. 60  (CH O)  2 30 2 C2H4O2  CH3COOH Ans. (d) Problem 9 : Complete combustion of 0.858 of compound X gives 2.63 g of CO2 and 1.28 g of H2O. The lowest molecular weight, X can have is (a) 43 (b) 86 (c) 129 (d) 172 Solution : % C  12  2.63 100  83.59 % H  44 0.858 2  1.28 18 0.858 100  16.57 Now. C : H  83.59 : 16.57  6.96 :16.57  1: 2.38 or 3 : 7 12 1  E.F. = C3H7 E.F. wt = 3 × 12 + 7 × 1 = 43 Ans. (a) Problem 10 : A hydrocarbon having empirical formula as CH2 has a density of 1.25 g/L. The molecular formula of the hydrocabron is (a) C2H2 (b) C2H4 (c) C2H6 (d) C3H8 Solution : 1 mole of hydrocarbon = wt. of 22.4 litres at NTP = 1.25 × 22.4 = 28  Mol. formula = CH  28  C H Ans. (b) 2 14 2 4 SECTION - III MULTIPLE CHOICE PROBLEMS Problem 1: Which of the following compounds may give blood red colouration while performing Lassaigne’s test for nitrogen ? (a) (NH2)2CO (b) (NH­2)2 C = S (c) p-NH2C6H4­SO3H (d) C6H5SO3H Solution : Both compounds contain N and S Ans. (b, c) Problem 2: The desiccants used for absorbing water during Liebig’s method for estimation of carbon and hydrogen are (a) Anhydrous CaCl2 (b) Anhydrous Na2SO4 (c) MgSO4 . 7H2O (d) Mg(ClO4)2 Solution : Anhydrous CaCl2 and Mg (ClO4)2 Ans. (a, d) Problem 3: Presence of halogen in a compound is tested by (a) lodoform test (b) Million’s test (c) Silver nitrate test (d) Beilstein test Solution : Silver nitrate test and Beilstein test Ans. (c, d) Problem 4: Which of the organic compounds will give white precipitate with AgNO3 ? (a) C6 H5 NHCl (b) NaCl (c) C6H5Cl (d) 2, 4, 6-trinitrochlorobenzene Solution : C6H5NH3+ Cl– and 2, 4, 6-Trinitrochlorobenzene will give white precipitate with AgNO3 Ans. (a, d) Problem 5: In Lassaigne’s test, the organic compound is at first fused with sodium metal. The sodium metal is used because (a) the melting point of sodium is low so it is easily fused with organic substances (b) sodium is very much effective to bring about destructive reductions of organic compounds forming ionic inorganic salts NaCN, Na2S and NaX (c) all sodium salts are soluble in water. (d) none of these Solution : a, b, c all are correct reason Ans. (a, b, c) Problem 6: Which of the following reactions occur during the detection of nitrogen in organic substances by Lassaigne’s test ? (a) Na + C + N  NaCN (b) FeSO4 + 6NaCN  Na4[Fe(CN)6] + Na2SO4 (c) 3Na4[Fe(CN)6] + 2Fe2(SO4)3  Fe3[Fe(CN)6]3 + 6Na2SO4 (d) (NH4)2SO4 + 2NaOH  2NH3 + Na2SO4 + 2H2O Solution : Reaction (d) occurs in kjeldahl’s method. Ans. (a, b, c) Problem 7: Which of the organic compound will give red colour is Lassaigne test : S (a) NaCNS (b) H2N C NH2 (c) O H2N C NH2 (d) H2N SO3Na Solution : Both compounds contain N and S Ans. (b, d) Problem 8: Organic compound A HNO3 / AgNO3 white ppt. A can be (a) NH4Cl (b) OH (c) Cl (d) CH2Cl Solution : Ans. (b, d) Problem 9: Detection of the chlorine is possible without preparing sodium extract in NO2 (a) O2N Cl NO2 (b) CHCl3 (c) CH2Cl (d) CH2 = CH – CH2Cl Solution : Ans. (a, c, d) Problem 10: Which will give colour with FeCl3 ? (a) OH (b) H3C OH C CH O C OC2H5 (c) OH (d) NH2 Solution : enol form will give colour with FeCl3 Ans. (a, b, c) MISCELLANEOUS PROBLEMS COMPREHENSION TYPE PROBLEMS Write up – I "Bonding in organic compounds are generally covalent, therefore, like inorganic compounds, no direct method is available for detection of elements. In sodium-fusion method covalent bonds of hetro atoms are broken by heating of organic compounds with sodium metal. This results in the formation of inorganic ions involving these elements, these ions can in turn be readily identified by inorganic qualitative methods." Problem 1 : To sodium fusion extract Cl2 water and CCl4 were added and shaken well. There is violet colour in the lower part (organic layer). This indicates presence of : (a) bromine (b) iodine (c) chlorine (d) bromine and iodine Solution : It indicates presence of iodine. Ans. (b) Problem 2 : Detection of elements in organic compound is done using sodium-fusion method which is also called. (a) Middleton’s fusion method (b) Lassaigne fusion method (c) Hofmann’s method (d) Hinsberg’s method Solution : Middleton’s fusion method. Ans. (b) Problem 3 : Red colour complex ion formed on adding FeCl3 to S.E. when N and S both are present in organic compound is : (a) [Fe(CN)6 ]4 (b) Fe(CNS)3 (c) [Fe(CNS)2 ] (d) [Fe(CN)6 ]3 Solution : 3NaCNS + FeCl3  Fe(CNS)3 + 3NaCl Ans. (b) Write up – II A mixture of two aromatic compound (A) and (B) was separated by dissolving in chloroform followed by extraction with aqueous KOH solution. The organic layer containing compound (A) when heated with alcoholic solution of KOH produced a compound (C), (C7H5N), associated with an unpleasant odour. The alkaline aqueous layer on the other hand when heated with chloroform and then acidified gave a mixture of two isomeric compound (D) and (E) of molecular formula C7H6O2. Problem 4 : Compound (A) extracted into organic (CHCl3) layer is NH2 CH2NH2 (a) (b) NHCH3 NH2 (c) (d) CH3 Ans. (a) Problem 5 : (A) has been tested above the unpleasant smell of the compound (C). (C) is …… formed from A by …… reaction. CN NC (a) , Reimer Tiemann (b) , Carbyl amine CN NC (c) , Carbyl amine (d) , Reimer Tiemann Ans. (b) Problem 6 : D and E are respectively : OH OH CHO (a) CHO (b) OH CHO OH CHO OH (c) Ans. (c) CHO OH CHO (d) CH3 OH OH CH3 OH CHO (Solution for 6, 7, and 9) NH2 NC  CHCl3  3KOH   3KCl  3H2O (A) (C) This reaction is called carbyl amine reaction. OH (B)  CHCl3  3KOH KCl  H2O OH (D) CHO + OH CHO (E) MATCHING TYPE PROBLEM 7. Column (I) Column (II) (a) Glycine (p) —NH2 (b) Salicyclic acid (q) —COOH (c) Aniline (r) —OH (d) Toluidine (s) —CH3 OH Sol. H2 NCH2COOH Glycine COOH salicyclic acid NH2 Aniline CH3 NH2 Toluidine Ans. (a) — (p, q) (b) — (q, r) (c) — (p) (d) — (p, s) ASSERTION-REASON TYPE PROBLEMS The question given below consist of an ASSERTION and the REASON. Use the following key for the appropriate answers (a) If both Assertion and Reason are correct and Reason is the correct explanation for Assertion (b) If both Assertion and Reason are correct and Reason is not the correct explanation for Assertion (c) If Assertion is correct but Reason is not correct. (d) If Assertion is incorrect but Reason is correct. 8. Assertion : A mixture of o-nitrophenol and p-nitrophenol can be separated by steam distillation Reason : o-Nitrophenol is steam volatile but p-nitrophenol is not though both are insoluble in water. Solution : R is the correct explanation of A. Ans. (a) 9. Assertion : Acetone (b.p. 329 K) and methyl alcohol (b.p. 338 K) are separated by fractional distillation Reason : Fractional distillation helps in separating two liquids from their mixture when their boiling points differ by 10º or so. Solution : R is the correct explanation of A. Ans. (a) 10. Assertion : Hydrazine contains nitrogen but does not give Lassaigne's test for nitrogen Reason : Hydrazine reacts with fused sodium to give H2 gas. Solution : Hydrazine does not contain carbon. Therefore, during fusion with Na metal, it does not form NaCN required for positive Lassaigne’s test. Ans. (b)