Purification & Characterisation of Organic Compounds Solved Problems

University:
Binghamton University
Course:
CHEM 231 | Organic Chemistry I
Academic year:

2022
Views:

498

Pages:

1
Author:

Lanrwst4tn

PURIFICATION & CHARACTERISATION OF ORGANIC COMPOUNDS SOLVED PROBLEMS Ex.1 Sol. Ex.2 Sol. Ex.3 Liquid which decompose below their normal boiling points can be distilled at lower temperature by (A) Increasing the pressure (B) Decreasing the pressure (C) Heating in water bath (D) Heating in sand bath Ans.(B) B.pt. gets lowered at low pressure. A bottle containing two immiscible liquids is given to you. They may be separated by using a (A) Fractionating coloumn (B) Separating funnel (C) Vacuum distillation (D) Steam distillation Ans.(B) Being immiscible liquids, they form different layers. Sol. In Lassaigne’s test for nitrogen, the blue colour is due to the formation of (A) Ferric ferrocyanide (B) Potassium ferrocyanide (C) Sodium ferrocyanide (D) Sodium cyanide Ans.(A) In Lassaigne’s test substance is heated strongly with sodium metal then water extract is boilded with alkaline FeSO 4 solution and after cooling FeCI3 solution and excess of HCI is added in it. If prussian blue or green ppt. is obtained then Nitrogen is confirmedNa + C + N  NaCN FeSO4 + 2NaOH  Fe(OH)2 + Na2SO4 6NaCN + Fe(OH)2  Na4[Fe(CN)6] + 2NaOH 3Na4[Fe(CN)6] + 4FeCl3  Fe 4[Fe(CN)6]3 + 12NaCl sodium ferrocyanide prussian blue or green (ferric ferrocyanide) In sodium fusion test of organic compounds, the nitrogen of an organic compound is converted to (A) Sodamide (B) Sodium cyanide (C) Sodium nitrite (D) Sodium nitrate Ans. (B) Na + C + N  fusion  NaCN Ex.5 Liebig method is used for the estimation of - Sol. Ex.4 Sol. Ex.6 (A) Nirogen (B) Sulphur (C) Carbon and Hydrogen (D) Halogens Ans. (C) Liebig method is used for the estimation of carbon and hydrogen. The quantitative determination of halogen in an organic compound is known as(A) Dumas method (B) Carius method (C) Kjeldahl method (D) Leibig method Ans.(B) Sol. Ex.7 Sol. Ex.8 Sol. The quantitative determination of halogen in an organic compound is called as carius method. 0.2475g of an organic compound gave on combustion 0.4950g of carbon dioxide and 0.2025 g of water. The percentage of carbon and hydrogen are(A) 54.54, 9.09 (B) 52.54, 8.09 (C) 120, 5.8 (D) None Ans. (A) Wt. of organic compound = 0.2475 g Wt. of CO2 produced = 0.4950 g Wt. of H2O produced = 0.2025 g Percentage of carbon Wt. of CO 2 12 = × x 100 Wt. of compound 44 12 0.4950 = × × 100 = 54.54 44 0.2475 Percentage of hydrogen Wt. of H2 O 2 = × × 100 Wt . of compound 18 2 0.2025 = × × 100 = 9.09 18 0.2475 0.257 g of an organic substance was heated with conc H2SO4 and then distilled with excess of strong alkali. The ammonia gas evolved was absorbed in 50 ml of N/ 10 HCl which required 23.2 ml of N/10 NaOH for neutralisation at the end of the p r ocess. The percentage of nitrogen in the compound is(A) 14.6 (B) 18.0 (C) 17.0 (D) 15.5 Ans.(A) N Volume of HCl taken = 50ml 10 N Volume of NaOH used for neutralisation 10 of unused acid = 23.2 ml Now N1 V1(NaOH) = N2V2(HCl) N N 23.2 ml of NaOH  23.2 ml of HCl 10 10 N  Volume of HCl unused = 23.2 ml 10  Volume of N/1 0 HC l requi red for neutralization of NH3 = 50 - 23.2 = 26.8 ml N 26.8 ml of HCl = 26.8 ml of N/10 NH3 10 1000 ml of 1N NH3 solution contains nitrogen = 14 g 26.8 ml of NH3 solution contains nitrogen 14 x 26.8 = 10 x 1000 Percentage of nitrogen 14 x 26.8 x 100 = = 14.6% 10 x 1000 x 0.257