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ESTIMATION

Explore problems on estimating carbon, hydrogen, nitrogen, bromine, sulfur percentages in organic compounds. Includes Dumas, Kjeldahl, Carius methods with detailed solutions and calculations.

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ESTIMATION

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  1. ESTIMATION

  2. Problem 12.20 On complete combustion, 0.246g of an organic compound gave 0.198gm of carbon dioxide and 0.1014 gm of water. Determine the percentage composition of carbon and hydrogen in the compound. Solution Percentage of carbon = 12 × 0.198 × 100 = 21.95% Percentage of hydrogen = 2 × 0.1014 × 100 =4.58% 18 × 0.246 44 × 0.246

  3. Problem 12.21 In Dumas method for estimation of nitrogen,0.3g of an organic compound gave 50ml of nitrogen collected at 300K temperature and 715mm pressure. Calculate the percentage composition of nitrogen in the compound. Aqueous tension at 300K = 15mm Solution Volume of nitrogen collected at 300k and 715mm pressure is 50ml. Actual pressure = 715-15 = 700mm Volume of nitrogen at STP=273 × 700 × 50 / 300 × 760 = 41.9ml 22.400ml of N2 at STP weighs = 28g 41.9ml of N2 weighs = 28 × 41.9g / 22400. Percentage of N2 = 28 × 41.9 ×100 / 22400 × 0.3 = 17.46%.

  4. Problem 12.22 During estimation of nitrogen present in an organic compound by Kjeldahl’s method., the ammonia evolved from 0.5g of the compound in Kjedahl’s estimation of nitrogen, neutralized 10mL of 1m H2SO4. Find out the percentage of nitrogen in the compound Solution 1m of 10mL H2SO4 =1m of 20ml NH3 1000 mL of 1M of ammonia contains 14g nitrogen. 20mL of 1M ammonia contains 14 × 20 / 1000 g nitrogen Percentage of nitrogen= 14 × 20 × 100 / 1000 × 0.5 = 56.0%.

  5. Problem 12.23 In carius method of estimation of halogen.0.15 gm of an organic compound, gave 0.12g of AgBr. Find out the percentage of bromine in the compound. Solution Molar mass of AgBr = 108 + 80=188g mol-1 1.188g AgBr contains 80g bromine. 0.12g AgBr contains 80 × 0.12 / 188 g bromine. Percentage of bromine =80 × 0.12 × 100 / 188 × 0.15 = 34.04%.

  6. Problem 12.24 In sulphur estimation, 0.157 g of an organic compound gave 0.4813 g of barium sulphate. What is the percentage of sulphur in the compound? Solution Molecular mass of BaSO4 = 137 + 32 + 64 = 233 g 233 g BaSO4 contains 32 g Sulphur 0.4813 g BaSO4 contains 32 × 0.4813 / 233 g Sulphur Percentage of Sulphur = 32 × 0.4813 × 100 / 233 × 0.157 = 42.10%

  7. Chapter 13 HYDROCARBONS

  8. Problem 13.1 Write structures of different chain isomers of alkanes corresponding to the molecular formula C6H14. Also write their IUPAC names Solution CH3-CH2-CH2-CH2-CH2-CH3 n-Hexane CH3-CH-CH2-CH2-CH32-Methylpentane CH3 CH3-CH2-CH-CH2-CH3 3-Methylpentane CH3-CH-CH-CH3 2,3-Dimethylbutane CH3-C-CH2-CH3 2,2-Dimethylbutane CH3 CH3 CH3 CH3 CH3

  9. Problem 13.2 Write structures of different isomeric alkyl groups corresponding to the molecular form C5H11. write IUPAC names of alcohols obtained by attachment of-OH groups at different carbons of the chains. Solution Structures of C5H11 groups • CH3-CH2-CH2-CH2-CH2- • CH3-CH-CH2-CH2-CH3 • CH3-CH2-CH-CH2-CH2 • CH3-CH-CH2-CH2- • CH3-CH2-CH-CH2- • CH3-C-CH2-CH3 • CH3-C-CH2- Corresponding alcohols CH3-CH2-CH2-CH2-CH2- CH3-CH-CH2-CH2-CH3 CH3-CH2-CH-CH2-CH2 CH3-CH-CH2-CH2- CH3-CH2-CH-CH2- CH3-C-CH2-CH3 CH3-C-CH2- Pentan-1-ol Pentan-2-ol Pentan-3-ol 3-Methylbutan-1-ol 2-Methylbutan-1-ol 2-Methylbutan-2-ol 2,2-Dimethyl propan-1-ol CH3 CH3 CH3 CH3 CH3 CH3 CH3 CH3 CH3 CH3

  10. Problem 13.3 Write IUPAC names of the following compounds • (CH3)3CCH2C(CH3)3 • (CH3)2C(C2H5)2 • Tetra-tert-butyl methane • 2,2,4,4- Tetramethylpentane • 3,3- Dimethylpentane • 3,3-Di-tert-butyl-2,2,4,4- Tetramehylpentane Solution

  11. Problem 13.4 • Write structures of the following compounds: • 3,4,4,5-Tetramethylheptane • 2,5-Dimethyhexane • Solution • CH3-CH2-CH-C-CH-CH-CH3 • CH3-CH-CH2-CH2-CH-CH3 CH3 CH3 CH3 CH3 CH3 CH3

  12. Problem 13.5 • Write structures for each the following compounds. Why are the given • names incorrect? Write correct IUPAC names. • 2-Ethylpentane • 5-Ethyl-3-methylheptane • Solution • CH3-CH-CH2-CH2-CH3 • CH3-CH2-CH-CH2-CH-CH2-CH3 Longest Chain is of 6 C atoms and not that of 5. hence, correct name is 3-Methylhexane. Numbering is to be started from the end which gives lower number to ethyl group. Hence, correct name is 3-ethyl-5-methylheptane C2H5 7 6 5 4 3 2 1 C2H5 CH3

  13. Problem 13.6 Sodium salt of which acid will be needed for the preparation of propane? Write chemical equation for the reaction. Solution Butanoic Acid CH3CH2CH2COONa + NaOH CH3CH2CH3 + Na2CO3 - + CaO

  14. Problem 13.7 • Write IUPAC names of the following compounds. • (CH3)2CH-CH=CH-CH2-CH • CH3-CH-CH • CH2=C(CH2CH2CH3)2 • (CH3)CH2CH2CH2 CH2CH3 • CH3-CHCH=C-CH2-CHCH3 • CH3 • Solution • 2,8- Dimethyl-3,6-decadiene • 1,3,5,7 Octatetraene • 2-n-Proplypent-1-ene • 4-Ethyl-2,6-Dimethyl-dec-4-ene Problem 13.8 Calculate no. of σ and π bonds in the above structures:- Solution σ bonds π bonds C2H5 • 33 2 • 7 4 • 23 1 • 41 1

  15. Problem 13.9 Write structures & IUPAC names of different structural isomers of alkenes corresponding to the C5H10. Solution CH2=CH-CH2-CH2-CH3 Pent 1-ene CH3-CH=CH-CH2-CH3 Pent 2-ene CH3-C=CH-CH3 2-Methylbut-2-ene CH3-CH-CH=CH2 3-Methylbut-1-ene CH2=C-CH2-CH3 2-Methylbut-1-ene CH3 CH3 CH3

  16. Problem 13.10 • Draw cis and trans isomers of the following compounds. Write their IUPAC names. • CHCl=CHCl • C2H5CCH3=CCH3C2H5 • Solution • C = C C=C • C=C C=C Cl H 1. H H H Cl Cl Cl 2. C2H5 CH3 CH3 CH3 C2H5 C2H5 CH3 C2H5

  17. Problem 13.11 • Which of the following compounds show cis-trans isomerism? • (CH3)2C = CH - C2H5 • CH2 = CBr2 • C6H5CH = CH - CH3 • CH3CH = CClCH3 • Solution • 3RD AND 4TH • In 1st and 2nd structures, two identical groups are attached to one of the doubly bonded carbon atoms

  18. Problem 13.12 Write IUPAC names of the products obtaine dby addition reactions of HBr to Hex-1-ene in absence of peroxide and In presence of peroxide Solution CH2=CH-CH2-CH2-CH2-CH3 + H-Br Hex-1-ene 2-Bromohexane CH3-CH-CH2-CH2-CH2-CH3 + H-Br CH2-CH2-CH2-CH2-CH2-CH3 1-Bromohexane No Peroxide CH3-CH-CH2-CH2-CH2-CH2-CH3 Br Peroxide

  19. Problem 13.13 Write structures of different isomers corresponding to the 5th member of alkyne series. Also write IUPAC names of all the isomers. What type of isomerism is exhibited by different pairs of isomers ? Solution 5th member of alkyne has the molecular formula C6H10, the possible isomers are: A)HC ≡ C-CH2-CH2-CH2-CH3Hex-1-yne B)CH3-C ≡ C-CH2-CH2-CH3Hex-2-yne C)CH3 CH2-C ≡ C-CH2–CH3Hex- 3yne D)HC ≡ C–CH-CH2-CH33–Methylpent-1-yne CH3 Position and chain isomerism shown by different parts. E)HC≡ C-CH2-CH-CH3 CH3 4-methylpent -1-yne F)CH3 –C C –CH –CH3 CH3 4-methylpent -2-yne CH3 G)HC C-C-CH3 CH3 3,3-dimethylbut- 1-yne

  20. NaOH(aq) Sodalime Problem 13.14 How will you convert ethanoic acid into benzene ? Solution CH3COOH CH3COONa CH4 C2H5CL C2H6 CH3Cl CH2=CH2 CH2Br CH2=CHBr CH=CH Na/ Dry Ether Wutz reaction Cl2 Br2 Alc. KOH CH2Br Red Hot Iron Tube 873 K

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