Chapter 24 Essay Example | Topics and Well Written Essays - 2500 words

Chapter 24 - Essay ExampleWhereas the aromatic changes are those mixs which overhear a cyclic carbon chain with alternating double bonds. A very important group of aromatic compounds includes benzene which is a cyclic compound with 6 carbon pieces and alternate double bonds.24.3 The hydrocarbons in which the adjacent carbon atoms have only one carbon-carbon bond are called sodding(a) hydrocarbons. These are called saturated compounds because all the four carbon valences are satisfied and no more hydrogen can be attached to the carbon atom. These hydrocarbons are represented by the formula CnH2n+2. Example of saturated hydrocarbon is ethane (C2H6).The hydrocarbons which contain one or more a double or triple bond in the compound are called unsaturated hydrocarbons. These compounds are called unsaturated because all the carbon valences are not satisfied by hydrogen atoms. The alkenes and alkynes are the examples of unsaturated compounds represented by the formula CnH2n and CnH2n- 2. Example of unsaturated hydrocarbon is ethane (C2H4).24.9 A carbon atom in a compound is called chiral, if the mirror image of this compound cannot be layered on itself. In a simplified way, if all the four substituent of a carbon atom are different, then such a carbon atom will be chiral. An example of such a compound is bromochlorofluoromethane.24.25 CH3CH(NH2)COOH and CH2(OH)CH(NH2)COOH are chiral amino acids as all the four substituent of the central carbon atom are different, so the... 24.27 Structural formula ofa) 3-methylhexaneb) 1,3,5-trichloro-cyclohexanec) 2,3-dimethylpentane d) 2-bromo-4-phenylpentanee) 3,4,5-trimethyloctane24.31 Structures area) 1-bromo-3-methylbenzeneb) 1-chloro-2-propylbenzenec) 1,2,4,5-tetramethylbenzene24.34 Lewis Structurea) Alcoholb) Etherc) Aldehyded) Ketonee) Carboxylic acidf) Esterg) Amine24.41 The products area) CH3-CH2-COOH + H2Ob) H2-C=CH-CH3c) Not Clear24.43 The possible isomers of C7H7Cl with a benzene ring area) b).c) d) 24.49 GivenDens ity of octane = 0.70 g/mlVolume of octane = 1 lt = 1000 ml wherefore mass of octane is 700 gAs the molecular weight of octane is 114.23 g/mole so the total moles of octane burnt are (700/114) = 6.13 molesWe know each mole of octane requires 12.5 moles of oxygen for complete combustion. Therefore for the combustion of 1 lt of octane, the oxygen required will be (12.5*6.13 =) 77.875 moles. It is known that multitude of 1 mole of a gas at 20oC is 24.04 lt. So the volume of oxygen required would be (77.875*24.04 =) 1872.2 lt. Given that oxygen is 22% of the communicate. So the total volume of air required is (1872.2*100/22 =) 8510 lt. Answer 8510 lt. of air is required for complete combustion of 1 lt. of octane. 24.53 The structural isomers of C4H8Cl2 are1. 1,1-dichlorobutane2. 1,2-dichlorobutaneThis is a chiral structure.3. 1,3-dichlorobutaneThis structure has a chiral carbon.4. 1,4-dichlorobutane5. 2,3-dichlorobutaneThis structure is chiral.6. 1,1-dichloro-2-methyl-propane7. 1,3-dic hloro-2-methyl-propane8. 1,2-dichloro-2-methyl-propane24.59 The structure ofa) Cyclopentane b)