Worksheet-01-Answers


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Worksheet-01-Answers

  • 21.1 What is Organic Chemistry (SB p.4)
  • Check Point 21-1
  • How was organic chemistry defined before 1800s?
  • Answer
  • (a) The knowledge of organic and inorganic compounds was raised during the 1780s. Scientists defined organic chemistry as the study of compounds that could be obtained from living organisms. They believed that the synthesis of organic compounds took place in living organisms only.
  • 12/27/19
  • Dr Seemal Jelani
  • 21.1 What is Organic Chemistry (SB p.4)
  • Check Point 21-1
  • (b) How is organic chemistry defined nowadays?
  • Back
  • Answer
  • (b) Nowadays, scientists have discovered that many organic compounds can be synthesized from inorganic substances. The updated definition of organic chemistry is the study of carbon compounds, except for carbon monoxide, carbon dioxide, carbonates, hydrogencarbonates, carbides and cyanides. These compounds have been traditionally classified under inorganic chemistry.
  • 12/27/19
  • Dr Seemal Jelani
  • 21.2 The Unique Nature of Carbon (SB p.5)
  • Why is carbon able to catenate?
  • Let's Think 1
  • Back
  • Answer
  • The ability to catenate of carbon is chiefly due to the high strength of the CC single bond (bond enthalpy of C  C single bond is 356 kJ mol-1).
  • 12/27/19
  • Dr Seemal Jelani
  • 21.2 The Unique Nature of Carbon (SB p.6)
  • Would you expect silicon, which is just below carbon in the Periodic Table, to catenate to form diverse molecular structures? Explain your answer.
  • Example 21-2
  • Answer
  • Silicon, unlike carbon, does not catenate to form diverse molecular structures. Carbon is able to catenate because carbon atoms have a relatively small atomic size. This enables a carbon atom to form strong covalent bonds with other carbon atoms. However, due to the greater atomic size of silicon, its ability to catenate is much lower than that of carbon.
  • Back
  • 12/27/19
  • Dr Seemal Jelani
  • 21.2 The Unique Nature of Carbon (SB p.7)
  • Check Point 21-2
  • Back
  • Answer
  • The electronic configuration of sulphur is 1s22s22p63s23p4. It has only two unpaired electrons. Its atomic size is larger than that of carbon. So it has a much lower tendency to catenate than carbon.
  • 12/27/19
  • Dr Seemal Jelani
  • 21.3 Classification of Organic Compounds (SB p.14)
  • Identify the functional group(s) in the following compounds:
  • (a)
  • Example 21-3A
  • Answer
  • Carbon-carbon double bond ( ) and chloro group (Cl)
  • 12/27/19
  • Dr Seemal Jelani
  • 21.3 Classification of Organic Compounds (SB p.14)
  • Identify the functional group(s) in the following compounds:
  • (b)
  • Example 21-3A
  • Answer
  • (b) Carbonyl group ( )
  • 12/27/19
  • Dr Seemal Jelani
  • 21.3 Classification of Organic Compounds (SB p.14)
  • Identify the functional group(s) in the following compounds:
  • (c)
  • Example 21-3A
  • Answer
  • (c) Amino group ( ) and carboxyl group ( )
  • Back
  • 12/27/19
  • Dr Seemal Jelani
  • 21.3 Classification of Organic Compounds (SB p.15)
  • To which homologous series does each of the following compounds belong?
  • (a)
  • Example 21-3B
  • Answer
  • (a) Ester
  • 12/27/19
  • Dr Seemal Jelani
  • 21.3 Classification of Organic Compounds (SB p.15)
  • To which homologous series does each of the following compounds belong?
  • (b)
  • Example 21-3B
  • Answer
  • (b) Amide
  • 12/27/19
  • Dr Seemal Jelani
  • 21.3 Classification of Organic Compounds (SB p.15)
  • To which homologous series does each of the following compounds belong?
  • (c)
  • Example 21-3B
  • Answer
  • (c) Acid anhydride
  • Back
  • 12/27/19
  • Dr Seemal Jelani
  • 21.3 Classification of Organic Compounds (SB p.15)
  • State whether each of the following pairs of compounds belongs to the same homologous series. Explain your answer.
  • (a)
  • Example 21-3C
  • Answer
  • (a) No, the first one is a carboxylic acid and the second one is an ester.
  • 12/27/19
  • Dr Seemal Jelani
  • 21.3 Classification of Organic Compounds (SB p.15)
  • State whether each of the following pairs of compounds belongs to the same homologous series. Explain your answer.
  • (b)
  • Example 21-3C
  • Answer
  • (b) Yes, both of them are alcohols.
  • 12/27/19
  • Dr Seemal Jelani
  • 21.3 Classification of Organic Compounds (SB p.15)
  • State whether each of the following pairs of compounds belongs to the same homologous series. Explain your answer.
  • (c)
  • Example 21-3C
  • Answer
  • (c) No, the first one is an amide and the second one is an amine.
  • Back
  • 12/27/19
  • Dr Seemal Jelani
  • 21.3 Classification of Organic Compounds (SB p.16)
  • Check Point 21-3
  • (a) Name the homologous series of organic compounds that contain oxygen atoms in their functional groups.
  • Answer
  • (a) Alcohol, ether, aldehyde, ketone, carboxylic acid, ester, acyl halide, amide and acid anhydride
  • 12/27/19
  • Dr Seemal Jelani
  • 21.3 Classification of Organic Compounds (SB p.16)
  • Check Point 21-3
  • Identify and name the functional groups in glucose which has the following structure.
  • Answer
  • (b) OH (hydroxyl group) and  O  (oxy group)
  • 12/27/19
  • Dr Seemal Jelani
  • 21.3 Classification of Organic Compounds (SB p.16)
  • Check Point 21-3
  • (c) Identify and name the functional groups in the following compounds:
  • Answer
  • Back
  • (c) Br (bromo), (aldehyde), (acyl chloride), (carbon-carbon double bond) groups
  • 12/27/19
  • Dr Seemal Jelani
  • 21.4 Factors Affecting the Physical Properties of Organic Compounds (SB p.18)
  • Why is oil immiscible with water?
  • Let's Think 2
  • Answer
  • Oil molecules do not have free OH groups, so they cannot form hydrogen bonds with water molecules.
  • Back
  • 12/27/19
  • Dr Seemal Jelani
  • 21.4 Factors Affecting the Physical Properties of Organic Compounds (SB p.20)
  • The relative molecular mass of glucose is 180.0, but it is soluble in water. Why?
  • Let's Think 3
  • Answer
  • Glucose molecules have OH groups, so they are able to form hydrogen bonds with water molecules. Therefore, glucose is soluble in water despite it has a high molecular mass.
  • Back
  • 12/27/19
  • Dr Seemal Jelani
  • 21.4 Factors Affecting the Physical Properties of Organic Compounds (SB p.20)
  • Despite the fact that butan-1-ol and ethoxyethane have the same relative molecular mass, they have very different boiling points. The boiling points of butan-1-ol and ethoxyethane are 117oC and 35oC respectively. Explain the difference.
  • Example 21-4A
  • Answer
  • 12/27/19
  • Dr Seemal Jelani
  • 21.4 Factors Affecting the Physical Properties of Organic Compounds (SB p.20)
  • Example 21-4A
  • There is an OH group in butan-1-ol. Thus, butan-1-ol molecules are able to form hydrogen bonds with one another and the energy required to separate butan-1-ol molecules would be much greater. Whereas for ethoxyethane, the attraction among the molecules is weak van der Waals’ forces only. The amount of energy required to break the forces would not be great. Therefore, the boiling point of ethoxyethane is lower than that of butan-1-ol.
  • Back
  • 12/27/19
  • Dr Seemal Jelani
  • 21.4 Factors Affecting the Physical Properties of Organic Compounds (SB p.21)
  • Example 21-4B
  • Explain why propan-1-ol is soluble in water but 1-chloropropane is insoluble in water.
  • Answer
  • The  OH group of propan-1-ol molecules enables it to form hydrogen bonds with water molecules. Thus it is soluble in water. Although 1-chloropropane is a polar molecule, it does not form hydrogen bonds with water molecules. So it is insoluble in water.
  • Back
  • 12/27/19
  • Dr Seemal Jelani
  • 21.4 Factors Affecting the Physical Properties of Organic Compounds (SB p.21)
  • Which molecule would have a higher boiling point, 1-bromobutane or 2-bromobutane? Why?
  • Let's Think 4
  • Answer
  • 1-bromobutane would have a higher boiling point. 1-bromobutane is a straight-chain molecule while 2-bromobutane is a branched-chain molecule. Straight-chain molecules have a greater surface area in contact with each other, so greater intermolecular forces exist among the molecules. Higher energy is required to break down the intermolecular forces among the molecules of 1-bromobutane.
  • Back
  • 12/27/19
  • Dr Seemal Jelani
  • 21.4 Factors Affecting the Physical Properties of Organic Compounds (SB p.22)
  • Example 21-4C
  • 1-Chlorobutane and 2-chloro-2-methylpropane have the same molecular mass, yet their melting points differ. The melting point of 1-chlorobutane is –123oC while that of 2-chloro-2-methylpropane is –27.1oC. Explain the difference.
  • Answer
  • 12/27/19
  • Dr Seemal Jelani
  • 21.4 Factors Affecting the Physical Properties of Organic Compounds (SB p.22)
  • Example 21-4C
  • Melting point is a measure of how efficient the molecules are packed together in the solid state instead of just comparing the van der Waals’ forces among molecules. Hence melting point is a function of the efficient packing of molecules but not the contact surface area. 1-Chlorobutane is a straight-chain molecule while 2-chloro-2-methylpropane is a branched-chain molecule. As 2-chloro-2-methylpropane is more spherical and symmetrical, its molecules are packed more efficiently in the solid state. 1-Chlorobutane is linear in shape and flattened, its packing in the solid state is not so efficient. Hence, it has a lower melting point than 2-chloro-2-methylpropane.
  • Back
  • 12/27/19
  • Dr Seemal Jelani
  • 21.4 Factors Affecting the Physical Properties of Organic Compounds (SB p.24)
  • Check Point 21-4
  • (a) What are the major factors that affect the physical properties of organic compounds?
  • Answer
  • (a) The physical properties of organic compounds are mainly affected by the structure of the functional groups, dipole moment of the molecule, the formation of hydrogen bonding between molecules, and the length of carbon chains of the molecule.
  • 12/27/19
  • Dr Seemal Jelani
  • 21.4 Factors Affecting the Physical Properties of Organic Compounds (SB p.24)
  • Check Point 21-4
  • The melting point and boiling point of pentane are –130oC and 36.3oC respectively while the melting point and boiling point of 2,2-dimethylpropane are –15.9oC and 9.5oC respectively. Account for the difference in melting point and boiling point between the two isomers.
  • Answer
  • 12/27/19
  • Dr Seemal Jelani
  • 21.4 Factors Affecting the Physical Properties of Organic Compounds (SB p.24)
  • Check Point 21-4
  • (b) Pentane is a straight-chain molecule, while 2,2-dimethylpropane is a branched-chain molecule. Straight-chain molecules have a greater surface area in contact with each other than branched-chain molecules. Straight-chain molecules are held together by stronger intermolecular forces. Therefore, pentane has a higher boiling point than 2,2-dimethylpropane. Molecules of 2,2-dimethylpropane are more spherical in shape and are packed more efficiently in the solid state. Molecules of pentane are linear in shape and flattened, so their packing in the solid state is not efficient. Since extra energy is required to break down the efficient packing of 2,2-dimethylpropane, 2,2-dimethylpropane has a higher melting point than pentane.
  • 12/27/19
  • Dr Seemal Jelani
  • 21.4 Factors Affecting the Physical Properties of Organic Compounds (SB p.24)
  • Check Point 21-4
  • (c) Which molecule, hexane or cyclohexane, would have a higher melting point? Explain your answer.
  • Answer
  • 12/27/19
  • Dr Seemal Jelani
  • 21.4 Factors Affecting the Physical Properties of Organic Compounds (SB p.24)
  • Check Point 21-4
  • (c) Cyclohexane has a higher melting point than hexane. Molecules of cyclohexane are more spherical in shape and are packed more eff iciently in the solid state. Molecules of hexane are linear in shape and flattened, so their packing in the solid state is not efficient. Since extra energy is required to break down the efficient packing of cyclohexane, cyclohexane has a higher melting point than hexane.
  • 12/27/19
  • Dr Seemal Jelani
  • 21.4 Factors Affecting the Physical Properties of Organic Compounds (SB p.24)
  • Check Point 21-4
  • (d) Arrange the following molecules in increasing order of boiling points. Explain your answer.
  • Answer
  • 12/27/19
  • Dr Seemal Jelani
  • 21.4 Factors Affecting the Physical Properties of Organic Compounds (SB p.24)
  • Check Point 21-4
  • (d) The boiling points increase in the order: butane < propanal < propan-1-ol
  • Molecules of butane are non-polar. Their molecules are held together by weak instantaneous dipole-induced dipole interactions. A relatively small amount of energy is required to separate the molecules in the process of boiling. Both propanal and propan-1-ol are polar molecules. Molecules of propanal are held together by relatively weak dipole-dipole interactions, while molecules of propan-1-ol are held together by intermolecular hydrogen bonds. Since the intermolecular forces present in molecules of propan-1-ol are stronger than those present in molecules of propanal, a larger amount of energy is required to separate the propan-1-ol molecules in the process of boiling.
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  • 12/27/19

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