ANSWERS
      

Tutorial 1:1












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Tutorial 1.2

Tutorial 1:3

Remember that the conformation of the molecule is not linear. For solid and liquid the most favoured (statistically) is a curled-up sphere and so intramolecular (between different parts of the same molecule) attractive interaction is more significant as compared with that in the gaseous state. Separation of molecules is dependent on intermolecular attraction rather than intramolecular attraction.

The boiling point for 2-methylpropane is -11.7�C, that is lower than butane. This is because it is more spherical and offers less surface area for intermolecular attractive interaction.     Back

Tutorial 1:4

Tutorial 1.5

Yes but it will involve the atoms oxygen and sulphur. The compounds will have different chemistry. So such compounds should be classified differently.     Back

Tutorial 2:1

Chain reaction. Initiation steps: (1) and (2). Propagation steps: (3) and (4). (Note that step (2) and (4) are similar and are labelled differently just to highlight the concept of a chain reaction mechanism.    Back

Tutorial 2:2

  • The concentration effect (otherwise known as the mass effect) and reactivity.
  • The bromine free radical is less relative as compared with chlorine. In general selectivity of a reaction increases with decrease in relativity.    Back

    • Tutorial 2:3

    Polypropylene has a secondary proton and so -CH2-CH(CH3)- should be more suceptable to attack by oxygen in sunlight. For all polyalkanes used outdoors it is a must to use compounds that would prevent such attacks of the proton. These are known as anti-ozonants of anti-oxidants.    Back

    Tutorial 2:4

    There are two possible molecules for sec-hexane CH3CH(CH3)CH2CH2CH3   and   CH3CH2CH(CH3)CH2CH3 and chemists would not know which molecule you are referring to. So it is wrong to use sec-hexane because it is not unique.    Back

    Tutorial 2:5

    The Cl� produced can knock into a methane molecule or its product methyl chloride. So along with methyl chloride you would expect to get methylene chloride (CH2Cl2: b.Pt. 40�C), chloroform (CH2Cl2: b.pt. 61�C) and carbon tetrachloride (b.pt. 77�C). To get the maximum yield of methyl chloride (b.pt. 24�C), your system must use a high methane/chlorine feed ratio and the methyl chloride must be removed continuously from the process before it reacts further to give the by-products.
    The hydrogen chloride can be isolated for sale. It should never, never be discharged into the atmosphere.        Answer

    Tutorial 2:6

    ALKANES
    ΔfHΘ / kJ mol‾�    		 Methane
    - 74.8    		 Ethane
    - 84.7    		 Propane
    - 103.8    		 Butane
    - 126.1    		 Pentane
    - 146.4    		 Hexane
    - 166.7
    Back

    Tutorial 2:7

    Tutorial 3:1

    CC double bond consist of a CC σ−bond + a CC π-bond. A π−bond is a bond formed by the overlap of two p-orbitals of the adjacent carbon atoms.    Back

    Tutorial 3:2

    Reacting an alkene with hydrogen chloride. Here only one alkyl chloride product will be obtained and so there is a great cost saving in the isolation process.    Back

    Tutorial 3:3

        H  Me
        │  │
    H−C=C−Me + H2SO4    		 →     H  Me
        │  │
    H−C−C−Me
        │ nbsp; +
        H    		  +  HSO4‾
    		→       Me
          │
    Me−C−
          │
          Me    		 HSO4
    		 H2O
    →    		       Me
          │
    Me−C−OH
          │
          Me    		  +  H2SO4

    Another name for isobutylene is 2−methyl−1−propene.    Back

    Tutorial 3:4

  • trans-2-chlorocyclohexanol. The chlorine will attack the π-bond followed by the reaction of the carbonium ion with the hydroxide anion.
  • trans-2-bromocyclohexyl methyl ether (trans-2-bromocyclohexyl-O-methyl)    Back

    • Tutorial 3:5

    The mechanism is as discussed for the syn addition. It is a concerted addition with the boron attacking the least substituted carbon owing to steric factor. Since it is a concerted reaction the addition is on the same side and so cis product could not be formed. Since the reaction did not proceed via a carbonium ion it will not follow the Markonikov rule.    Back

    Tutorial 3:6







    		You will find that it is not possible. So they are enantiomers.    Back

    Tutorial 3:7

    Poly(2-methyl-2-butene) or poly(cis-1,4-isoprene). Poly(cis-1,4-isoprene) is the more widely used.    Back

    Tutorial 3:8

    There are four stereosisomers possible.
    3-methylcyclohex-3-ene-exo-carbaldehyde, 3-methylcyclohex-3-ene-endo-carbaldehyde,
    3-methylcyclohex-2-ene-exo-carbaldehyde, and 3-methylcyclohex-2-ene-endo-carbaldehyde.
    The endo poducts are obtained when the acrolein approaches the isoprene with the aldehyde group on the same side of the diene. (Imagine that there is a partition along the cc double bond of the acrolein.) Our diagram shows the acrolein approaching the isoprene with the aldehyde group on the other side of the diene to give the exo isomer. The endo and exo isomers are enantiomers.        Back

    Tutorial 5:1

    "Terminal alkyne is a nucleophile in the nucleophilic substitution of alkyl halide" means terminal alkyne can become a nucleophile to attack alkyl halide, while "acetylene can undergo nucleophilic addition" means acetylene can be attacked by a nucleophile which led to the addition of groups onto acetylene?    Back

    Tutorial 6:1

    Reacting an alkene with hydrogen chloride. Here only one alkyl chloride product will be obtained and so there is a great cost saving in the isolation process.    Back

    Tutorial 6:2

    Owing to the electron-donating-inductive effect ClCH2−C(�)H(C6H5) is more stable than
    ClCH(C6H5)−C(�)H2.        Back

    Tutorial 6:3

    In2 → 2 In� (1)
    In�   +   CH2=CHX → InCH2C(�)HX (2)
    InCH2C(�)HX   +   CH2=CHX → InCH2CHX−CH2C(�)HX (3)

    where X=Cl or CH3COO−. The polymer obtained would be -[CH2CHCl]n-, poly(vinyl chloride) or PVC, and -[CH2CH(OC(O)CH3)]n-, poly(vinyl acetate). PVC is used as imitation leather for bags, shoes, etc. Poly(vinyl acetate) is used from making gramophone records to emulsion paints and adhesives.    Back

    Tutorial 6:4







    		 H+
    ◄`►






    		− H2O
    ◄`►





    		H+
    ◄`►





    118.2 g of pinocal should yield 100.2 if the conversion is complete (100%). Or 30 g should yield 25.4 g. So yield is (17.8/25.4)x100% = 70%.

    If the pinacolone can be removed as it is formed this should drive the reaction forward. The pinacolone is an organic compound while the other products are water miscible. One setup that can do this is as shown. B is graduated like a burette. The distillate should collect in B and when it is full the bottom portion should overflow back into the reaction flask.

    Measure 23 ml of pinacol into a 250 ml round-bottom flask followed by 130 ml of 6N sulphuric acid. Reflux the mixture to about 120�C. Monitor the volume of organic liquid collected in B. When the volume is constant it is assumed that the reaction is over. Dry the product and purify by distillation. Collect the 103 - 107�C portion.    Back

    Tutorial 6:5

    The initial reaction will produce the desired vinyl chloride. When there are both acetylene and vinyl chloride present in the system, the HCl acid is more likely to attack the vinyl chloride rather than the acetylene since the alkyl carbonium ion is preferred. So we end up with a high amount of 1,1-dichloroethylene. We would need another process to remove a molecule of HCl from 1,1-dichloroethylene to give vinyl chloride.    Back

    Tutorial 6:6

    When an anion reacts with a proton it is labelled a base. When it attacks and bonds with a carbon it is labelled a nucleophile.      Back
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