Answers to Topic 16 Test Yourself Questions

Answers to Topic 16 Test yourself questions

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© G. Hill and A. Hunt 2008 Edexcel Chemistry for AS CD-ROM

1a)pentane

b)1,2-dichloropentane

c)2-bromopentane

3a)Poly(ethene), CH2=CH2, ethene

b)Poly(chloroethene), CH2=CHCl, chloroethene

4CH3CHBrCH3 + OH–→

CH2=CHCH3 + H2O + Br–

5a)Chlorine in the presence of ultraviolet light.

b)Heat with iodine and red phosphorus.

c)Heat with alcoholic ammonia.

d)Heat with aqueous potassium hydroxide.

6Heat with a limited amount of a solution of potassium dichromate(vi) in dilute sulfuric acid and distil off the product as it forms.

7a)CH3CH2CHOHCH3 + [O] →

CH3CH2COCH3 + H2O

b)CH3CH2CH2CH2CH2OH + 2[O] →

CH3CH2CH2CH2COOH + H2O

8a)Acidify the mixture with nitric acid and then add silver nitrate solution.

b)A bright yellow precipitate of silver iodide forms in the presence of iodide ions.

9Covalent bonds have to break to produce free radicals. This takes energy because the bond energies of most covalent bonds are relatively high. The energy to break the bonds comes either from the kinetic energy of fast moving molecules as they collide or from photons of UV light.

11Alkanes are relatively inert. The reaction with halogens produce more reactive halogenoalkanes which are themselves useful and can also be converted to other valuable chemicals.

12a)Chloromethane and hydrogen chloride

b)Methyl radicals form during the free radical chain reaction. A possible termination step is two methyl radicals joining to form an ethane molecule (see Section 6.3 in the Student’s Book).

c)Further substitution of chlorine for hydrogen is possible when chloromethane reacts with chorine atoms in the reaction mixture.

13See Figure 5.17 in Section 5.6 of the Student’s Book. The bond breaking in chloromethane is similar to that in bromomethane. The chlorine atom gains an electron from carbon when the bond breaks heterolytically so that it becomes a chloride ion. This leaves a positive charge on the carbon atom.

14Free radicals: Cl, Br, CH3

Electrophiles: Cl2, HBr, Br2, H+

Nucleophiles: H2O, Br–, OH–, NH3

16a)Cis–trans (geometric) isomerism

b)Cis and trans

17Free rotation about the single bond in
1,2-dibromoethane means that there is no possibility of distinguishing cis and trans isomers. The ring in 1,2-dibromocyclohexane prevents rotation about the C−C bond.

18a)A bromide ion is set free during the first step of electrophilic addition. This makes it possible for it to approach the C+ carbon atom from the other side of the molecule.

b)The relatively large size of bromine atoms means that, during the second step, it is easier for the larger bromide ion to combine with the C+ carbon atom from the other side of the molecule.

20A C−I bond is less polar than a C−Cl bond. A C−I bond is weaker than a C−Cl bond. The trend correlates better with bond strength.

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