A Data Booklet may be used.
CommonCH4 and Cl2 react together in the presence of ultraviolet light. Which statements are correct?
Source: 2022 HCI J1 Promo P1 Q18
1,1-dimethylcyclopentane undergoes free-radical substitution with limited bromine in the presence of ultraviolet light. A mixture of mono-brominated products is obtained.
Assuming that the reaction occurs at the same rate at every carbon atom, what is the total number of mono-brominated products (ignore stereoisomers) obtained?
Source: 2022 HCI J1 Promo P1 Q19
Use of the Data Booklet is relevant to this question. Chlorofluorocarbons (CFCs) were widely used as refrigerants, solvents and propellants. However, the manufacture of CFCs has been phased out, as they undergo homolytic fission by ultraviolet radiation in the stratosphere, which contributes to ozone depletion. Which radical could result from the irradiation of CHFClCF2Cl?
Source: VJC 2012 Prelim P1 Q17, via EJC Alkanes Topical Revision, Q2
What are the products from the following reaction? ĊH3 + CH3CH2Cl → ?
Source: DHS 2013 Prelim P1 Q13, via EJC Alkanes Topical Revision, Q6
CommonCrude oil consists of a mixture of hydrocarbons, namely alkanes, cycloalkanes and various aromatic hydrocarbons.
Source: MJC 2018 Promo Section B Q1(a) and (b), via EJC Alkanes Topical Revision, Section B Q6
Octane has a total of 18 isomers. Isomers of octane may be distinguished by the number of mono-brominated products formed upon reaction with bromine under ultraviolet light. Draw an isomer of octane which reacts with limiting bromine to give
Source: RJC H2 Chemistry Tutorial 11 (Alkanes), Q7(b)(i)
Source: 2024 JPJC J2 Prelim P2 Q2(a)
Source: RJC H2 Chemistry Tutorial 11 (Alkanes), Q3
A3 only[1]
B3[1]
There are a total of 6 primary hydrogens and 8 secondary hydrogens. Substitution of any of them gives a mono-brominated product. However, due to an internal plane of symmetry, mono-substitution of hydrogen on carbon-2 or carbon-5 gives the same product. The case is similar for carbon-3 or carbon-4, and for the 2 primary methyl groups on carbon-1. Hence there will be a total of 3 different mono-substituted products.
A•CHFCF2Cl[1]
Homolytic fission of the C–Cl bond occurs, as the C–Cl bond is the weakest, compared with the C–F and C–H bonds.
ACH4 + CH3ĊHCl[1]
In this free radical substitution reaction, there is a higher probability of the methyl radical abstracting an H atom from chloroethane to form methane, than of it colliding with the C atom, which is sterically hindered by H atoms. Hence, option A is preferred to options C and D.
In addition, for option D, the formation of a hydrogen radical is highly unlikely and unfavourable. Option B will not happen, as there are too many bonds broken and formed simultaneously.
Free radical substitution.
Initiation: [1]
Propagation: [1]
Termination (any one): [1]
Both compounds have simple molecular structures. 2-bromobutane has a straight chain, but 2-bromo-2-methylpropane has a branched chain.[1]
Extent of surface area of contact, extent of distortion of the electron cloud, and so extent of the (weak) intermolecular instantaneous dipole–induced dipole attractions are greater for 2-bromobutane than for 2-bromo-2-methylpropane.[1]
More energy is required to overcome the instantaneous dipole–induced dipole attractions in 2-bromobutane than in 2-bromo-2-methylpropane. Hence, the boiling point of 2-bromobutane is higher than that of 2-bromo-2-methylpropane.[1]
The catalyst in the catalytic converter is a heterogeneous catalyst. The reactant molecules adsorb onto the surface of the metal catalyst by forming weak, temporary bonds.[1]
This weakens the bonds in the reactant molecules and so provides an alternative reaction pathway of lower activation energy. The surface concentration of the reactants also increases. The product molecules then desorb from the catalyst surface, making the surface available for the adsorption of new reactant molecules.[1]
A: CH3CCl2CH3 (gives only CH3CCl2CH2Cl)[1]
B: ClCH2CH2CH2Cl (gives ClCH2CHClCH2Cl and Cl2CHCH2CH2Cl)[1]
C: CH3C*HClCH2Cl (gives CH3CHClCHCl2, CH3CCl2CH2Cl and ClCH2CHClCH2Cl)[1]
D: CH3CH2CHCl2 (gives CH3CH2CCl3, CH3CHClCHCl2 and ClCH2CH2CHCl2)[1]
E: CH3C*HClCHCl2 (optically active)[1]