NJC Alkane 2023 Tutorial Answers

1 HYDROCARBONS - ALKANES AND CYCLOALKANES TUTORIAL Suggested answers Structure and naming 1 Give the IUPAC name for each compound. Identify any chiral center and draw its stereoisomers, if any. (a) CH3CH2CH2CH(CH3)CH2CH2CH3 4-methylheptane; no chiral C (b) CH3CH(CH3)CH(CH3)CH2CH3 2,3-dimethylpentane CH(CH3)2 C CH2CH3CH3 H enantiomers CH(CH3)2 C CH3 H CH3CH2 (c) 1,3-dimethylcyclohexane (2 chiral centres with internal mirror plane; actual no of stereoisomers = 3) non-superimposable mirror images C C CH3 HCH3 H C C CH3 H CH3 H ; superimposable mirror images (identical structures) C C H HCH3 CH3 C C H H CH3 CH3 2 Give the structural formula of the following alkanes. (a) 4-ethyl-3,4-dimethylheptane C CH H H C H H C CH3 H C CH2 CH3 C H H CH3 C H H H H H (b) a saturated compound with molecular formula C5H10 (cannot be alkene; must be cycloalkane - any of the structures below)

2 Physical Properties 3 Crude oil consists mainly of alkanes, the saturated hydrocarbons. The alkanes in crude oil can be separated because they have different boiling points. The table below shows the boiling points of some alkanes. Alkane Boiling point / oC Mr Butane 0 58 Pentane 36 72 Hexane 69 86 2-methylbutane 28 72 2,2-dimethylpropane 10 72 3-methylpentane ? 86 2,3-dimethylbutane 58 86 (a) State what is meant by the term saturated hydrocarbon. A compound containing: - only C and H atoms where all C atoms form 4 single covalent bonds each. - only single bonds, no multiple bonds (b) (i) Explain the trend in boiling points of the straight chain alkanes. boiling point increases from butane to hexane, due to: - increasing Mr; hence size of electron cloud increases - increasing electron cloud size leads to more easily induced dipole/ greater ease of distortion - strength of intermolecular id-id interactions increases, - bp increases as more E is required to break the increasingly stronger IMF (ii) Explain the difference in the boiling points of the three isomers with Mr = 72. - decreasing order of bp: pentane > 2-methylbutane > 2,2-dimethylpropane - increasing branching of isomers leads to smaller surface area of contact between molecules - hence weaker intermolecular id-id interactions - less E needed to break the IMF and hence decreasing bp in branched compounds (iii) With reference to your answer in b(ii), suggest a possible value for the boiling point of 3-methylpentane. Any value: 58 oC - 69 oC (higher than 2,3-dimethylbutane but lower than hexane)

3 Combustion of hydrocarbon 4 What is the volume of CO2 formed when y cm3 of a 50:50 mixture of methane and propane is completely burnt? A 2y cm3 B 5𝑦 2 cm3 C 4y cm3 D 5y cm3 CH4 + 2O2 → CO2 + 2H2O V / cm3 ½y ½y C3H8 + 5O2 → 3CO2 + 4H2O V / cm3 ½y 2 3y Total vol of CO2 produced = 2y cm3 Ans: A 5 When 15 cm3 of a gaseous hydrocarbon A were burned in 100 cm3 of oxygen, the final gaseous mixture contained 60 cm 3 of carbon dioxide and 10 cm 3 of unreacted oxygen. [All gaseous volumes measured under identical conditions.] What is the formula of hydrocarbon A? A C3H6 B C3H8 C C4H8 D C4H10 Let X be CxHy CxHy + (x + 𝒚 𝟒)O2 → xCO2 + 𝒚 𝟐H2O Vinitial / cm3 15 100 0 - Vfinal / cm3 0 10 60 - Reacting vol 15 90 60 - Vol ratio 1 6 4 By Avagadro’s law: V ∝ n at constant p and T x = 4; x + 𝒚 𝟒 = 6, y = 8 Ans: C 6 (2016 P3 Q1c) Carbon monoxide is a product of the incomplete combustion of hydrocarbons in internal combustion engines. Write an equation for the incomplete combustion of octane, C 8H18, giving CO 2 and CO in a 3:1 molar ratio. C8H18 + 23 2 O2 → 6CO2 + 2CO + 9H2O Free Radical Substitution 7 (2016 P3 Q4d) Alkanes are very unreactive. (i) Suggest two reasons why this is the case. (ii) However, alkanes such as propane, C3H8, do react with oxygen and with chlorine. For each of these reactions, • State the conditions under which the reaction is carried out. • Name the type of reaction which takes place, • Write an equation for the reaction.

4 (i) The C−C (BE = 350 kJ mol−1) and C−H bonds (BE = 410 kJ mol−1) are strong. Larger amount of energy is required for reaction to occur. The C−C and C−H bonds are non polar, thus they do not attract electrophiles and nucleophiles. (ii) C3H8 + 5O2 → 3CO2 + 4H2O Conditions: Burn propane in excess O2 Type of reaction: Combustion C3H8 + Cl2 → CH3CH2CH2Cl + HCl Conditions: Cl2(g), UV light Type of reaction: Free radical substitution Note: Mechanism for free radical substitution is not required. 8 Which explains why chlorine gas reacts readily with ethane in the presence of ultraviolet light? A UV light breaks the C-H bonds in ethane. B UV light increases the temperature of the mixture. C UV light splits the chlorine molecules into chlorine atoms. Note: In initiation step, homolytic fission of chlorine molecules into chlorine radicals (which is chlorine atoms) take place in presence of UV light. D UV light breaks up the chlorine molecules into chloride ions. Ans: C 9 Which statement best explains why a high yield of 2-bromobutane is not usually obtained when butane and bromine react in the presence of UV light? A Bromine reacts with butane too vigorously. B Bromine can replace any hydrogen atom in butane. When monobromination take place in presence of limiting Br 2, 1-bromobutane and 2-bromobutane can be formed. Also, poly-substitution can also take place if excess Br2 is used. C The second and third carbon atoms of butane are too strongly electrophilic. D The second and third carbon atoms of butane are too strongly nucleophilic. Ans: B 10 Which is a propagation step in the reaction between propane and bromine when they are irradiated with ultraviolet light? A CH3CH2CH2Br + Br• → CH3CH2CH2 + Br2 B CH3CH2CH2 + Br• → CH3CH2CH2Br C CH3CH2CHBr + Br2 → CH3CH2CHBr2 + Br• D CH3CH2CH2 + Br2 → CH3CH2CHBr + HBr Every propagation step should involve consumption of one radical & production of another radical. Option D is incorrect. Option B is Termination step between 2 radicals. Option A should generate an organic radical and HBr. It is not energetically favorable to break C—Br bond (280 kJ mol—1) and form a weaker Br—Br bond (193 kJ mol—1). Ans: C • • • •

5 11 An alkane J is reacted with chlorine gas in the presence of ultraviolet light to form only two monochlorinated alkanes in an approximate molar ratio of 6 : 1. Which could be J? A Cl2, UV lightH C CH3 H CH3 H C CH3 Cl CH3 H C CH3 H CH2Cl 2 6 1 3 B Cl2, UV lightCH3 C CH3 H CH3 CH3 C CH3 Cl CH3 CH3 C CH3 H CH2Cl 1 9 C Cl2, UV light 12 8 CH3CH2 C CH2CH3 CH2CH3 CH2CH3 CH3CH2 C CH2CH3 CH2CH3 CH2CH2Cl CH3CH2 C CH2CH3 CH2CH3 CHClCH3 3 2 D Cl2, UV light 2 1 CH3 C CH3 H C CH3 H CH3 CH3 C CH3 H C CH3 Cl CH3 CH3 C CH3 H C CH3 H CH2Cl 12 6 Ans: D

6 12 Propane reacts with chlorine gas in the presence of UV light. Which is true about the reaction? A The maximum number of mono-chlorinated structural isomers with formula C3H7Cl is 2. True. 1-chloropropane and 2-chloropropane B C6H12 is present in a small quantity in the product. False. The radical formed in propagation is C3H7 radical Termination involving two C3H7 radical gives C6H14. C Homolytic fission occurs only in the initiation step. False. Homolytic fission also occur in propagation steps. Propagation steps involve homolytic fission that generate free radical. D HCl is formed in the termination step. False. HCl is only formed in propagation step. Ans: A 13. It is found by experiment that during free radical substitution, primary, secondary and tertiary hydrogen atoms are replaced by chlorine atoms at different rates, as shown in the following table. reaction relative rate RCH3 ⎯→ RCH2Cl 1 R2CH