CHEMISTRY 2 SK025 2025/2026CHAPTER 5 HYDROCARBONSSK 1TUTORIAL 5A: ALKANES1. Classifya) unsaturated c) saturatedb) saturated d) unsaturated2. IUPAC namea) 3-ethyl-2-methylpentane d) 5-ethyl-1,2,3-trimethylcyclohexaneb) 2-methylbutane e) 4-isopropyl-3,6-dimethyloctane c) 3-isopropyl-1-methyl-1-propylcyclopentane3. Draw structural formulaa)Clc)BrBrb)4. a) methane < propane < pentane < hexaneincreasing order of boiling points• Molecular weight @ number of carbons of hexane > pentane > butane > propane.• Molecular size of hexane > pentane > butane > propane.• Strength of van der Waals forces between molecules of hexane > pentane > butane > propane.• Hence, energy needed to overcome the van der Waals forces betweenmolecules of hexane > pentane > butane > propaneb) 2,2-dimethylpropane < 2-methylbutane < pentaneincreasing order of boiling points• 2,2-dimethylpropane, 2-methylbutane and pentane have the same molecular weight @number of carbon atoms.• Number of branches in 2,2-dimethylpropane > 2-methylbutane, while pentaneis a straight chain alkane.• Molecular surface area of 2,2-dimethylpropane < 2-methylbutane < pentane.• Strength of van der Waals forces between molecules of 2,2-dimethylpropane < 2-methylbutane < pentane.• Therefore, energy needed to overcome the van der Waals forces betweenmolecules of 2,2-dimethylpropane < 2-methylbutane < pentane. 5. Butane soluble in benzene but insoluble in water,Reason:• Benzene is a non-polar solvent. Butane is soluble in benzene because butane and benzene have the same type of intermolecular forces which is Van der Waals forces, so they attract each other.• Water is a polar solvent. Butane is insoluble in water because it is a non-polar compound and cannot form hydrogen bond with water.
CHEMISTRY 2 SK025 2025/2026CHAPTER 5 HYDROCARBONSSK 26. Chemical equationa) In excess oxygen:2CH3CH2CH2CH3(g) + 13O2(g) ⎯→ 8CO2(g) + 10H2O(g)b) In limited oxygen:2CH3CH2CH2CH3(g) + 9O2(g) ⎯→ 8CO(g) + 10H2O(g)c) Very limited oxygen:CH3CH2CH2CH3(g) + O2(g) ⎯→ 4C(s) + 5H2O(g)7. a) Structural formulae of monosubstituted major and minor product formed.i. CH3CH2Br iv.Brii. major product : CH3CHBrCH3 minor product : CH3CH2CH2Briii. major product:H3C C CH3BrCH3 minor product:CH3CHCH2BrCH3b) i. Equationii. Reaction mechanismInitiation step: Propagation step:
CHEMISTRY 2 SK025 2025/2026CHAPTER 5 HYDROCARBONSSK 3Termination step: iii. Observation of the reaction:The reddish-brown colour of bromine is decolourised.iv. Observation in the dark:The reddish-brown colour of bromine is remain unchanged.
CHEMISTRY 2 SK025 2025/2026CHAPTER 5 HYDROCARBONSSK 4TUTORIAL 5B: ALKENES1. IUPAC namea) 2-methyl-2-pentene d) 1-ethyl-2-isopropylcyclopenteneb) 3,5-dimethylcyclohexene e) 4-ethyl-2,7-nonadienec) cis-3-heptene2. Structural formulaea)C CH3CH3C CH2CH3CH3c)C CH3CCH2ClHHb)C CCH2H HH3C CH2CH2CH2CH3d)Cl Cl3. trans-3-pentenea)C CH3CCH2CH3HHb) Correct name: trans-2-penteneReason: • The chain is numbered from the end nearer to the double bond. • The position of the double bond is designated using the lower number of the two doubly bonded carbon atoms.2-methylcyclohexenea)CH3b) Correct name: 1-methylcyclohexeneReason:• The position of the double bond is designated using the lower number of the two doubly bonded carbon atoms cis-2,3-dimethyl-2-pentenea) b) Correct name: 2,3-dimethyl-2-penteneReason:• Does not show any cis-trans isomerism because one of C at restricted rotation site attached to same group of atoms.4. a) State Saytzeff’s ruleSayzteff’s rule states that an elimination reaction will give an alkene with the most highly substituted double bond as the major product. b) Structural formulae of major and minor productCH3CH CCH3CH3 CH2CCH2CH3 H3CMajor product Minor product
CHEMISTRY 2 SK025 2025/2026CHAPTER 5 HYDROCARBONSSK 5Mechanism for the formation of major product:Step 1:H3C COHCH2CH3CH3 H3C CO+CH2CH3CH3HH+ H O SO3H + O SO3H-Step 2:H3C CO+CH2CH3CH3HHC CH2CH3H3C CH3+O HH+Step 3:C CCH3H3C CH3HH+O HH+H3C C C CH3CH3H+ O+HHH5. a) State Markovnikov’s ruleMarkovnikov’s rule states that in the addition of HX to an alkene or alkyne, the hydrogen atom adds to the carbon – carbon double bond that already has the greater number of hydrogens atoms.b) i. EquationH3C C CH2CH3 HBrH3C C CH3CH3Brii. Type of reactionElectrophilic addition reaction.iii. Reaction mechanismStep 1:H3C C CH2CH3+ H3C C CH3CH3H Brslow++ Br-Step 2:H3C C CH3CH3+ fastBr-H3C C CH3CH3Br
CHEMISTRY 2 SK025 2025/2026CHAPTER 5 HYDROCARBONSSK 66. Complete reaction schemea) Alkene to vicinal dihalide and halohydrinb) Alkene to haloalkane (Markovnikov and anti-Markovnikov) and alcohol.c) Oxidation of alkened) Alkene to oxidize and haloalkane
CHEMISTRY 2 SK025 2025/2026CHAPTER 5 HYDROCARBONSSK 7e) Alkene to oxidize and alcohol7. State reagents and conditiona) R1: Cl2, CH2Cl2, uv c) R6: HBr, H2O2R2: KOH, ethanol, reflux R7: H2O, H+R3: Br2, CH2Cl2 R8: KMnO4, H+, ∆b) R4: conc. H2SO4, ∆ R9: KMnO4, HO-, coldR5: i. O3 ii. Zn, H2O8. Outline the synthesisa)b)9. Chemical test to distinguish between 2-butene and butaneTest Reaction with bromine in inert solventReagent & ConditionsReagent: Br2 in CH2Cl2Conditions: In the darkCompound butane 2-buteneObservation Reddish brown colour of bromine is remain unchanged.Reddish brown colour of bromine is decolourised.Equation
CHEMISTRY 2 SK025 2025/2026CHAPTER 5 HYDROCARBONSSK 810. Suggest structural formula and write chemical equation.11. (a) Monomer: The smallest and simplest repeating unit of a polymer. (b) Copolymer: The polymers that is made up of two or more different monomers. (c) Cross-linked polymer: The polymer contains branches that connects linear polymer chain. 12. Natural polymers are polymers that are found naturally such as rubber, protein, carbohydrates. Synthetic polymers are polymers that are prepared chemically from monomers. Example: nylon, plastic, synthetic rubber, Kevlar, Dacron. 13.
CHEMISTRY 2 SK025 2025/2026CHAPTER 5 HYDROCARBONSSK 914.MEKA 51. naminga) 2-methylbutane d) 1-buteneb) 1-cyclopentyl-3-methylcyclohexane e) trans-2-hexenec) 3-bromo-1-methylcyclobutene f) 1-chloro-1,3,5-trimethylcyclohexane2. drawa)CH3H3Ce)CH3f) H2C=CHCH2CHBrCHBrCH2CH3b) g)c)Clh)d)ClHHCl3. a) Free radical substitution mechanism+ Cl2uv CH Cl CH4 3 + HClInitiation Step:Cl Cl uv Cl + ClPropagation Step:C + Cl HCl +HHHH H CHH
CHEMISTRY 2 SK025 2025/2026CHAPTER 5 HYDROCARBONSSK 10H CHH+ C ClHHHCl Cl + ClTermination Step:Cl + Cl Cl2 H CHH+ Cl C ClHHHH CHH+ C HHHHC CHHHHHb) electrophilic addition mechanismCH3CH2C CH2CH3HBrCH3CH2C CH3CH3BrStep 1:CH3CH2C CH2CH3CH3CH2C CH3CH3+ H Brslow ++ Br -Step 2:CH3CH2C CH3CH3++ fast Br -CH3CH2C CH3CH3Br4. a) Sayzteff’s rule states that the major product in elimination is the most stable alkenes which has the most highly substituted double bond. b) A: B:CH3CH2Mechanism:Step 1:+ H O+HH + H OHfastOCH3HHO+CH3HHH
CHEMISTRY 2 SK025 2025/2026CHAPTER 5 HYDROCARBONSSK 11Step 2:+ H OHslowO+CH3HHHC+CH3HStep 3:+ H OH+ H O+HHfastC+CH3HCH35. chemical test to differentiate cyclopentane(alkane) and cyclopentene(alkene).Test: Bromine testReagent and conditions: Br2, CH2Cl2dark placeObservation and equation:cyclopentaneReddish brown colour of bromine remain unchanged.cyclopenteneReddish brown colour of bromine decolourised.Br2, CH2Cl2dark placeBrBr6. Structure of A to H.A: B: C:D: E: F: G:H:H3C C CH3CH3OHH C C CH2Br 3CH3OHH3C CCH3O+ CO2 + H2OH3C C CH2OHCH3OHH C C CH2Br 3CH3HH3C C CH3CH3HH3C C CH3CH3OHH C C CH2Br 3CH3H
CHEMISTRY 2 SK025 2025/2026CHAPTER 5 HYDROCARBONSSK 127. (a) i. Basic molecular units that can joined to many others to form a polymer.ii. Polymer that are built from the same type of monomer units. iii. Polymers that are built by different monomers. (b)8. (a) Addition polymerization is the addition reaction in which unsaturated monomers are joined together by covalent bonds to form a polymer without elimination of a small molecule.(b) Polymer Monomer(i) ???? = ??2(ii) ??2 = ???? = ??2(iii) ??2 = ?(??)?? = ??2
CHEMISTRY 2 SK025 2025/2026CHAPTER 5 HYDROCARBONSSK 13KUMBE 51. naminga) 2-chloro-5-methylheptaneb) 3-methylpentanec) 5-isopropyl-5,6-dimethyl-2-heptened) 2-isopropyl-1-pentene e) 1-bromo-3-(1-chloroethyl)-1-methylcyclopentane f) 4-bromo-4-chloro-5-isopropylcyclohexene2. a) free radical substitution reaction+hClCl2Initiation Step:uvCl Cl Cl + ClPropagation step:C HH+ Cl HCl +CHCH+ Cl Cl + ClCClHTermination Step:Cl + Cl Cl ClH+HH+ClClH
CHEMISTRY 2 SK025 2025/2026CHAPTER 5 HYDROCARBONSSK 14b) i. X is cycloalkane because it undergoes free radical substitution with bromine when heated with bromine at very high temperature. ii. X:iii.+ Br2BrCH2Cl2high temperature3. 3-methylhexane < heptane < octane• Octane has the highest boiling point because it has the highest molecular weight. • The strength of van der Waals forces between octane molecules are strongest.• 3-methylhexane and heptane have the similar molecular weight. 3-methylhexane is branched alkane, while heptane is a straight chain alkane. • Hence, the molecular surface area of 3-methylhexane is smaller than heptane. • The strength of van der Waals forces between molecules of 3-methylhexane is weakercompared to heptane. • The energy to overcome the van der Waals forces between molecules of octane heptane 3-methylhexane.4. Alkane is generally unreactiveReason:Alkane is a saturated hydrocarbon where all C atoms are bonded with four others C atoms.5. a) A: c) C:H3C C CHCH3CH3b) B:CH3or CH26. dehydrohalogenation of 2-bromo-3-methylbutaneCH3C CH2CH3CH3BrMajor product Minor productA: B: A is the major product because it is most highly substituted double bond.CH3C CH CH3CH3H2C CCH2CH3CH3
CHEMISTRY 2 SK025 2025/2026CHAPTER 5 HYDROCARBONSSK 157. Product of methylcyclobutenea) d)IBrBrb) e)CCH2CH2COHOCH3c) f)BrCCH2CH2COHOOCH38. (a)(b)(c) Addition polymerisation(d) Peroxide(e) Piping, floor tiles, clothing, toys, wire covering