JPJC 2026 Alkane Assignment Answer

2026 JPJC J1 H2 Chemistry (9476) 1 [Turn Over JURONG PIONEER JUNIOR COLLEGE 2026 JC1 H2 Chemistry (9476) Topical Assignment Topical 9: Alkanes (Suggested Answers) Name : __________________________________________ 18 Class : 26S ___ Date : ____________ Bond Linkage −1 / NA Duration : 30 minutes 1 2-bromopropane, CH3CHBrCH3, reacts with chlorine to form a mono-chlorinated compound which exhibits stereoisomerism. (i) Write a balanced equation for the reaction. CH3CHBrCH3 + Cl2 ⟶ CH2ClCHBrCH3 + HCl [1m] [1] (ii) State the conditions required for this reaction and the type of reaction undergone. Conditions: limited Cl2 (or excess 2-bromopropane), UV light [1m] Type of reaction: free radical substitution [1m] [2] (iii) Describe the mechanism for the above reaction. ✓1initiation step: Cl2 ⟶ 2Cl (Note: If arrow is shown, it must be .) ✓1propagation step: CH3CHBrCH3 + Cl ⟶ CH2CHBrCH3 + HCl CH2CHBrCH3 + Cl2 ⟶ CH2ClCHBrCH3 + Cl ✓1termination step: (any two equations) 2 Cl ⟶ Cl2 2 CH2CHBrCH3 ⟶ Cl + CH2CHBrCH3 ⟶ CH2ClCHBrCH3 6(✓): [3m] ; 4−5(✓): [2m] ; 2−3(✓): [1m] ; Minus one (✓) if the three steps are not labelled. [3] (✓) (✓) (✓) 2(✓) any two equations • • • • • • • • CH2CHBrCH3 CH2CHBrCH3 • UV

2026 JPJC J1 H2 Chemistry (9476) 2 (iv) State the type of stereoisomerism the product exhibited and draw appropriate diagrams to illustrate the stereoisomers. Type of stereoisomerism: Enantiomerism [1m] [2m] 1 mark for each correct enantiomer [3] (v) Explain why the reaction does not require prolonged radiation of ultraviolet light. Once the reaction has been initiated, the chlorine radical / C l• is regenerated during the propagation/ propagating step. [1m] Hence, prolonged radiation is not required. [1] 2 An alkane A, C 4H10, when reacted with bromine, forms only two isomeric monobromo compounds, B and C, with the formula C4H9Br. Compound B exhibits stereoisomerism but not compound C. (i) Draw the structure of A, B and C. structure of A structure of B structure of C C C C CH H H H H H H H H H [1m] C C C CH H H H Br H H H H H* [1m] C C C CH H Br H H H H H H H [1m] [3] (ii) Suggest, with a reason, the approximate ratio in which B and C were formed. Assuming that the substitution process is purely random , there are 2 types of replaceable H atoms in pentane and hence, 2 different mono –brominated compounds are formed. To form B, there are 4 possible H atoms to be substituted To form C, there are 6 possible H atoms to be substituted. Hence, the ratio of B : C is 2 : 3 (i.e. simplest ratio of 4 : 6). [1m] Ecf from structures of B and C in (i) [2] [Total: 15] [1m] non-superimposable mirror images C Br H3C * * CH3 C H Br CH2Cl H CH2Cl (✓1)*- labelling of all steps correctly (✓2) – balanced eqn + UV for initiation step; if arrow drawn must be (✓3) - balanced eqn for step 1 of propagation + correct radical formed with “•” on C (✓4) - balanced eqn for step 2 of propagation (✓5) (✓6) - for each of any 2 possible eqn of termination

2026 JPJC J1 H2 Chemistry (9476) 3 [Turn Over 3 Which of the following represents a valid propagation step in the reaction between ethane and chlorine? Which statements about the process are correct? A CH3CH3 + Cl• ⟶ CH3CH2Cl + H• Incorrect. Recall: H• radical is not formed. Reason: The formation of a H –Cl bond is more exothermic, therefore, more energetically preferred, than the formation of a C–Cl bond in CH3CH2Cl. Preferred: CH3CH3 + Cl• → CH3CH2 + HCl See Alkanes notes Pg 12. B •CH2CH3 + Cl• ⟶ CH3CH2Cl Incorrect. This is a termination step since it involves the combination of two radicals to form a molecule. C CH3CH2Cl + Cl• ⟶ •CH2CH2Cl + HCl Correct. The formation of H–Cl bond over a C–Cl bond is energetically preferred. (Contrast this with statement A.) D CH3CH2Cl + H• ⟶ CH3CH2• + HCl Incorrect. H• radical is not formed during the reaction. 4 The molecule, Q, with molecular formula C6H14 has the following structural formula Q CH3CH2CH(CH3)CH2CH3 Q reacts with chlorine to form monochloro compounds C6H13Cl. How many possible constitutional isomers, each with formula C6H13Cl, could be formed by Q? A 3 B 4 C 5 D 6 C C C CH H H H H H H H C H H C HH H H Types of H to be replaced: a b c b a d There are 4 types of replaceable hydrogen atoms (labelled a-d) in compound Q, giving rise to 4 possible constitutional isomers. •

2026 JPJC J1 H2 Chemistry (9476) 4 5 Propane undergoes free -radical substitution when mixed with chlorine and exposed to ultra -violet light. Which compounds are possible products from this reaction? 1 CH3CH2CH2Cl 2 CH2ClCH2CH2Cl 3 CH3CH2CH2CH2CH2CH3 4 CH3CH2CH2CH2CH3 A 1, 2, 3 and 4 B 1, 2 and 3 C 1 and 4 only D 2 and 3 only 1. CH3CH2CH2Cl This can be formed from propagation step or termination step. 2. CH2ClCH2CH2Cl This can be formed from further chlorination of CH3CH2CH2Cl. 3. CH3CH2CH2CH2CH2CH3 This can be formed from the termination step through combining two CH 3CH2CH2 radicals. Recall that trace amount of by – products with number of carbon atoms twice as many as the reactant can be found. 4. CH3CH2CH2CH2CH3 The structure of this compound suggests that it can be formed from combining CH 3CH2CH2 and CH 2CH3 radicals but CH 2CH3 radical is not formed in the reaction. • CH3CH2CH2 + Cl2 → CH3CH2CH2Cl + Cl• CH3CH2CH2 + Cl• → CH3CH2CH2Cl • • • • •