RI 2025 Arenes Tutorial (ans)

-1- RAFFLES INSTITUTION YEAR 5 H2 CHEMISTRY 2025 Tutorial 13 – Arenes ______________________________________________________________________________ Suggested Solutions to Practice Questions 1 Electrophile Major Organic Product(s) (i) Br+ Br Br -CH3 group is 2,4-directing (ii) (iii) NO2+ CO2H NO2 -COOH group is 3-directing (iv) O O (v) O O

-2- 2(a) Reaction Reagents and conditions Type of reaction Unknown Reactant / Product I CH3CH2Cl, AlCl3 electrophilic substitution -- II H2, Ni catalyst reduction A: CH=CH2 III KMnO4(aq), dilute H2SO4, heat / heat under reflux oxidation C : COOH IV limiting Br2, uv light, room temperature free radical substitution -- V Br2, AlBr3 electrophilic substitution -- VI conc. HNO3, conc. H2SO4, 30 °C electrophilic substitution -- VII Br2 in CCl4, dark electrophilic addition B : Br Br 2(b) 2(c) Positional isomers of D: isomer D1 isomer D2 D1 is more likely to be formed along with D . This is because the – CH2CH3 group is 2,4-directing (and not 3-directing). CH2CH3 CO2H + CO2 + 2H2O+ 6[O] CH2CH3Br Br CH2CH3

-3- 2(d) (i) Reaction IV : Br Br2, uv room temp. Reaction VII : Br Br2 dark Br Reaction of alkyl side chain vs alkene. Reaction IV requires energy (in the form of uv light) to be provided while reaction VII does not. Compare regions of high / low electron density • Reaction IV involves the reaction of the alkyl group which does not possess any regions of high or low electron density. • Reaction VII involves reaction at the highly electron rich C=C which polarises the Br 2 electron cloud to generate the Br δ+ electrophilic site / attracts the Br 2 electrophile. Comparing bonds broken • The C–H bonds broken in Reaction IV are relatively strong. Hence, uv light is needed to generate the reactive bromine radicals which make the substitution reaction thermodynamically feasible. • The slow step of Reaction VII involves breaking of the relatively weak π bond of C=C. (ii) Reaction V : Br2, AlBr3 Reaction VII : Br Br2 dark Br Br Reaction V requires the use of a Lewis acid catalyst while reaction VII does not. In reaction V, the conversion of ethylbenzene to D involves disrupting the resonance stabilisation due to the delocalisa tion of the 6 π electrons of benzene. Hence an electrophile stronger than Br 2 is required for the electrophilic substitution. A Lewis acid such as AlBr3 is required since it reacts with Br2 to generate the stronger Br+ electrophile. However, in reaction VII, the addition of Br 2 across the C=C bond does not disrupt any resonance stabilisation of the benzene ring. Also, the electron rich C=C bond of the alkene is able t o polarise the Br 2 electron cloud to generate the Brδ+ electrophilic site. Hence, no lewis acid is required.

-4- 2(e) Reaction I + CH3CH2Cl AlCl3 CH2CH3 + HCl Electrophilic Substitution CH 3CH2Cl + AlCl3 ⇌ CH3CH2+ + [AlCl4]– Step 1: CH2CH3 slow H CH2CH3 Step 2: H CH2CH3 CH2CH3 + HCl + AlCl3 fast[AlCl4]+ Reaction V Electrophilic Substitution Br2 + AlBr3 ⇌ Br+ + [AlBr4]– Step 1: Br+ H Br slow CH2CH3 CH2CH3 Step 2: H Br CH2CH3 + fast + HBr + AlBr3[AlBr4] CH2CH3 Br CH2CH3 CH2CH3 Br + HBr + Br2 FeBr3 AlBr3

-5- Reaction VI + HNO3 + H2OO2N CH2CH3CH2CH3 conc. HNO3 conc. H2SO4 heat Electrophilic Substitution 2H2SO4 + HNO3 ⇌ NO2+ + 2HSO4– + H3O+ Step 1: NO2+ H O2Nslow CH2CH3 CH2CH3 Step 2: + HSO4 fast + H2SO4O2N CH2CH3H O2N CH2CH3 3 The product is bromobenzene. Since C l is more electronegative than Br , Cl attracts the bonding electrons to itself when the Br–Cl bond cleaves heterolytically to react with AlCl3 to form Br + and AlCl4–. Br+ acts as the electrophile which attacks benzene to form bromobenzene. 4(a)(i) The –CH3 substituent is a 2,4-directing group and directs the substitution of –SO3H on the ring at the 2- and 4-positions with respect to it. H, however, has the –SO3H at the 3-position with respect to –CH3. [1] Comments Students are required to provide an explanation, instead of just stating that –CH3 is 2,4-directing (which is given in the Data Booklet). 4(a)(ii) The –CH3 group poses steric hindrance to the approach of the electrophile at the 2-position [1] during the reaction. Hence, substitution by –SO3H at the 2-position with respect to –CH3 occurs less than the 4-position, resulting in a lower concentration of G than J. 30 °C

-6- 4(b)(i) + S O O O H slow electrophile H SO3H intermediate fast SO3H + H+ (displayed structure must be shown) [3] Comments In the displayed structure of the electrophile, the positive charge is on the S atom (since C–S bond is formed in the product). Hence, the slow step involves the movement of 2 π electrons from the benzene ring to the S atom of the electrophile, forming a σ bond between S and one C atom of the benzene ring. 4(b)(ii) Benzene does not undergo addition reactions with fuming sulfuric acid as the addition product will not have the delocalisation of six π electrons [1], leading to the loss of aromaticity and resonance stabilisation [1]. 4(b)(iii) In decreasing order of reactivity with fuming sulfuric acid: methylbenzene > benzene > K methylbenzene : The –CH3 group is an electron-donating group which increases the electron density of the benzene ring, making it more susceptible to electrophilic attack by SO3H+ than an unsubstituted benzene ring. [1] K: The –COCH3 group is an electron-withdrawing group which decreases the electron density of the benzene ring, making it less susceptible to electrophilic attack by SO3H+ than an unsubstituted benzene ring. [1] 5(a) Step 1 Step 2 NO2 Cl NO2 Reagents and conditions Step 1: conc. HNO3, conc. H2SO4, 50 °C Step 2 : Cl2, AlCl3

-7- 5(b) Step 1 Step 2 Step 3 CH2Cl Br CH3 CH3 Br Reagents and conditions Step 1: CH3Cl, AlCl3 Step 2 : Br2, AlBr3 Step 3 : limited Cl2, uv light, room temp. 5(c) Step 1 Step 2 Step 3 Step 4 CO2H NO2 Br CH3 CH3 Br CO2H Br Reagents and conditions Step 1: CH3Cl, AlCl3 Step 2 : Br2, AlBr3 Step 3 : KMnO4(aq), H2SO4(aq), heat / heat under reflux Step 4: conc. HNO3, conc. H2SO4, heat / heat under reflux 6(a) Test Add two drops of KMnO4 acidified with dilute sulfuric acid to each compound in a test- tube, and heat each reaction mixture in a hot water bath. (Note: DO NOT heat under reflux) Observation • For methylbenzene, there is decolourisation of purple KMnO4(aq). • For cyclohexane, there is no decolourisation of purple KMnO4(aq). 6(b) Test Add Br2 to each compound in separate test-tubes and irradiate each mixture with uv light at room temperature. Observation • For cyclohexane, there is decolourisation of reddish-brown Br2. • For benzene, there is no decolourisation of reddish-brown Br2.

-8- OR Test Add concentrated HNO 3 and concentrated H2SO4 to each compound in a test tube, and heat each reaction in a water bath maintained at 5 0 oC (for about 10 min). Then pour each mixture into a small beaker containing some water. Observation • For benzene, pale yellow oily droplets of nitrobenzene will be observed at the base of the beaker. • For cyclohexane, pale yellow oily droplets will not be observed. 6(c) Test Add two drops of KMnO4 acidified with dilute sulfuric acid to each compound in a test- tube, and heat each reaction mixture in a hot water bath. (Note: DO NOT heat under reflux) Observation • For , there is decolourisation of purple KMnO4(aq) and efferverscence of CO2(g) which gives a whit