7 Benzene and aromatic compounds Answers

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Practice: pages 161 163 1 Answer is D. Methyl takes precedence over nitro and, therefore, automatically takes position 1, which doesn t have to be included in the name. [1] 2 Answer is C.

1 Practice: pages Answer is D. Methyl takes precedence over nitro and, therefore, automatically takes position 1, which doesn t have to be included in the name. [1] 2 Answer is C. If Kekulé was correct it would undergo electrophilic addition reactions and not electrophilic substitution reactions. [1] 3 Answer is C. Conditions have to be anhydrous (non-aqueous). [1] 4 Answer is A. Phenol only reacts with reactive metals like Na or with alkali. Mg(OH) 2 (aq) is alkaline but the products would be magnesium phenoxide and water. [1] 5 Answer is B. Benzene requires nitration and alkylation. The nitro group is 3-directing, whereas the methyl group would be 2, 4 and 6-directing. [1] 6 Answer is A. All three are correct. [1] 7 Answer is A. All three are correct. [1] 8 Answer is C. 2 & 3 are correct both would generate the correct electrophile CH 3 C + =CHCH 3. [1] 9 Answer is B. 1 & 2 are correct. 1 reacts by electrophilic substitution at the ring, 2 reacts by radical substitution on the side chain. [1] 10 Answer is D. Phenol is 2, 4-directing, nitro is 3-directing, but the amine group is 2, 4-directing. [1] 11 a) Benzene is less reactive with electrophiles than alkenes because, due to the delocalised electrons, the electron density in benzene is lower than the electron density in alkenes.

b) Concentrated HNO 3 and concentrated H 2 SO 4 are refluxed with benzene at a temperature of 55 60 C.

catalyst c) Describe the conditions needed for the nitration of benzene to form nitrobenzene, and outline the mechanism of

12 a) A 1,2-dimethylbenzene B 1,3-dimethylbenzene C 1,3,52-trimethylbenzene D 2-nitromethylbenzene E

2 b) Concentrated HNO 3 and concentrated H 2 SO 4 are refluxed with benzene at a temperature of C. HNO 3 + H 2 SO 4 HSO H 2 O + NO 2 formation of electrophile substitution at ring H + + HSO 4 H 2 SO 4 regeneration of catalyst c) Describe the conditions needed for the nitration of benzene to form nitrobenzene, and outline the mechanism of the reaction using curly arrows where appropriate. 12 a) A 1,2-dimethylbenzene B 1,3-dimethylbenzene C 1,3,52-trimethylbenzene D 2-nitromethylbenzene E 2,4,6-trichloromethylbenzene F 4-methylphenol [6] b) C 7 H 8 O [1] c) C 4 H 5 [1] d) A = 1,4-dimethylbenzene, B = ethylbenzene e) C 9 H O 2 9CO 2 + 6H 2 O [1] 13 a) Empirical formula is C 4 H 5. b) Molecular formula is C 8 H 10. c) [8]

14 a) i) Aluminium chloride, iron(iii) chloride or iron.

Nucleophilic substitution reaction [1] 15 a) The molecular formula and the empirical formula of naphthalene is C 10 H 8.

electronegative than I. [1] b) A halogen carrier is used to form an electrophile.

3 14 a) i) Aluminium chloride, iron(iii) chloride or iron. [1] ii) AlCl 3 + Cl 2 AlCl 4 + Cl + [1] AlCl 4 + H + AlCl 3 + HCl [1] b) i) (Radical) addition reaction [1] ii) 1,2,3,4,5,6-hexachlorocyclohexane [1] c) i) [1] ii) Nucleophilic substitution reaction [1] 15 a) The molecular formula and the empirical formula of naphthalene is C 10 H 8. b) i) Electrophilic [1] substitution [1] ii) HNO 3 + H 2 SO 4 HSO 4 + NO H 2 O [1] HSO 4 + H + H 2 SO 4 [1] 16 a) The product is C 6 H 5 I because Cl is more electronegative than I. [1] b) A halogen carrier is used to form an electrophile. [1] c) AlCl 3 + I Cl AlCl 4 + l + [1] AlCl 4 + H + AlCl 3 + HCl [1] 17 Phenol reacts with NaOH(aq)because it is acidic [1] but benzene and ethanol do not react with NaOH as they are not acidic. [1] 18 a) Type of reaction: benzene and phenol electrophilic substitution cyclohexene electrophilic addition [1] Reagents and conditions (if any) benzene halogen carrier is needed [1] phenol and cyclohexene don t require a halogen carrier

Organic product and observations benzene bromobenzene is formed, Br 2 will be decolourised phenol 2,4,6-tribromophenol is formed, Br 2 will be decolourised and a white precipitate will form

electron density [1] such that it polarises the bromine more readily than benzene.

4 Organic product and observations benzene bromobenzene is formed, Br 2 will be decolourised phenol 2,4,6-tribromophenol is formed, Br 2 will be decolourised and a white precipitate will form cyclohexene 1,2-dibromocyclohexane, Br 2 will be decolourised b) benzene and phenol phenol reacts faster because a lone pair of electrons from the O is delocalised into the ring [1], increasing the electron density [1] such that it polarises the bromine more readily than benzene. [1] benzene and cyclohexene cyclohexene reacts faster because the C=C has high electron density [1] and can polarise the Br 2 [1], whereas in benzene the electron density is low because the electrons are delocalised around the ring. [1] 19 a) Dilute nitric acid [1] 20 b) The OH group is activating [1] and directs to the 2,4 and 6 positions. [1] c) The OH group is activating and directs to the 2,4 and 6 positions for mono-substitution the 2 and 6 positions are equivalent [1], so there are two possible of attack, compared to attack at the 4 position. [1] From the equation, 1 mol TNT produces 7 mols of CO 2 (g) mols of N 2 2 (g) = 8 1 mols 2 of gas (this assumes that the H 2 O(l) was produced. [1] molar mass of TNT = 227 g mol 1 hence moles of TNT used = 9.08/227 = mol [1] mols of gas = = mol [1] 2 volume of gas = 8.16 dm 3 = cm 3 [1] 21 a) mols of NaOH = /1000 = 0.01 = mols of TCP in 100 cm 3 solution [1] molar mass of TCP = g mol 1 [1] mass of TCP in 100 cm 3 solution = = g (1.98 g) [1] b) Concentration of TCP = = mol dm 3 [1]

5 22 Methyl groups are activating and 2,4,6-directing. [1] Nitro groups are deactivating and 3,5-directing. [1] If nitration occurred first, the product would be 3-nitromethylbenzene. [1] [4] Challenge 23 a) [4] b) c)

6 Carbonium ions are unstable and often don t exist long enough to collide and react. [1] Anything that stabilises the carbonium ion increases the chance that it will exist long enough to collide and react. [1] or Carbonium ions are stabilised by the inductive effect of adjacent alkyl groups [1], which release electrons along the σ-bond. The primary carbonium ion is stabilised by the adjacent ethyl, C 2 H 5 group [1], while the secondary carbonium ion is stabilised by both adjacent methyl groups. [1]This makes the secondary carbonium ion more stable and, therefore, more likely to exist long enough to collide and react. [1] 24 a) All nitro phenols contain six C. [1] Therefore, 1 mol picric acid gives 6 mol CO 2 [1] and, hence, 1.5 mol of H 2 O. [1] Hence, picric acid contains six C and three H: C 6 H 3 N y O z. [1] One of the three H is in the OH of the phenol, so there are only two H on the ring. [1] Hence, there must be three NO 2 on the ring [1], so the molecular formula is C 6 H 3 N 3 O 7 [1], which has a molar mass of 229 g mol 1 [1]. % N = {(3 x14)/229} 100 = [1] b) 2,4,6-trinitrophenol [1], because the OH group is 2,4,6-directing. [1] c) C 6 H 3 N 3 O O 4 2 6CO H 2 2 O N 2 2 or any multiple, such as 4C 6 H 3 N 3 O O 2 24CO 2 + 6H 2 O + 6N 2

Acceptable sequence of stages are: chlorination. nitration, reduction, chlorination nitration. nitration, chlorination, reduction, reduction

Question er Mark Guidance 1 (a) (i) Response requires three stages Acceptable sequence of stages are: chlorination nitration, reduction, chlorination nitration nitration, chlorination, reduction, reduction