Benzene and Aromatic Compounds: Quiz
Seven questions on the structure and stability of benzene, naming aromatic compounds and electrophilic substitution, with explanations.
Benzene and Aromatic Compounds: Quiz
7 questions
All C–C bonds are 139 pm, between C–C (154 pm) and C=C (134 pm), because the π electrons are delocalized.
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Answer: All six are the same length, between a single and a double bond
All C–C bonds are 139 pm, between C–C (154 pm) and C=C (134 pm), because the π electrons are delocalized.
An electrophile replaces a hydrogen, keeping the stable aromatic ring intact.
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Answer: Electrophilic substitution
An electrophile replaces a hydrogen, keeping the stable aromatic ring intact.
Bromine adds across the C=C of cyclohexene. Benzene does not react without a catalyst, because addition would destroy its delocalized ring.
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Answer: Only cyclohexene decolourizes it
Bromine adds across the C=C of cyclohexene. Benzene does not react without a catalyst, because addition would destroy its delocalized ring.
Expected for 3 C=C: 3 × (−120) = −360 kJ/mol. Actual −208 kJ/mol. Difference: 360 − 208 = 152 kJ/mol.
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Answer: 152 kJ/mol
Expected for 3 C=C: 3 × (−120) = −360 kJ/mol. Actual −208 kJ/mol. Difference: 360 − 208 = 152 kJ/mol.
Sulfuric acid helps form the nitronium ion, NO₂⁺, the electrophile that attacks the ring (about 50 °C).
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Answer: Concentrated HNO₃ and concentrated H₂SO₄
Sulfuric acid helps form the nitronium ion, NO₂⁺, the electrophile that attacks the ring (about 50 °C).
C₆H₅– is the phenyl group, so C₆H₅NH₂ is phenylamine (common name aniline).
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Answer: Phenylamine
C₆H₅– is the phenyl group, so C₆H₅NH₂ is phenylamine (common name aniline).
7.80 g ÷ 78.11 g/mol = 0.09986 mol; × 123.11 g/mol = 12.3 g (1 : 1). 9.83 g is the mass at 80 % yield.
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Answer: 12.3 g
7.80 g ÷ 78.11 g/mol = 0.09986 mol; × 123.11 g/mol = 12.3 g (1 : 1). 9.83 g is the mass at 80 % yield.