Vidaara.orgClass 11 · Chemistry
CodeVID-C11-13-T3-01
Assignment — Aromatic Hydrocarbons
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- All questions are compulsory.
- Section A carries 1 mark each, Section B 2 marks, Section C 3 marks and Section D 5 marks.
- Show all working for Sections B, C and D. Only final answers are given at the end — for full solutions, raise your doubts with your teacher.
Section A — Multiple Choice Questions
5 × 1 = 5 marks
1.
The number of delocalised π electrons in benzene is:
- A.2
- B.4
- C.6
- D.8
2.
Benzene prefers substitution over addition because it is:
- A.highly unsaturated
- B.aromatic and stable
- C.non-planar
- D.ionic
3.
The catalyst used in the halogenation of benzene is:
- A.Ni
- B.Lindlar's catalyst
- C.anhydrous FeCl3
- D.Pt
4.
An ortho/para directing, activating group is:
- A.–NO2
- B.–COOH
- C.–NH2
- D.–SO3H
5.
Friedel–Crafts acylation of benzene with CH3COCl/AlCl3 gives:
- A.toluene
- B.acetophenone
- C.benzoic acid
- D.nitrobenzene
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
State Huckel's rule and apply it to benzene.
7.
Why are all the carbon–carbon bonds in benzene of equal length?
8.
Write the reagents for the sulphonation of benzene and name the product.
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Explain why –NO2 is a meta-directing, deactivating group while –OH is an ortho/para directing, activating group.
10.
Describe the mechanism of nitration of benzene in two steps.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Discuss the structure of benzene in terms of the Kekulé model, resonance and the orbital (MO) picture, and state how this stability is reflected in its reactions.
Answer Key
Section A — Multiple Choice Questions
- (C) 6
- (B) aromatic and stable
- (C) anhydrous FeCl3
- (C) –NH2
- (B) acetophenone
Section B — Short Answer (2 marks)
- A cyclic, planar, fully conjugated ring is aromatic if it has (4n+2) π electrons (n = 0,1,2,...). Benzene has 6 π electrons, so 4n+2 = 6 gives n = 1; hence benzene is aromatic.
- Benzene is a resonance hybrid of two Kekulé structures, so the π electrons are delocalised over the ring; every C–C bond has the same partial double-bond character and the same length (139 pm).
- Benzene heated with fuming sulphuric acid (oleum, containing SO3) gives benzenesulphonic acid, C6H5SO3H; the reaction is reversible.
Section C — Short Answer (3 marks)
- –NO2 withdraws electron density (−I, −M), so it makes the ring electron-poor (deactivating) and destabilises the arenium ion most at the ortho/para positions, leaving meta as the favoured site. –OH donates electron density through its lone pair (+M), making the ring electron-rich (activating) and stabilising the arenium ion at the ortho/para positions.
- Step 1: conc. HNO3/H2SO4 form NO2+, which is attacked by the benzene π cloud to give the resonance-stabilised arenium ion (sigma complex). Step 2: HSO4- removes the proton from the carbon bearing NO2, restoring aromaticity and giving nitrobenzene.
Section D — Long Answer (5 marks)
- Kekulé proposed a six-membered ring with alternating single and double bonds, but this cannot explain why all six C–C bonds are equal (139 pm) or why benzene resists addition. Resonance treats benzene as a hybrid of two equivalent Kekulé structures with delocalised π electrons; its measured stability exceeds the calculated value by the resonance energy (~150 kJ mol-1). In the orbital/MO picture each carbon is sp2 hybridised and planar, and the six p-orbitals overlap to form continuous π clouds above and below the ring holding six delocalised electrons (satisfying Huckel's 4n+2 rule with n=1). This stability makes benzene prefer electrophilic substitution — which preserves the aromatic sextet — over addition, which would destroy it.
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