Vidaara.orgClass 11 · Chemistry
CodeVID-C11-13-CH-01
Chapter Assignment — Hydrocarbons
Name: ____________________
Roll No.: __________
Date: ____________
General Instructions
- This is a full-length test covering the whole chapter — every topic is included.
- 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.
Section A — Multiple Choice Questions
6 × 1 = 6 marks
1.
The general formula of an alkene is:
- A.CnH2n
- B.CnH2n+2
- C.CnH2n-2
- D.CnHn
2.
The least stable conformation of ethane is the:
- A.staggered
- B.eclipsed
- C.gauche
- D.anti
3.
Markovnikov addition of HCl to propene gives:
- A.1-chloropropane
- B.propan-1-ol
- C.1,2-dichloropropane
- D.2-chloropropane
4.
A reagent that decolourises with an alkene but not an alkane is:
- A.Baeyer's reagent
- B.dilute HCl
- C.AlCl3
- D.NaCl solution
5.
Aromaticity requires (4n+2) π electrons; benzene satisfies this with n equal to:
- A.0
- B.1
- C.2
- D.3
6.
The meta-directing group among the following is:
- A.–OH
- B.–CH3
- C.–NO2
- D.–NH2
Section B — Short Answer (2 marks)
4 × 2 = 8 marks
7.
State Markovnikov's rule and give the product of propene + HCl.
8.
Why does benzene undergo substitution rather than addition?
9.
Distinguish a terminal alkyne from an alkene using one chemical test.
10.
What are the conditions for nitration of benzene and the electrophile involved?
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
11.
Explain the free-radical mechanism of the monochlorination of methane, naming the three stages.
12.
Account for the directive influence of –CH3 and –NO2 groups on the benzene ring.
Section D — Long Answer (5 marks)
2 × 5 = 10 marks
13.
Describe the two-step mechanism of electrophilic aromatic substitution, using nitration of benzene as the example, and explain the role of the arenium ion.
14.
Contrast the preparation and characteristic reactions of alkanes, alkenes and alkynes with one example each, and explain why their reactivities differ.
Answer Key
Section A — Multiple Choice Questions
- (A) CnH2n
- (B) eclipsed
- (D) 2-chloropropane
- (A) Baeyer's reagent
- (B) 1
- (C) –NO2
Section B — Short Answer (2 marks)
- On addition of HX to an unsymmetrical alkene, H adds to the carbon bearing more H atoms (more stable carbocation) and X to the more substituted carbon. Propene + HCl → 2-chloropropane.
- Benzene's six delocalised π electrons give it a large resonance stabilisation; substitution preserves this aromatic sextet, whereas addition would destroy it, so substitution is strongly favoured.
- A terminal alkyne gives a white precipitate (silver acetylide) with ammoniacal AgNO3 because of its acidic ≡C-H, whereas an alkene gives no precipitate.
- Concentrated HNO3 with concentrated H2SO4 (about 330 K); the attacking electrophile is the nitronium ion NO2+.
Section C — Short Answer (3 marks)
- Initiation: Cl2 ⟶[hν] 2 Cl·. Propagation: Cl· + CH4 → CH3· + HCl, then CH3· + Cl2 → CH3Cl + Cl·. Termination: two radicals combine, e.g. CH3· + Cl· → CH3Cl.
- –CH3 donates electron density (+I/hyperconjugation), activating the ring and stabilising the arenium ion at ortho/para positions, so it is an o/p director. –NO2 withdraws electrons (−I, −M), deactivating the ring and destabilising the o/p arenium ions most, so the electrophile enters at meta.
Section D — Long Answer (5 marks)
- Step 1 (slow): conc. HNO3/H2SO4 generate NO2+. The benzene π cloud attacks NO2+ to form a resonance-stabilised carbocation, the arenium ion (sigma complex), in which the positive charge is delocalised over three ring carbons and aromaticity is temporarily lost. Step 2 (fast): a base (HSO4-) removes the proton from the sp3 carbon bearing NO2, restoring the aromatic sextet and giving nitrobenzene. The arenium ion is the key high-energy intermediate; its relative stability at different ring positions controls the rate and the orientation (directive influence) of substitution.
- Alkanes (CnH2n+2) are prepared by hydrogenation of alkenes, Wurtz reaction or decarboxylation; their typical reaction is free-radical substitution, e.g. CH4 + Cl2 ⟶[hν] CH3Cl + HCl. Alkenes (CnH2n) are made by dehydrohalogenation (alc. KOH) or dehydration of alcohols; they undergo electrophilic addition, e.g. CH2=CH2 + Br2 → CH2BrCH2Br. Alkynes (CnH2n-2) come from CaC2+H2O or double dehydrohalogenation; they add reagents like alkenes and, if terminal, show acidic behaviour, e.g. HC≡CH + 2[Ag(NH3)2]+ → AgC≡CAg. Alkanes are least reactive because they have only strong, non-polar σ bonds; alkenes and alkynes are more reactive because their exposed, electron-rich π bonds attract electrophiles, so they react readily by addition.
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