Vidaara.orgClass 11 · Chemistry
CodeVID-C11-13-T2-01
Assignment — Alkenes & Alkynes
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 hybridisation of carbon in ethyne (acetylene) is:
- A.sp3
- B.sp2
- C.sp
- D.sp3d
2.
But-2-ene exists as cis and trans forms because of:
- A.chain isomerism
- B.restricted rotation about C=C
- C.free rotation
- D.resonance
3.
Acetylene is prepared in the lab from:
- A.sodium acetate
- B.calcium carbide and water
- C.ethanol
- D.methane
4.
Markovnikov addition of HCl to propene gives:
- A.1-chloropropane
- B.2-chloropropane
- C.propan-1-ol
- D.1,2-dichloropropane
5.
Ethene polymerises to give:
- A.polystyrene
- B.PVC
- C.polythene
- D.Teflon
Section B — Short Answer (2 marks)
3 × 2 = 6 marks
6.
State Markovnikov's rule with an example.
7.
How is cis-but-2-ene obtained from but-2-yne?
8.
Why is the ≡C–H of a terminal alkyne acidic?
Section C — Short Answer (3 marks)
2 × 3 = 6 marks
9.
Explain the peroxide effect in the addition of HBr to propene and write the product.
10.
Describe ozonolysis and use it to deduce the structure of an alkene that gives propanone and methanal.
Section D — Long Answer (5 marks)
1 × 5 = 5 marks
11.
Compare the addition of HBr to propene with and without peroxide, and outline two chemical tests that distinguish an alkene from an alkane.
Answer Key
Section A — Multiple Choice Questions
- (C) sp
- (B) restricted rotation about C=C
- (B) calcium carbide and water
- (B) 2-chloropropane
- (C) polythene
Section B — Short Answer (2 marks)
- When an unsymmetrical reagent HX adds to an unsymmetrical alkene, the H attaches to the carbon already bearing more hydrogens (forming the more stable carbocation), and X to the more substituted carbon. Example: propene + HBr → 2-bromopropane.
- Partial hydrogenation of but-2-yne over Lindlar's catalyst (Pd poisoned with quinoline/BaSO4) adds H2 to the same face, giving cis-but-2-ene.
- The hydrogen is bonded to an sp carbon, which has high s-character and is strongly electronegative; it holds the bonding electrons close, so the H can leave as H+, making the alkyne weakly acidic.
Section C — Short Answer (3 marks)
- In the presence of peroxides, HBr adds by a free-radical chain. A Br radical adds to the terminal carbon to give the more stable 2° carbon radical, then abstracts H from HBr. So Br ends up on C-1: the product is 1-bromopropane (anti-Markovnikov). It is seen only with HBr, not HCl or HI.
- Ozonolysis treats the alkene with O3 to form an ozonide, then Zn/H2O cleaves it into two carbonyl fragments at the double bond. Propanone (CH3)2C=O and methanal HCHO recombine at the cleaved bond to give (CH3)2C=CH2, i.e. 2-methylprop-1-ene (isobutylene).
Section D — Long Answer (5 marks)
- Without peroxide the reaction is ionic and follows Markovnikov (Br on the more substituted carbon, 2-bromopropane) via the more stable 2° carbocation. With peroxide it is a free-radical chain following anti-Markovnikov (Br on terminal carbon, 1-bromopropane) via the more stable 2° radical; the effect is limited to HBr. Tests for unsaturation: (i) bromine water (orange) is decolourised by an alkene but not an alkane; (ii) Baeyer's reagent (cold dilute alkaline KMnO4, purple) is decolourised, forming a diol, with an alkene only.
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