CBSE Class 11 Chemistry Hydrocarbons Worksheet Set 05

Hydrocarbons Worksheet

Question 1. Give the IUPAC names of (i) CH₃–CH(CH₃)–CH₂–CH₃, (ii) CH₃–CH=C(CH₃)–CH₃, (iii) CH₃–C≡C–CH₂–CH₃ and (iv) CH₃–C₆H₄–CH₃ (the 1,2-isomer).
Answer:
(i) The longest chain has four carbons with a methyl group on C-2: 2-methylbutane.
(ii) The longest chain containing the double bond has four carbons. Numbering from the right end gives the double bond at C-2 and a methyl group on C-2: 2-methylbut-2-ene.
(iii) A five-carbon chain with the triple bond starting at C-2: pent-2-yne.
(iv) A benzene ring with two adjacent methyl groups: 1,2-dimethylbenzene (o-xylene).
In simple words: Find the longest chain that holds the multiple bond, number it so that the multiple bond gets the lowest number and then name the branches.

Teacher's Note:
1. The double or triple bond gets the lowest locant before substituents do.
2. Write the locants of the substituents in front of the name, separated by commas and hyphens.

Question 2. What is the Wurtz reaction? Write the equation for the preparation of ethane and explain why it is not a good method for preparing an unsymmetrical alkane such as propane.
Answer:
The Wurtz reaction is the treatment of an alkyl halide with sodium metal in dry ether, which gives a higher alkane with double the number of carbon atoms.
2CH₃Br + 2Na → CH₃–CH₃ + 2NaBr (dry ether).
If two different alkyl halides are used, three alkanes are formed: R–R, R'–R' and R–R'. For example CH₃Br and C₂H₅Br give ethane, butane and propane.
These alkanes have close boiling points and are difficult to separate, so the yield of the required alkane is poor. So the method is useful only for symmetrical alkanes (R–R); an alkane such as propane, which cannot be made from two identical groups, can only be obtained together with other alkanes.
In simple words: Two alkyl groups join through sodium in dry ether; mixed halides give a mixture.

Teacher's Note:
1. Name the reagent: Na and dry ether.
2. A mixture of products makes the method suitable only for symmetrical alkanes.

Question 3. Explain the free-radical chlorination of methane with the three steps of the mechanism.
Answer:
Initiation: Light or heat splits the Cl–Cl bond homolytically: Cl₂ → 2Cl·
Propagation: Cl· + CH₄ → CH₃· + HCl and then CH₃· + Cl₂ → CH₃Cl + Cl·. The chlorine radical regenerated in the second step repeats the cycle. Further substitution gives CH₂Cl₂, CHCl₃ and CCl₄.
Termination: Radicals combine and the chain stops: Cl· + Cl· → Cl₂; CH₃· + CH₃· → C₂H₆; CH₃· + Cl· → CH₃Cl.
The reaction needs light, heat or a radical initiator, and the overall reaction is CH₄ + Cl₂ → CH₃Cl + HCl.
In simple words: Light makes chlorine radicals, the radicals carry the chain forward by two repeated steps, and the reaction stops when two radicals meet.

Teacher's Note:
1. Radicals are made in initiation, used and remade in propagation, and destroyed in termination.
2. Write the dot on the atom that has the unpaired electron.

Question 4. Draw and explain the staggered and eclipsed conformations of ethane. Which one is more stable and why? What is the energy difference?
Answer:
Conformations are the different spatial arrangements of atoms that arise from rotation about a C–C single bond.
In the staggered form the hydrogens on the two carbons are as far apart as possible (dihedral angle 60°). In the eclipsed form the hydrogens on one carbon are directly in front of those on the other (dihedral angle 0°).
The staggered form is more stable because the C–H bond electron clouds are farthest apart, so the repulsion (torsional strain) is least. In the eclipsed form the repulsion is greatest.
The energy difference is about 12.5 kJ/mol. The barrier is so small that the conformations cannot be separated at room temperature, and rotation is almost free.
In simple words: Ethane twists about its C–C bond; the staggered form has the least repulsion and so the lowest energy.

Teacher's Note:
1. Newman projections show the molecule viewed along the C–C bond.
2. Staggered: 60°. Eclipsed: 0°.

Question 5. Why does but-2-ene show geometrical isomerism but but-1-ene does not? Draw the two isomers of but-2-ene (in words) and state which has the higher boiling point.
Answer:
Rotation about a C=C double bond is restricted, because rotating would break the π bond. If each doubly bonded carbon carries two different groups, two different spatial arrangements are possible.
In CH₃–CH=CH–CH₃ each double-bond carbon carries H and CH₃. In cis-but-2-ene the two CH₃ groups are on the same side of the double bond, and in trans-but-2-ene they are on opposite sides.
In but-1-ene (CH₂=CH–CH₂CH₃) one doubly bonded carbon carries two H atoms, so there is only one arrangement.
The cis isomer has the higher boiling point (277 K against 274 K for trans). In the cis isomer the bond dipoles add up to give a small net dipole moment, while in the trans isomer they cancel, so the cis isomer has stronger dipole–dipole attractions.
In simple words: A double bond cannot rotate; two different groups on each carbon give cis and trans forms.

Teacher's Note:
1. The cis isomer is polar and the trans isomer is almost non-polar.
2. The trans isomer is usually the more stable one and has the higher melting point.

Question 6. Write the product of the addition of HBr to propene (i) in the absence and (ii) in the presence of benzoyl peroxide. Name the rule or effect in each case.
Answer:
(i) CH₃CH=CH₂ + HBr → CH₃CHBrCH₃ (2-bromopropane). This follows Markovnikov's rule: H adds to the carbon with more hydrogens, so the Br goes to the carbon with fewer hydrogens.
(ii) CH₃CH=CH₂ + HBr (peroxide) → CH₃CH₂CH₂Br (1-bromopropane). This is the peroxide effect (Kharasch effect), an anti-Markovnikov addition through a free-radical mechanism.
The peroxide effect occurs with HBr only, not with HCl or HI.
In simple words: Without peroxide, Br goes to the middle carbon; with peroxide it goes to the end carbon.

Teacher's Note:
1. Peroxide makes a Br· radical which attacks the carbon with more hydrogens to give the more stable secondary radical.
2. Remember that HCl and HI do not show the peroxide effect.

Question 7. Complete the following reactions: (a) CH₃CH=CH₂ + H₂O (H⁺), (b) CH₂=CH₂ + cold dilute alkaline KMnO₄, (c) CH₃CH=CH₂ + O₃ followed by Zn/H₂O.
Answer:
(a) Acid-catalysed hydration follows Markovnikov's rule: CH₃CH=CH₂ + H₂O → CH₃CH(OH)CH₃ (propan-2-ol).
(b) Hydroxylation gives a diol: CH₂=CH₂ + H₂O + [O] → HOCH₂CH₂OH (ethane-1,2-diol), and the pink colour of KMnO₄ disappears.
(c) Ozonolysis cleaves the double bond: CH₃CH=CH₂ → CH₃CHO + HCHO (ethanal and methanal).
In simple words: Water adds as H and OH, KMnO₄ adds two OH groups, and ozone breaks the double bond into two carbonyl compounds.

Teacher's Note:
1. Ozonolysis: each =CHR carbon becomes an aldehyde group and each =CR₂ carbon becomes a ketone.
2. Cold dilute KMnO₄ gives a diol and hot acidic KMnO₄ cleaves the double bond.

Question 8. How is ethyne prepared in the laboratory? Why is it called acidic in nature? Give one chemical test that distinguishes ethyne from ethene.
Answer:
Preparation: Calcium carbide reacts with water: CaC₂ + 2H₂O → HC≡CH + Ca(OH)₂.
Acidic character: In ethyne the carbon is sp hybridised, with 50% s-character, so it is more electronegative than sp² or sp³ carbon and holds the C–H electrons more tightly. The hydrogen is therefore easily lost as H⁺, and ethyne can react with sodium (giving sodium ethynide and H₂) or with ammoniacal silver nitrate.
Test: Ethyne gives a white precipitate of silver acetylide with ammoniacal AgNO₃: HC≡CH + 2[Ag(NH₃)₂]⁺ → AgC≡CAg + 2NH₄⁺ + 2NH₃. Ethene gives no precipitate.
In simple words: The more s-character a carbon has, the more easily its hydrogen leaves as a proton, so ethyne is much more acidic than ethene or ethane.

Teacher's Note:
1. Acidity order of the C–H bond: ≡C–H > =C–H > –C–H.
2. Both ethene and ethyne decolourise bromine water, so use the silver test to tell them apart.

Question 9. How will you convert (a) ethyne into benzene and (b) ethyne into ethanal?
Answer:
(a) Pass ethyne through a red hot iron tube at 873 K: 3HC≡CH → C₆H₆ (cyclic polymerisation).
(b) Hydration with dilute H₂SO₄ and HgSO₄ at 333 K: HC≡CH + H₂O → [CH₂=CHOH] → CH₃CHO. The unstable vinyl alcohol (enol) rearranges to the aldehyde.
In simple words: Three ethynes join to make a ring, and added water gives an aldehyde through an unstable enol.

Teacher's Note:
1. Higher alkynes add water to give ketones.
2. Do not forget the HgSO₄ catalyst.

Question 10. Describe the structure of benzene. Why are all C–C bonds equal in length (139 pm)? State Hückel's rule.
Answer:
Benzene, C₆H₆, is a planar, regular hexagon. Each carbon is sp² hybridised: it forms two C–C σ bonds and one C–H σ bond, and all bond angles are 120°.
The unhybridised p orbital on each carbon overlaps sideways with the p orbitals of both neighbours, so the six π electrons are delocalised over the entire ring (above and below the plane). The two Kekulé structures are resonance forms and the real molecule is a resonance hybrid.
Because of this delocalisation each C–C bond has partial double-bond character and the length is 139 pm, between a single (154 pm) and a double bond (134 pm).
Hückel's rule: a planar, cyclic, fully conjugated molecule is aromatic if it has (4n + 2) π electrons, where n = 0, 1, 2, ... Benzene has 6 π electrons (n = 1).
In simple words: Benzene is a flat ring with six π electrons shared by all six carbons, so every bond is the same length and the ring is unusually stable.

Teacher's Note:
1. Draw the two Kekulé structures with a double-headed arrow between them.
2. The resonance energy of benzene is the reason it prefers substitution over addition.

Question 11. Explain, with the example of toluene and nitrobenzene, the meaning of ortho/para-directing and meta-directing groups. Write the product of nitration in each case.
Answer:
Ortho/para-directing groups (–CH₃, –OH, –NH₂, –OCH₃ and, though deactivating, the halogens) increase the electron density at the ortho and para positions, so the electrophile enters there. Toluene on nitration gives o-nitrotoluene and p-nitrotoluene.
Meta-directing groups (–NO₂, –CN, –CHO, –COR, –COOH, –SO₃H) withdraw electrons by resonance and reduce the density most at ortho and para positions, so the electrophile enters the meta position. Nitrobenzene on further nitration (fuming HNO₃ / conc. H₂SO₄) gives mainly 1,3-dinitrobenzene.
Activating groups also make the ring more reactive than benzene, and deactivating groups make it less reactive.
In simple words: Electron-donating groups send the new group to the ortho and para positions; electron-withdrawing groups send it to the meta position.

Teacher's Note:
1. Halogens are the exception: deactivating but o/p-directing.
2. A meta-directing group always deactivates the ring.

Question 12. Write the equations for the Friedel–Crafts alkylation and acylation of benzene using chloromethane and ethanoyl chloride. What is the catalyst, and why does nitrobenzene not undergo these reactions?
Answer:
Alkylation: C₆H₆ + CH₃Cl → C₆H₅CH₃ + HCl (anhydrous AlCl₃), giving toluene.
Acylation: C₆H₆ + CH₃COCl → C₆H₅COCH₃ + HCl (anhydrous AlCl₃), giving acetophenone.
The catalyst is the Lewis acid anhydrous AlCl₃, which generates the electrophile (CH₃⁺ or CH₃CO⁺).
Nitrobenzene does not undergo these reactions because the strongly electron-withdrawing –NO₂ group deactivates the ring so much that it does not attack the weak electrophile, and AlCl₃ also complexes with the nitro group.
In simple words: An alkyl or acyl group replaces a ring hydrogen when AlCl₃ is present; a strongly deactivated ring cannot take part.

Teacher's Note:
1. Acylation gives a ketone; alkylation gives an alkylbenzene.
2. Use anhydrous AlCl₃; moisture destroys it.

Question 13. Why does p-xylene give only one mononitro derivative while o-xylene gives two and m-xylene gives three?
Answer:
The number of products equals the number of different kinds of ring hydrogens (positions open to attack).
In p-xylene (1,4-dimethylbenzene) all four ring hydrogens are equivalent by symmetry, so only 2-nitro-1,4-dimethylbenzene forms.
In o-xylene there are two types (positions 3 and 4), giving 3-nitro and 4-nitro-1,2-dimethylbenzene.
In m-xylene there are three types (positions 2, 4 and 5), giving three isomers (2-, 4- and 5-nitro-1,3-dimethylbenzene).
In simple words: Count the different kinds of hydrogens on the ring; each kind gives a separate product.

Teacher's Note:
1. Draw the plane or axis of symmetry to find equivalent hydrogens.
2. Here the methyl groups direct to ortho/para, but every open position still gives a product.

Question 14. Calculate (a) the volume of oxygen at STP needed for the complete combustion of 4.4 g of propane and (b) the volume of CO₂ and the mass of water formed. (C = 12, H = 1, O = 16; molar volume at STP = 22.4 L)
Answer:
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O
Molar mass of C₃H₈ = 3 × 12 + 8 × 1 = 44 g/mol, so moles of propane = 4.4 ÷ 44 = 0.1 mol.
(a) O₂ needed = 5 × 0.1 = 0.5 mol, which is 0.5 × 22.4 = 11.2 L.
(b) CO₂ formed = 3 × 0.1 = 0.3 mol = 0.3 × 22.4 = 6.72 L. Water formed = 4 × 0.1 = 0.4 mol = 0.4 × 18 = 7.2 g.
In simple words: Balance the equation first; the mole ratio then gives the volumes and masses.

Teacher's Note:
1. Complete combustion of a hydrocarbon gives CO₂ and H₂O only.
2. One mole of any gas occupies 22.4 L at STP.