CBSE Class 11 Chemistry Hydrocarbon Notes Set 05

These Class 11 CBSE Chemistry notes for Chapter 9 Hydrocarbons, part one, cover the classification of hydrocarbons, IUPAC nomenclature and isomerism, the preparation and reactions of alkanes, ethane conformations, and the structure, geometrical isomerism, preparation and addition reactions of alkenes. Four solved examples show how to apply Markovnikov's rule, the peroxide effect and ozonolysis.

Important Notes and Formulas

Classification of Hydrocarbons

Hydrocarbons are compounds of carbon and hydrogen only. Saturated hydrocarbons may be open chain or cyclic (cycloalkanes, C_nH_2n). The same names are used for the two series, so check the suffix: -ane, -ene, -yne.

Alkanes: Nomenclature and Isomerism

1. Select the longest continuous carbon chain; this gives the parent name (meth-, eth-, prop-, but-, pent-, hex-, hept-, oct-, non-, dec-).
2. Number the chain from the end that gives the lowest locants to the substituents (branches).
3. Name the branches as alkyl groups (methyl, ethyl, propyl) and arrange them alphabetically; use di-, tri- for repeated groups, e.g. 2,4-dimethylhexane.
4. Alkanes show chain isomerism only: C₄H₁₀ has 2 isomers (n-butane and 2-methylpropane), C₅H₁₂ has 3 (pentane, 2-methylbutane and 2,2-dimethylpropane) and C₆H₁₄ has 5.

Preparation of Alkanes

Wurtz reaction gives an alkane with double the carbon number of the halide. Using two different halides gives a mixture of three alkanes, so it is suitable only for symmetrical alkanes. Decarboxylation gives an alkane with one carbon less than the acid.

Physical Properties of Alkanes

1. C₁ to C₄ are gases, C₅ to C₁₇ are colourless liquids, and the higher members are waxy solids at room temperature.
2. Alkanes are non-polar, so they are insoluble in water but soluble in non-polar solvents. Their densities are less than that of water.
3. The boiling point rises regularly with molar mass (about 20–30 K per CH₂ group).
4. Among isomers the branched chain has the lower boiling point, because its smaller surface area gives weaker van der Waals forces: n-pentane 309 K, isopentane 301 K and neopentane 283 K.

Chemical Properties of Alkanes

Alkanes are fairly unreactive (they were called paraffins, meaning little affinity) because the C–C and C–H bonds are strong and non-polar. They react under special conditions.

1. Halogenation (substitution): CH₄ + Cl₂ → CH₃Cl + HCl in diffused light, UV or at 573 K. Further substitution gives CH₂Cl₂, CHCl₃ and CCl₄. The order of reactivity of halogens is F₂ > Cl₂ > Br₂ > I₂; iodination is reversible and needs an oxidising agent such as HIO₃. The order of replacement of H is 3° > 2° > 1°.
2. Mechanism (free radical chain): initiation (Cl₂ → 2Cl·), propagation (Cl· + CH₄ → CH₃· + HCl; CH₃· + Cl₂ → CH₃Cl + Cl·) and termination (e.g. CH₃· + Cl· → CH₃Cl).
3. Combustion: C_nH_2n+2 + (3n+1)/2 O₂ → nCO₂ + (n+1)H₂O, with release of heat; e.g. C₄H₁₀ + 13/2 O₂ → 4CO₂ + 5H₂O.
4. Controlled oxidation: 2CH₄ + O₂ → 2CH₃OH (Cu, 523 K, 100 atm); CH₄ + O₂ → HCHO + H₂O (Mo₂O₃); 2CH₃CH₃ + 3O₂ → 2CH₃COOH + 2H₂O (Cu, 523 K, 100 atm); 2(CH₃)₃CH + O₂ → 2(CH₃)₃COH (KMnO₄).
5. Isomerisation: n-butane → isobutane (anhydrous AlCl₃ / HCl, heat).
6. Reaction with steam: CH₄ + H₂O → CO + 3H₂ (Ni, 1273 K); this is a source of dihydrogen.
7. Aromatisation: n-hexane → benzene + 4H₂ (Cr₂O₃ on Al₂O₃ or V₂O₅, 773 K, 10–20 atm); n-heptane gives toluene.
8. Pyrolysis (cracking): a higher alkane splits into smaller alkanes, alkenes and hydrogen on heating, e.g. C₁₂H₂₆ → C₇H₁₆ + C₅H₁₀ + other products.

Conformations of Alkanes (Ethane)

Rotation about a C–C σ bond produces different spatial arrangements of atoms, called conformations (or conformers). Two extreme forms of ethane are shown: the staggered form (hydrogens as far apart as possible) and the eclipsed form (hydrogens of the two carbons directly opposite each other).

1. In the staggered conformation the dihedral angle between C–H bonds on the two carbons is 60°; repulsion between the electron clouds of the bonds is minimum, so this form is the most stable.
2. In the eclipsed conformation the dihedral angle is 0° and repulsion is maximum, so it is the least stable.
3. The difference in energy (torsional strain) is about 12.5 kJ/mol. This is small, so the conformations rapidly interconvert at room temperature and cannot be isolated.
4. Other representations: the sawhorse projection (viewed from an oblique angle) and the Newman projection (viewed along the C–C axis, front carbon at the centre of the circle, rear carbon as the circle).

Alkenes: Structure, Nomenclature and Isomerism

1. Alkenes have the general formula C_nH_2n; the C=C carbons are sp² hybridised. One C=C consists of one σ bond (sp²–sp² overlap) and one π bond (sideways overlap of unhybridised p orbitals). The geometry about each double-bond carbon is trigonal planar (bond angle about 120°).
2. C=C bond length is 134 pm and bond enthalpy is 681 kJ/mol, against 154 pm and 348 kJ/mol for C–C. The π bond is weaker than the σ bond, so alkenes are reactive, mainly in addition reactions.
3. Nomenclature: choose the longest chain that contains the double bond, name it with the suffix -ene and number from the end that gives the lowest number to C=C, e.g. but-2-ene and 2-methylbut-2-ene.
4. Structural isomerism: chain isomerism (but-1-ene and 2-methylpropene) and position isomerism (but-1-ene and but-2-ene).
5. Geometrical (cis-trans) isomerism: restricted rotation about the C=C gives two arrangements when each double-bond carbon has two different groups. In the cis isomer the identical groups are on the same side and in the trans isomer on opposite sides. cis-But-2-ene (b.p. 277 K) has a net dipole moment and a higher boiling point than trans-but-2-ene (274 K), which is non-polar; the trans isomer has the higher melting point and is more stable.

Preparation of Alkenes

Properties of Alkenes

Alkenes are colourless and weakly polar. C₂ to C₄ are gases, C₅ to C₁₇ are liquids and the higher ones are solids. They are insoluble in water, but soluble in non-polar solvents. The boiling point rises with chain length.

1. Addition of H₂: CH₂=CH₂ + H₂ → CH₃CH₃ (Ni, Pt or Pd).
2. Addition of halogens: CH₂=CH₂ + Br₂ → BrCH₂CH₂Br. The reddish-orange colour of Br₂ in CCl₄ is discharged; this is a test for unsaturation.
3. Addition of HX (X = Cl, Br, I): CH₂=CH₂ + HBr → CH₃CH₂Br. For unsymmetrical alkenes Markovnikov's rule applies: the negative part of the reagent goes to the double-bond carbon with fewer hydrogen atoms (CH₃CH=CH₂ + HBr → CH₃CHBrCH₃). Reactivity of HX: HI > HBr > HCl.
4. Peroxide effect (Kharasch effect): in the presence of peroxide, HBr adds in the anti-Markovnikov way by a free radical mechanism: CH₃CH=CH₂ + HBr → CH₃CH₂CH₂Br. HCl and HI do not show this effect.
5. Addition of H₂SO₄ and water: CH₂=CH₂ + H₂SO₄ → CH₃CH₂OSO₃H; the acid-catalysed hydration of propene gives propan-2-ol (Markovnikov).
6. Oxidation with cold dilute alkaline KMnO₄ (Baeyer's reagent): CH₂=CH₂ → HOCH₂CH₂OH (glycol), and the pink colour disappears; this is also a test for unsaturation. Hot acidified KMnO₄ or K₂Cr₂O₇ cleaves the double bond to give ketones and/or carboxylic acids (and CO₂ from =CH₂).
7. Ozonolysis: the alkene reacts with O₃ to form an ozonide, which is decomposed by Zn and water to give aldehydes and/or ketones: =CHR gives an aldehyde and =CR₂ gives a ketone, e.g. CH₃CH=CH₂ → CH₃CHO + HCHO.
8. Combustion: CH₂=CH₂ + 3O₂ → 2CO₂ + 2H₂O. Polymerisation: n CH₂=CH₂ → (–CH₂–CH₂–)n polythene (high pressure, catalyst).

Solved Examples

Example 1. Write the IUPAC names of all the chain isomers of C₅H₁₂ and say which one has the lowest boiling point.
Answer:
There are three isomers: pentane (CH₃CH₂CH₂CH₂CH₃), 2-methylbutane (CH₃CH(CH₃)CH₂CH₃) and 2,2-dimethylpropane (C(CH₃)₄).
The boiling point falls with increased branching (a more compact, near-spherical molecule has a smaller surface area). So 2,2-dimethylpropane (neopentane, b.p. 283 K) has the lowest boiling point.

Example 2. How many conformations are possible for ethane? Which is more stable, and what is the energy difference?
Answer:
Rotation about the C–C bond gives an infinite number of conformations; the two extreme ones are the staggered and eclipsed forms.
The staggered form is more stable because the C–H bonds on the two carbons are as far apart as possible, so the repulsion is least.
The eclipsed form is higher in energy by about 12.5 kJ/mol (torsional strain).

Example 3. Write the products when propene reacts with HBr (i) in the absence and (ii) in the presence of benzoyl peroxide. Explain the difference.
Answer:
(i) CH₃CH=CH₂ + HBr → CH₃CHBrCH₃ (2-bromopropane); the ionic mechanism follows Markovnikov's rule. H⁺ adds to the CH₂ carbon to give the more stable secondary carbocation CH₃CH⁺CH₃, and Br⁻ then attacks.
(ii) In the presence of peroxide a Br· radical is produced and adds to the CH₂ carbon to give the more stable secondary radical CH₃CH·CH₂Br, which abstracts H from HBr. The product is CH₃CH₂CH₂Br (1-bromopropane).
The first is the Markovnikov addition and the second is the anti-Markovnikov addition (peroxide effect). The effect is seen only with HBr.

Example 4. An alkene on ozonolysis (followed by Zn/H₂O) gives propanone and ethanal. Identify the alkene and write the reaction.
Answer:
Join the two carbonyl carbons through a double bond: propanone (CH₃)₂C=O and ethanal CH₃CH=O give (CH₃)₂C=CHCH₃.
The alkene is 2-methylbut-2-ene.
(CH₃)₂C=CHCH₃ + O₃ → ozonide → (CH₃)₂C=O + CH₃CHO (Zn/H₂O).