Class 11 Chemistry Chapter 9: Hydrocarbons worksheet with answer key

Answer: (b) C₈H₁₈

Alkanes have the general formula CₙH₂ₙ₊₂; for n = 8 this is C₈H₁₈. C₇H₁₄ and C₉H₁₈ fit CₙH₂ₙ and C₆H₁₀ fits CₙH₂ₙ₋₂.

Answer: (d) 2,2-dimethylbutane

In 2,2-dimethylbutane, CH₃–C(CH₃)₂–CH₂–CH₃, carbon 2 is bonded to four carbon atoms, so it is quaternary. The other three have at most tertiary carbons (bonded to three carbons).

Answer: (a) ĊH₃ + Cl–Cl → CH₃–Cl + Ċl

A propagation step consumes one free radical and produces another, so the chain continues: the methyl radical takes a chlorine atom and releases Ċl. Homolysis of Cl₂ is initiation; the two radical–radical combinations remove radicals and are termination steps.

Answer: (c) the alkane is formed by joining two alkyl free radicals, so it has at least two carbon atoms

At the anode carboxylate ions lose electrons, give alkyl free radicals and CO₂, and two radicals combine. Even the smallest radical, ĊH₃, gives CH₃–CH₃; methane is not formed.

Answer: (b) butane < 2,2-dimethylpropane < 2-methylbutane < pentane

Among the C₅H₁₂ isomers branching lowers the boiling point: pentane 309.1 K, 2-methylbutane 300.9 K, 2,2-dimethylpropane 282.5 K. Butane (molar mass 58 u) boils at 272.4 K, below even the most branched pentane, because the larger molecular mass outweighs the effect of branching here.

Molar masses: CH₄ = 16 g mol⁻¹, C₄H₁₀ = 58 g mol⁻¹. Heat per gram: methane 890 / 16 = 55.6 kJ g⁻¹; butane 2875.84 / 58 = 49.6 kJ g⁻¹. Methane releases more heat per gram, although butane releases more per mole.

• both molar masses and the two values per gram (1)
• methane releases more per gram (1)

(i) Cl–Cl + AlCl₃ → Cl⁺ + [AlCl₄]⁻; electrophile Cl⁺, the chloronium ion. (ii) CH₃COCl + AlCl₃ → CH₃–C⁺=O + [AlCl₄]⁻; electrophile CH₃C⁺O, the acylium ion. In both the Lewis acid AlCl₃ takes a chloride ion.

• Cl⁺ with its equation (1)
• acylium ion CH₃C⁺O with its equation (1)

Rotation about the C–C single bond of ethane is hindered only by about 12.5 kJ mol⁻¹, which collisions at ordinary temperature supply, so the conformations keep changing into one another. Rotation about C=C would require the sideways overlap of the p orbitals, the π bond, to break; rotation is restricted, so the cis and trans forms do not interconvert and can be separated.

• small barrier in ethane: conformers interconvert (1)
• restricted rotation about C=C (π bond must break) (1)

(i) Heat with soda lime (NaOH + CaO): CH₃CH₂COONa + NaOH →(CaO, Δ) CH₃–CH₃ + Na₂CO₃ (decarboxylation, one carbon fewer). (ii) Chlorinate the ethane in light: CH₃–CH₃ + Cl₂ →(hν) CH₃CH₂Cl + HCl. (iii) Wurtz reaction on chloroethane: 2CH₃CH₂Cl + 2Na →(dry ether) CH₃CH₂CH₂CH₃ + 2NaCl.

• soda lime decarboxylation to ethane (1)
• Cl₂ in light to chloroethane (1)
• Na in dry ether to butane (1)

(i) CH₃CH₂CH(OSO₂OH)CH₃, butan-2-yl hydrogen sulphate (sec-butyl hydrogen sulphate). (ii) CH₃CH₂CH(OH)CH₃, butan-2-ol. (iii) CH₃CH₂CHBr–CH₂Br, 1,2-dibromobutane. In (i) and (ii) the negative part (–OSO₂OH, –OH) goes to C2, the carbon with fewer hydrogens: Markovnikov. In (iii) both added atoms are the same, so the rule does not apply.

• (i) butan-2-yl hydrogen sulphate (1)
• (ii) butan-2-ol (1)
• (iii) 1,2-dibromobutane and the Markovnikov statement (1)

Only one 1,2-dibromobenzene is known, whereas a fixed Kekulé structure would allow one with a double bond and one with a single bond between the two Br-bearing carbons. X-ray diffraction shows all six C–C bonds equal, 139 pm, between 154 pm (single) and 133 pm (double). In the resonance picture each p orbital overlaps equally with both neighbours, so the six π electrons are delocalised over the ring: every C–C bond is identical, hence one ortho isomer and one bond length.

• only one 1,2-dibromobenzene (1)
• all C–C bonds 139 pm (1)
• delocalisation makes all bonds equivalent (1)