Class 12 Chemistry Chapter 6: Haloalkanes and Haloarenes worksheet with answer key

Answer: (c) SO₂ and HCl, both gases

R–OH + SOCl₂ → R–Cl + SO₂ + HCl. Both by-products are gases that escape, leaving the alkyl chloride pure. H₃PO₃ comes from PCl₃, POCl₃ from PCl₅.

Answer: (b) 2

The molecule is symmetrical: all twelve CH₃ hydrogens are equivalent and the two CH hydrogens are equivalent. Two types of hydrogen give two monochlorides: ClCH₂CH(CH₃)CH(CH₃)₂ and (CH₃)₂CClCH(CH₃)₂.

Answer: (a) 1-chloro-1-methylcyclohexane

By Markovnikov's rule H adds to the ring CH of the double bond (the carbon with more hydrogens) and Cl to the carbon carrying the methyl group, giving the tertiary chloride 1-chloro-1-methylcyclohexane. (d) would need Cl₂, not HCl.

Answer: (b) a gem-dihalide, 1,1-dichlorobutane

Both chlorine atoms sit on the same (terminal) carbon, so the compound is geminal (an alkylidene halide, butylidene chloride); its IUPAC name is 1,1-dichlorobutane. Vicinal would need the two halogens on adjacent carbons.

Answer: (c) a tertiary alkyl halide

The carbon carrying chlorine is bonded to three ethyl groups, i.e. to three other carbon atoms, so the halide is tertiary (3°). The carbon is sp³, so it is not vinylic.

Step 1: diazotise 4-chloroaniline with NaNO₂ and HBr at 273–278 K, giving 4-chlorobenzene diazonium bromide, ClC₆H₄N₂⁺Br⁻. Step 2: warm the diazonium salt with cuprous bromide, Cu₂Br₂; –N₂⁺ is replaced by –Br with loss of N₂ (Sandmeyer), giving 1-bromo-4-chlorobenzene.

• NaNO₂ + HBr, 273–278 K, diazonium salt (1)
• Cu₂Br₂, Sandmeyer, product named (1)

β-hydrogens are on the two methyl groups of C-2 and on C-3. Removing H from a C-2 methyl gives CH₂=C(CH₃)CH(CH₃)₂, 2,3-dimethylbut-1-ene. Removing the C-3 hydrogen gives (CH₃)₂C=C(CH₃)₂, 2,3-dimethylbut-2-ene. By Zaitsev's rule the major product is 2,3-dimethylbut-2-ene, with four alkyl groups on the double-bond carbons.

• both alkenes drawn (1)
• 2,3-dimethylbut-2-ene major, by Zaitsev (1)

C-3 is asymmetric: it carries H, Br, an ethyl group and a propyl group, four different groups. The two enantiomers have the same melting point, boiling point and refractive index; they differ only in rotating plane-polarised light, by equal amounts in opposite directions.

• C-3, with H, Br, C₂H₅ and C₃H₇ (1)
• two shared properties and opposite rotation (1)

(i) Heat propan-2-ol with red phosphorus and bromine (PBr₃ formed in situ): 3(CH₃)₂CHOH + PBr₃ → 3(CH₃)₂CHBr + H₃PO₃. (ii) Heat with NaI in dry acetone (Finkelstein): (CH₃)₂CHBr + NaI → (CH₃)₂CHI + NaBr↓. (iii) Heat with alcoholic KOH (β-elimination): (CH₃)₂CHI + KOH → CH₃CH=CH₂ + KI + H₂O.

• red P/Br₂ (PBr₃) or constant-boiling HBr (1)
• NaI, dry acetone (1)
• alcoholic KOH, heat, propene (1)

Step 1 (slow, reversible): (CH₃)₂CBrCH₂CH₃ ⇌ (CH₃)₂C⁺CH₂CH₃ + Br⁻; the energy to break C–Br comes from solvation of the ions by water. Step 2 (fast): (CH₃)₂C⁺CH₂CH₃ + H₂O → (CH₃)₂C(OH)CH₂CH₃ + H⁺, giving 2-methylbutan-2-ol. Rate = k[(CH₃)₂CBrCH₂CH₃], first order: only the halide is in the slow step. The reaction is fast because the carbocation is tertiary and stabilised by three alkyl groups.

• slow ionisation to the tertiary carbocation (1)
• fast attack, 2-methylbutan-2-ol (1)
• first-order rate law, tertiary cation stability (1)

X is 2-chlorobutane, CH₃CHClCH₂CH₃, the only C₄H₉Cl with an asymmetric carbon: C-2 carries H, Cl, CH₃ and C₂H₅. With alcoholic KOH it can lose a β-H from C-3 or C-1; by Zaitsev's rule Y is but-2-ene, CH₃CH=CHCH₃, and but-1-ene is the minor product. X + Mg → CH₃CH(MgCl)CH₂CH₃; this with water gives CH₃CH₂CH₂CH₃, butane, which confirms the unbranched four-carbon chain.

• X = 2-chlorobutane, asymmetric C-2 (1)
• Y = but-2-ene major, but-1-ene minor (1)
• Grignard then water gives butane (1)