CBSE Class 12 Chemistry Halo alkanes and Haloarene Questions Answers

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Halo alkanes and Haloarene

HALOALKANES AND HALOARENES

HALOALKANES AND HALOARENES — QUESTIONS AND ANSWERS

SET 1

One-mark questions

Give the IUPAC name of the following organic compound: CH₃-CH=C(CH₃)-CH(Br)-CH₃

Answer: 4-Bromo-3-methylpent-2-ene.

Write the structural formula of 4-chloropent-2-ene.

Answer: CH₃-CH=CH-CH(Cl)-CH₃

Arrange the following halides in order of increasing S_N2 reactivity: CH₃Cl, CH₃Br, CH₃CH₂Cl, (CH₃)₂CHCl.

Answer: (CH₃)₂CHCl < CH₃CH₂Cl < CH₃Cl < CH₃Br

What is the order of reactivity of different alkyl halides in nucleophilic substitution reactions?

Answer: RI > RBr > RCl.

An alkyl halide C₄H₉Cl is optically active. What is its structure?

Answer: 2-Chlorobutane, CH₃-C*H(Cl)-CH₂-CH₃ (the starred carbon is the chiral carbon).

Which type of solvents are generally used to carry out S_N1 reactions?

Answer: Polar protic solvents.

Identify the chiral and achiral molecules in the following pair of compounds: A: CH₃CH(Br)CH₂CH₃; B: CH₃CH₂CH₂CH₂Br.

Answer: A (2-bromobutane) is chiral and B (1-bromobutane) is achiral.

Give two uses of iodoform (CHI₃).

Answer: (1) As an antiseptic. (2) As a photosensitiser in the emulsion of AgBr used to make photographic film.

Write a chemical reaction in which the iodide ion displaces the diazonium group from a diazonium salt.

Answer:

C₆H₅N₂⁺Cl⁻ + KI → C₆H₅I + KCl + N₂

How will you convert 2-bromopropane to 1-bromopropane?

Answer: Dehydrohalogenation with alcoholic KOH gives propene; addition of HBr in the presence of peroxide (anti-Markovnikov addition) then gives 1-bromopropane.

CH₃-CHBr-CH₃ → CH₃-CH=CH₂ (alc. KOH)

CH₃-CH=CH₂ + HBr → CH₃-CH₂-CH₂Br (peroxide)

How will you convert aniline into chlorobenzene?

Answer:

C₆H₅NH₂ + NaNO₂ + HCl → C₆H₅N₂⁺Cl⁻ + NaCl + 2H₂O (273–278 K)

C₆H₅N₂⁺Cl⁻ → C₆H₅Cl + N₂ (CuCl/HCl)

Name the iodine-containing hormone, the deficiency of which causes goitre.

Answer: Thyroxine.

Name the synthetic halogen compound which is used in the treatment of malaria.

Answer: Chloroquine.

Which isomer of C₄H₉Br will have the lowest boiling point?

Answer: tert-Butyl bromide, (CH₃)₃C-Br (the most branched isomer has the smallest surface area).

Write the IUPAC name of DDT.

Answer: 1,1,1-Trichloro-2,2-bis(4-chlorophenyl)ethane.

Why is sulphuric acid not used during the reaction of alcohols with KI?

Answer: It first converts KI into HI and then oxidises the HI to I₂.

Out of CH₃Br and CH₃Cl, which will have the higher boiling point and why?

Answer: CH₃Br. The greater the molecular mass, the greater the van der Waals forces of attraction, and the higher the boiling point.

Which one of the following has the highest dipole moment? (i) CH₂Cl₂ (ii) CHCl₃ (iii) CCl₄

Answer: CH₂Cl₂.

Two-mark questions

Define the following terms: (i) ambident nucleophile; (ii) chirality.

Answer: (i) A nucleophile having two nucleophilic centres, e.g. the cyanide group, which can attack through C (–CN) or through N (–NC). (ii) An object which is not superimposable on its mirror image is said to be chiral; the property of being chiral is called chirality.

Write a short note on the Sandmeyer reaction.

Answer: A primary aromatic amine is diazotised with NaNO2 and HCl at 273–278 K and the diazonium salt is then treated with CuCl/HCl (or CuBr/HBr) to give the aryl halide.

NaNO₂ + HCl → NaCl + HNO₂ (273–278 K)

C₆H₅NH₂ + HNO₂ + HCl → C₆H₅N₂⁺Cl⁻ + 2H₂O

C₆H₅N₂⁺Cl⁻ → C₆H₅Cl + N₂ (CuCl/HCl)

Write the structures of the main products: (i) chlorination of benzene in the presence of UV light; (ii) propene treated with HBr in the presence of benzoyl peroxide.

Answer:

(i) C₆H₆ + 3Cl₂ → C₆H₆Cl₆ (UV light; benzene hexachloride)

(ii) CH₃-CH=CH₂ + HBr → CH₃-CH₂-CH₂-Br (peroxide)

Complete the following reactions: (i) C₆H₅N₂Cl + KI; (ii) CH₂=CH₂ + Br₂ in CCl₄.

Answer:

(i) C₆H₅N₂Cl + KI → C₆H₅I + KCl + N₂

(ii) CH₂=CH₂ + Br₂ → CH₂Br-CH₂Br (in CCl₄)

Explain why haloarenes are much less reactive than haloalkanes towards nucleophilic substitution reactions.

Answer: (i) Due to resonance the C–Cl bond acquires partial double-bond character. (ii) In haloarenes the carbon bearing the halogen is sp² hybridised, so the C–X bond is shorter and stronger.

Write a short note on (i) the Wurtz reaction; (ii) the Wurtz–Fittig reaction.

Answer: (i) Alkyl halides react with sodium in dry ether to give higher alkanes. (ii) An alkyl halide and an aryl halide react with sodium in dry ether to give an alkylbenzene.

R-X + 2Na + X-R → R-R + 2NaX (dry ether)

R-X + 2Na + X-C₆H₅ → R-C₆H₅ + 2NaX (dry ether)

How will you convert (i) ethyl chloride into ethyl alcohol; (ii) ethyl chloride to ethane?

Answer:

(i) C₂H₅Cl + KOH(aq) → C₂H₅OH + KCl

(ii) C₂H₅Cl + 2[H] → C₂H₆ + HCl (LiAlH₄)

Alkyl halides are insoluble in water though they contain a polar C–X bond. Why?

Answer: Haloalkanes cannot form hydrogen bonds with water molecules, and at the same time they cannot break the hydrogen bonds present among water molecules.

Give one test to distinguish between (i) chloroform and carbon tetrachloride; (ii) methanol and ethanol.

Answer: (i) Carbylamine test: chloroform, when heated with aniline and alcoholic KOH, gives the offensive smell of phenyl isocyanide; carbon tetrachloride does not. (ii) Iodoform test: ethanol gives a yellow precipitate of iodoform, methanol does not.

C₆H₅NH₂ + CHCl₃ + 3KOH(alc.) → C₆H₅NC + 3KCl + 3H₂O

CH₃CH₂OH + 4I₂ + 6NaOH → CHI₃ + HCOONa + 5NaI + 5H₂O

CH₃OH + NaOI (NaOH + I₂) → no yellow precipitate

Write a short note on (i) the Finkelstein reaction; (ii) the Swarts reaction.

Answer: (i) Finkelstein reaction: an alkyl chloride or bromide is converted into the alkyl iodide by treatment with NaI in dry acetone. (ii) Swarts reaction: an alkyl chloride or bromide is converted into the alkyl fluoride by heating with a metallic fluoride such as AgF.

CH₃-CH₂-Cl + NaI → CH₃-CH₂-I + NaCl (acetone)

CH₃-CH₂-Cl + AgF → CH₃-CH₂-F + AgCl

Three-mark questions

Explain why: (a) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride; (b) the Grignard reagent should be prepared under anhydrous conditions; (c) chloroform is stored in dark brown bottles.

Answer: (a) The sp² carbon of chlorobenzene is more electronegative than the sp³ carbon of cyclohexyl chloride (and the C–Cl bond is also shortened by resonance), so the C–Cl dipole is smaller. (b) Grignard reagents are very reactive and react readily with water: RMgX + H₂O → RH + Mg(OH)X. (c) In the presence of air and sunlight chloroform is oxidised to the poisonous gas phosgene: CHCl₃ + 1/2 O₂ → COCl₂ + HCl.

What happens when (a) ethyl chloride is treated with alcoholic KOH; (b) an alcohol reacts with thionyl chloride?

Answer:

(a) C₂H₅Cl + KOH(alc) → CH₂=CH₂ + H₂O + KCl

(b) R-OH + SOCl₂ → R-Cl + SO₂ + HCl

How will you convert (a) chlorobenzene into toluene; (b) chlorobenzene into phenol; (c) ethyl bromide into diethyl ether?

Answer:

(a) C₆H₅Cl + 2Na + CH₃Cl → C₆H₅CH₃ + 2NaCl (dry ether)

(b) C₆H₅Cl + NaOH → C₆H₅ONa (623 K, high pressure)

C₆H₅ONa + H⁺ → C₆H₅OH (H₂O/H⁺)

(c) C₂H₅Br + NaOC₂H₅ → C₂H₅-O-C₂H₅ + NaBr

Give a chemical test to distinguish between the following pairs of compounds: (i) ethyl chloride and ethyl bromide; (ii) chlorobenzene and benzyl chloride.

Answer: Boil each with aqueous KOH, acidify with dilute HNO₃ and add AgNO₃ solution. (i) Ethyl chloride gives a white precipitate of AgCl (soluble in NH₄OH); ethyl bromide gives a pale yellow precipitate of AgBr (partially soluble in NH₄OH). (ii) Benzyl chloride gives a white precipitate of AgCl; chlorobenzene does not react with aqueous KOH, so there is no precipitate.

C₂H₅Cl + KOH(aq) → C₂H₅OH + KCl

KCl + AgNO₃ → AgCl (white ppt.) + KNO₃

C₂H₅Br + KOH(aq) → C₂H₅OH + KBr

KBr + AgNO₃ → AgBr (pale yellow ppt.) + KNO₃

C₆H₅CH₂Cl + KOH(aq) → C₆H₅CH₂OH + KCl

C₆H₅Cl + KOH(aq) → no reaction

Give reasons: (i) The boiling point of an alkyl bromide is higher than that of the alkyl chloride. (ii) Alkyl halides are better solvents than aryl halides. (iii) The haloalkanes used as solvents in industry are chloro compounds rather than bromo compounds.

Answer: (i) Because of the higher magnitude of van der Waals forces in the alkyl bromide than in the alkyl chloride. (ii) Due to the greater polarity of alkyl halides. (iii) The C–Cl bond is more polar than the C–Br bond, so chloro compounds are the better solvents (and they are cheaper).

Answer the following: (i) What effect should the resonance of vinyl chloride, CH₂=CH-Cl ↔ ⁻CH₂-CH=Cl⁺, have on its dipole moment? (ii) Iodoform is obtained by the reaction of acetone with hypoiodite but not with iodide ion. (iii) Vinyl chloride is hydrolysed more slowly than ethyl chloride.

Answer: (i) It decreases the dipole moment of vinyl chloride relative to ethyl chloride. (ii) To prepare iodoform from acetone positive iodine (I⁺) is required; it can be supplied by IO⁻ but not by I⁻, so hypoiodite is used. (iii) Due to resonance there is double-bond character between carbon and chlorine, which makes the bond stronger.

Write the structure of the major organic product in each of the following reactions: (i) CH₃CH₂CH₂Cl + NaI in acetone; (ii) CH₃CH₂CH₂OH + SOCl₂; (iii) CH₃CH₂CH=CH₂ + HBr in the presence of peroxide.

Answer:

(i) CH₃-CH₂-CH₂-Cl + NaI → CH₃-CH₂-CH₂-I + NaCl

(ii) CH₃-CH₂-CH₂-OH + SOCl₂ → CH₃-CH₂-CH₂-Cl + SO₂ + HCl

(iii) CH₃-CH₂-CH=CH₂ + HBr → CH₃-CH₂-CH₂-CH₂-Br

Give uses of the following: CCl₄, DDT, chloroform.

Answer: CCl₄: as a solvent. DDT: as an insecticide. Chloroform: as an anaesthetic in surgery (it was formerly used for this purpose).

Distinguish between S_N1 and S_N2 reactions.

Answer:

Five-mark questions

What happens when (a) n-butyl chloride is treated with alcoholic KOH; (b) bromobenzene is treated with Mg in the presence of dry ether; (c) ethyl chloride is treated with aqueous KOH; (d) ethyl bromide is treated with Na in the presence of dry ether; (e) methyl chloride is treated with KCN?

Answer:

(a) CH₃-CH₂-CH₂-CH₂-Cl + KOH(alc) → CH₃-CH₂-CH=CH₂ + KCl + H₂O

(b) C₆H₅Br + Mg → C₆H₅MgBr (dry ether)

(c) C₂H₅Cl + KOH(aq) → C₂H₅OH + KCl

(d) 2C₂H₅Br + 2Na → CH₃-CH₂-CH₂-CH₃ + 2NaBr (dry ether)

(e) CH₃Cl + KCN → CH₃CN + KCl

A primary alkyl halide A, C₄H₉Br, reacted with alcoholic KOH to give compound B. B reacted with HBr to give C, which is an isomer of A. When A was reacted with Na metal it gave a compound D, C₈H₁₈, that was different from the compound obtained when n-butyl bromide reacted with sodium. Give the structural formulae of A and write the equations for all the reactions.

Answer: A = (CH₃)₂CH-CH₂Br (1-bromo-2-methylpropane); B = (CH₃)₂C=CH₂ (2-methylpropene); C = (CH₃)₃C-Br (2-bromo-2-methylpropane); D = (CH₃)₂CH-CH₂-CH₂-CH(CH₃)₂ (2,5-dimethylhexane).

A → B: (CH₃)₂CH-CH₂Br + KOH(alc) → (CH₃)₂C=CH₂ + KBr + H₂O

B + HBr → C: (CH₃)₂C=CH₂ + HBr → (CH₃)₃C-Br

2A + 2Na → D: 2(CH₃)₂CH-CH₂Br + 2Na → (CH₃)₂CH-CH₂-CH₂-CH(CH₃)₂ + 2NaBr

Write short notes on (i) the Fittig reaction; (ii) Friedel–Crafts alkylation; (iii) Friedel–Crafts acylation; (iv) the Gattermann reaction; (v) the carbylamine reaction.

Answer:

(i) C₆H₅Cl + 2Na + C₆H₅Cl → C₆H₅-C₆H₅ + 2NaCl (dry ether)

(ii) C₆H₆ + CH₃Cl → C₆H₅CH₃ + HCl (anhydrous AlCl₃)

(iii) C₆H₆ + CH₃COCl → C₆H₅COCH₃ + HCl (anhydrous AlCl₃)

(iv) C₆H₅N₂Cl → C₆H₅Cl + N₂ (Cu/HCl)

(v) CH₃CH₂NH₂ + CHCl₃ + 3KOH(alc.) → CH₃CH₂NC + 3KCl + 3H₂O

Give reasons: (i) Benzyl chloride undergoes S_N1 reactions faster than cyclohexylmethyl chloride. (ii) p-Dichlorobenzene has a higher melting point than o-dichlorobenzene. (iii) Out of chlorobenzene and chloromethane, which is more reactive towards nucleophilic substitution? (iv) Thionyl chloride is preferred for preparing alkyl chlorides from alcohols. (v) The iodide ion is a better nucleophile than the bromide ion.

Answer: (i) In benzyl chloride the carbocation formed after the loss of Cl⁻ is stabilised by resonance. (ii) The symmetry of p-dichlorobenzene lets it fit better in the crystal lattice than o-dichlorobenzene. (iii) Chloromethane, being an alkyl halide, is more reactive. (iv) The by-products SO₂ and HCl are gases and escape, leaving the alkyl chloride in an almost pure state. (v) Because of its bigger size (greater polarisability) and lower electronegativity.

SET 2

One-mark questions

What happens when tert-butyl alcohol is treated with Cu at 573 K?

Answer: Dehydration takes place and isobutene is formed.

(CH₃)₃C-OH → (CH₃)₂C=CH₂ + H₂O (Cu, 573 K)

Arrange the following halides in order of increasing S_N2 reactivity: CH₃Cl, CH₃Br, CH₃CH₂Cl, (CH₃)₂CHCl.

Answer: (CH₃)₂CHCl < CH₃CH₂Cl < CH₃Cl < CH₃Br. (As the size of the alkyl group increases, S_N2 reactivity decreases.)

Alkyl halides, though polar, are immiscible with water. Why?

Answer: Alkyl halides cannot form hydrogen bonds with water molecules and hence are insoluble in water.

Grignard reagents should be prepared under anhydrous conditions. Why?

Answer: Grignard reagents have a very strong affinity for H⁺ ions. In the presence of water they abstract H⁺ from water and form alkanes. To prevent this, they must be prepared under anhydrous conditions.

Which of the following two compounds would react faster by the S_N2 pathway: 1-bromobutane or 2-bromobutane?

Answer: 1-Bromobutane. The reactivity in S_N2 depends upon the extent of steric hindrance; 1-bromobutane is a 1° alkyl halide and 2-bromobutane is a 2° alkyl halide, which has more steric hindrance.

Allyl chloride is more reactive than n-propyl chloride towards nucleophilic substitution. Explain.

Answer: Allyl chloride readily ionises to give the resonance-stabilised allyl carbocation, which readily combines with OH⁻ ions to form allyl alcohol.

Explain why chlorobenzene is more difficult to hydrolyse than ethyl chloride.

Answer: The lone pair of electrons on chlorine in chlorobenzene takes part in resonance with the benzene ring, so the C–Cl bond acquires partial double-bond character. This C–Cl bond is stronger than the pure single C–Cl bond in ethyl chloride, so chlorobenzene is difficult to hydrolyse.

R–Cl is hydrolysed to R–OH slowly, but the reaction is rapid if a catalytic amount of KI is added to the reaction mixture. Why?

Answer: The iodide ion is a powerful nucleophile (and a good leaving group), so it reacts rapidly with R–Cl to form R–I, which is then easily hydrolysed.

KI → K⁺ + I⁻

R-Cl + I⁻ → R-I + Cl⁻

Why are haloalkanes more reactive than haloarenes?

Answer: In haloarenes there is partial double-bond character between carbon and halogen due to resonance, which makes them less reactive.

Why do haloalkanes undergo nucleophilic substitution whereas haloarenes undergo electrophilic substitution?

Answer: In haloalkanes the halogen is electronegative, so the carbon atom becomes slightly positive and is easily attacked by nucleophiles. In haloarenes, because of resonance, the ring carbon atoms become slightly negative (electron rich) and are attacked by electrophiles.

Why is sulphuric acid not used during the reaction of alcohols with KI?

Answer: The HI formed is oxidised to I₂ by concentrated sulphuric acid, which is an oxidising agent.

p-Dichlorobenzene has a higher melting point than the o- and m-isomers. Why?

Answer: p-Dichlorobenzene is symmetrical, fits into the crystal lattice more readily and so has the higher melting point.

Although chlorine is an electron-withdrawing group, it is ortho and para directing in electrophilic aromatic substitution reactions. Why?

Answer: Chlorobenzene is a resonance hybrid with negative charge at the ortho and para positions, so electrophilic substitution takes place at those positions because of the +R effect. The +R effect dominates over the –I effect (the –I effect only makes the ring less reactive overall).

Explain why vinyl chloride is unreactive in nucleophilic substitution reactions.

Answer: Because of the double-bond character between C and Cl (resonance), the C–Cl bond is difficult to break.

Arrange the following compounds according to their reactivity towards nucleophilic substitution with CH₃ONa: 4-nitrochlorobenzene, 2,4-dinitrochlorobenzene, 2,4,6-trinitrochlorobenzene.

Answer: 2,4,6-Trinitrochlorobenzene > 2,4-dinitrochlorobenzene > 4-nitrochlorobenzene.

Arrange in order of boiling points: (a) bromobenzene, bromoform, chloromethane, dibromomethane; (b) 1-chloropropane, isopropyl chloride, 1-chlorobutane.

Answer: (a) Chloromethane < dibromomethane < bromoform < bromobenzene. (b) Isopropyl chloride < 1-chloropropane < 1-chlorobutane.

Predict the order of reactivity in S_N1: C₆H₅CH₂Br, (C₆H₅)₂CHBr, C₆H₅CH(CH₃)Br, C₆H₅C(CH₃)(C₆H₅)Br.

Answer: C₆H₅C(CH₃)(C₆H₅)Br (3°) > (C₆H₅)₂CHBr (2°) > C₆H₅CH(CH₃)Br (2°) > C₆H₅CH₂Br (1°).

Why is vinyl chloride less reactive than ethyl chloride?

Answer: Due to the sp² hybridisation of the carbon and the resonance in vinyl chloride, the C–Cl bond is stronger.

Chloroform is stored in a dark-coloured, sealed bottle. Why?

Answer: In air and light it is oxidised to the poisonous gas phosgene.

Two-mark questions

The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols, but in the presence of alcoholic KOH alkenes are the major products. Explain.

Answer: In aqueous solution KOH is almost completely ionised to give OH⁻ ions, which are strong nucleophiles and bring about substitution to form alcohols. In aqueous solution the OH⁻ ions are also highly solvated (hydrated), which reduces their basic character, so they fail to abstract a hydrogen from the β-carbon of the alkyl halide. An alcoholic solution of KOH contains alkoxide (RO⁻) ions, which are a much stronger base than OH⁻, and preferentially abstract a hydrogen from the β-carbon to form the alkene.

R-X + KOH(aq) → R-OH + KX

C₂H₅O⁻ + CH₃CH₂Cl → CH₂=CH₂ + C₂H₅OH + Cl⁻

p-Dichlorobenzene has a higher melting point and lower solubility than those of the o- and m-isomers. Discuss.

Answer: The p-isomer, being more symmetrical, fits closely in the crystal lattice and so has stronger intermolecular forces of attraction than the o- and m-isomers. Since the crystal lattice must break during melting or dissolution, a larger amount of energy is needed to melt or dissolve the p-isomer.

Haloalkanes react with KCN to form alkyl cyanides as the major product, while AgCN forms isocyanide as the chief product. Explain.

Answer: KCN is ionic and provides cyanide ions in solution. Although both C and N can donate an electron pair, the attack takes place mainly through carbon, since the C–C bond formed is more stable than the C–N bond. AgCN is mainly covalent and nitrogen is free to donate its electron pair, forming the isocyanide as the main product.

Which compound will react faster in an S_N2 reaction with OH⁻? (a) CH₃Br or CH₃I; (b) (CH₃)₃C-Cl or CH₃Cl.

Answer: (a) CH₃I reacts faster (iodide is a better leaving group). (b) CH₃Cl reacts faster (the 3° halide is sterically hindered).

How can the following conversions be carried out? (a) Propene to propan-1-ol; (b) 1-bromopropane to 2-bromopropane.

Answer:

(a) CH₃-CH=CH₂ + HBr → CH₃-CH₂-CH₂-Br (peroxide; anti-Markovnikov addition)

CH₃-CH₂-CH₂-Br + KOH(aq) → CH₃-CH₂-CH₂-OH + KBr (aq. KOH, heat; nucleophilic substitution)

(b) CH₃-CH₂-CH₂-Br → CH₃-CH=CH₂ + HBr (alc. KOH, heat; dehydrohalogenation)

CH₃-CH=CH₂ + HBr → CH₃-CHBr-CH₃ (Markovnikov addition)

Tert-butyl chloride reacts with aqueous NaOH by the S_N1 mechanism while n-butyl chloride reacts by the S_N2 mechanism. Why?

Answer: Heterolytic cleavage of the C–Cl bond in tert-butyl chloride gives a highly stable 3° carbocation, which favours S_N1; moreover the bulky 3° halide has steric hindrance, which does not allow S_N2. In n-butyl chloride (1°) the carbon of the C–Cl bond is less crowded, so the nucleophile can attack from the back side, forming the transition state; the low steric hindrance favours S_N2.

Why are alkyl halides generally not prepared in the laboratory by free radical halogenation of alkanes?

Answer: (i) It gives a mixture of isomeric monohalogenated products whose boiling points are so close that they cannot be separated easily. (ii) Polyhalogenation may also take place, which makes the mixture more complex and difficult to separate.

Three-mark questions

Why does the preparation of an aryl iodide by electrophilic substitution require the presence of an oxidising agent? Why can an aryl fluoride not be prepared by this method?

Answer: Iodination is reversible and requires an oxidising agent (HNO₃ etc.) to oxidise the HI formed during iodination, which promotes the forward reaction. Fluoro compounds cannot be prepared by this method because of the very high reactivity of fluorine.

Why are aryl halides extremely less reactive towards nucleophilic substitution?

Answer: (i) Resonance effect: due to resonance the C–Cl bond acquires partial double-bond character. (ii) Difference in hybridisation: in haloarenes the carbon attached to the halogen is sp² hybridised, while it is sp³ in haloalkanes; the C–X bond in the sp² case is shorter and stronger and difficult to break. (iii) Instability of the phenyl cation. (iv) Possible repulsion between the incoming nucleophile and the electron-rich arene.

Complete the reactions: (a) CH₃OCH₃ + PCl₅; (b) C₂H₅OCH₃ + HCl; (c) (C₂H₅)₂O + HCl.

Answer:

(a) CH₃OCH₃ + PCl₅ → 2CH₃Cl + POCl₃

(b) C₂H₅OCH₃ + HCl → CH₃Cl + C₂H₅OH

(c) (C₂H₅)₂O + HCl → C₂H₅Cl + C₂H₅OH

Distinguish between the following pairs of organic compounds: (i) CCl₄ and CHCl₃; (ii) chlorobenzene and benzyl chloride.

Answer: (i) Carbylamine test: chloroform gives the offensive smell of an isocyanide when heated with a primary amine and alcoholic KOH; CCl₄ does not. (ii) Boil each with aqueous NaOH, acidify with dilute HNO₃ and add AgNO₃: benzyl chloride gives a white precipitate of AgCl, chlorobenzene gives none.

How are the following conversions carried out? (i) Benzyl chloride to benzyl alcohol; (ii) ethylmagnesium chloride to propan-1-ol; (iii) C₆H₅N₂Cl to chlorobenzene.

Answer:

(i) C₆H₅CH₂Cl + KOH(aq) → C₆H₅CH₂OH + KCl

(ii) C₂H₅MgCl + HCHO → CH₃CH₂CH₂OMgCl

CH₃CH₂CH₂OMgCl + H₂O → CH₃CH₂CH₂OH + Mg(OH)Cl

(iii) C₆H₅N₂Cl → C₆H₅Cl + N₂ (Cu/HCl)

Write the formula of the main product formed in the following reactions: (a) (CH₃)₂CH-Cl + Na in dry ether; (b) CH₃Br + AgF (heat); (c) CH₃CH₂Br + NaI in dry acetone.

Answer:

(a) 2(CH₃)₂CH-Cl + 2Na → (CH₃)₂CH-CH(CH₃)₂ + 2NaCl

(b) CH₃Br + AgF → CH₃F + AgBr

(c) CH₃CH₂Br + NaI → CH₃CH₂I + NaBr

Five-mark questions

(a) Give the IUPAC name of CH₃CH₂-C(CH₃)(Br)-C(CH₃)(Br)-CH₂CH₂-OH. (b) Complete the reactions: (i) C₆H₅ONa + C₂H₅Cl; (ii) CH₃CH₂CH₂OH + SOCl₂; (iii) The preparation of a haloalkane from an alkane and a halogen using UV light is the least preferred method. Explain.

Answer: (a) 3,4-Dibromo-3,4-dimethylhexan-1-ol. (b)(iii) It gives a mixture of isomeric (primary, secondary, tertiary) mono- and polyhalogenated products, which are difficult to separate.

(i) C₆H₅ONa + C₂H₅Cl → C₆H₅-O-C₂H₅ + NaCl

(ii) CH₃CH₂CH₂OH + SOCl₂ → CH₃CH₂CH₂Cl + SO₂ + HCl

(a) p-Dichlorobenzene has a higher melting point and lower solubility than the o- and m-isomers. Explain. (b) Which has the higher boiling point: 1-chloropentane or 2-chloro-2-methylbutane? Give reasons. (c) Why is chloroform not used as an anaesthetic nowadays?

Answer: (a) The p-isomer, being more symmetrical, fits better in the crystal lattice and has stronger intermolecular forces of attraction than the o- and m-isomers; a large amount of energy is therefore needed to melt or dissolve it. (b) 1-Chloropentane. On branching, the surface area and hence the van der Waals forces of attraction decrease. (c) On oxidation it forms the poisonous gas phosgene.

(a) p-Nitrochlorobenzene undergoes nucleophilic substitution faster than chlorobenzene. Explain, giving the resonating structures as well. (b) Allyl chloride is more reactive than n-propyl chloride towards nucleophilic substitution. Why? (c) Give the IUPAC name of C₆H₅CH₂Cl.

Answer: (a) In this reaction a carbanion intermediate is formed. In p-nitrochlorobenzene this carbanion is stabilised by resonance, because the negative charge of the ring can be delocalised onto the oxygen atoms of the nitro group at the para position; in chlorobenzene there is no such stabilisation. (b) In allyl chloride the carbocation formed is stabilised by resonance, while the carbocation formed from n-propyl chloride is less stable, so allyl chloride is more reactive. (c) Chlorophenylmethane (benzyl chloride).

(a) Why do alcohols have higher boiling points than haloalkanes of the same molecular mass? (b) Convert: (i) benzene to aniline; (ii) benzene to diphenyl; (iii) benzene to p-chlorotoluene.

Answer: (a) Alcohols are capable of forming intermolecular hydrogen bonds. (b)

(i) C₆H₆ + Cl₂ → C₆H₅Cl + HCl (FeCl₃)

C₆H₅Cl + NH₃ → C₆H₅NH₂ + HCl (Cu₂O, heat, pressure)

(ii) C₆H₆ + Cl₂ → C₆H₅Cl + HCl (FeCl₃)

2C₆H₅Cl + 2Na → C₆H₅-C₆H₅ + 2NaCl (dry ether)

(iii) C₆H₆ + CH₃Cl → C₆H₅CH₃ + HCl (anhydrous AlCl₃)

C₆H₅CH₃ + Cl₂ → p-CH₃C₆H₄Cl + HCl (FeCl₃; the o-isomer is also formed and is separated)

     • i) Ambidient nucleophile (ii) Chirality

Q2.Write short note on sandmeyer reaction?

Q3.Write the structures of main products:

Q6.Write short note on:

(i).Wurtz reaction

• Three Marks Questions:

     1.Explain why:

(a).Dipole moment of cholorobenzene is lower than cyclohexyl chloride.

(b).Chlorobenzene to phenol

(c).Ethyl bromide to diethyl ether.