CBSE Class 12 Chemistry Haloalkanes and Haloarenes Important Questions Set 04

Very Short Answer Questions (1 Mark)

Question 1. Give one chemical test to distinguish between chlorobenzene and benzyl chloride.
Answer:
AgNO₃ test: on warming with aqueous/alcoholic AgNO₃, benzyl chloride gives a white precipitate of AgCl.
Chlorobenzene does not give a precipitate, because its C–Cl bond is not broken under these conditions.

Teacher's Note:
1. This test separates halides that ionise easily (alkyl, benzyl, allyl) from those that do not (aryl, vinyl).

Question 2. Why does iodoform show antiseptic properties?
Answer:
Iodoform shows antiseptic properties due to the free iodine that it liberates (when it comes in contact with skin), not due to iodoform itself.

Teacher's Note:
1. Because of its unpleasant smell, iodoform has largely been replaced by other iodine-based antiseptics.

Question 3. The presence of nitro group (–NO₂) at ortho or para positions increases the reactivity of haloarenes towards nucleophilic substitution reactions. Explain.
Answer:
The –NO₂ group is strongly electron withdrawing. When present at ortho or para position, it withdraws electron density from the ring and the negative charge of the carbanion intermediate formed by attack of the nucleophile is delocalised onto the nitro group by resonance.
This stabilises the intermediate and increases the reactivity. At the meta position no such resonance stabilisation is possible.

Teacher's Note:
1. Draw the resonance forms of the intermediate: only for o/p-NO₂ does a form appear with the negative charge on the nitro oxygen.

Question 4. For the preparation of alkyl chlorides from alcohols, thionyl chloride (SOCl₂) is preferred. Give reason.
Answer:
R–OH + SOCl₂ → R–Cl + SO₂↑ + HCl↑
The by-products SO₂ and HCl are gases and escape, leaving pure alkyl chloride. Hence SOCl₂ is preferred.

Teacher's Note:
1. The key word in the answer is "gaseous by-products", which means no separation step is needed.

Question 5. Arrange the following compounds according to their reactivity towards nucleophilic substitution reaction with CH₃ONa:
4-nitrochlorobenzene, 2,4-dinitrochlorobenzene, 2,4,6-trinitrochlorobenzene
Answer:
2,4,6-Trinitrochlorobenzene > 2,4-dinitrochlorobenzene > 4-nitrochlorobenzene

Teacher's Note:
1. Each additional nitro group at an ortho or para position further stabilises the negatively charged intermediate.

Question 6. Why is sulphuric acid not used during the reaction of alcohols with KI?
Answer:
It is because HI formed will get oxidised to I₂ by concentrated sulphuric acid, which is an oxidising agent.

Teacher's Note:
1. Phosphoric acid (H₃PO₄) is used instead because it is non-oxidising.

Question 7. p-Dichlorobenzene has higher m.p. than those of ortho and meta isomers. Why?
Answer:
p-Dichlorobenzene is symmetrical; it fits into the crystal lattice more readily and has a higher melting point.

Teacher's Note:
1. Melting point depends on packing in the solid, so symmetry matters; boiling points of the three isomers are nearly the same.

Question 8. Write the structure of
(i) 2, 4-dinitrochlorobenzene
(ii) 1-bromo-4-chlorobut-2-ene
(iii) 3-bromo-2-methylprop-1-ene (CBSE 2017)
Answer:
(i) 2, 4-Dinitrochlorobenzene (Cl at C-1, NO₂ groups at C-2 and C-4):
(ii) ¹BrCH₂–²CH=³CH–⁴CH₂Cl   1-Bromo-4-chlorobut-2-ene
(iii) Br–³CH₂–²C(CH₃)=¹CH₂   3-Bromo-2-methylprop-1-ene

Teacher's Note:
1. Draw the parent chain first and number it from the end nearer the double bond, then attach each substituent at its locant.

Question 9. (i) Out of the two compounds shown below, which is an example of allylic halide ?
(ii) Out of C₆H₅CHCl₂ and C₆H₅CH₂CH₂Cl, which is an example of a benzylic halide ? (CBSE 2017)

Answer:
(i) The first compound, in which X is attached to the sp^3 carbon next to the C=C double bond, is an allylic halide while the second compound, in which X is attached to a doubly bonded carbon, is a vinylic halide.
(ii) Compounds in which the halogen is attached to an sp³⁻hybridized carbon atom next to the benzene ring (C₆H₅–C<, C₆H₅–CH< or C₆H₅–CH₂–) are called benzylic halides. Thus, out of C₆H₅CHCl₂ and C₆H₅CH₂CH₂Cl, C₆H₅CHCl₂ is a benzylic halide.

Teacher's Note:
1. Allylic and benzylic carbons are the sp³ carbons directly next to the C=C or the ring; a halogen on the double-bond carbon itself is vinylic.

Question 10. Convert propanoic acid to bromoethane ?
Answer:
This can be carried out by Hunsdiecker reaction
CH₃CH₂COOH (Propanoic acid) \( \xrightarrow{\text{AgNO}_3,\ -\text{HNO}_3} \) CH₃CH₂COOAg (Silver propanoate)
CH₃CH₂COOAg \( \xrightarrow{\text{Br}_2\text{/boiling CCl}_4} \) CH₃CH₂Br (Bromoethane) + AgBr + CO₂   (Hunsdiecker reaction)

Teacher's Note:
1. Hunsdiecker reaction removes one carbon (as CO₂): a C₃ acid gives a C₂ bromide.

Question 11. How will you obtain 1-bromopropane from propene ?
Answer:
With HBr in presence of peroxides
CH₃CH=CH₂ \( \xrightarrow{\text{HBr, peroxides}} \) CH₃CH₂CH₂Br

Teacher's Note:
1. Peroxide effect (anti-Markovnikov addition) works only with HBr, not with HCl or HI.

Question 12. How will you bring about the conversion : methyl bromide to methyl iodide ?
Answer:
CH₃–Br (Methyl bromide) + NaI \( \xrightarrow{\text{Dry acetone}} \) CH₃–I (Methyl iodide) + NaBr   (Finkelstein reaction)

Teacher's Note:
1. The reaction goes forward because NaBr (or NaCl) is insoluble in dry acetone and precipitates out.

Question 13. What happens when chlorine is passed through boiling toluene in the presence of sunlight ?
Answer:
Free radical substitution occurs at the methyl group giving first benzyl chloride, then benzal dichloride and finally benzotrichloride.
C₆H₅CH₃ (Toluene) + Cl₂ \( \xrightarrow{383\ \text{K},\ h\nu,\ -\text{HCl}} \) C₆H₅CH₂Cl (Benzyl chloride)
\( \xrightarrow{383\ \text{K},\ h\nu,\ -\text{HCl}} \) C₆H₅CHCl₂ (Benzal dichloride) \( \xrightarrow{383\ \text{K},\ h\nu,\ -\text{HCl}} \) C₆H₅CCl₃ (Benzotrichloride)

Teacher's Note:
1. Light and heat give side-chain (free radical) substitution; a Lewis acid like FeCl₃ in the dark gives ring substitution instead.

Question 14. n-Butyl bromide has higher boiling point than t-butyl bromide. Give reasons. (CBSE 2015)
Answer:
For the same alkyl group, the b.p. decreases as the branching increases or the surface area decreases. Since n-butyl bromide has no branching while t-butyl bromide has two branches, therefore, b.p. of n-butyl bromide is higher than that of t-butyl bromide.
CH₃CH₂CH₂CH₂–Br: no branching, surface area is more and hence has higher b.p. (i.e., 375 K)
(CH₃)₃C–Br: two branches at α-carbon, surface area is less and hence has lower b.p. (i.e., 346 K)

Teacher's Note:
1. Among isomeric haloalkanes, more branching means a more compact, nearly spherical molecule with weaker van der Waals forces and a lower boiling point.

Question 15. Arrange the following in order of increasing boiling point :
(i) CH₃CH₂CH₂CH₂Br   (ii) (CH₃)₃CBr   (iii) (CH₃)₂CHCH₂Br.
Answer:
The b.p. increases as the branching decreases (or the surface area increases), i.e.,
(CH₃)₃CBr < (CH₃)₂CHCH₂Br < CH₃CH₂CH₂CH₂Br.

Teacher's Note:
1. These are isomers, so molecular mass is the same; only the degree of branching decides the order.

Question 16. Out of ethyl bromide and ethyl chloride which has higher boiling point and why ? (CBSE 2007)
Answer:
Due to bigger size of Br over Cl, the van der Waals forces of attraction are stronger in ethyl bromide than in ethyl chloride. Therefore, ethyl bromide has higher boiling point than ethyl chloride.

Teacher's Note:
1. For the same alkyl group, b.p. rises in the order R–F < R–Cl < R–Br < R–I as size and mass of the halogen increase.

Short Answer Questions (2 Marks)

Question 17. Write short note on:
(i) Sandmeyer reaction
(ii) Finkelstein reaction
Answer:
(i) Sandmeyer reaction: When a freshly prepared solution of benzenediazonium chloride is treated with cuprous chloride (in HCl) or cuprous bromide (in HBr), the diazonium group is replaced by –Cl or –Br.
C₆H₅N₂⁺Cl⁻ –[CuCl/HCl]→ C₆H₅Cl + N₂;   C₆H₅N₂⁺Cl⁻ –[CuBr/HBr]→ C₆H₅Br + N₂
(ii) Finkelstein reaction: Alkyl iodides are prepared by heating alkyl chlorides or bromides with NaI in dry acetone.
R–X + NaI –[dry acetone]→ R–I + NaX (X = Cl, Br)
NaCl or NaBr formed is insoluble in dry acetone and precipitates out, which drives the reaction forward.

Teacher's Note:
1. In Finkelstein reaction always mention "dry acetone" and why it is used – examiners look for the precipitation of NaCl/NaBr.

Question 18. Name the reagents used to convert:
(i) 2-chloropropane to 2-nitropropane
(ii) Chloroethane to n-butane
Answer:
(i) AgNO₂: (CH₃)₂CHCl + AgNO₂ → (CH₃)₂CHNO₂ + AgCl
(ii) Na/dry ether (Wurtz reaction): 2C₂H₅Cl + 2Na → C₄H₁₀ + 2NaCl

Teacher's Note:
1. AgNO₂ gives nitroalkanes (attack through N), while KNO₂ gives mainly alkyl nitrites (attack through O).

Question 19. Draw structure of monohalo product in each of the following:
(i) Cyclohexanol –[SOCl₂]→
(ii) C₆H₅–CH₂CH=CH₂ + HBr –[peroxide]→
Answer:
(i) C₆H₁₁–Cl (chlorocyclohexane)
(ii) C₆H₅–CH₂CH₂CH₂Br (1-bromo-3-phenylpropane)

Teacher's Note:
1. SOCl₂ replaces –OH by –Cl directly; the ring itself is not affected.

Question 20. Why are haloalkanes more reactive than haloarenes?
Answer:
(i) In haloarenes, there is partial double bond character between carbon and halogen due to resonance effect, which makes them less reactive.
(ii) In benzene, carbon is sp² hybridised, which is smaller in size than the sp³ carbon present in haloalkanes. So the C–Cl bond in aryl halides is shorter and stronger.

Teacher's Note:
1. These two reasons are the standard two-mark answer; add the resonance structures of chlorobenzene if a third mark is offered.

Question 21. When an alkyl halide is treated with ethanolic solution of KCN, the major product is alkyl cyanide, whereas if alkyl halide is treated with AgCN, the major product is alkyl isocyanide. Explain.
Answer:
KCN is predominantly ionic and provides cyanide ions in solution. The cyanide ion is an ambident nucleophile, but the attack takes place mainly through the carbon atom, since the C–C bond formed is more stable than a C–N bond. Hence alkyl cyanide is the major product: R–X + KCN → R–CN + KX
AgCN is mainly covalent (Ag–C bond), so the carbon is not free and the attack takes place through the lone pair on nitrogen, forming isocyanide as the major product: R–X + AgCN → R–NC + AgX

Teacher's Note:
1. Link the product to the nature of the reagent: ionic KCN → R–CN, covalent AgCN → R–NC.

Question 22. The treatment of alkyl chlorides with aqueous KOH leads to the formation of alcohols but in presence of alcoholic KOH alkenes are the major products. Explain.
Answer:
In aqueous KOH, OH⁻ is the nucleophile which replaces the halogen (a substitution reaction):
R–X + KOH → R–OH + KX
Whereas in alcoholic KOH, C₂H₅O⁻ ion is produced, which is a strong base; hence β-elimination takes place to form an alkene:
C₂H₅OH + KOH → C₂H₅O⁻ + K⁺ + H₂O
CH₃CH₂–Cl + alcoholic KOH → CH₂=CH₂ + KCl + H₂O

Teacher's Note:
1. OH⁻ in water is heavily solvated and acts mainly as a nucleophile; ethoxide is a stronger base and removes a β-hydrogen.

Question 23. Explain why vinyl chloride is unreactive in nucleophilic substitution reaction.
Answer:
Vinyl chloride is unreactive in nucleophilic substitution reaction because of the double bond character of the C–Cl bond, which is difficult to break.
The lone pair on Cl is delocalised by resonance: H₂C=CH–Cl: ↔ H₂C⁻–CH=Cl⁺

Teacher's Note:
1. Show the resonance structure with the curved-arrow shift of the Cl lone pair; it is what earns the mark.

Question 24. Explain why alkyl halides are generally not prepared in the laboratory by free radical halogenation of alkanes.
Answer:
Free radical halogenation is not a suitable method for laboratory synthesis of alkyl halides because of the following two reasons:
(i) It gives a mixture of isomeric monohalogenated products whose boiling points are so close that they cannot be easily separated in the laboratory.
(ii) Polyhalogenation may also occur to some extent thereby making the mixture more complex and hence more difficult to separate.

Teacher's Note:
1. In the lab, alkyl halides are made from alcohols (SOCl₂, PCl₅, HX/ZnCl₂) or from alkenes, which give a single product.

Question 25. Wurtz reaction fails in case of tert-alkyl halides. Explain.
Answer:
tert-Alkyl halides prefer to undergo dehydrohalogenation in presence of a strong base such as Na metal instead of undergoing Wurtz reaction as explained below:
(CH₃)₃C–Br (tert-butyl bromide) + 2Na → (CH₃)₃C⁻Na⁺ (tert-butylsodium) + Na⁺Br⁻
(CH₃)₃C⁻Na⁺ + H–CH₂–C(CH₃)₂–Br →(−NaBr) (CH₃)₃CH (isobutane) + CH₂=C(CH₃)–CH₃ (isobutylene)
Thus, only 1° and 2° alkyl halides undergo Wurtz reaction while 3° alkyl halides prefer to undergo dehydrohalogenation to form alkenes.

Teacher's Note:
1. The bulky tert-butyl carbanion acts as a base and removes a β-H rather than attacking the crowded 3° carbon.

Question 26. Complete the following reaction:
CH₃–CH=CH₂ →(HBr, peroxide) X →(NaI, Acetone) Y
or How will you convert propene to 1-iodopropane? (CBSE 2016)
Answer:
CH₃–CH=CH₂ (propene) →(HBr, peroxide; anti-Markovnikov addition) CH₃CH₂CH₂–Br [1-bromopropane (X)]
CH₃CH₂CH₂–Br →(NaI, acetone; Finkelstein reaction) CH₃CH₂CH₂–I [1-iodopropane (Y)]

Teacher's Note:
1. Finkelstein reaction works because NaBr (or NaCl) formed is insoluble in dry acetone and precipitates, driving the reaction forward.

Question 27. CHF₃ is less acidic than CHCl₃. Explain.
Answer:
Due to stronger −I-effect of F than Cl, CHF₃ should be more acidic than CHCl₃. But actually reverse is true.
This is due to the reason that :CCl₃⁻ left after the removal of a proton from CHCl₃ is stabilized by resonance due to the presence of d-orbitals on Cl but :CF₃⁻ left after the removal of a proton from CHF₃ is not stabilized by resonance due to the absence of d-orbitals on F.

Teacher's Note:
1. Acidity is decided by the stability of the conjugate base; here the extra stabilisation of :CCl₃⁻ outweighs the −I effect of F.

Question 28. What will be the major organic product of the following reaction?
C₆H₅C₂H₅ →((i) Br₂, hν; (ii) NaCN)
Answer:
When ethylbenzene is treated with Br₂ in presence of light, the bromine free radical formed during the reaction, abstracts the α (but not β-)hydrogen from the ethyl group to form a resonance stabilized benzylic radical (I) which then reacts with Br₂ to form 1-bromo-1-phenylethane (II). This on treatment with NaCN undergoes nucleophilic substitution to form 2-phenylpropanenitrile (III).
C₆H₅–CH₂–CH₃ (ethylbenzene) →(Br·, −HBr) C₆H₅–C·H–CH₃ (I) →(Br–Br, −Br·) C₆H₅–CHBr–CH₃ [1-bromo-1-phenylethane (II)]
C₆H₅–CHBr–CH₃ →(Na⁺CN⁻, −NaBr) C₆H₅–CH(CN)–CH₃ [2-phenylpropanenitrile (III)]

Teacher's Note:
1. Side-chain halogenation of alkylbenzenes always occurs at the benzylic (α) carbon because the benzylic radical is resonance stabilised.

Question 29. Arrange the following halides in order of increasing S_N2 reactivity:
CH₃Cl, CH₃Br, CH₃CH₂Cl, (CH₃)₂CHCl.
Answer:
As the size of the alkyl group increases, S_N2 reactivity decreases. Further, C–Br bond being weaker is easier to break than C–Cl bond. Therefore, the overall increasing S_N2 reactivity follows the order:
(CH₃)₂CHCl < CH₃CH₂Cl < CH₃Cl < CH₃Br

Teacher's Note:
1. Two factors decide S_N2 rate: steric crowding at the carbon (methyl fastest) and leaving-group ability (I > Br > Cl > F).

Short Answer Questions (3 Marks)

Question 30. Answer the following:
(i) Haloalkanes easily dissolve in organic solvents, why?
(ii) What is known as racemic mixture? Give example.
(iii) Of the two bromo derivatives, C₆H₅CH(CH₃)Br and C₆H₅CH(C₆H₅)Br, which one is more reactive in S_N1 substitution reaction and why?
Answer:
(i) Haloalkanes dissolve in organic solvents because the new intermolecular attractions between haloalkane and solvent molecules are of about the same strength as the attractions being broken in the separate haloalkane and solvent.
(ii) An equimolar mixture of a pair of enantiomers (d- and l-forms) is called a racemic mixture. It is optically inactive because the rotation due to one isomer is cancelled by the other. Example: (±)-butan-2-ol, formed by S_N1 hydrolysis of optically active 2-bromobutane.
(iii) C₆H₅CH(C₆H₅)Br is more reactive, because the carbocation (C₆H₅)₂CH⁺ it forms is stabilised by resonance with two phenyl rings, while C₆H₅C⁺HCH₃ is stabilised by only one.

Teacher's Note:
1. A racemic mixture shows zero optical rotation even though it contains chiral molecules – this is called external compensation.

Question 31. Answer the following:
(i) What is meant by chirality of a compound? Give an example.
(ii) Which one of the following compounds is more easily hydrolysed by KOH and why?
CH₃CHClCH₂CH₃ or CH₃CH₂CH₂Cl
(iii) Which one undergoes S_N2 substitution reaction faster and why?
1-iodopentane or 1-chloropentane
Answer:
(i) A compound (object) which is non-superimposable on its mirror image is said to be chiral, and this property is called chirality. Example: butan-2-ol, CH₃CH(OH)CH₂CH₃, or 2-chlorobutane, whose C-2 carries four different groups.
(ii) CH₃CHClCH₂CH₃ (2-chlorobutane) is more easily hydrolysed, because it is a secondary halide and forms a more stable secondary carbocation than the primary carbocation from CH₃CH₂CH₂Cl.
(iii) 1-Iodopentane reacts faster by S_N2, because the C–I bond is weaker than the C–Cl bond and I⁻ is a better leaving group than Cl⁻.

Teacher's Note:
1. Molecules with exactly one chiral carbon are always chiral; with two or more, check for a plane of symmetry (meso forms).

Question 32. Complete the following reactions:
(i) CH₃CH₂OH –[SOCl₂]→ A –[KCN]→ B
(ii) (CH₃)₂CHBr + Na →
(iii) CH₃CH₂Cl –[AgNO₃]→
Answer:
(i) A = CH₃CH₂Cl; B = CH₃CH₂CN
CH₃CH₂OH + SOCl₂ → CH₃CH₂Cl + SO₂ + HCl;   CH₃CH₂Cl + KCN → CH₃CH₂CN + KCl
(ii) 2(CH₃)₂CHBr + 2Na –[dry ether]→ (CH₃)₂CH–CH(CH₃)₂ + 2NaBr (2,3-dimethylbutane, Wurtz reaction)
(iii) CH₃CH₂Cl + AgNO₃ → CH₃CH₂ONO₂ + AgCl↓ (white precipitate)

Teacher's Note:
1. In Wurtz reaction, the product is found by joining two alkyl groups at the carbons that carried the halogen.

Question 33. How can the following conversions be carried out?
(i) But-1-ene to n-butyl iodide
(ii) Tert-butyl bromide to isobutyl bromide
(iii) Ethanol to but-1-yne
Answer:
(i) CH₃CH₂CH=CH₂ –[HBr, peroxide]→ CH₃CH₂CH₂CH₂Br –[NaI, dry acetone]→ CH₃CH₂CH₂CH₂I
(ii) (CH₃)₃C–Br –[alc. KOH, Δ]→ (CH₃)₂C=CH₂ –[HBr, peroxide]→ (CH₃)₂CH–CH₂Br
(iii) CH₃CH₂OH –[conc. H₂SO₄, 443 K]→ CH₂=CH₂ –[Br₂]→ BrCH₂CH₂Br –[alc. KOH, then NaNH₂]→ HC≡CH
HC≡CH –[NaNH₂]→ HC≡C⁻Na⁺;   CH₃CH₂OH –[PBr₃]→ CH₃CH₂Br
HC≡C⁻Na⁺ + CH₃CH₂Br → HC≡C–CH₂CH₃ (but-1-yne) + NaBr

Teacher's Note:
1. Moving Br from a tertiary to a primary carbon is a two-step trick: eliminate, then add HBr in presence of peroxide.

Question 34. Write short notes on:
(i) Wurtz–Fittig reaction
(ii) Fittig reaction
(iii) Dehydrohalogenation reaction
Answer:
(i) Wurtz–Fittig reaction: A mixture of an alkyl halide and an aryl halide gives an alkylarene when treated with sodium in dry ether.
C₆H₅Cl + CH₃Cl + 2Na –[dry ether]→ C₆H₅CH₃ + 2NaCl
(ii) Fittig reaction: Aryl halides when treated with sodium in dry ether give diaryls.
2C₆H₅Cl + 2Na –[dry ether]→ C₆H₅–C₆H₅ + 2NaCl
(iii) Dehydrohalogenation: When a haloalkane having a β-hydrogen is heated with alcoholic KOH, a molecule of HX is eliminated (H from the β-carbon and X from the α-carbon) and an alkene is formed (β-elimination). When more than one alkene is possible, the more substituted alkene is the major product (Saytzeff rule).
CH₃CH₂CH(Br)CH₃ –[alc. KOH, Δ]→ CH₃CH=CHCH₃ (major, 81%) + CH₃CH₂CH=CH₂ (19%)

Teacher's Note:
1. Wurtz: R + R; Fittig: Ar + Ar; Wurtz–Fittig: R + Ar – all with Na in dry ether.

Question 35. An organic compound ‘A’ having molecular formula C₄H₈ on treatment with dil. H₂SO₄ gives another compound ‘B’. B on treatment with conc. HCl and anhy. ZnCl₂ gives ‘C’. C on treatment with sodium ethoxide gives back ‘A’. Identify the compounds. Write the equations involved.
Answer:
A = (CH₃)₂C=CH₂ (2-methylpropene); B = (CH₃)₃C–OH (2-methylpropan-2-ol); C = (CH₃)₃C–Cl (2-chloro-2-methylpropane)
(CH₃)₂C=CH₂ + H₂O –[dil. H₂SO₄]→ (CH₃)₃C–OH
(CH₃)₃C–OH + HCl –[anhy. ZnCl₂]→ (CH₃)₃C–Cl + H₂O
(CH₃)₃C–Cl + C₂H₅ONa → (CH₃)₂C=CH₂ + C₂H₅OH + NaCl

Teacher's Note:
1. With a strong base like ethoxide, tertiary halides undergo elimination rather than substitution.

Question 36. What happens when:
(i) 1-bromopropane reacts with metallic sodium.
(ii) Bromoethane is treated with caustic potash.
(iii) Iodomethane is treated with ammonia.
Answer:
(i) n-Hexane is formed (Wurtz reaction): 2CH₃CH₂CH₂Br + 2Na –[dry ether]→ CH₃(CH₂)₄CH₃ + 2NaBr
(ii) With aqueous KOH, ethanol is formed: C₂H₅Br + KOH(aq) → C₂H₅OH + KBr. With alcoholic KOH, ethene is formed: C₂H₅Br + KOH(alc) → CH₂=CH₂ + KBr + H₂O
(iii) Methylamine is formed: CH₃I + NH₃ → CH₃NH₂ + HI. With excess CH₃I, further alkylation gives (CH₃)₂NH, (CH₃)₃N and finally (CH₃)₄N⁺I⁻.

Teacher's Note:
1. Aqueous KOH means substitution (alcohol); alcoholic KOH means elimination (alkene) – always state which one you assume.

Question 37. Identify A, B and C:
2-propanol –[SOCl₂]→ A –[Mg, dry ether]→ B –[H₂O]→ C
Answer:
A = (CH₃)₂CHCl (2-chloropropane)
B = (CH₃)₂CHMgCl (isopropylmagnesium chloride)
C = CH₃CH₂CH₃ (propane)
(CH₃)₂CHMgCl + H₂O → CH₃CH₂CH₃ + Mg(OH)Cl

Teacher's Note:
1. Water simply replaces MgX by H, so the Grignard route converts an alcohol into the corresponding alkane.

Question 38. Account for the following:
(i) A small amount of ethyl alcohol is added to CHCl₃ stored for use as an anaesthetic.
(ii) After using CCl₄ as a fire extinguisher inside a closed space, the space is thoroughly ventilated.
(iii) When 2-chloro-3-methylbutane is treated with alcoholic potash, 2-methylbut-2-ene is the main product.
Answer:
(i) Chloroform is slowly oxidised by air in light to poisonous phosgene: 2CHCl₃ + O₂ → 2COCl₂ + 2HCl. Ethyl alcohol converts any harmful COCl₂ formed into harmless ethyl carbonate: COCl₂ + 2C₂H₅OH → (C₂H₅O)₂CO + 2HCl
(ii) At the high temperature of the fire, CCl₄ vapour reacts with water vapour to form poisonous phosgene: CCl₄ + H₂O → COCl₂ + 2HCl. The space is ventilated to sweep out the COCl₂.
(iii) According to Saytzeff rule, the more substituted alkene is the major product of dehydrohalogenation:
(CH₃)₂CH–CHCl–CH₃ –[alc. KOH]→ (CH₃)₂C=CH–CH₃ (major) + (CH₃)₂CH–CH=CH₂ (minor)

Teacher's Note:
1. In Saytzeff questions, count alkyl groups on the C=C carbons; the alkene with more of them is the major product.

Question 39. How will you distinguish between:
(i) Vinyl chloride and ethyl chloride
(ii) Chlorobenzene and cyclohexyl chloride
(iii) Ethyl chloride and ethyl bromide
Answer:
(i) Boil each with aqueous KOH, acidify with dil. HNO₃ and add AgNO₃ solution. Ethyl chloride gives a white precipitate of AgCl; vinyl chloride does not, since its C–Cl bond has partial double bond character.
(ii) By the same test: cyclohexyl chloride gives a white precipitate of AgCl, while chlorobenzene gives no precipitate.
(iii) By the same test: ethyl chloride gives a white precipitate (AgCl) soluble in ammonium hydroxide, while ethyl bromide gives a pale yellow precipitate (AgBr) sparingly soluble in ammonium hydroxide.

Teacher's Note:
1. Always acidify with dil. HNO₃ before adding AgNO₃, otherwise excess KOH precipitates brown Ag₂O and spoils the test.

Question 40. Explain the following:
(i) The dipole moment of chloroethane is higher than that of chlorobenzene.
(ii) Although haloalkanes are polar in character, yet they are insoluble in water.
(iii) Vinyl chloride is unreactive in nucleophilic substitution reactions.
Answer:
(i) In chlorobenzene, the C–Cl carbon is sp² hybridised (more electronegative than sp³ carbon) and the lone pair of Cl is delocalised into the ring (+R effect). Both reduce the polarity of the C–Cl bond, so its dipole moment is lower than that of chloroethane.
(ii) To dissolve in water, energy is needed to break the hydrogen bonds between water molecules. Haloalkanes cannot form hydrogen bonds with water, and the energy released in forming new attractions between haloalkane and water molecules is less than that required. Hence they are insoluble.
(iii) In vinyl chloride, the lone pair on Cl is in resonance with the C=C bond: CH₂=CH–Cl ↔ ⁻CH₂–CH=Cl⁺. The C–Cl bond gets partial double bond character and the carbon is sp² hybridised, so the bond is shorter, stronger and difficult to break.

Teacher's Note:
1. The same two ideas – resonance and sp² hybridisation – explain the low reactivity and low dipole moment of both vinyl and aryl halides.

Long Answer Questions (5 Marks)

Question 41. Complete the following reactions:
(i) C₆H₆ –[Cl₂/Fe]→ X –[CuCN, pyridine]→ Y –[H⁺, H₂O]→
(ii) C₂H₄ –[HBr]→ X –[aq. KOH]→ Y –[I₂, NaOH]→ Z
(iii) CH₃CH₂Br –[AgCN]→ A
(iv) 3-Ethylpent-2-ene –[Br₂/H₂O]→ B
Answer:
(i) X = C₆H₅Cl (chlorobenzene); Y = C₆H₅CN (benzonitrile); final product = C₆H₅COOH (benzoic acid)
(ii) X = CH₃CH₂Br; Y = CH₃CH₂OH; Z = CHI₃ (iodoform, yellow precipitate)
(iii) A = CH₃CH₂NC (ethyl isocyanide)
(iv) CH₃CH=C(C₂H₅)₂ + Br₂/H₂O → CH₃CH(Br)–C(OH)(C₂H₅)₂; B = 2-bromo-3-ethylpentan-3-ol (bromohydrin; OH goes to the more substituted carbon)

Teacher's Note:
1. Ethanol gives the iodoform test because it is oxidised to ethanal, which has the CH₃CO– group.

Question 42. Account for the following:
(i) Sulphuric acid is not used during the reaction of alcohols with KI.
(ii) p-Methoxybenzyl bromide reacts faster than p-nitrobenzyl bromide with sodium ethoxide to form an ether product.
(iii) Organic halogen compounds used as solvents in industry are chlorides rather than bromides and iodides.
(iv) Wurtz reaction fails in case of tert-alkyl halides.
(v) Alkyl halides are insoluble in water though they contain a polar C–X bond.
(vi) Use of CHCl₃ as anaesthetic is not preferred.
Answer:
(i) H₂SO₄ is an oxidising agent; it oxidises the HI formed to I₂, so no alkyl iodide is obtained: 2KI + H₂SO₄ → K₂SO₄ + 2HI;   2HI + H₂SO₄ → I₂ + SO₂ + 2H₂O. A non-oxidising acid such as H₃PO₄ is used instead.
(ii) The reaction proceeds through a benzylic carbocation (S_N1). The electron-donating –OCH₃ group (+R effect) stabilises the p-methoxybenzyl carbocation, whereas the electron-withdrawing –NO₂ group destabilises the p-nitrobenzyl carbocation. Hence p-methoxybenzyl bromide reacts faster.
(iii) Chlorides are cheaper and more easily available, and being less reactive (stronger C–Cl bond) they are more stable than bromides and iodides. Hence chlorides are used as solvents.
(iv) Tert-alkyl halides are sterically hindered and, with sodium, they undergo elimination to form alkenes instead of coupling. Hence Wurtz reaction fails.
(v) Alkyl halides cannot form hydrogen bonds with water. The energy released in forming new attractions between alkyl halide and water molecules is less than the energy needed to break the hydrogen bonds in water; hence they are insoluble.
(vi) In air and light chloroform is oxidised to poisonous phosgene (COCl₂). Moreover, chloroform vapour depresses the central nervous system and can damage the liver and kidneys. Hence safer anaesthetics are used.

Teacher's Note:
1. For "account for" questions, give the reason and, where possible, the equation – the equation often carries one of the marks.