CBSE Class 12 Chemistry Haloalkanes and Haloarenes Important Questions Set 03

Very Short Answer Questions (1 Mark)

Question 1. In how many of the following pairs does the second compound react faster than the first in S_N1 reaction with OH⁻? (Answer is an integer from 0 to 9.)
(i) 3-chloropentane, 1-chloropentane   (ii) CH₃I, CH₃Br
(iii) (CH₃)₃CBr, (CH₃)₂CHBr   (iv) 3-chloropentane, 2-chloro-2-methylpropane
(v) CH₃Cl, (CH₃)₂CHCl   (vi) (CH₃)₂CHBr, CH₃CH₂Br
Answer:
Answer: 2
S_N1 reactivity depends on carbocation stability (3° > 2° > 1° > CH₃) and, for the same alkyl group, on the leaving group (I > Br > Cl).
(i) 2° vs 1° – first is faster. (ii) I⁻ is a better leaving group – first is faster. (iii) 3° vs 2° – first is faster.
(iv) 2° vs 3° – second is faster. (v) CH₃ vs 2° – second is faster. (vi) 2° vs 1° – first is faster.
So the second compound is faster only in pairs (iv) and (v): 2.

Teacher's Note:
1. Check both factors in each pair: the class of carbon (carbocation stability) and the leaving group.

Question 2. Give IUPAC name of:
CH₃–CH₂–CH(Br)–C(CH₃)(Br)–CH₂Cl
Answer:
1-Chloro-2,3-dibromo-2-methylpentane

Teacher's Note:
1. Number the chain from the end nearer the first substituent (CH₂Cl end) so the locant set {1, 2, 2, 3} is lowest.

Question 3. Identify A and B in the following process:
CH₃CH₂Cl –[NaCN]→ A –[Reduction, Ni/H₂]→ B
Answer:
A : CH₃–CH₂–CN (propanenitrile); B : CH₃CH₂CH₂NH₂ (propan-1-amine)

Teacher's Note:
1. The –CN route adds one carbon: a two-carbon halide gives a three-carbon amine on reduction.

Question 4. Draw the structure of 4-bromo-3-methylpent-2-ene.
Answer:
CH₃–CH=C(CH₃)–CH(Br)–CH₃

Teacher's Note:
1. Write the five-carbon chain first, put the C2=C3 double bond, then place CH₃ on C-3 and Br on C-4.

Question 5. Why should Grignard reagent be prepared under anhydrous conditions?
Answer:
Grignard reagents are highly reactive and react with water (even moisture) to form alkanes:
RMgX + H₂O → RH + Mg(OH)X
Therefore, they are prepared under anhydrous conditions.

Teacher's Note:
1. Any compound with an acidic H (water, alcohols, NH₃, terminal alkynes) destroys a Grignard reagent in the same way.

Question 6. Chloroform is stored in dark coloured and sealed bottles. Why?
Answer:
Chloroform is slowly oxidised by air in the presence of light to an extremely poisonous gas, phosgene (carbonyl chloride):
2CHCl₃ + O₂ –[light]→ 2COCl₂ + 2HCl
Dark coloured bottles cut off light and sealing (filling completely) keeps out air.

Teacher's Note:
1. Both conditions are needed – light AND oxygen; dark bottles stop one, sealing stops the other.

Question 7. An alkyl halide having molecular formula C₄H₉Cl is optically active. What is its structure?
Answer:
CH₃–CH(Cl)–CH₂–CH₃ (2-chlorobutane)
C-2 is attached to four different groups (H, Cl, CH₃, C₂H₅), so it is chiral.

Teacher's Note:
1. Of the four C₄H₉Cl isomers, only 2-chlorobutane has a chiral carbon.

Question 8. An organic compound ‘A’ on treatment with KCN gave B which on hydrolysis with dil. HCl gave acetic acid. Identify A.
Answer:
A : CH₃Cl (chloromethane)
CH₃Cl + KCN → CH₃CN (B) + KCl;   CH₃CN + 2H₂O + HCl → CH₃COOH + NH₄Cl

Teacher's Note:
1. Work backwards: the acid has one more carbon than the halide, because the CN carbon becomes the COOH carbon.

Question 9. Write IUPAC name of iodoform.
Answer:
Triiodomethane (CHI₃)

Teacher's Note:
1. Haloforms are CHX₃: chloroform, bromoform and iodoform are trichloro-, tribromo- and triiodomethane.

Question 10. Which one of the following two substances undergoes S_N1 reaction faster and why?
3-chloropentane or 1-chloropentane
Answer:
3-Chloropentane (a 2° halide) undergoes S_N1 reaction faster.
It forms a secondary carbocation, which is more stable than the primary carbocation formed from 1-chloropentane.

Teacher's Note:
1. S_N1 rate follows carbocation stability: 3° > 2° > 1°.

Question 11. Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms isocyanides as the chief product. Explain.
Answer:
KCN is predominantly ionic and provides cyanide ions in solution. Both C and N can donate electrons, but the attack takes place mainly through carbon, since the C–C bond formed is more stable than the C–N bond. Hence alkyl cyanide is the main product.
AgCN is mainly covalent in nature, so the carbon is bonded to Ag and only the nitrogen lone pair is free for attack. Hence isocyanide is the main product.

Teacher's Note:
1. The cyanide ion is ambident; which end attacks depends on whether the reagent is ionic (KCN) or covalent (AgCN).

Question 12. Write the IUPAC name of the following compound:
CH₂=C(CH₃)–CH₂Br
Answer:
3-Bromo-2-methylprop-1-ene

Teacher's Note:
1. The double bond gets the lowest number; Br is then on C-3 and CH₃ on C-2.

Question 13. Arrange the following in order of their increasing reactivity in nucleophilic substitution reactions:
CH₃F, CH₃I, CH₃Br, CH₃Cl
Answer:
CH₃F < CH₃Cl < CH₃Br < CH₃I
The C–X bond strength decreases from C–F to C–I, so the halide ion leaves more easily.

Teacher's Note:
1. For the same alkyl group, reactivity is set by the leaving group: I⁻ is the best and F⁻ the poorest.

Question 14. Allyl chloride is more reactive than n-propyl chloride towards nucleophilic substitution reaction. Explain why?
Answer:
Allyl chloride ionises to give the allyl carbocation, CH₂=CH–CH₂⁺, which is stabilised by resonance:
CH₂=CH–CH₂⁺ ↔ ⁺CH₂–CH=CH₂
n-Propyl chloride gives a primary carbocation which has no such stabilisation. Hence allyl chloride is more reactive.

Teacher's Note:
1. Allylic and benzylic halides are reactive by both S_N1 (stable cation) and S_N2 (stabilised transition state) routes.

Question 15. Complete the reaction:
Cyclohexene + Br₂ –[heat or UV light]→
Answer:
Cyclohexene + Br₂ –[heat or UV light]→ 3-bromocyclohexene + HBr
Under these conditions free radical substitution takes place at the allylic carbon.

Teacher's Note:
1. In the dark at room temperature Br₂ adds across C=C; with heat or light it substitutes the allylic H instead.

Question 16. How will you convert 2-bromopropane into 1-bromopropane?
Answer:
CH₃–CH(Br)–CH₃ –[alc. KOH, Δ]→ CH₃–CH=CH₂ –[HBr, peroxide]→ CH₃–CH₂–CH₂Br

Teacher's Note:
1. Moving a halogen to the terminal carbon: eliminate to the alkene, then add HBr anti-Markovnikov (peroxide effect).

Short Answer Questions (2 Marks)

Question 17. Complete the following reactions:
(i) C₆H₅N₂Cl + KI →
(ii) CH₂=CH₂ + Br₂ –[CCl₄]→
Answer:
(i) C₆H₅N₂Cl + KI → C₆H₅I + KCl + N₂↑ (iodobenzene)
(ii) CH₂=CH₂ + Br₂ –[CCl₄]→ BrCH₂–CH₂Br (1,2-dibromoethane)

Teacher's Note:
1. Iodobenzene is made from the diazonium salt with KI alone – no copper(I) salt is needed, unlike for chloro- and bromobenzene.

Question 18. Carry out the following conversions in not more than two steps:
(i) Toluene to benzyl alcohol
(ii) Benzyl alcohol to phenylethanenitrile
Answer:
(i) C₆H₅CH₃ –[Cl₂, hv]→ C₆H₅CH₂Cl –[aq. KOH]→ C₆H₅CH₂OH
(ii) C₆H₅CH₂OH –[SOCl₂]→ C₆H₅CH₂Cl –[KCN (alc.)]→ C₆H₅CH₂CN

Teacher's Note:
1. Use light (not FeCl₃) to chlorinate the side chain of toluene; with FeCl₃ the ring is chlorinated instead.

Question 19. Give reasons:
(i) Boiling point of alkyl bromide is higher than alkyl chloride.
(ii) Alkyl halides are better solvents than aryl halides.
Answer:
(i) Due to the higher magnitude of van der Waals forces in alkyl bromides, as bromine is larger and more polarisable than chlorine and the molecular mass is higher.
(ii) The C–X bond is more polar in haloalkanes than in haloarenes (where it has partial double bond character), so alkyl halides dissolve polar and non-polar organic substances better.

Teacher's Note:
1. For the same alkyl group, boiling point rises in the order RF < RCl < RBr < RI.

Question 20. Which of the following compounds would undergo S_N1 reaction faster and why?
(A) C₆H₁₁–CH₂Cl (chloromethylcyclohexane)   (B) C₆H₅–CH₂Cl (benzyl chloride)
Answer:
(B) benzyl chloride undergoes S_N1 reaction faster.
It forms the benzyl carbocation C₆H₅CH₂⁺, which is stabilised by resonance with the benzene ring. Compound (A) would form a primary carbocation which has no resonance stabilisation.

Teacher's Note:
1. Both are primary halides; the difference comes only from the ring being aromatic (resonance) in B and saturated in A.

Question 21. Identify and indicate the presence of centre of chirality, if any, in the following molecules. How many stereoisomers are possible for those containing chiral centre?
(i) 1,2-dichloropropane
(ii) 3-bromopent-1-ene
Answer:
(i) CH₃–C*H(Cl)–CH₂Cl: C-2 is the chiral centre (attached to H, Cl, CH₃ and CH₂Cl). Number of stereoisomers = 2 (a pair of enantiomers).
(ii) CH₂=CH–C*H(Br)–CH₂CH₃: C-3 is the chiral centre (attached to H, Br, CH=CH₂ and C₂H₅). Number of stereoisomers = 2.

Teacher's Note:
1. With n chiral centres the maximum number of stereoisomers is 2ⁿ; one centre always gives a pair of enantiomers.

Question 22. Convert:
(i) Benzene to m-nitrochlorobenzene
(ii) Benzene to diphenyl
Answer:
(i) C₆H₆ –[conc. HNO₃ + conc. H₂SO₄]→ C₆H₅NO₂ –[Cl₂, anhy. FeCl₃]→ m-ClC₆H₄NO₂
(ii) C₆H₆ –[Cl₂, anhy. FeCl₃]→ C₆H₅Cl;   2C₆H₅Cl + 2Na –[dry ether]→ C₆H₅–C₆H₅ + 2NaCl (Fittig reaction)

Teacher's Note:
1. Order of steps matters: nitrate first, because –NO₂ is meta-directing; chlorinating first would give o/p products.

Question 23. What happens when:
(i) Propene is treated with HBr in presence of peroxide.
(ii) Benzene is treated with methyl chloride in presence of AlCl₃.
Answer:
(i) Anti-Markovnikov addition (peroxide effect) gives 1-bromopropane: CH₃CH=CH₂ + HBr –[peroxide]→ CH₃CH₂CH₂Br
(ii) Friedel–Crafts alkylation gives toluene: C₆H₆ + CH₃Cl –[anhy. AlCl₃]→ C₆H₅CH₃ + HCl

Teacher's Note:
1. The peroxide effect works only with HBr, not with HCl or HI.

Question 24. Alkyl iodides develop colouration on long standing, particularly in light. Explain.
Answer:
The C–I bond is weak. On standing, particularly in light, alkyl iodides decompose and liberate iodine, which gives the colour:
2R–I –[light]→ R–R + I₂

Teacher's Note:
1. C–I is the weakest carbon–halogen bond, which is also why alkyl iodides are the most reactive alkyl halides.

Question 25. Tert-butyl bromide reacts with aq. NaOH by S_N1 mechanism while n-butyl bromide reacts by S_N2 mechanism. Why?
Answer:
Tert-butyl bromide forms a stable tertiary carbocation, (CH₃)₃C⁺, and the three bulky methyl groups hinder backside attack of the nucleophile. Hence it follows S_N1.
n-Butyl bromide would form an unstable primary carbocation, but there is very little steric hindrance at the carbon bearing Br, so the nucleophile can attack from the back side. Hence it follows S_N2.

Teacher's Note:
1. Two factors decide the mechanism: carbocation stability favours S_N1, low steric hindrance favours S_N2.

Question 26. Identify the products:
(i) m-Bromonitrobenzene (1-bromo-3-nitrobenzene) + Mg –[dry ether]→
(ii) CH₃–CH(Br)–CH₃ –[alc. KOH]→ A –[HBr, H₂O₂]→ B
Answer:
(i) m-O₂N–C₆H₄–MgBr (3-nitrophenylmagnesium bromide, a Grignard reagent)
(ii) A = CH₃–CH=CH₂ (propene); B = CH₃CH₂CH₂Br (1-bromopropane, anti-Markovnikov addition in presence of peroxide)

Teacher's Note:
1. H₂O₂ over the arrow with HBr signals the peroxide (Kharasch) effect – Br goes to the terminal carbon.

Question 27. (i) Arrange the following halides in order of increasing S_N1 reactivity:
CH₃Cl, CH₃Br, CH₃CH₂Cl, (CH₃)₂CHCl
(ii) Which out of 1-bromobutane and 2-bromobutane would react faster by S_N2 pathway and why?
Answer:
(i) CH₃Cl < CH₃Br < CH₃CH₂Cl < (CH₃)₂CHCl (S_N1 reactivity increases with stability of the carbocation; for the same alkyl group, Br⁻ is a better leaving group than Cl⁻).
(ii) 1-Bromobutane reacts faster by S_N2, because it is a primary halide and offers less steric hindrance to the backside attack of the nucleophile than the secondary 2-bromobutane.

Teacher's Note:
1. When the alkyl groups differ, carbocation stability outweighs the leaving group; compare leaving groups only for the same alkyl group.

Question 28. Identify the products:
C₆H₆ –[CH₃Cl, anhy. AlCl₃]→ A –[Cl₂ (1 mole), hv]→ B –[aq. KOH]→ C –[HBr]→ D
Answer:
A = C₆H₅CH₃ (toluene)
B = C₆H₅CH₂Cl (benzyl chloride)
C = C₆H₅CH₂OH (benzyl alcohol)
D = C₆H₅CH₂Br (benzyl bromide)

Teacher's Note:
1. "1 mole" of Cl₂ tells you only one side-chain H is replaced; excess Cl₂ would give C₆H₅CHCl₂ and C₆H₅CCl₃.

Question 29. Carry out the following conversions:
(i) But-1-ene to n-butyl iodide
(ii) Isopropyl alcohol to iodoform
Answer:
(i) CH₃CH₂CH=CH₂ –[HBr, peroxide]→ CH₃CH₂CH₂CH₂Br –[NaI, dry acetone]→ CH₃CH₂CH₂CH₂I
(ii) (CH₃)₂CHOH + 4I₂ + 6NaOH → CHI₃↓ + CH₃COONa + 5NaI + 5H₂O (iodoform reaction; yellow precipitate)

Teacher's Note:
1. Isopropyl alcohol gives iodoform because it is first oxidised by I₂/NaOH to acetone, which has the CH₃CO– group.

Question 30. An organic compound A reacts with PCl₅ to give compound B. Compound B reacts with Na/ether to give n-butane. What are compounds A and B?
Answer:
A = C₂H₅OH (ethanol), B = C₂H₅Cl (chloroethane)
C₂H₅OH + PCl₅ → C₂H₅Cl + POCl₃ + HCl
2C₂H₅Cl + 2Na –[dry ether]→ CH₃CH₂CH₂CH₃ + 2NaCl (Wurtz reaction)

Teacher's Note:
1. Wurtz reaction doubles the alkyl group, so a symmetrical alkane with 4 C comes from a 2-carbon halide.

Short Answer Questions (3 Marks)

Question 31. How can the following conversion be carried out:
(i) Aniline to bromobenzene
(ii) Chlorobenzene to 2-chloroacetophenone
(iii) Chloroethane to butane [CBSE OD 2015]
Answer:
(i) C₆H₅NH₂ (Aniline) ⟶^{NaNO₂ + HCl, 273–278 K} C₆H₅N₂⁺Cl⁻ (Benzene diazonium chloride)
C₆H₅N₂⁺Cl⁻ ⟶^CuBr/HBr C₆H₅Br (Bromobenzene) + N₂
(ii) C₆H₅Cl (Chlorobenzene) + CH₃COCl ⟶^{Anhy. AlCl₃} o-ClC₆H₄COCH₃ (2-Chloroacetophenone) + HCl
(iii) 2CH₃CH₂Cl (Chloroethane) + 2Na ⟶^{dry ether} CH₃CH₂CH₂CH₃ (Butane) + 2NaCl

Teacher's Note:
1. In Friedel-Crafts acylation of chlorobenzene the para isomer (4-chloroacetophenone) is actually formed in larger amount; the ortho isomer is separated from the mixture.

Question 32. What happens when
(i) Chlorobenzene is treated with Cl₂/FeCl₃,
(ii) Ethyl chloride is treated with AgNO₂,
(iii) 2-bromopentane is treated with alcoholic KOH?
Write the chemical equations in support of your answer. [CBSE OD 2015]
Answer:
(i) C₆H₅Cl + Cl₂ ⟶^{Anhy. FeCl₃} p-C₆H₄Cl₂ (1,4-dichlorobenzene) + o-C₆H₄Cl₂ (1,2-dichlorobenzene) + HCl
(ii) CH₃CH₂Cl + AgNO₂ → CH₃CH₂NO₂ + AgCl
(iii) CH₃CH₂CH₂CH(Br)CH₃ + KOH (alc.) → CH₃CH₂CH = CHCH₃ + KBr + H₂O

Teacher's Note:
1. In (iii), pent-2-ene is the major product (Saytzeff rule); pent-1-ene forms only in small amount.

Question 33. Draw the structures of the major monohalo product for each of the following reactions:
(i) p-Cl–C₆H₄–CH₂–CH₃ ⟶^{Br₂, heat}
(ii) 1-Methylcyclohexene + HBr → ?
(iii) p-HO–C₆H₄–CH₂OH ⟶^{HCl, heat} ? [CBSE Foreign Set-1, 2, 3 2017]
Answer:
(i) p-Cl–C₆H₄–CH(Br)–CH₃
(ii) 1-Bromo-1-methylcyclohexane (Br and CH₃ on the same ring carbon)
(iii) p-HO–C₆H₄–CH₂Cl

Teacher's Note:
1. In (iii) only the alcoholic (benzylic) OH is replaced; the phenolic OH does not react with HCl because its C–O bond has partial double bond character.

Question 34. (i) Account for the following:
(a) Electrophilic substitution reactions in haloarenes occur slowly.
(b) Haloalkanes, though polar, are insoluble in water.
(ii) Arrange the following compounds in increasing order of reactivity towards S_N2 displacement:
2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane [CBSE Comptt. OD Set-1, 2, 3 2017]
Answer:
(i) (a) Due to −I effect of X, the ring is partially deactivated.
(b) They fail to form hydrogen bonds with water / more energy is required to break hydrogen bonds in water than is released when new attractions are set up.
(ii) 2-Bromo-2-methylbutane < 2-Bromopentane < 1-Bromopentane

Teacher's Note:
1. For S_N2 the order is 1° > 2° > 3°; write the class under each name before arranging them.

Question 35. (i) Write the structural formula of A, B, C and D in the following sequence of reaction:
CH₃–CH(Cl)–CH₃ ⟶^{alc. KOH} A ⟶^{HBr, Peroxide} B ⟶^{NaI, dry ether} C ⟶^{Mg, dry ether} D
(ii) Illustrate Sandmeyer's reaction with the help of a suitable example. [CBSE Comptt. Delhi Set-1, 2, 3 2017]
Answer:
(i) A: CH₃–CH = CH₂
B: CH₃–CH₂–CH₂Br
C: CH₃–CH₂–CH₂I
D: CH₃–CH₂–CH₂MgI
(ii) C₆H₅N₂⁺X⁻ ⟶^{CuX/HX or CuCN/KCN} C₆H₅–X (Aryl halide) + N₂,   X = Cl, Br, CN
e.g., C₆H₅N₂⁺Cl⁻ ⟶^{Cu₂Cl₂/HCl} C₆H₅Cl + N₂

Teacher's Note:
1. Track the carbon skeleton step by step: elimination, anti-Markovnikov addition, Finkelstein exchange, then Grignard formation – the halogen ends up on the terminal carbon.

Question 36. Write the major monohalo product(s) in each of the following reactions:
(i) Cyclohexene + Br₂ ⟶^{UV light} ?
(ii) C₆H₅–CH₂–CH = CH₂ + HBr ⟶^Peroxide ?
(iii) p-HO–C₆H₄–CH₂OH + HCl ⟶^heat ? [CBSE OD Set-2 2016]
Answer:
(i) Cyclohexene + Br₂ ⟶^{UV light} 3-Bromocyclohexene + HBr (Br replaces an allylic H, next to the C=C)
(ii) C₆H₅–CH₂–CH = CH₂ + HBr ⟶^Peroxide C₆H₅–CH₂–CH₂–CH₂–Br
(iii) p-HO–C₆H₄–CH₂OH + HCl ⟶^Δ p-HO–C₆H₄–CH₂–Cl + H₂O

Teacher's Note:
1. Br₂ in UV light gives allylic substitution (via the stable allylic radical), not addition across the C=C.

Question 37. (a) Identify the chiral molecule in the following pair:
(i) (CH₃)₂CH–CH(OH)–CH₃    (ii) (CH₃)₂CH–CH(OH)–CH(CH₃)₂
(b) Write the structure of the product when chlorobenzene is treated with methyl chloride in the presence of sodium metal and dry ether.
(c) Write the structure of the alkene formed by dehydrohalogenation of 1-bromo-1-methylcyclohexane with alcoholic KOH. [CBSE Delhi/OD 2018]
Answer:
(a) (i) (CH₃)₂CH–C*H(OH)–CH₃ is chiral.
(b) Wurtz-Fittig reaction: C₆H₅Cl + CH₃Cl + 2Na ⟶^{Dry ether} C₆H₅CH₃ (Toluene) + 2NaCl
(c) 1-Methylcyclohexene (major, double bond in the ring) + Methylenecyclohexane (minor, exocyclic C=CH₂)

Teacher's Note:
1. In (a)(ii) the carbinol carbon carries two identical isopropyl groups, so it cannot be a chiral centre.

Question 38. Which one of the following compounds will undergo faster hydrolysis reaction by S_N1 mechanism? Justify your answer.
C₆H₅CH₂Cl or CH₃CH₂CH₂Cl [CBSE SQP 2018-2019]
Answer:
C₆H₅CH₂Cl will undergo S_N1 reaction faster. The carbocation formed by C₆H₅CH₂Cl gets stabilized through resonance.
Greater the stability of carbocation, greater will be its ease of formation from the respective halide.
The positive charge of the benzyl carbocation C₆H₅CH₂⁺ is delocalised over the CH₂ carbon and the two ortho and the para carbons of the ring (four resonance structures), whereas CH₃CH₂CH₂⁺ is an unstable primary carbocation.

Teacher's Note:
1. Benzyl and allyl halides are exceptions to the 'primary means S_N2' rule: their carbocations are resonance-stabilised, so they also react fast by S_N1.

Question 39. A compound is formed by the substitution of two chlorine atoms for two hydrogen atoms in propane. Write the structures of the isomers possible. Give the IUPAC name of the isomer which can exhibit enantiomerism. [CBSE SQP 2018-2019]
Answer:
CH₃–CH(Cl)–CH₂Cl
ClCH₂–CH₂–CH₂Cl
CH₃–CCl₂–CH₃
CH₃–CH₂–CHCl₂
The following isomer will exhibit enantiomerism: CH₃–C*H(Cl)–CH₂Cl
IUPAC name: 1, 2-Dichloropropane.

Teacher's Note:
1. Write the isomers systematically: 1,1-, 1,2-, 1,3- and 2,2-dichloropropane – only 1,2- has a carbon with four different groups.

Question 40. Write the product(s) formed when
(i) 2-Bromopropane undergoes dehydrohalogenation reaction.
(ii) Chlorobenzene undergoes nitration reaction.
(iii) Methylbromide is treated with KCN. [CBSE Comptt. Delhi/OD 2018]
Answer:
(i) Propene: CH₃–CHBr–CH₃ + KOH (alc.) → CH₃–CH = CH₂ + KBr + H₂O
(ii) 4-Nitrochlorobenzene and 2-nitrochlorobenzene: C₆H₅Cl + HNO₃ ⟶^{conc. H₂SO₄} p-ClC₆H₄NO₂ (major) + o-ClC₆H₄NO₂ + H₂O
(iii) Methylcyanide / Ethanenitrile: CH₃Br + KCN → CH₃CN + KBr

Teacher's Note:
1. Cl is o,p-directing, so nitration never gives the meta product as the major one; the para isomer dominates for steric reasons.

Question 41. Rearrange the compounds of each of the following sets in order of reactivity towards S_N2 displacement:
(i) 2-bromo-2-methylbutane, 1-bromopentane, 2-bromopentane
(ii) 1-bromo-3-methylbutane, 2-bromo-2-methylbutane, 2-bromo-3-methylbutane
(iii) 1-bromobutane, 1-bromo-2,2-dimethylpropane, 1-bromo-2-methylbutane
Answer:
(i) 1-Bromopentane > 2-bromopentane > 2-bromo-2-methylbutane (1° > 2° > 3°)
(ii) 1-Bromo-3-methylbutane > 2-bromo-3-methylbutane > 2-bromo-2-methylbutane (1° > 2° > 3°)
(iii) 1-Bromobutane > 1-bromo-2-methylbutane > 1-bromo-2,2-dimethylpropane (all 1°; steric hindrance increases with branching at the β-carbon)
S_N2 reactivity decreases as steric hindrance to the backside attack of the nucleophile increases.

Teacher's Note:
1. In set (iii) all are primary – look one carbon further: methyl groups on the β-carbon still crowd the backside approach.

Long Answer Questions (5 Marks)

Question 42. How would you bring about the following conversions:
(i) Propene to 2-bromopropane
(ii) Bromoethane to propanoic acid
(iii) 1-Chloropropane to 1-propanol
(iv) Ethanol to chloroethane
(v) 1-Iodopropane to propene
Answer:
(i) CH₃CH=CH₂ + HBr → CH₃CH(Br)CH₃ (Markovnikov addition)
(ii) CH₃CH₂Br –[KCN (alc.)]→ CH₃CH₂CN –[H⁺/H₂O, hydrolysis]→ CH₃CH₂COOH
(iii) CH₃CH₂CH₂Cl + KOH(aq) → CH₃CH₂CH₂OH + KCl
(iv) CH₃CH₂OH + SOCl₂ –[pyridine]→ CH₃CH₂Cl + SO₂ + HCl
(v) CH₃CH₂CH₂I + KOH(alc) –[Δ]→ CH₃CH=CH₂ + KI + H₂O

Teacher's Note:
1. Count carbons first: if the product has one more carbon (as in ii), the CN⁻ route is almost always the answer.

Question 43. What happens when: (Give chemical reactions)
(i) Cyclohexanol is treated with thionyl chloride
(ii) p-Hydroxybenzyl alcohol is heated with HCl
(iii) Ethyl bromide is refluxed with NaI in acetone
(iv) Ethyl bromide is treated with mercurous fluoride
(v) Chlorobenzene is subjected to hydrolysis
Answer:
(i) Chlorocyclohexane is formed: C₆H₁₁OH + SOCl₂ → C₆H₁₁Cl + SO₂ + HCl
(ii) Only the benzylic –OH is replaced: p-HO–C₆H₄–CH₂OH + HCl → p-HO–C₆H₄–CH₂Cl + H₂O
(iii) Ethyl iodide is formed (Finkelstein reaction): C₂H₅Br + NaI –[acetone]→ C₂H₅I + NaBr↓
(iv) Ethyl fluoride is formed (Swarts reaction): 2C₂H₅Br + Hg₂F₂ → 2C₂H₅F + Hg₂Br₂
(v) Phenol is formed when chlorobenzene is heated with aqueous NaOH at 623 K and 300 atm, followed by acidification:
C₆H₅Cl + NaOH –[623 K, 300 atm]→ C₆H₅ONa –[H⁺]→ C₆H₅OH

Teacher's Note:
1. Finkelstein makes iodides (NaI), Swarts makes fluorides (AgF, Hg₂F₂, CoF₂, SbF₃) – do not mix up the two.