CBSE Class 12 Chemistry Haloalkanes and Haloarenes Important Questions Set 02

Very Short Answer Questions (1 Mark)

Question 1. Write the IUPAC name of the following:
CH₂ = C(CH₃)–CH₂Br [CBSE Delhi 2012]
Answer:
3-bromo-2-methyl prop-1-ene

Teacher's Note:
1. The double bond must get the lowest locant, so numbering starts from the =CH₂ end, which puts Br on C₃.

Question 2. Out of CH₃–CH(CH₃)–CH₂–Cl and CH₃–CH₂–CH(CH₃)–Cl, which is more reactive towards S_N1 reaction and why? [CBSE Delhi 2016]
Answer:
CH₃–CH₂–CH(CH₃)–Cl is more reactive as being secondary halide it forms more stable carbocation intermediate than primary halides.

Teacher's Note:
1. In an S_N1 question, the word 'why' always needs the carbocation stability argument, not just the class of halide.

Question 3. Out of (A) and (B), which is an example of allylic halide? [CBSE OD Set-1 2017]

Answer:
(A), in which X is attached to the sp³ carbon next to the C=C double bond, is the allylic halide (3-halocyclohexene).

Teacher's Note:
1. Allylic: X on the carbon next to C=C. Vinylic: X directly on a C=C carbon.

Question 4. Out of (A) and (B), which is an example of vinylic halide? [CBSE OD Set-2 2017]

Answer:
(B), in which X is attached directly to a doubly bonded (sp²) carbon, is the vinylic halide (1-halocyclohexene).

Teacher's Note:
1. Vinylic halides behave like aryl halides: the C–X bond has partial double bond character and is hard to break.

Question 5. Which would undergo S_N2 reaction faster in the following pair and why?
CH₃–CH₂–Br and CH₃–C(CH₃)(Br)–CH₃ [CBSE Delhi Set-1, 2, 3 2015]
Answer:
CH₃–CH₂–Br would undergo S_N2 reaction faster as it has less steric hindrance than tert-butyl bromide, because it is a primary halide / (1°) halide.

Teacher's Note:
1. S_N2 order is the reverse of S_N1 order: 1° > 2° > 3°, because the nucleophile attacks from the back side.

Question 6. Amongst the isomeric alkanes of molecular formula C₅H₁₂, identify the one that on photochemical chlorination yields a single monochloride. [CBSE SQP 2016]
Answer:
Neopentane or 2, 2-Dimethylpropane, C(CH₃)₄.

Teacher's Note:
1. A single monochloride means all hydrogens are equivalent – in neopentane all 12 H are on identical CH₃ groups.

Question 7. In the following pair of halogen compounds, which will react faster by S_N1 mechanism?
(CH₃)₃C–Cl and CH₃CH₂CH(Cl)CH₂CH₃ [CBSE OD 2014]
Answer:
The 3° alkyl halides are most reactive, because the intermediate carbocation formed in their case is most stable in S_N1 mechanism. The most stable intermediate is formed at faster rate.
(CH₃)₃C–Cl (3° alkyl halide) reacts faster than CH₃CH₂CH(Cl)CH₂CH₃ (2° alkyl halide).

Teacher's Note:
1. Tertiary carbocations are stabilised by the +I effect and hyperconjugation of three alkyl groups.

Question 8. How can methyl bromide be preferentially converted to methyl isocyanide? [CBSE Comptt. Delhi 2014]
Answer:
CH₃Br + alc. AgCN → CH₃NC + AgBr
(Methyl isocyanide)

Teacher's Note:
1. AgCN is mainly covalent, so it attacks through nitrogen and gives isocyanide; ionic KCN attacks through carbon and gives cyanide.

Question 9. Identify the compound that on hydrogenation produces an optically active compound from the following compounds:
(A) CH₂=CH–CH(Br)–CH₂CH₃
(B) CH₃–CH=CH–CH(Br)–CH₃ [CBSE SQP 2018-19]
Answer:
B
On hydrogenation, (B) gives CH₃CH₂CH₂CH(Br)CH₃ (2-bromopentane), which has a chiral carbon; (A) gives CH₃CH₂CH(Br)CH₂CH₃ (3-bromopentane), which is not chiral.

Teacher's Note:
1. Both starting compounds are chiral – what matters is whether the chiral centre survives after the C=C becomes an ethyl group.

Question 10. Predict the major product formed when sodium ethoxide reacts with tert.Butyl chloride. [CBSE Comptt. Delhi/O.D. 2018]
Answer:
2-Methylprop-1-ene / isobutene
(CH₃)₃C–Cl + CH₃CH₂ONa → (CH₃)₂C=CH₂ + CH₃CH₂OH + NaCl

Teacher's Note:
1. A strong base with a tertiary halide gives elimination, not Williamson ether synthesis – a frequent trap in exams.

Question 11. Which of the following two reactions is S_N2 and why? [CBSE OD Set-2 2016]

Answer:
The reaction (i) is S_N2 as it shows inversion of configuration.

Teacher's Note:
1. In (i) the three groups C₂H₅, H and CH₃ flip to the other side like an umbrella in the wind (Walden inversion); in (ii) they keep their positions, i.e., retention.

Teacher's Note:
1. Wurtz: R + R; Fittig: Ar + Ar; Wurtz-Fittig: Ar + R – the name tells you which halides are joined.

Question 13. Write the IUPAC name of the given compound: C₆H₅–CH₂–CH₂–OH [CBSE OD 2016]
Answer:
2-Phenylethanol.

Teacher's Note:
1. The OH-bearing carbon is C₁ of the ethanol chain, so the phenyl group sits on C₂.

Question 14. Write the IUPAC name of the given compound: [CBSE Delhi 2016]

Answer:
2,4,6-Tribromoaniline / 2,4,6-Tribromobenzenamine.

Teacher's Note:
1. The carbon bearing NH₂ is C₁, so the bromines at the two ortho and the para positions get locants 2, 4 and 6.

Question 15. Draw the structure of major monohalo product in the following reaction:
C₆H₆ + Cl₂ ⟶^{Fe, dark} ? [CBSE Comptt. Delhi 2012]
Answer:
C₆H₆ + Cl₂ ⟶^{Fe, dark} C₆H₅Cl + HCl
(Chlorobenzene)

Teacher's Note:
1. Fe reacts with Cl₂ to form FeCl₃, the Lewis acid that actually produces the Cl⁺ electrophile.

Question 16. Out of chlorobenzene and benzyl chloride, which one gets easily hydrolysed by aqueous NaOH and why? [CBSE Delhi/OD 2018]
Answer:
Benzyl chloride; due to resonance, stable benzyl carbocation is formed.
Benzyl chloride gets easily hydrolysed by aqueous NaOH as chlorobenzene possesses partial double bond character in the Cl–C bond. The lone pairs delocalised in the ring strengthens Cl–C bond reducing its reactivity. Whereas benzyl chloride undergoes S_N1 reaction to form stable benzyl carbocation.

Teacher's Note:
1. Both compounds contain a benzene ring, but only in benzyl chloride is Cl on an sp³ carbon – that is the deciding difference.

Short Answer Questions (2 Marks)

Question 17. Write the equations for the preparation of 1-bromobutane from:
(i) 1-butanol
(ii) but-1-ene [CBSE Comptt. Delhi 2016]
Answer:
(i) CH₃CH₂CH₂CH₂–OH ⟶^HBr CH₃CH₂CH₂CH₂Br
(ii) CH₃CH₂CH = CH₂ ⟶^HBr/Peroxide CH₃CH₂CH₂CH₂Br

Teacher's Note:
1. Without peroxide, HBr would add by Markovnikov's rule and give 2-bromobutane – always write the peroxide condition.

Question 18. Which one of the following compounds is more reactive towards S_N2 reaction and why?
CH₃CH(Cl)CH₂CH₃ or CH₃CH₂CH₂Cl [CBSE Comptt. Delhi/OD 2018]
Answer:
CH₃CH₂CH₂Cl, due to primary halide which has less steric hindrance.
In S_N2 reaction, attack of nucleophile takes place from backward direction. Primary alkyl halide is the least sterically hindered among primary, secondary and tertiary alkyl halides. While in secondary halides presence of bulky –CH₃ group around causes steric hindrance for nucleophilic attack. Therefore, primary alkyl halides are more reactive towards S_N2 reaction.

Teacher's Note:
1. For a 2-mark 'why', mention both the backside attack and the crowding of the transition state.

Question 19. Which alkyl halide from the following pair is (i) Chiral and (ii) undergoes S_N1 reaction faster?
(a) (CH₃)₃CBr    (b) CH₃CH₂CHBrCH₃ [CBSE Comptt. Set-2 2017]
Answer:
(i) (b) is chiral.
(ii) (a)

Teacher's Note:
1. (CH₃)₃CBr has three identical CH₃ groups on the C–Br carbon, so it cannot be chiral, but it gives the most stable 3° carbocation.

Question 20. (i) Allyl chloride can be distinguished from Vinyl chloride by NaOH and silver nitrate test. Comment.
(ii) Alkyl halide reacts with Lithium aluminium hydride to give alkane. Name the attacking reagent which will bring out this change. [CBSE SQP 2017]
Answer:
(i) Vinyl chloride does not respond to NaOH and silver nitrate test because of partial double bond character due to resonance. Allyl chloride is easily hydrolysed by NaOH to give Cl⁻ ions, which form a white precipitate of AgCl with silver nitrate.
(ii) Hydride ion / H⁻

Teacher's Note:
1. Allyl chloride is reactive because the allyl carbocation is resonance-stabilised; vinyl chloride is unreactive because its C–Cl bond is strengthened by resonance.

Question 21. Account for the following:
(i) The C–Cl bond length in chlorobenzene is shorter than that in CH₃–Cl.
(ii) Chloroform is stored in closed dark brown bottles. [CBSE Delhi 2013]
Answer:
(i) In chlorobenzene each carbon atom is sp² hybridised and due to resonance there is a partial double bond character, so bond length is short.
(ii) In the presence of air and sunlight, chloroform get oxidised to phosgene (Carbonyl chloride – COCl₂) a poisonous compound. Thus, to avoid the formation of phosgene, chloroform is stored in dark brown bottles filled up to brim.
CHCl₃ + ½O₂ → COCl₂ + HCl

Teacher's Note:
1. Two effects shorten the C–Cl bond in chlorobenzene: partial double bond character from resonance, and the smaller size of the sp² carbon.

Question 22. Chlorobenzene is extremely less reactive towards a nucleophilic substitution reaction. Give two reasons for the same. [CBSE Delhi 2013]
Answer:
The two reasons are as follows:
(i) Resonance effect: The electron pair of chlorine atom is involved in conjugation with the π-electrons of the benzene ring. As a result, electrons of C–Cl bond get delocalized and a partial double bond character develops in the bond and hence, it becomes difficult for the nucleophile to cleave the C–Cl bond.
(ii) Increased electron density: A repulsion is suffered by the nucleophile due to increased electron density on the benzene ring which prohibits the nucleophile to make a close access for the attack on the molecule.

Teacher's Note:
1. Other accepted reasons are the sp² hybridised carbon (shorter, stronger C–Cl bond) and the instability of the phenyl cation; any two well-explained reasons earn full marks.

Question 23. Write a test to detect the presence of double bond in a molecule. [NCERT Exemplar]
Answer:
Presence of a double bond in a molecule can be detected by bromine water test and Baeyer's test. The molecules with a double bond decolourise bromine water and aqueous KMnO₄.
>C = C< + Br₂ (brown) → >C(Br)–C(Br)< (colourless)
>C = C< ⟶^{aq. KMnO₄ (purple)} >C(OH)–C(OH)< (colourless)

Teacher's Note:
1. Baeyer's reagent is cold, dilute, alkaline KMnO₄; the purple colour disappears and a brown precipitate of MnO₂ forms.

Question 24. (i) Draw the structures of major monohalo products in each of the following reactions:
(a) C₆H₁₁–CH₂OH ⟶^PCl₅
(b) C₆H₅–CH₂–CH = CH₂ + HBr →
(ii) Which halogen compound in each of the following pairs will react faster in S_N2 reaction:
(a) CH₃Br or CH₃I
(b) (CH₃)₃C–Cl or CH₃–Cl [CBSE Delhi 2014]
Answer:
(i) (a) C₆H₁₁–CH₂OH + PCl₅ → C₆H₁₁–CH₂–Cl + POCl₃ + HCl
(b) C₆H₅–CH₂–CH = CH₂ + HBr → C₆H₅–CH₂–CH(Br)–CH₃ (major)
(ii) (a) CH₃I will react faster in S_N2 reaction because CH₃I has a better leaving group than CH₃Br.
(b) CH₃Cl will react faster in S_N2 reaction because 1° alkyl halide undergoes faster reaction than 3° alkyl halide.

Teacher's Note:
1. Without peroxide, H adds to the CH₂ end and Br goes to the carbon with fewer hydrogens (Markovnikov's rule). Compare with SA Q₁(ii), where peroxide reverses this.

Question 25. How will you carry out the following conversion:
(i) 2-Bromopropane to 1-bromopropane
(ii) Benzene to p-chloronitrobenzene [CBSE Comptt. OD Set-1, 2, 3 2017]
Answer:
(i) CH₃–CH(Br)–CH₃ ⟶^{alc. KOH} CH₃–CH = CH₂ ⟶^{HBr, Peroxide} CH₃–CH₂–CH₂–Br
(ii) C₆H₆ + Cl₂ ⟶^{Fe, dark} C₆H₅Cl ⟶^{HNO₃, conc. H₂SO₄} p-ClC₆H₄NO₂

Teacher's Note:
1. Order matters in (ii): chlorinate first, because Cl is o,p-directing; nitrating first would put Cl meta to NO₂.

Question 26. Although chlorine is an electron withdrawing group, yet it is ortho-, para-directing in electrophilic aromatic substitution reactions. Explain why it is so? [CBSE Delhi 2012]
Answer:
Although chlorine is an electron withdrawing group, yet it is ortho-para directing in electrophilic aromatic substitution because when chlorine is present in benzene ring, it releases electron by resonance whereas it acts as withdrawing group only through inductive effect.
By inductive effect, chlorine atom destabilizes the intermediate carbocation formation but by resonance, chlorine atom stabilises the intermediate carbocation and effect is more at o- and p-positions. Resonance effect opposes inductive effect. Inductive is stronger than resonance effect due to which reactivity is controlled by inductive effect and orientation by resonance.
When E⁺ attacks at the ortho- or para-position, one resonance structure of the intermediate carbocation has the positive charge on the carbon carrying Cl, where the lone pair of Cl can stabilise it (C=Cl⁺). This extra stabilisation is not possible for attack at the meta-position.

Teacher's Note:
1. Remember the one-line summary: −I effect controls reactivity (deactivation), +R effect controls orientation (o/p).

Question 27. Haloalkanes undergo nucleophilic substitution whereas haloarenes undergo electrophilic substitution. Explain. [CBSE Comptt. Delhi 2012]
Answer:
In haloalkanes, the halogen atom is attached to carbon atom. As the halogen atom is more electronegative than carbon, the bond between carbon and halogen is polar in character: C^δ+–X^δ−
Due to presence of partial positive charge on carbon atom, the nucleophiles can attack on electron deficient carbon thereby resulting in the displacement of weaker nucleophile, thus typical reactions of alkyl halides are nucleophilic substitution reactions:
R:X + :Z⁻ → R:Z + X⁻   (Z⁻ = nucleophile, X⁻ = halide ion)
Whereas in haloarenes, as the halogen atom attached to benzene ring releases electron towards benzene ring (+R effect) thereby the ortho and para positions become site of attack for electrophiles more than nucleophiles.

Teacher's Note:
1. Think of where the electron density is: the δ+ carbon of R–X attracts nucleophiles, while the electron-rich π ring of Ar–X attracts electrophiles.

Question 28. Give chemical tests to distinguish between the following pairs of compounds:
(i) Benzyl chloride and chlorobenzene
(ii) Chloroform and carbon tetrachloride [CBSE Comptt. OD 2013]
Answer:
(i) On adding sodium hydroxide and silver nitrate to both the compounds, benzyl chloride forms white precipitate but chlorobenzene does not form white precipitate.
C₆H₅CH₂Cl + NaOH → C₆H₅CH₂OH + NaCl
NaCl + AgNO₃ → AgCl(s) + NaNO₃   (white ppt.)
(ii) On heating chloroform and carbon tetrachloride with aniline and ethanolic potassium hydroxide separately chloroform forms pungent smelling isocyanide but carbon tetrachloride does not form this compound.
C₆H₅NH₂ + CHCl₃ + 3KOH ⟶^Δ C₆H₅NC (Phenyl isocyanide) + 3KCl + 3H₂O

Teacher's Note:
1. A distinguishing test must state what each compound does – the observation for one and 'no change' for the other.

Question 29. Discuss the role of Lewis acids in the preparation of aryl bromides and chlorides in the dark. [NCERT Exemplar]
Answer:
Direct chlorination or bromination of benzene or other aromatic hydrocarbon gives chloroarenes and bromoarenes. These reactions are carried out in the presence of Lewis acids such as ferric or aluminium halides (FeCl₃, AlCl₃) in the dark, at ordinary temperatures (310-320 K). The reaction is electrophilic aromatic substitution. The halogen itself is not a strong electrophile for the attack of benzene ring. The Lewis acid promotes the electrophilic aromatic substitution by polarizing the halogen molecule. Actually, it generates a better electrophile for the attack of benzene ring.
C₆H₅CH₃ + X₂ ⟶^{Fe/FeCl₃, dark} o-XC₆H₄CH₃ + p-XC₆H₄CH₃
Mechanism: Cl–Cl + FeCl₃ → FeCl₄⁻ + Cl⁺
C₆H₆ + Cl⁺ → [C₆H₆Cl]⁺ (arenium ion) ⟶^FeCl₄− C₆H₅Cl + FeCl₃ + HCl

Teacher's Note:
1. 'In the dark' is essential: in light, free radical addition or side-chain halogenation would compete with ring substitution.

Question 30. Diphenyls are potential threat to the environment. How are these produced from arylhalides? [NCERT Exemplar]
Answer:
Diphenyls such as p,p'-dichlorodiphenyl trichloroethane (DDT) are posing serious environment problems because of its chemical stability and its fat solubility. Its residue accumulates in environment and its long-term effects could be highly dangerous.
Diphenyls can be prepared from aryl halides by the following two methods:
(i) Fittig reaction: 2C₆H₅–Cl + 2Na ⟶^Ether C₆H₅–C₆H₅ (Diphenyl) + 2NaCl
(ii) Ullmann biaryl synthesis: 2C₆H₅–I + 2Cu ⟶^Heat C₆H₅–C₆H₅ (Diphenyl) + 2CuI

Teacher's Note:
1. Fittig uses sodium in dry ether; Ullmann uses copper powder on heating – both join two aryl groups through a C–C bond.

Short Answer Questions (3 Marks)

Question 31. Give reasons for the following observations:
(i) p-dichlorobenzene has higher melting point than those of o- and m- isomers.
(ii) Haloarenes are less reactive than haloalkanes towards nucleophilic substitution reaction.
(iii) The treatment of alkyl chloride with aqueous KOH leads to the formation of alcohol but in the presence of alcoholic KOH, alkene is the major product. [CBSE 2019 (56/4/1)]
Answer:
(i) It is due to the greater symmetry of para-isomer that fits in the crystal better as compared to ortho and meta-isomers.
(ii) As C–X bond in aryl halide acquires a partial double bond character due to resonance while the C–X bond in alkyl halide is a pure single bond.
(iii) Alkoxide ion present in alcoholic KOH, is not only a strong nucleophile but also a strong base so preferentially eliminate a molecule of HCl from alkyl halide to form alkenes.

Teacher's Note:
1. In aqueous KOH the OH⁻ ion is heavily hydrated and acts mainly as a nucleophile; in alcohol the alkoxide ion acts mainly as a base.

Question 32. (i) Why are alkyl halides insoluble in water?
(ii) Why is butan-1-ol optically inactive but butan-2-ol is optically active?
(iii) Although chlorine is an electron withdrawing group, yet it is ortho, para directing in electrophilic aromatic substitution reactions. Why? [CBSE (F) 2015]
Answer:
(i) This is due to the inability of alkyl halide molecule to form intermolecular hydrogen bonds with water molecules.
(ii) CH₃–CH₂–C*H(OH)–CH₃, due to presence of a chiral carbon butan-2-ol is an optically active compound. Butan-1-ol, CH₃CH₂CH₂CH₂OH, has no chiral carbon, so it is optically inactive.
(iii) As the weaker resonance (+R) effect of Cl which stabilise the carbocation formed tends to oppose the stronger inductive (−I) effect of Cl which destabilise the carbocation at ortho and para positions and makes deactivation less for ortho and para positions.

Teacher's Note:
1. For (iii), the +R effect pushes electron density to the o- and p-positions only, so those positions are deactivated least.

Question 33. Give reasons:
(i) n-Butyl bromide has higher boiling point than t-butyl bromide.
(ii) Racemic mixture is optically inactive.
(iii) The presence of nitro group (–NO₂) at o/p positions increases the reactivity of haloarenes towards nucleophilic substitution reactions. [CBSE Delhi 2015; 2019 (56/4/1)]
Answer:
(i) n-Butyl bromide being a straight chain alkyl halide has larger surface area than tert-butyl bromide. Larger the surface area, larger the magnitude of the van der Waal’s forces and hence higher is the boiling point.
(ii) A racemic mixture contains the two enantiomers d and l in equal proportions. As the rotation due to one enantiomer is cancelled by equal and opposite rotation of another enantiomer, therefore, it is optically inactive.
(iii) The presence of NO₂ group at o/p position in haloarenes helps in the stabilisation of resulting carbanion by −R and −I effects and hence increases the reactivity of haloarenes towards nucleophilic substitution reactions.

Teacher's Note:
1. Note the contrast for (ii): a racemic mixture is inactive by external compensation, whereas a meso compound is inactive by internal compensation.

Question 34. Give reasons:
(i) C–Cl bond length in chlorobenzene is shorter than C–Cl bond length in CH₃–Cl.
(ii) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(iii) S_N1 reactions are accompanied by racemization in optically active alkyl halides. [CBSE Delhi 2016]
Answer:
(i) In chlorobenzene each carbon atom is sp² hybridised due to resonance there is a partial double bond character, so bond length is short.
(ii) In chlorobenzene, carbon to which chlorine is attached is sp² hybridised and is more electronegative than the corresponding carbon in cyclohexyl chloride which is sp³ hybridised. So the dipole moment is lower in chlorobenzene. In chlorobenzene, −I and +R effect oppose each other while in the other only −I effect is the only contributing factor resulting in higher dipole moment of cyclohexyl chloride.
(iii) In S_N1 reaction, carbocation intermediate formed is a planar molecule which will lead to form d- and l- products. Hence, racemization occurs.

Teacher's Note:
1. For (iii), stress that the nucleophile can attack the planar carbocation equally from both faces – that is why a 50:50 mixture forms.

Question 35. Explain the following:
(i) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(ii) Alkyl halides, though polar, are immiscible with water.
(iii) Grignard's reagents should be prepared under anhydrous conditions. [CBSE Comptt. OD Delhi 2012]
Answer:
(i) In chlorobenzene, carbon to which chlorine is attached is sp² hybridised and is more electronegative than the corresponding carbon in cyclohexyl chloride which is sp³ hybridised. So the dipole moment is lower in chlorobenzene.
(ii) Alkyl halides though polar are immiscible with water because less energy is released when new attractions are set up between the alkyl halide and the water molecules as they are not as strong as the original hydrogen bonds in the water as a result the solubility of alkyl halides in water is low.
(iii) Grignard reagents should be prepared under anhydrous conditions because Grignard reagents are highly reactive. In the presence of moisture they react to give alkanes.
RMgX + H₂O → R–H + Mg(OH)X

Teacher's Note:
1. For (ii), the key comparison is energy: breaking water's hydrogen bonds costs more than is regained from haloalkane–water attractions.

Question 36. What happens when:
(i) CH₃–Cl is treated with aqueous KOH?
(ii) CH₃–Cl is treated with KCN?
(iii) CH₃–Br is treated with Mg in the presence of dry ether? [CBSE Delhi 2016]
Answer:
(i) CH₃–Cl + KOH(aq) → CH₃OH + KCl   (Methanol)
(ii) CH₃–Cl + KCN → CH₃CN + KCl   (Methyl cyanide)
(iii) CH₃Br + Mg ⟶^{Dry ether} CH₃MgBr   (Methyl magnesium bromide)

Teacher's Note:
1. Name the product as well as writing the equation; aqueous KOH gives substitution while alcoholic KOH would give elimination (not possible for CH₃Cl).

Question 37. Give the IUPAC name of the product formed when:
(i) 2-Methyl-1-bromopropane is treated with sodium in the presence of dry ether.
(ii) 1-Methyl cyclohexene is treated with HI.
(iii) Chloroethane is treated with silver nitrite. [SQP 2017]
Answer:
(i) 2,5-Dimethyl hexane.
(ii) 1-Methyl-1-iodocyclohexane
(iii) Nitroethane
Detailed Answer:
(i) 2 (CH₃)₂CH–CH₂Br + 2Na ⟶^{Dry ether} (CH₃)₂CH–CH₂–CH₂–CH(CH₃)₂ + 2NaBr
2-Methyl-1-bromopropane → 2,5-Dimethyl hexane
(ii) 1-Methylcyclohexene + HI → 1-Methyl-1-iodocyclohexane (I adds to the ring carbon carrying CH₃, by Markovnikov's rule)
(iii) C₂H₅Cl + AgNO₂ → C₂H₅NO₂ + AgCl

Teacher's Note:
1. AgNO₂ is covalent and attacks through N to give nitroalkanes, whereas KNO₂ is ionic and attacks through O to give alkyl nitrites.

Question 38. Give the IUPAC names of the following compounds:
(i) CH₃–CH(Br)–CH₂–CH₃
(ii)(iii) CH₂ = CH–CH₂–Cl [CBSE Comptt. OD 2015]
Answer:
(i) 2-Bromobutane
(ii) 1,3-Dibromobenzene
(iii) 3-Chloropropene

Teacher's Note:
1. In (iii) the double bond gets the lowest locant, so the chain is numbered from the =CH₂ end and Cl is on C₃, not C₁.

Question 39. Name the following according to IUPAC system:
(i) CH₃–CH(OH)–CH₂–CH₃
(ii)(iii) CH₃–C(CH₃)₂–CH₂–Cl [CBSE Comptt. Delhi 2015]
Answer:
(i) Butan-2-ol
(ii) 2-Bromotoluene (1-bromo-2-methylbenzene)
(iii) 1-Chloro-2,2-dimethylpropane

Teacher's Note:
1. (CH₃)₃CCH₂Cl is neopentyl chloride; the longest chain has only three carbons, so it is named as a substituted propane.

Question 40. How do you convert:
(i) Chlorobenzene to biphenyl,
(ii) Propene to 1-iodopropane,
(iii) 2-bromobutane to but-2-ene. [CBSE OD 2016]
Answer:
(i) Chlorobenzene to biphenyl: When two chlorobenzene combine with sodium metal in the presence of dry ether it forms biphenyl.
2C₆H₅Cl + 2Na ⟶^{dry ether} C₆H₅–C₆H₅ (Biphenyl) + 2NaCl
(ii) Propene to 1-iodopropane:
CH₃CH = CH₂ ⟶^HBr/Peroxide CH₃CH₂CH₂Br ⟶^NaI/acetone CH₃CH₂CH₂I
(iii) 2-bromobutane to but-2-ene:
CH₃–CH₂–CH(Br)–CH₃ ⟶^{Alc. KOH} CH₃–CH = CH–CH₃ (But-2-ene, 80%) + CH₃–CH₂–CH = CH₂ (But-1-ene, 20%)

Teacher's Note:
1. In (iii) quote Saytzeff's rule: the more substituted alkene, but-2-ene, is the major product.

Question 41. Write the major product(s) in the following:
(i) p-O₂N–C₆H₄–CH₂–CH₃ ⟶^{Br₂, UV light} ?
(ii) 2CH₃–CH(Cl)–CH₃ ⟶^{Na, dry ether} ?
(iii) CH₃–CH₂–Br ⟶^AgCN ? [CBSE OD 2016]
Answer:
(i) p-O₂N–C₆H₄–CH₂CH₃ (4-Ethylnitrobenzene) ⟶^{Br₂, UV light} p-O₂N–C₆H₄–CH(Br)–CH₃ (1-(1-Bromoethyl)-4-nitrobenzene)
(ii) 2CH₃–CH(Cl)–CH₃ ⟶^{Na, dry ether} (CH₃)₂CH–CH(CH₃)₂ (2,3-dimethyl butane)
(iii) CH₃CH₂Br ⟶^AgCN C₂H₅NC (Carbylaminoethane / Ethyl isocyanide)

Teacher's Note:
1. Under UV light Br substitutes at the benzylic carbon because the benzylic free radical is resonance-stabilised.

Long Answer Questions (5 Marks)

Question 42. Some alkylhalides undergo substitution whereas some undergo elimination reaction on treatment with bases. Discuss the structural features of alkyl halides with the help of examples which are responsible for this difference. [NCERT Exemplar]
Answer:
Primary alkyl halides prefer to undergo substitution reactions by S_N2 mechanism. A transition state is formed in which carbon is bonded to nucleophile and finally halogen atom is pushed out.
CH₃–Cl + OH⁻ (aq. KOH, weak base) → [HO···CH₃···Cl]⁻ (Transition state) → HO–CH₃ + Cl⁻
Tertiary alkyl halides prefer to undergo elimination reactions due to the formation of stable carbocations. 3° alkyl halides follow S_N1 mechanism by forming tertiary carbocations. Now, if the reagent used is a weak base then substitution occurs while if it is a strong base then instead of substitution elimination occurs.
(CH₃)₃C–Cl → (CH₃)₃C⁺ + Cl⁻
(CH₃)₃C⁺ + OH⁻ → (CH₃)₃C–OH
Here the reagent used i.e., aq. KOH is a weak base, so substitution takes place.
(CH₃)₃C–Cl ⟶^{alc. KOH} (CH₃)₃C⁺ ⟶^{OH−, −H+} (CH₃)₂C = CH₂ + H₂O   (2-methylprop-1-ene)
Here the reagent used i.e., alc. KOH is a strong base, so elimination competes over substitution and alkene is formed.
Secondary alkyl halide can undergo substitution or elimination depending on type of solvent and temperature conditions.

Teacher's Note:
1. Use the pair 'aqueous KOH → substitution, alcoholic KOH → elimination' and link it to the class of the halide; that structure-plus-reagent argument is what the examiner looks for.

Question 43. Some halogen containing compounds are useful in daily life. Some compounds of this class are responsible for exposure of flora and fauna to more and more of UV light which causes destruction to a great extent. Name the class of these halocompounds. In your opinion, what should be done to minimise harmful effects of these compounds. [NCERT Exemplar]
Answer:
There are a number of halogen containing compounds such as chloroform, iodoform, carbon tetrachloride, DDT, benzene hexachloride, freons, etc, which are useful in our daily life. But these compounds have adverse effects on health of humans, animals and other living beings. Compounds like freons containing chloroflorocarbons cause the depletion of ozone layer. Consequently, UV rays reach the earth which harm flora and fauna.
(i) Tetrachloromethane: In contact with air, carbon tetrachloride rises to the atmosphere and depletes the ozone layer. Depletion of the ozone layer increases human exposure to UV rays leading to increased skin cancer, eye diseases and disorders, and possible disruption of the immune system. These UV rays cause damages to plants, and reduction of plankton populations in the ocean's euphotic zone.
(ii) Freons: Freon-113 remains in the air long enough to reach the upper atmosphere. It provides chlorine atoms which damage the ozone layer. Because of this depletion, UV rays enter in our atmosphere and become responsible for the damage to great extent.
(iii) p-p'-dichlorodiphenyltrichloroethane (DDT): It is non-biodegradable so gets deposited in fatty tissues. If ingestion continues for a long time, DDT builds up within the animal and affect the reproductive system.
To minimize the harmful effect of these compounds the use of equipments like, air-conditioner, refrigerator, aerosol that produce such compounds should be discouraged. Government has tightened laws on use of DDT. Such rules should be strictly followed.

Teacher's Note:
1. Name the class clearly – chlorofluorocarbons (freons) – before discussing effects; a 5-mark answer also needs concrete measures to reduce their use.