CBSE Class 12 Chemistry Haloalkanes and Haloarenes Important Questions Set 06

Very Short Answer Questions (1 Mark)

Question 1. Write a chemical reaction to illustrate Saytzeff's rule.
Answer:
Saytzeff rule states that if the structure of an alkyl halide is such that it can undergo dehydrohalogenation in two or more different ways to give a mixture of alkenes, then the alkene which is most highly substituted is the major product. For example,
CH₃CHBrCH₂CH₃ (2-Bromobutane) \( \xrightarrow{\text{Alc. KOH},\ \Delta} \) CH₃CH=CHCH₃ (But-2-ene, more stable, 81%) + CH₃CH₂CH=CH₂ (But-1-ene, less stable, 19%)

Teacher's Note:
1. Count the alkyl groups on the C=C carbons: the alkene with more alkyl groups (more substituted) is the Saytzeff (major) product.

Question 2. How do you convert : 2-bromobutane to but-2-ene ? (CBSE 2016)
Answer:
2-Bromobutane on heating with alc. KOH undergoes dehydrohalogenation to give but-2-ene as the major product in accordance with Saytzeff rule.
CH₃–CH(Br)–CH₂CH₃ (2-Bromobutane) \( \xrightarrow{\text{KOH (alc.)},\ \Delta,\ (-\text{HBr})} \) CH₃–CH=CH–CH₃ (But-2-ene, major product) + CH₃CH₂CH=CH₂ (But-1-ene, minor product)

Teacher's Note:
1. Write the reagent as alcoholic KOH; aqueous KOH would give butan-2-ol instead.

Question 3. Write the structure of the alkene formed by dehydrohalogenation of 1-bromo-1-methylcyclohexane with alcoholic KOH. (CBSE 2018)
Answer:
1-Bromo-1-methylcyclohexane \( \xrightarrow{\text{Alc. KOH},\ -\text{HBr}} \) 1-Methylcyclohexene (major, Saytzeff product) + Methylenecyclohexane (minor)

Teacher's Note:
1. The ring hydrogens on C-2 give the trisubstituted endocyclic alkene, which is more substituted than the exocyclic C=CH₂, so it is the major product.

Question 4. Name the alkyl halide which can be used to prepare methane and ethane in single steps.
Answer:
Methyl iodide (CH₃–I).
CH₃–I (Methyl iodide) + HI \( \xrightarrow{\text{Red P},\ \Delta\ (\text{Reduction})} \) CH₄ (Methane) + I₂
2 CH₃–I (Methyl iodide) + 2Na \( \xrightarrow{\text{Dry ether}\ (\text{Wurtz reaction})} \) CH₃–CH₃ (Ethane) + 2NaI

Teacher's Note:
1. Reduction keeps the number of carbons; Wurtz reaction doubles it, which is why one halide can give both C₁ and C₂ alkanes.

Question 5. What is plane polarized light ?
Answer:
A beam of light which has vibrations in only one plane is called plane polarised light.

Teacher's Note:
1. Ordinary light vibrates in all planes; a Nicol prism (or polaroid) lets through only the vibrations in one plane.

Question 6. What is an asymmetric or chiral carbon ?
Answer:
A carbon which is attached to four different atoms/groups is called asymmetric or chiral carbon. For example, the carbon atom in BrC*HCl or CH₃C*HOHCH₂CH₃.

Teacher's Note:
1. Check all four groups on the carbon; two identical groups (for example two CH₃ or two H) mean it is not chiral.

Question 7. What is the condition to be satisfied for a compound to be chiral ?
Answer:
Its mirror image should be non-superimposable.

Teacher's Note:
1. A molecule with a plane of symmetry is achiral even if it has chiral carbons (for example meso compounds).

Question 8. What type of isomerism is shown by lactic acid. (Karnataka Board 2012)
Answer:
Lactic acid contains one chiral carbon since it is attached to four different atoms/groups H, OH, CH₃ and COOH. Therefore, lactic acid shows optical isomerism and thus exists in two stereoisomers (i.e., enantiomers) which are non-superimposable mirror images of each other.

Teacher's Note:
1. Lactic acid is CH₃CH(OH)COOH; the central carbon carries H, OH, CH₃ and COOH, so it has one chiral centre and two enantiomers.

Question 9. What do prefixes (+), (–) and (±) before an organic compound mean ?
Answer:
Prefixes (+), (–) and (±) imply dextrorotatory, laevorotatory and racemic modification respectively.

Teacher's Note:
1. (+)/(−) are found by experiment with a polarimeter; they have no fixed link with the D/L or R/S labels.

Question 10. What is known as a racemic mixture ? Give one example. (CBSE 2011)
or Racemic mixture is optically inactive. Give reason. (CBSE 2015)
Answer:
A 50 : 50 mixture of two enantiomers of any optically active compound is called a racemic mixture. It is always optically inactive since rotation caused by the molecules of one enantiomer is exactly cancelled by equal and opposite rotation caused by the molecules of the other enantiomer. For example, an equimolar mixture of (+)-2-bromobutane and (–)-2-bromobutane is called a racemic mixture.

Teacher's Note:
1. A racemic mixture is inactive due to external compensation; a meso compound is inactive due to internal compensation.

Question 11. S_N1 reactions are accompanied by racemization in optically active halides. (CBSE 2016)
Answer:
In S_N1 reactions, carbocations are the intermediates. Since carbocations are planar species, therefore, attack of the nucleophile on them can occur from both the sides (front and rear) with almost equal ease giving a 50 : 50 mixture of the two enantiomers, i.e., a racemic mixture is obtained.

Teacher's Note:
1. Link the three words in your answer: planar carbocation, attack from both faces, racemic mixture.

Question 12. What is the lowest molecular mass alkane that is chiral ? Is there another alkane of the same molecular formula which is also chiral ? If so, then give its structure also.
Answer:
The lowest molecular mass alkane which is chiral must have a chiral carbon which is attached to four different atoms/groups of lowest molecular mass, i.e., H, CH₃, CH₃CH₂ and CH₃CH₂CH₂. Therefore, the alkane is 3-methylhexane.
Since the isomeric form of n-propyl group is isopropyl group, therefore, another alkane which is chiral is 2, 3-dimethylpentane
CH₃CH₂–C*H(CH₃)–CH₂CH₂CH₃   3-Methylhexane
CH₃CH₂–C*H(CH₃)–CH(CH₃)–CH₃   2, 3-Dimethylpentane

Teacher's Note:
1. Both are C₇H₁₆; the chiral carbon must carry H, methyl, ethyl and a three-carbon group (propyl or isopropyl).

Question 13. An acid of molecular formula, C₅H₁₀O₂ is optically active. What is its structure ?
Answer:
CH₃CH₂–C*H(CH₃)–COOH   2-Methylbutanoic acid

Teacher's Note:
1. The chiral carbon here carries H, CH₃, C₂H₅ and COOH; the other C₅ acids lack such a carbon.

Question 14. A carboxylic acid of the formula, C₃H₅O₂Br is optically active. What is its structure ?
Answer:
CH₃–C*H(Br)–COOH   2-Bromopropanoic acid.

Teacher's Note:
1. 3-Bromopropanoic acid, BrCH₂CH₂COOH, has no chiral carbon, so it is the optically inactive isomer.

Question 15. Give the structure of lowest molecular mass alcohol which is chiral ?
Answer:
The lowest molecular mass alcohol which is chiral is butan-2-ol, i.e., CH₃CH₂–C*H(OH)–CH₃.

Teacher's Note:
1. Propan-2-ol is not chiral because its central carbon carries two identical CH₃ groups.

Question 16. Give the structure of an optically active hydrocarbon (C₆H₁₂) which on catalytic hydrogenation gives an optically inactive compound (C₆H₁₄).
Answer:
CH₂=CH–C*H(CH₃)–CH₂CH₃ (3-Methylpent-1-ene, Optically active) \( \xrightarrow{\text{H}_2\text{/Ni}} \) CH₃CH₂–CH(CH₃)–CH₂CH₃ (3-Methylpentane, Optically inactive)

Teacher's Note:
1. On hydrogenation the vinyl group becomes a second ethyl group, so the carbon now carries two identical groups and loses its chirality.

Short Answer Questions (2 Marks)

Question 17. Write four structural isomers of compound having molecular formula C₄H₉Br. (Hr. Board 2012, 2013)
Answer:
The four structural isomers of C₄H₉Br are :
(i) CH₃CH₂CH₂CH₂Br   1-Bromobutane (n-butyl bromide, 1°)
(ii) CH₃CH₂CH(Br)CH₃   2-Bromobutane (sec-butyl bromide, 2°)
(iii) (CH₃)₂CHCH₂Br   1-Bromo-2-methylpropane (isobutyl bromide, 1°)
(iv) (CH₃)₃CBr   2-Bromo-2-methylpropane (tert-butyl bromide, 3°)

Teacher's Note:
1. Two isomers come from the straight chain and two from the branched chain; only 2-bromobutane is chiral.

Question 18. Explain why free radical bromination of n-butane yields 2- bromobutane as the major product.
Answer:
Free radical bromination of n-butane can give two products : 1-bromobutane (via the 1° butyl radical) and 2-bromobutane (via the 2° sec-butyl radical).
CH₃CH₂CH₂CH₃ + Br₂ \( \xrightarrow{h\nu\ \text{or}\ \Delta} \) CH₃CH₂CH(Br)CH₃ (2-Bromobutane, major) + CH₃CH₂CH₂CH₂Br (1-Bromobutane, minor) + HBr
Since the stability of free radicals follows the order 3° > 2° > 1°, the 2° free radical is formed more easily than the 1° free radical. Further, bromine atom is less reactive and hence more selective; it abstracts preferentially the 2° H-atom. Therefore, 2-bromobutane is the major product.

Teacher's Note:
1. Bromination is far more selective than chlorination because the Br radical is less reactive; it mostly removes the easiest (2° or 3°) hydrogen.

Question 19. Write the major product of the following reactions :
(i) CH₃CH₂CH₂OH + SOCl₂ → (Hr. Board 2011)
(ii) CH₃CH = CH₂ + HBr \( \xrightarrow{\text{Peroxide}} \) (HP Board 2013)
(iii) CH₃CH = CH₂ + HBr \( \xrightarrow{\text{Absence of peroxide}} \) (Hr. Board 2013)
(iv) CH₃CH₂CH₂CH₂Cl + NaI \( \xrightarrow{\text{Acetone, heat}} \) (Hr. Board 2011)
Answer:
(i) CH₃CH₂CH₂OH + SOCl₂ → CH₃CH₂CH₂Cl (1-Chloropropane) + SO₂↑ + HCl↑
(ii) CH₃CH = CH₂ + HBr \( \xrightarrow{\text{Peroxide}} \) CH₃CH₂CH₂Br (1-Bromopropane; anti-Markovnikov addition)
(iii) CH₃CH = CH₂ + HBr \( \xrightarrow{\text{Absence of peroxide}} \) CH₃CH(Br)CH₃ (2-Bromopropane; Markovnikov addition)
(iv) CH₃CH₂CH₂CH₂Cl + NaI \( \xrightarrow{\text{Acetone, heat}} \) CH₃CH₂CH₂CH₂I (1-Iodobutane) + NaCl↓ (Finkelstein reaction)

Teacher's Note:
1. Peroxide changes the product only for HBr; with HCl or HI Markovnikov's rule is followed even in the presence of peroxide.

Question 20. What are alkyl halides? How is n-propyl bromide prepared from propylene? (Maharashtra Board 2012)
Answer:
Alkyl halides are the halogen derivatives of alkanes obtained by replacing one hydrogen atom of an alkane by a halogen atom. Their general formula is C_nH_2n+1X or R–X, where X = F, Cl, Br or I, e.g., CH₃Cl, C₂H₅Br, etc.
n-Propyl bromide is prepared by the addition of HBr to propylene in presence of peroxides (anti-Markovnikov addition or peroxide effect) :
CH₃CH=CH₂ (Propylene) + HBr \( \xrightarrow{\text{Peroxide}} \) CH₃CH₂CH₂Br (n-Propyl bromide)

Teacher's Note:
1. Without peroxide the same reaction gives isopropyl bromide; mention the peroxide clearly over the arrow.

Question 21. Give reasons in one or two sentences for the following :
(i) Alkyl halides are better solvents than aryl halides.
(ii) Organic halogen compounds used as solvents in industry are chlorides rather than bromides and iodides.
Answer:
(i) Alkyl halides are more polar (higher dipole moment) than aryl halides and are also more volatile; hence they dissolve a large number of organic compounds and can be easily removed after use. Therefore, alkyl halides are better solvents than aryl halides.
(ii) Organic alkyl chlorides are widely used as solvents in industry because chlorides are not only more stable but are also more volatile (and cheaper) than bromides and iodides.

Teacher's Note:
1. A good industrial solvent should dissolve the substance, be stable and evaporate easily; chlorides meet all three needs best.

Question 22. Why do alkyl halides show nucleophilic substitution reactions ?
Answer:
Due to greater electronegativity of halogen than carbon, the C–X bond in alkyl halides is polar; the carbon atom carries a partial positive charge (C^δ+–X^δ−) and the halogen a partial negative charge.
Therefore, a nucleophile (an electron-rich species) attacks this electron-deficient carbon and the halogen leaves as halide ion (a good leaving group). Thus, alkyl halides undergo nucleophilic substitution reactions, e.g.,
R–X + Nu⁻ → R–Nu + X⁻

Teacher's Note:
1. Two things make this possible: the δ+ carbon that attracts the nucleophile and the halide ion that is a stable leaving group.

Question 23. How will you convert ethyl bromide to
(i) ethane
(ii) ethoxyethane? (J & K Board 2012)
Answer:
(i) By reduction with zinc and dil. HCl (or H₂/Pd) :
C₂H₅Br + 2[H] \( \xrightarrow{\text{Zn/HCl}} \) C₂H₆ (Ethane) + HBr
(ii) By heating with sodium ethoxide (Williamson synthesis) :
C₂H₅Br + C₂H₅ONa → C₂H₅–O–C₂H₅ (Ethoxyethane) + NaBr

Teacher's Note:
1. Williamson synthesis is an S_N2 reaction; it works best with a 1° alkyl halide like ethyl bromide.

Question 24. How will you convert an alkyl halide into an alkene and an alkane? (Hr. Board 2012)
Answer:
Alkene : By heating with alcoholic KOH (dehydrohalogenation), e.g.,
CH₃CH₂Br + KOH (alc.) \( \xrightarrow{\Delta} \) CH₂=CH₂ + KBr + H₂O
Alkane : By reduction with Zn/HCl, or with HI and red P, or by Wurtz reaction, e.g.,
CH₃CH₂Br + 2[H] \( \xrightarrow{\text{Zn/HCl}} \) CH₃CH₃ + HBr ;   2CH₃CH₂Br + 2Na \( \xrightarrow{\text{Dry ether}} \) CH₃CH₂CH₂CH₃ + 2NaBr

Teacher's Note:
1. Reduction gives an alkane with the same number of carbons; Wurtz reaction gives one with double the number.

Teacher's Note:
1. When asked for 'two differences', give matched points on the same basis (e.g., steps vs steps, kinetics vs kinetics).

Question 26. How would you differentiate between S_N1 and S_N2 mechanisms of substitution reactions ? Give one example of each. (CBSE 2010)
Answer:
S_N1 : It occurs in two steps through a carbocation intermediate; rate depends only on the concentration of alkyl halide (Rate = k[RX]); optically active halides give racemic mixture; 3° halides react fastest. Example : hydrolysis of tert-butyl bromide with aqueous KOH :
(CH₃)₃C–Br + KOH(aq) → (CH₃)₃C–OH + KBr
S_N2 : It occurs in one step through a transition state; rate depends on the concentration of both alkyl halide and nucleophile (Rate = k[RX][Nu⁻]); inversion of configuration occurs; CH₃X and 1° halides react fastest. Example : hydrolysis of methyl bromide with aqueous KOH :
CH₃–Br + KOH(aq) → CH₃–OH + KBr

Teacher's Note:
1. Choose a 3° halide as the S_N1 example and CH₃X or a 1° halide as the S_N2 example.

Question 27. Explain the steps involved in the mechanism of hydrolysis of tertiary butyl bromide using aqueous potassium hydroxide. (Karnataka Board 2012)
Answer:
Hydrolysis of tert-butyl bromide occurs by S_N1 mechanism in two steps :
Step 1. (Slow, rate-determining) The C–Br bond ionizes to give a 3° carbocation and bromide ion :
(CH₃)₃C–Br \( \xrightarrow{\text{Slow}} \) (CH₃)₃C⁺ (tert-Butyl carbocation) + Br⁻
Step 2. (Fast) The nucleophile OH⁻ attacks the planar carbocation to give tert-butyl alcohol :
(CH₃)₃C⁺ + OH⁻ \( \xrightarrow{\text{Fast}} \) (CH₃)₃C–OH (tert-Butyl alcohol)
Since the slow step involves only the alkyl halide, Rate = k[(CH₃)₃CBr], i.e., it is a first order reaction.

Teacher's Note:
1. The 3° carbocation is stabilised by +I effect and hyperconjugation of three CH₃ groups; this is why the first step is possible.

Question 28. (a) The nucleophilic substitution of primary alkyl chlorides with sodium acetate is catalysed by sodium iodide.
(b) p-Methoxybenzyl bromide reacts faster than p-nitrobenzyl bromide with ethanol to form an ether product.
Answer:
(a) In presence of NaI, alkyl chlorides form alkyl iodides which being more reactive catalyse the nucleophilic substitution reaction (I⁻ is a better nucleophile and also a better leaving group than Cl⁻; it is regenerated in the second step).
R–Cl + I⁻ → R–I + Cl⁻ ;   R–I + CH₃COO⁻ → CH₃COOR + I⁻
(b) The reaction with ethanol occurs by S_N1 mechanism. p-Methoxybenzyl cation is stabilized by the +R- effect of the methoxy group while p-nitrobenzyl cation is destabilized by the –R-effect of the NO₂ group. Therefore, p-methoxybenzyl bromide reacts faster.

Teacher's Note:
1. Electron-donating groups at the para position speed up S_N1 of benzyl halides; electron-withdrawing groups slow it down.

Question 29. Racemic mixture is optically inactive. Give reasons. (CBSE 2015)
Answer:
A racemic mixture contains the two enantiomers in equal amounts (50 : 50). The rotation of plane polarised light caused by the molecules of one enantiomer is exactly cancelled by the equal and opposite rotation caused by the molecules of the other enantiomer. Hence, a racemic mixture is optically inactive (external compensation).

Teacher's Note:
1. Write 'external compensation' for a racemic mixture and 'internal compensation' for a meso compound.

Question 30. How is chlorobenzene prepared from :
(i) benzene and
(ii) aniline or benzenediazonium chloride ?
Answer:
(i) By electrophilic substitution of benzene with chlorine in presence of a Lewis acid catalyst such as anhydrous FeCl₃ in dark at room temperature :
C₆H₆ + Cl₂ \( \xrightarrow{\text{Anhyd. FeCl}_3,\ \text{dark}} \) C₆H₅Cl + HCl
(ii) Aniline is first diazotised with NaNO₂ and HCl at 273–278 K, and the benzenediazonium chloride so formed is treated with cuprous chloride dissolved in HCl (Sandmeyer reaction) :
C₆H₅NH₂ \( \xrightarrow{\text{NaNO}_2 + \text{HCl},\ 273\text{--}278\ \text{K}} \) C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\text{Cu}_2\text{Cl}_2\text{/HCl}} \) C₆H₅Cl + N₂

Teacher's Note:
1. Without a Lewis acid and in light, chlorine adds to benzene (giving C₆H₆Cl₆) instead of substituting.

Short Answer Questions (3 Marks)

Question 31. Write the major product of the following reactions:
(a) CH₂=CHBr →(AgCN, alcohol)
(b) C₆H₆ + CH₃–C(CH₃)₂–CH₂–Br →(Anhyd. AlCl₃)
(c) C₆H₅CH₃ (toluene) + CH₃–CH(CH₃)–CH₂Br →(Anhyd. AlCl₃) C
Answer:
(a) Since vinyl halides are unreactive towards nucleophilic substitution reactions, therefore, CH₂=CHBr does not react with AgCN.
CH₂=CHBr →(AgCN, alcohol) No reaction
(b) The initially formed less stable 1° carbocation (I) undergoes 1,2-methyl shift to give the more stable 3° carbocation (II) which then attacks the benzene ring to give F.C. alkylation product (III),
(CH₃)₃C–CH₂–Br (neopentyl bromide) →(AlCl₃, −AlCl₄⁻) (CH₃)₃C–C⁺H₂ [1° carbocation (I), less stable] →(1,2-methyl shift) (CH₃)₂C⁺–CH₂CH₃ [3° carbocation (II), more stable]
(CH₃)₂C⁺–CH₂CH₃ + C₆H₆ →(−H⁺) C₆H₅–C(CH₃)₂–CH₂CH₃ [2-methyl-2-phenylbutane (III)]
(c) The initially formed less stable 1° carbocation (I) undergoes rearrangement by 1,2-hydride shift to form more stable 3° carbocation (II) which then attacks the toluene ring at the less hindered p-position to form p-tert-butyltoluene.
(CH₃)₂CH–CH₂Br (isobutyl bromide) →(Anhyd. AlCl₃, −AlCl₃Br⁻) (CH₃)₂CH–C⁺H₂ [isobutyl carbocation (I), 1°, less stable] →(1,2-hydride shift) (CH₃)₃C⁺ [tert-butyl carbocation (II), 3°, more stable]
(CH₃)₃C⁺ + CH₃–C₆H₅ →(−H⁺) p-CH₃–C₆H₄–C(CH₃)₃ [p-tert-butyltoluene, 1-tert-butyl-4-methylbenzene]

Teacher's Note:
1. Friedel–Crafts alkylation goes through carbocations, so 1° alkyl halides usually give rearranged (more branched) products.

Question 32. Identify and indicate the presence of centre of chirality, if any, in the following molecules. How many stereoisomers are possible for each?
(i) 2-Aminobutane (ii) 3-Bromopent-1-ene (iii) 1,2-Dichloropropane (iv) 3-Methyl-1-pentene.
Answer:
(i) CH₃CH₂–C*H(NH₂)–CH₃ (2-aminobutane): one chiral carbon atom; two stereoisomers which are enantiomers.
(ii) CH₂=CH–C*HBr–CH₂CH₃ (3-bromopent-1-ene): one chiral carbon atom; two stereoisomers which are enantiomers.
(iii) ClCH₂–C*HCl–CH₃ (1,2-dichloropropane): one chiral carbon atom; two stereoisomers which are enantiomers.
(iv) CH₃CH₂–C*H(CH₃)–CH=CH₂ (3-methyl-1-pentene): one chiral carbon atom; two stereoisomers which are enantiomers.
(Chiral carbon atoms are marked with an asterisk.)

Teacher's Note:
1. With n chiral carbons a molecule can have at most 2ⁿ stereoisomers; one chiral carbon always gives one pair of enantiomers.

Question 33. p-Chloronitrobenzene undergoes nucleophilic substitution faster than chlorobenzene. Explain giving the resonating structures as well.
Answer:
Attack of nucleophile on p-chloronitrobenzene gives carbanion (I) which is stabilized by π-electrons of the benzene ring as well as by the −R-effect and the −I-effect of the —NO₂ group as shown below:
p-Cl–C₆H₄–NO₂ + :Nu⁻ →(slow) carbanion (I), in which C-1 carries both Cl and Nu (sp³) and the negative charge is spread by resonance over C-2, C-4 and C-6 of the ring;
when the charge is on C-4 (the carbon bearing NO₂), it is further delocalised onto the NO₂ group, giving a structure with C=N⁺(O⁻)O⁻, i.e., the negative charge resides on the electronegative oxygen atoms.
Carbanion (I) →(fast, −Cl⁻) p-Nu–C₆H₄–NO₂ (substitution product)
On the other hand, attack of the nucleophile on chlorobenzene gives carbanion (II) which is stabilized only by the π-electrons of the benzene ring as shown below:
C₆H₅Cl + :Nu⁻ →(slow) carbanion (II), negative charge delocalised only over C-2, C-4 and C-6 of the ring (three resonance structures) →(fast, −Cl⁻) C₆H₅–Nu (substitution product)
Due to greater stability of the carbanion (I) over carbanion (II), p-chloronitrobenzene undergoes nucleophilic substitution faster than chlorobenzene.

Teacher's Note:
1. The key extra resonance structure is the one with the negative charge on the oxygen atoms of NO₂; it is possible only when NO₂ is ortho or para to Cl.

Question 34. Give the IUPAC names of the following :
(i) CH₃CH₂–C(CH₃)(Br)–CH(Br)–CH₂–Cl
(ii) CH₃–C(C₂H₅)(Cl)–CH₂–C(C₂H₅)(Cl)–CH₃
(iii) CH₂ = CHCH₂Br (CBSE 2011)
(iv) (CH₃)₃CCH₂Br (CBSE 2011)
(v) CH₃–CH(Br)–CH₂–CH(Cl)–CH₃ (CBSE 2013)
(vi) CH₃–CH(Cl)–CH₂–CH=CH₂ (CBSE 2013)
(vii) CH₃–C(CH₃)₂–CH(Cl)–CH₃ (CBSE 2013)
(viii) CH₃CH = CH–C(CH₃)₂–Br (CBSE 2013, 2019)
(ix) (CH₃)₂CHCH(Cl)CH₃ (CBSE 2013)
Answer:
(i) 2, 3-dibromo-1-chloro-3-methylpentane, (ii) 3,5-Dichloro-3, 5-dimethylheptane, (iii) 3-Bromoprop-1-ene, (iv) 1-Bromo-2, 2-dimethylpropane, (v) 2-bromo-4-chloropentane, (vi) 4-chloropent-1-ene, (vii) 3-chloro-2, 2-dimethylbutane, (viii) 4-bromo-4-methylpent-2-ene, (ix) 2-chloro-3-methylbutane

Teacher's Note:
1. Prefixes are written in alphabetical order (bromo before chloro), ignoring multiplying prefixes like di- and tri-.
2. In (ii) the ethyl groups extend the main chain, so the parent is heptane, not pentane.

Question 35. Explain the following :
(i) Although haloalkanes are polar in character yet they are insoluble in water.
or Alkyl halides are insoluble in water though they contain a polar C–X bond.
(ii) The boiling point of bromoethane is higher than that of chloroethane.
(iii) n-Butyl bromide has higher boiling point than t-butyl bromide. (CBSE 2015)
(iv) The dipole moment of chloroethane is higher than that of chlorobenzene.
Answer:
(i) To dissolve in water, energy is required to overcome the attractions between haloalkane molecules and to break the hydrogen bonds between water molecules. Since haloalkanes cannot form hydrogen bonds with water, less energy is released when new attractions are set up between haloalkane and water molecules. Hence haloalkanes, though polar, are insoluble in water.
(ii) Due to bigger size and higher molecular mass of Br over Cl, the van der Waals forces of attraction are stronger in bromoethane than in chloroethane. Therefore, bromoethane has higher b.p. than chloroethane.
(iii) For the same alkyl group, b.p. decreases as the branching increases (surface area decreases). n-Butyl bromide has no branching while t-butyl bromide is highly branched, therefore, n-butyl bromide (b.p. 375 K) has higher b.p. than t-butyl bromide (b.p. 346 K).
(iv) In chloroethane, Cl is attached to a sp³⁻hybridized carbon while in chlorobenzene it is attached to a sp²⁻hybridized carbon which, due to greater s-character, is more electronegative and has less tendency to release electrons to Cl. Further, due to resonance, electrons of Cl are delocalised into the ring. As a result, C–Cl bond in chlorobenzene is less polar. Hence dipole moment of chloroethane (2.05 D) is higher than that of chlorobenzene (1.69 D).

Teacher's Note:
1. For solubility questions, always compare energy needed (to break H-bonds of water) with energy released (new attractions formed).

Question 36. How do the products differ when ethyl bromide reacts separately with :
(i) Aqueous KOH and alcoholic KOH (ii) KCN and AgCN (iii) KNO₂ and AgNO₂ ?
Answer:
(i) With aqueous KOH, ethyl bromide undergoes substitution to give ethyl alcohol while with alcoholic KOH, it undergoes dehydrohalogenation (elimination) to give ethene.
C₂H₅Br + KOH(aq) → C₂H₅OH (Ethanol) + KBr ;   C₂H₅Br + KOH(alc) \( \xrightarrow{\Delta} \) CH₂=CH₂ (Ethene) + KBr + H₂O
(ii) KCN is ionic and attack occurs through the C-atom of CN⁻ giving ethyl cyanide (propanenitrile), while AgCN is mainly covalent and attack occurs through N-atom giving ethyl isocyanide.
C₂H₅Br + KCN → C₂H₅–C≡N (Ethyl cyanide) + KBr ;   C₂H₅Br + AgCN → C₂H₅–N⁺≡C⁻ (Ethyl isocyanide) + AgBr
(iii) KNO₂ is ionic and attack occurs through O-atom of nitrite ion giving ethyl nitrite, while AgNO₂ is covalent and attack occurs through N-atom giving nitroethane.
C₂H₅Br + KONO → C₂H₅–O–N=O (Ethyl nitrite) + KBr ;   C₂H₅Br + AgNO₂ → C₂H₅–NO₂ (Nitroethane) + AgBr

Teacher's Note:
1. CN⁻ and NO₂⁻ are ambident nucleophiles: ionic K salts give cyanide/nitrite, covalent Ag salts give isocyanide/nitro compound.

Question 37. Outline the preparation of following compounds using a nucleophilic substitution reaction.
(a) CH₃C ≡ CCH₂CH₃
(b) CH₃CH₂CH₂N(CH₃)₂
(c) C₆H₅CH₂OCOCH₃
(d) CH₃CH₂CH₂CH₂NO₂
(e) CH₃CH₂CH₂CN
(f) C₆H₅CH₂N⁺≡C:⁻
(g) CH₃CH₂–O–N = O
Answer:
(a) CH₃C≡C⁻Na⁺ (Sodium propynide) + CH₃CH₂Br → CH₃C≡CCH₂CH₃ + NaBr
(b) CH₃CH₂CH₂Br + HN(CH₃)₂ (Dimethylamine) → CH₃CH₂CH₂N(CH₃)₂ + HBr
(c) C₆H₅CH₂Cl + CH₃COO⁻Ag⁺ (Silver acetate) → C₆H₅CH₂OCOCH₃ (Benzyl acetate) + AgCl
(d) CH₃CH₂CH₂CH₂Br + AgNO₂ → CH₃CH₂CH₂CH₂NO₂ + AgBr
(e) CH₃CH₂CH₂Br + KCN \( \xrightarrow{\text{aq. ethanol}} \) CH₃CH₂CH₂CN + KBr
(f) C₆H₅CH₂Cl + AgCN → C₆H₅CH₂N⁺≡C:⁻ (Benzyl isocyanide) + AgCl
(g) CH₃CH₂Br + KONO (Potassium nitrite) → CH₃CH₂–O–N=O (Ethyl nitrite) + KBr

Teacher's Note:
1. In each case pick the alkyl halide that supplies the alkyl part and the reagent that supplies the nucleophile; keep the halide primary to avoid elimination.

Question 38. With the help of chemical equations, show how will you convert :
(i) 1-propanol to 2-bromopropane.
(ii) 2-bromopropane to 1-bromopropane ?
(iii) 1-bromopropane to 2-bromopropane
(iv) propanone to iodoform.
(v) isopropyl chloride to n-propyl chloride?
(vi) 1-chlorobutane to 1-iodobutane (CBSE 2019)
Answer:
(i) CH₃CH₂CH₂OH \( \xrightarrow{\text{Conc. H}_2\text{SO}_4,\ 443\ \text{K}} \) CH₃CH=CH₂ \( \xrightarrow{\text{HBr}} \) CH₃CH(Br)CH₃
(ii) CH₃CH(Br)CH₃ \( \xrightarrow{\text{alc. KOH},\ \Delta} \) CH₃CH=CH₂ \( \xrightarrow{\text{HBr, peroxide}} \) CH₃CH₂CH₂Br
(iii) CH₃CH₂CH₂Br \( \xrightarrow{\text{alc. KOH},\ \Delta} \) CH₃CH=CH₂ \( \xrightarrow{\text{HBr (no peroxide)}} \) CH₃CH(Br)CH₃
(iv) CH₃COCH₃ + 3I₂ + 4NaOH → CHI₃ (Iodoform) + CH₃COONa + 3NaI + 3H₂O
(v) CH₃CH(Cl)CH₃ \( \xrightarrow{\text{alc. KOH},\ \Delta} \) CH₃CH=CH₂ \( \xrightarrow{\text{HBr, peroxide}} \) CH₃CH₂CH₂Br \( \xrightarrow{\text{aq. KOH}} \) CH₃CH₂CH₂OH \( \xrightarrow{\text{SOCl}_2} \) CH₃CH₂CH₂Cl
(vi) CH₃CH₂CH₂CH₂Cl + NaI \( \xrightarrow{\text{Dry acetone}} \) CH₃CH₂CH₂CH₂I + NaCl (Finkelstein reaction)

Teacher's Note:
1. To shift a halogen between C-1 and C-2, go through the alkene: eliminate with alc. KOH, then add HBr with or without peroxide.
2. HCl shows no peroxide effect, so 1-chloropropane is reached through 1-bromopropane and propan-1-ol.

Question 39. Discuss briefly the mechanism of S_N1 and S_N2 reactions. (Uttarakhand Board 2012)
Answer:
S_N2 mechanism : It is a one-step (concerted) bimolecular reaction. The nucleophile attacks the carbon carrying the halogen from the side opposite to the halogen (back side attack). In the transition state the C-atom is partly bonded to both Nu and X; as the C–Nu bond forms, the C–X bond breaks, and the configuration is inverted (Walden inversion). Rate = k[RX][Nu⁻].
HO⁻ + CH₃–Br → [HO^δ−···CH₃···Br^δ−] (Transition state) → HO–CH₃ + Br⁻
S_N1 mechanism : It occurs in two steps. In the first slow (rate-determining) step, the alkyl halide ionizes to give a carbocation and a halide ion. In the second fast step, the nucleophile attacks the planar carbocation from either side. Rate = k[RX].
(CH₃)₃C–Br \( \xrightarrow{\text{Slow}} \) (CH₃)₃C⁺ + Br⁻ ;   (CH₃)₃C⁺ + OH⁻ \( \xrightarrow{\text{Fast}} \) (CH₃)₃C–OH

Teacher's Note:
1. Remember the molecularity in the name: S_N2 rate depends on both RX and Nu⁻, S_N1 rate on RX only.

Question 40. Arrange the following compounds in increasing order of S_N1 reactivity.
(a) ClCH₂CH = CHCH₂CH₃, CH₃C(Cl) = CHCH₂CH₃, CH₃CH = CHCH₂CH₂Cl
(b) CH₃CH₂Br, CH₂ = CHCH(Br)CH₃, CH₂ = CHBr, CH₃CH(Br)CH₃
(c) (CH₃)₃CCl, C₆H₅C(CH₃)₂Cl, (CH₃)₂CHCl, CH₃CH₂CH₂Cl
Answer:
(a) CH₃C(Cl) = CHCH₂CH₃ < CH₃CH = CHCH₂CH₂Cl < ClCH₂CH = CHCH₂CH₃
(b) CH₂ = CHBr < CH₃CH₂Br < CH₃CH(Br)CH₃ < CH₂ = CHCH(Br)CH₃
(c) CH₃CH₂CH₂Cl < (CH₃)₂CHCl < (CH₃)₃CCl < C₆H₅C(CH₃)₂Cl

Teacher's Note:
1. Rank by the carbocation formed: vinylic (does not form) < 1° < 2° < 3°, with allylic and benzylic cations getting extra resonance stability.

Question 41. Explain why : (a) Neopentyl bromide undergoes nucleophilic substitution reaction very slowly.
(b) 3-Bromocyclohexene is more reactive than 4- bromocyclohexene in hydrolysis with aqueous NaOH.
(c) tert-Butyl chloride reacts with aqueous sodium hydroxide by S_N1 mechanism while n-butyl chloride reacts by S_N2 mechanism.
Answer:
(a) Neopentyl bromide, (CH₃)₃C–CH₂Br, is a 1° alkyl halide and hence does not form a stable carbocation; thus it cannot react by S_N1. It cannot react easily by S_N2 either because the bulky tert-butyl group on the β-carbon causes large steric hindrance to the back side attack of the nucleophile. Therefore, it undergoes nucleophilic substitution very slowly.
(b) 3-Bromocyclohexene is an allylic halide. On ionization it gives an allylic carbocation which is stabilized by resonance. 4-Bromocyclohexene gives a 2° carbocation which is not stabilized by resonance. Therefore, 3-bromocyclohexene is more reactive than 4-bromocyclohexene in hydrolysis.
(c) tert-Butyl chloride on ionization gives a stable 3° carbocation and is too hindered for back side attack; hence it reacts by S_N1. n-Butyl chloride would give an unstable 1° carbocation but has very little steric hindrance; hence it reacts by S_N2 mechanism.

Teacher's Note:
1. Neopentyl halides are the classic exception: primary, yet slow by both mechanisms.

Long Answer Questions (5 Marks)

Question 42. (a) Alkyl halides are amongst the most reactive of the organic compounds. Why?
(b) The treatment of alkyl halides with alcoholic KOH leads to the formation of alkenes. Justify. (Pb. Board 2012)
Answer:
(a) Due to greater electronegativity of halogen than carbon, the C–X bond in alkyl halides is polar, the carbon carrying a partial positive charge (C^δ+–X^δ−). Therefore, nucleophiles readily attack this carbon. Further, the C–X bond (except C–F) is weaker than C–H and C–C bonds and the halide ion is a good leaving group. Hence, alkyl halides readily undergo nucleophilic substitution, elimination, reduction and reactions with metals, and are amongst the most reactive organic compounds.
(b) Alcoholic KOH contains alkoxide ions (RO⁻), which are strong bases (KOH + C₂H₅OH ⇌ C₂H₅O⁻K⁺ + H₂O). The strong base removes a β-hydrogen as a proton and simultaneously the halogen leaves as halide ion, forming a double bond (β-elimination or dehydrohalogenation). For example,
CH₃CH₂Br + KOH (alc.) \( \xrightarrow{\Delta} \) CH₂=CH₂ + KBr + H₂O
CH₃CH(Br)CH₂CH₃ + KOH (alc.) \( \xrightarrow{\Delta} \) CH₃CH=CHCH₃ (major, Saytzeff product) + CH₂=CHCH₂CH₃ (minor)
In aqueous KOH, OH⁻ ions are highly hydrated and act mainly as nucleophiles, giving alcohols instead.

Teacher's Note:
1. Remember: aqueous KOH → substitution (alcohol); alcoholic KOH → elimination (alkene).

Question 43. What are haloarenes ? How are they classified ? Give one method each for the preparation of nuclear and side chain substituted haloarenes.
Answer:
Haloarenes : The halogen derivatives of aromatic hydrocarbons are called haloarenes, i.e., compounds in which a halogen atom is attached to an aromatic ring or to its side chain.
Classification : (i) Nuclear halogen derivatives (aryl halides) : the halogen is directly attached to an sp^2 carbon of the benzene ring, e.g., chlorobenzene (C₆H₅Cl), o-chlorotoluene.
(ii) Side chain halogen derivatives (aralkyl halides) : the halogen is attached to an sp^3 carbon of the side chain, e.g., benzyl chloride (C₆H₅CH₂Cl), benzal chloride (C₆H₅CHCl₂), benzotrichloride (C₆H₅CCl₃). Depending on the number of halogen atoms, haloarenes are also classified as mono-, di-, tri- and poly-halogen compounds.
Preparation of nuclear haloarene : By electrophilic substitution in presence of a Lewis acid in dark :
C₆H₆ + Cl₂ \( \xrightarrow{\text{Anhyd. FeCl}_3,\ \text{dark}} \) C₆H₅Cl + HCl
(Also from diazonium salts by Sandmeyer reaction : C₆H₅N₂⁺Cl⁻ \( \xrightarrow{\text{Cu}_2\text{Cl}_2\text{/HCl}} \) C₆H₅Cl + N₂)
Preparation of side chain substituted haloarene : By passing chlorine into boiling toluene in presence of light (free radical substitution) :
C₆H₅CH₃ + Cl₂ \( \xrightarrow{h\nu,\ 383\ \text{K}} \) C₆H₅CH₂Cl (Benzyl chloride) + HCl

Teacher's Note:
1. Same reagents, different conditions: Lewis acid in dark gives ring substitution; light/heat gives side-chain substitution.