CBSE Class 12 Chemistry Haloalkanes and Haloarenes Important Questions Set 05

Very Short Answer Questions (1 Mark)

Question 1. Which alkyl halide has the highest density and why ?
Answer:
CH₃I ; because of its lowest carbon content and heaviest halogen i.e. I.

Teacher's Note:
1. Density rises with the mass of the halogen and falls as the number of carbon atoms increases.

Question 2. Haloalkanes dissolve easily in organic solvents, why ? (CBSE 2011)
Answer:
Haloalkanes dissolve easily in organic solvents of low polarity such as petroleum ether, benzene, diethyl ether, chloroform, carbon tetrachloride, etc. because the new forces of attraction set up between haloalkane and solvent molecules are of the same strength as the forces of attraction being broken, i.e., existing between haloalkane – haloalkane and solvent – solvent molecules.

Teacher's Note:
1. Contrast with water: haloalkanes cannot form H-bonds with water, so the energy released is less than that needed to break water's H-bonds and they are insoluble.

Question 3. Organic halogen compounds used in industry as solvents are chlorides rather than bromides and iodides. Explain.
Answer:
Organic alkyl chlorides are widely used as solvents in industry because chlorides are not only more stable but are also more volatile than bromides and iodides.

Teacher's Note:
1. Chlorides are also cheaper; the stronger C–Cl bond is the reason for their greater stability.

Question 4. Which is a better nucleophile, a bromide ion or an iodide ion ?
Answer:
Because of bigger size and lower electronegativity, iodide ion can donate a pair of electrons more easily than bromide ion and hence iodide ion is a better nucleophile than bromide ion.

Teacher's Note:
1. In protic solvents nucleophilicity of halide ions is I⁻ > Br⁻ > Cl⁻ > F⁻, the reverse of their basicity.

Question 5. Ethyl iodide undergoes S_N2 reaction faster than ethyl bromide. (CBSE 2013)
Answer:
The rate of S_N2 reactions depends upon the leaving ability of the leaving group. Since I⁻ is a better leaving group than Br⁻, therefore, ethyl iodide undergoes S_N2 reaction faster than ethyl bromide.

Teacher's Note:
1. The C–I bond is the weakest carbon–halogen bond, so it breaks most easily in the transition state.

Question 6. What happens when CH₃–Br is treated with KCN ? (CBSE 2013, 2018 S)
Answer:
Ethanenitrile or Acetonitrile is formed.
CH₃–Br (Bromomethane) + KCN \( \xrightarrow{\text{S}_\text{N}2} \) CH₃–CN (Ethanenitrile) + KBr

Teacher's Note:
1. KCN is ionic and attacks through carbon, giving a nitrile (one carbon added); AgCN would give an isocyanide.

Question 7. What happens when ethyl chloride is treated with aqueous KOH ? (CBSE 2013)
Answer:
Ethyl chloride undergoes hydrolysis to form ethyl alcohol.
CH₃CH₂–Cl (Ethyl chloride) + KOH (aq) \( \xrightarrow{\text{S}_\text{N}2} \) CH₃CH₂–OH (Ethyl alcohol) + KCl

Teacher's Note:
1. Aqueous KOH gives substitution (alcohol); alcoholic KOH gives elimination (alkene). Note the solvent in every question.

Question 8. Which compound in each of the following pairs will react faster in S_N2 reaction with OH⁻ ?
(a) CH₃Br or CH₃I
(b) CH₂ = CHBr or CH₂ = CHCH₂Br.
Answer:
(a) Since I⁻ ion is a better leaving group than Br⁻ ion, therefore, iodides are more reactive than bromides. In other words, CH₃I is more reactive than CH₃Br.
(b) Reactivity in S_N2 reactions among other things depends upon the magnitude of the +ve charge on the carbon atom carrying the halogen. Now the sp²⁻carbon atom of the CH₂ group carrying the Br atom in CH₂=CHCH₂Br carries more +ve charge due to –I-effect of the Br atom and the CH₂=CH– group as compared to sp²⁻carbon atom of CH group in CH₂=CHBr due to electron-donating resonance effect.
CH₂=CH–Br: ↔ ⁻:CH₂–CH=Br⁺:
As a result, CH₂ = CHCH₂Br will react faster than CH₂ = CHBr with OH⁻ ions.

Teacher's Note:
1. Vinyl halides are very unreactive towards nucleophiles because resonance gives the C–X bond partial double bond character.

Question 9. In each of the following pairs of compounds, identify the compound which will undergo S_N1 reaction faster ?
(i) (CH₃)₃C–Cl and CH₃CH₂CH(Cl)CH₃
(ii) Chlorobenzene (C₆H₅Cl) and benzyl chloride (C₆H₅CH₂Cl) (CBSE 2018)
Answer:
(i) The reactivity in S_N1 reactions depends upon the stability of the intermediate carbocation which an alkyl halide gives on ionization. Thus,
(CH₃)₃C–Cl (3° Alkyl chloride) \( \xrightarrow{\text{Ionization},\ -\text{Cl}^-} \) (CH₃)₃C⁺ (3° carbocation, more stable)
CH₃CH₂CH(Cl)CH₃ (2° Alkyl chloride) \( \xrightarrow{\text{Ionization},\ -\text{Cl}^-} \) CH₃CH₂C⁺HCH₃ (2° carbocation, less stable)
Since 3° carbocations are more stable than 2° carbocations, therefore, (CH₃)₃C–Cl will react faster than CH₃CH₂CH(Cl)CH₃ in S_N1 reactions.
(ii) Benzyl chloride (C₆H₅CH₂Cl) readily forms benzyl cation (C₆H₅CH₂⁺) which is stabilized by resonance. In contrast, chlorobenzene (C₆H₅Cl) does not form phenyl cation (C₆H₅⁺) because it is highly unstable. Thus, benzyl chloride undergoes S_N1 reaction faster than chlorobenzene.

Teacher's Note:
1. For S_N1 questions always compare the carbocations formed: 3° > 2° > 1°, and benzylic/allylic cations are extra stable due to resonance.

Question 10. Which one of the following reacts faster in an S_N1 reaction and why ? (CBSE 2008)

Answer:
The reactivity in S_N1 reactions depends upon the stability of the intermediate carbocation which an alkyl halide gives on ionization. Thus,
CH₃CH₂CH(Cl)CH₂CH₃ (2° Alkyl halide) \( \xrightarrow{\text{Ionization}} \) CH₃CH₂C⁺HCH₂CH₃ (2° carbocation, more stable) + Cl⁻ ;
CH₃CH₂CH₂CH₂Cl (1° Alkyl halide) \( \xrightarrow{\text{Ionization}} \) CH₃CH₂CH₂CH₂⁺ (1° carbocation, less stable) + Cl⁻
Since 2° carbocation is more stable than 1° carbocation, therefore, CH₃CH₂CH(Cl)CH₂CH₃ (3-chloropentane) will react faster than CH₃CH₂CH₂CH₂Cl (1-chlorobutane) in S_N1 reaction.

Teacher's Note:
1. Identify each halide as 1°, 2° or 3° from the skeletal formula first: count the carbons attached to the C bearing Cl.

Question 11. A solution of KOH hydrolyses CH₃CHClCH₂CH₃ and CH₃CH₂CH₂CH₂Cl. Which one of these is more easily hydrolysed ? (CBSE 2010, 2012)
Answer:
Under aqueous conditions, substitution occurs by S_N1 mechanism since carbocation intermediates are stabilized by solvation. Further, the reactivity in S_N1 reactions depends upon the stability of carbocations.
CH₃–CHCl–CH₂CH₃ \( \xrightarrow{\text{Ionization}} \) CH₃–C⁺H–CH₂CH₃ + Cl⁻ (2° carbocation, more stable)
CH₃CH₂CH₂CH₂–Cl \( \xrightarrow{\text{Ionization}} \) CH₃CH₂CH₂C⁺H₂ + Cl⁻ (1° carbocation, less stable)
Now since CH₃CHClCH₂CH₃ upon ionization gives the more stable 2° carbocation while CH₃CH₂CH₂CH₂Cl on ionization gives the less stable 1° carbocation, therefore, CH₃CHClCH₂CH₃ hydrolyses faster than CH₃CH₂CH₂CH₂Cl.

Teacher's Note:
1. Polar protic solvents such as water favour S_N1 by solvating both the carbocation and the leaving halide ion.

Question 12. In the pair, (CH₃)₃CCl and CH₃Cl, CH₃Cl will react faster in S_N2 reaction with OH⁻. Explain. (CBSE 2010 S)
Answer:
In S_N2 reactions, reactivity depends upon the steric hindrance. In (CH₃)₃CCl, the α-carbon carrying the Cl atom is attached to three CH₃ groups but in CH₃–Cl, the carbon carrying the Cl atom is attached to three hydrogen atoms. Since CH₃ group is much bigger than H-atoms, therefore, (CH₃)₃CCl suffers greater steric hindrance to nucleophilic attack by OH⁻ than CH₃Cl and hence CH₃Cl will react faster than (CH₃)₃CCl in S_N2 reaction with OH⁻.
CH₃–Cl, Methyl chloride (carbon carrying Cl atom is attached to three small H-atoms, no steric hindrance, more reactive)
(CH₃)₃C–Cl, tert-Butyl chloride (carbon carrying Cl atom is attached to three bigger CH₃ groups, more steric hindrance, less reactive)

Teacher's Note:
1. S_N2 reactivity order: CH₃X > 1° > 2° > 3°, because the nucleophile must reach the carbon from the side opposite to the halogen.

Question 13. Which one undergoes S_N2 substitution reaction faster and why? (CBSE 2012)

Answer:
Since I⁻ ion is a better leaving group than Cl⁻ ion, therefore, the iodo compound (first compound) reacts faster than the chloro compound (second compound) in S_N2 substitution reactions.

Teacher's Note:
1. When the carbon skeleton is identical, compare only the leaving groups: R–I > R–Br > R–Cl.

Question 14. Of the two bromoderivatives C₆H₅CH(CH₃)Br and C₆H₅CH(C₆H₅)Br, which one is more reactive in S_N1 substitution reaction and why? (CBSE 2011)
Answer:
In S_N1 reactions, reactivity depends upon the stability of the intermediate carbocation. Thus,
C₆H₅–CH(Br)–CH₃ \( \xrightarrow{\text{Ionization},\ -\text{Br}^-} \) C₆H₅–C⁺H–CH₃ (2° carbocation (I), less stable)
C₆H₅–CH(Br)–C₆H₅ \( \xrightarrow{\text{Ionization},\ -\text{Br}^-} \) C₆H₅–C⁺H–C₆H₅ (2° carbocation (II), more stable)
The carbocation (II) derived from C₆H₅CH(C₆H₅)Br is stabilized by +R-effect of the two C₆H₅ groups but the carbocation (I) derived from C₆H₅CH(CH₃)Br is stabilized by +R-effect of one C₆H₅ group and +I-effect of the CH₃ group. Since +R-effect is much stronger than +I-effect, therefore, carbocation (II) is more stable than carbocation (I). Thus, C₆H₅CH(C₆H₅)Br is more reactive than C₆H₅CH(CH₃)Br in S_N1 substitution reaction.

Teacher's Note:
1. Each extra phenyl group on the cationic carbon spreads the positive charge over another ring, so diphenylmethyl cations are much more stable than 1-phenylethyl cations.

Question 15. Which compound in each of the following pairs will react faster by S_N1 reactions in aqueous KOH. Give reasons.
(a) CH₃CH₂–Br or CH₃CH₂–Cl
(b) CH₃CH₂CH₂CH₂–Br or (CH₃)₃C–Br (Raj. Board 2012, CBSE 2019)
Answer:
(a) S_N1 reactions occur by carbocation intermediates. Since, CH₃CH₂Br or CH₃CH₂Cl give the same carbocation, CH₃CH₂⁺, therefore, stability of the carbocation does not affect the rate of the reaction. Further, since Br⁻ ion is a better leaving group than Cl⁻ ion, therefore, CH₃CH₂–Br will react faster than CH₃CH₂–Cl with aq. KOH by S_N1 mechanism.
(b) S_N1 reactions occur by carbocation intermediates
CH₃CH₂CH₂CH₂–Br \( \xrightarrow{\text{Ionization}} \) CH₃CH₂CH₂C⁺H₂ (1° carbocation) + Br⁻
(CH₃)₃C–Br \( \xrightarrow{\text{Ionization}} \) (CH₃)₃C⁺ (3° carbocation) + Br⁻
Since, 3° carbocation is more stable than 1° carbocation, therefore, (CH₃)₃C–Br will react faster than CH₃CH₂CH₂CH₂–Br with aq. KOH by S_N1 mechanism.

Teacher's Note:
1. If the carbocation is the same, the leaving group decides the rate; if the leaving group is the same, carbocation stability decides it.

Question 16. What is the reagent used for dehydrohalogenation of an alkyl halide? (Karnataka Board 2012)
Answer:
Alcoholic KOH.
For example,
CH₃–CH₂–Br (Bromoethane) \( \xrightarrow{\text{Alc KOH, }\Delta\text{}} \) CH₂=CH₂ (Ethene) + KBr + H₂O   (Dehydrohalogenation)

Teacher's Note:
1. In alcoholic KOH the stronger base RO⁻ (alkoxide) is present, which favours removal of a β-hydrogen (elimination) over substitution.

Short Answer Questions (2 Marks)

Question 17. Predict the order of reactivity of the following compounds in S_N1 reactions:
chlorocyclohexane, 1-chloro-1-methylcyclohexane, 1-bromo-1-methylcyclohexane, 1-iodo-1-methylcyclohexane
Answer:
The first compound is a 2° alkyl halide while all others are 3° alkyl halides. Since 3° alkyl halides are more reactive than 2° alkyl halides in S_N1 reactions, therefore, first compound is the least reactive.
Further, reactivity increases in the order: chloride < bromide < iodide. Thus, the increasing order of reactivity in S_N1 reactions is the same in which they are listed above:
chlorocyclohexane < 1-chloro-1-methylcyclohexane < 1-bromo-1-methylcyclohexane < 1-iodo-1-methylcyclohexane

Teacher's Note:
1. For S_N1, compare the carbocation first (3° > 2°); when the carbocation is the same, the better leaving group (I > Br > Cl) decides.

Question 18. Write down the structure of the product of the following reaction.
(CH₃)₃C–Br →(C₂H₅ONa, C₂H₅OH) ?
Answer:
3° Alkyl halides on treatment with a strong base undergo dehydrohalogenation to form alkenes, i.e.,
CH₃–C(CH₃)₂–Br + C₂H₅O⁻Na⁺ → CH₃–C(CH₃)=CH₂ (2-methylpropene) + C₂H₅OH + Na⁺Br⁻

Teacher's Note:
1. With a strong, bulky-enough base, a 3° halide gives elimination, not the ether; to make tert-butyl ethyl ether use (CH₃)₃CONa + C₂H₅Br.

Question 19. Differentiate between chiral and achiral molecules.
Answer:
A chiral molecule is one that is not superimposable on its mirror image. These molecules contain one asymmetric or chiral carbon atom. For example, butan-2-ol.
An achiral molecule, on the other hand, is one which is superimposable on its mirror image. For example, propan-2-ol.

Teacher's Note:
1. Quick test: a carbon with four different groups (C-2 of butan-2-ol) makes the molecule chiral; in propan-2-ol C-2 carries two identical CH₃ groups.

Question 20. Identify the chiral molecule in the following pair:
(CH₃)₂CH–CH(OH)–CH₃ and (CH₃)₂CH–CH(OH)–CH(CH₃)₂ (CBSE 2018)
Answer:
(CH₃)₂CH–C*H(OH)–CH₃ has a chiral centre because it is attached to four different atoms/groups and hence is a chiral molecule.
(CH₃)₂CH–CH(OH)–CH(CH₃)₂ is a symmetrical molecule since C₃ is attached to two similar groups and hence this molecule is achiral.

Teacher's Note:
1. Check each carbon carrying OH: in the second molecule both sides of C-3 are isopropyl groups, so it cannot be a stereocentre.

Question 21. Identify the compound that on hydrogenation produces an optically active compound from the following compounds:
(A) CH₂=CH–C*HBr–CH₂CH₃ (one enantiomer of 3-bromopent-1-ene)
(B) CH₃–CH=CH–C*HBr–CH₃ (one enantiomer of 4-bromopent-2-ene)
Answer:
CH₂=CH–C*HBr–CH₂CH₃ [optically active (A)] →(H₂/Ni, hydrogenation) CH₃CH₂–CHBr–CH₂CH₃ (has a plane of symmetry, no chiral centre; optically inactive)
CH₃–CH=CH–C*HBr–CH₃ [optically active (B)] →(H₂/Ni, hydrogenation) CH₃CH₂CH₂–C*HBr–CH₃ (contains a chiral centre; optically active)
Thus, compound (B) on hydrogenation produces optically active compound.

Teacher's Note:
1. Hydrogenation does not touch the C–Br stereocentre; the question is whether that carbon still has four different groups after the C=C becomes an ethyl or propyl group.

Question 22. What is meant by chirality of a compound. Give one example. (CBSE 2012)
Answer:
A compound is said to be chiral if it exists in two stereoisomeric forms which are non-superimposable mirror images of each other. This property of non-superimposability is called chirality and the two compounds which are non-superimposable mirror images of each other are called enantiomers.

For example, butan-2-ol is a chiral compound because it exists in two stereoisomers which are non-superimposable mirror images of each other.

Teacher's Note:
1. Your example must have a carbon with four different groups; butan-2-ol (H, OH, CH₃, C₂H₅) is the standard one.

Question 23. What are enantiomers? Draw the structures of the possible enantiomers of 3-methylpent-1-ene.
Answer:
Stereoisomers which are non-superimposable mirror images of each other are called enantiomers. The enantiomers of 3-methylpent-1-ene are:

Teacher's Note:
1. When drawing enantiomers, keep two groups in the same place and draw the exact mirror image across a dashed mirror line.

Question 24. Distinguish between enantiomers and diastereomers. (Tamil Nadu Board 2012)
Answer:
Stereoisomers which are non-superimposable mirror images of each other are called enantiomers while stereoisomers which are not mirror images of each other are called diastereomers. For example, tartaric acid exists in three stereoisomeric forms I, II and III

Structures I and II are non-superimposable mirror images of each other and hence are enantiomers. But structures I and III (or II and III) are not mirror images and hence are diastereomers.

Teacher's Note:
1. Enantiomers have identical physical properties (except sign of rotation); diastereomers differ in m.p., b.p. and solubility.

Question 25. (±)-2-Butanol is optically inactive. Give reasons. (CBSE 2013, 2022)
Answer:
(±)-2-Butanol contains equal amounts of the two enantiomers, i.e., (+)-2-butanol and (−)-2-butanol. Whereas (+)-2-butanol rotates the plane of polarized light toward right, (−)-2-butanol rotates the plane of polarized light towards left but to the same extent. The overall rotation is zero and hence (±)-2-butanol is optically inactive.

Teacher's Note:
1. Use the term “racemic mixture” and “external compensation” in the answer; that is what examiners look for.

Question 26. Optically active 2-iodobutane on treatment with NaI in acetone gives a product which does not show optical activity. Explain.
Answer:
Optically active 2-iodobutane on treatment with NaI in acetone undergoes racemization and hence the product does not show optical activity as explained below:
I⁻ + H–C(CH₃)(CH₂CH₃)–I [2-iodobutane (I), say dextrorotatory] ⇌ I–C(CH₃)(CH₂CH₃)–H [2-iodobutane (II), laevorotatory] + I⁻
When optically active 2-iodobutane (I) is treated with I⁻ ion, it undergoes Walden inversion (S_N2 reaction) to give 2-iodobutane (II) which is the enantiomer of (I). Now 2-iodobutane (II) undergoes Walden inversion to give enantiomer (I). As a result of repetition of these two Walden inversions, ultimately a 50 : 50 mixture of the two enantiomers is obtained. In other words, optically active 2-iodobutane undergoes racemization.

Teacher's Note:
1. Here the nucleophile and the leaving group are the same (I⁻), so each S_N2 step simply converts one enantiomer into the other until a racemic mixture results.

Question 27. Which alkyl halide from the following pair would you expect to react more rapidly by an S_N2 mechanism?
CH₃CH₂–CH(CH₃)–Br or CH₃–C(CH₃)₂–Br (CBSE 2019)
Answer:
Reactivity in S_N2 reactions depends upon the steric hindrance at α-carbon to which the halogen is attached. Since the steric hindrance increases in the order 1° < 2° < 3°, alkyl halides, therefore, reactivity increases in the reverse order: 1° > 2° > 3° alkyl halides.
In other words CH₃CH₂–CHBr–CH₃ being 2° alkyl bromide reacts faster than the 3° alkyl bromide, (CH₃)₃C–Br.

Teacher's Note:
1. Both compounds are isomers (C₄H₉Br), so only the degree of the carbon bearing Br decides the S_N2 rate.

Question 28. Which one of the following will undergo faster hydrolysis reaction by S_N1 mechanism?
C₆H₅CH₂Cl or CH₃CH₂CH₂Cl
Justify your answer. (CBSE Sample Paper 2019-20)
Answer:
C₆H₅–CH₂Cl [benzyl chloride (I)] →(ionization, −Cl⁻) C₆H₅–C⁺H₂ [benzyl cation, stabilized by strong resonance effect]
CH₃CH₂CH₂–Cl →(ionization, −Cl⁻) CH₃CH₂–C⁺H₂ [n-propyl cation, stabilized by weak +I-effect of CH₃CH₂ group]
Now benzyl chloride gives benzyl cation which is stabilized by strong resonance effect while n-propyl chloride is stabilized by weak +I effect of the CH₃CH₂ group. As a result, benzyl cation is more stable than n-propyl cation, therefore, benzyl chloride undergoes hydrolysis much faster than n-propyl chloride.

Teacher's Note:
1. Benzylic and allylic halides behave like 3° halides in S_N1 even though they are 1°, because their carbocations are resonance stabilised.

Question 29. Out of the various possible isomers of C₇H₇Cl containing a benzene ring, suggest the structure with the weakest C–Cl bond.
Answer:
Four isomers are possible; their structures are: (i) C₆H₅CH₂Cl (benzyl chloride), (ii) o-chlorotoluene, (iii) m-chlorotoluene, (iv) p-chlorotoluene (CH₃–C₆H₄–Cl).
Just as in chlorobenzene, the C–Cl bond in o-, m- and p-chlorotoluenes, has some double bond character due to resonance. In contrast, in benzyl chloride, the C–Cl bond is a pure single bond. Hence, out of the four isomers, the C–Cl bond is the weakest in benzyl chloride. Alternatively, the cleavage of C–Cl bond in benzyl chloride gives benzyl cation which is stabilized by resonance.
C₆H₅CH₂–Cl →(ionization, −Cl⁻) C₆H₅CH₂⁺ (benzyl cation): the positive charge is delocalised by resonance from the CH₂ carbon to the ortho and para positions of the ring.
On the other hand, cleavage of C–Cl bond in o-, m- or p-chlorotoluenes gives tolyl cation which is highly unstable. As a result, the energy of activation for the cleavage of C–Cl bond in benzyl chloride is much smaller than that of C–Cl bond in o-, m- or p-chlorotoluenes. Consequently, the C–Cl bond is the weakest in benzyl chloride.

Teacher's Note:
1. Halogen on a side-chain sp³ carbon (benzylic) behaves like an alkyl halide; halogen on the ring carbon behaves like an aryl halide.

Question 30. (i) Complete the following, giving the structures of the principal organic products.
CH₃–C₆H₄–I (p-) + Cu + heat →
(ii) What would be the major products in the following reactions?
p-O₂N–C₆H₄–F →(NaOMe)
Answer:
(i) p-Iodotoluene when heated with copper undergoes Ullmann biaryl synthesis/reaction to give 4,4′-dimethyldiphenyl.
2 p-CH₃–C₆H₄–I (p-iodotoluene) + 2Cu →(Δ) p-CH₃–C₆H₄–C₆H₄–CH₃-p (4,4′-dimethyldiphenyl) + 2CuI
(ii) F is activated towards nucleophilic substitution by the electron-withdrawing effect of the NO₂ group. Therefore, nucleophilic attack by CH₃O⁻ first gives resonance stabilized carbanion (I) which then loses a F⁻ ion to give p-nitroanisole.
p-O₂N–C₆H₄–F + ⁻OCH₃ →(slow) carbanion (I) [C-1 carries F and OCH₃] →(fast, −F⁻) p-O₂N–C₆H₄–OCH₃ (p-nitroanisole)

Teacher's Note:
1. In activated aryl halides fluoride is replaced fastest, because the slow step is the attack of the nucleophile, which the highly electronegative F speeds up.

Short Answer Questions (3 Marks)

Question 31. (i) Which will have a higher boiling point? 1-chloroethane or 2-chloro-2-methylbutane. Give reason.
(ii) p-Chloronitrobenzene undergoes nucleophilic substitution faster than chlorobenzene. Explain giving resonating structures as well.
Answer:
(i) 2-Chloro-2-methylbutane has the higher boiling point. It has a much larger molecular mass and size (C₅H₁₁Cl against C₂H₅Cl), so the van der Waals forces between its molecules are stronger.
(ii) When OH⁻ attacks C-1 (carrying Cl) of p-chloronitrobenzene, a carbanion intermediate is formed. Its negative charge is delocalised over C-2, C-4 and C-6 of the ring; the form with the charge on C-4 (which carries –NO₂) passes it on to the nitro group:
Resonating structures of the intermediate: (I) negative charge on C-2; (II) negative charge on C-4; (III) negative charge on C-6; (IV) C-4=N⁺(O⁻)–O⁻, i.e. the charge on an oxygen of the nitro group.
Structure (IV), with the charge on the electronegative oxygen, gives extra stability to the intermediate. No such structure is possible for chlorobenzene, so p-chloronitrobenzene reacts faster.

Teacher's Note:
1. In the exam, draw the ring with sp³ C-1 carrying both OH and Cl, and show the charge moving to C-2, C-4, C-6 and then onto the NO₂ oxygen.

Question 32. (i) What are ambident nucleophiles? Explain with an example.
(ii) Convert ethyl bromide to diethyl ether.
(iii) What are freons?
Answer:
(i) Nucleophiles which have two nucleophilic sites through which they can attack are called ambident nucleophiles. Example: cyanide ion, [:C≡N:]⁻, can attack through carbon to give alkyl cyanide (R–CN) or through nitrogen to give alkyl isocyanide (R–NC). Nitrite ion (attack through O gives R–ONO, through N gives R–NO₂) is another example.
(ii) Williamson synthesis: C₂H₅Br + C₂H₅ONa → C₂H₅–O–C₂H₅ + NaBr
(iii) Freons are chlorofluorocarbon compounds of methane and ethane, e.g. CCl₂F₂ (freon-12). They are unreactive, non-toxic, non-corrosive and easily liquefiable gases, used as refrigerants and aerosol propellants. In the stratosphere they break down to free radicals that deplete the ozone layer.

Teacher's Note:
1. Freon-12 is CCl₂F₂ – a frequently asked formula.

Question 33. A hydrocarbon ‘A’ (C₄H₈) adds HBr in accordance with Markovnikov’s rule to give compound ‘B’ which on hydrolysis with aqueous alkali forms tertiary alcohol ‘C’ (C₄H₁₀O). Identify A, B and C.
Answer:
A = (CH₃)₂C=CH₂ (2-methylpropene)
B = (CH₃)₃C–Br (2-bromo-2-methylpropane)
C = (CH₃)₃C–OH (2-methylpropan-2-ol)
(CH₃)₂C=CH₂ + HBr → (CH₃)₃CBr;   (CH₃)₃CBr + KOH(aq) → (CH₃)₃COH + KBr

Teacher's Note:
1. The only C₄ tertiary alcohol is tert-butyl alcohol – working backwards from it fixes A and B.

Question 34. (i) Which isomer of C₄H₉Cl will have the lowest boiling point?
(ii) Predict the alkenes that would be formed by dehydrohalogenation with sodium ethoxide in ethanol, and identify the major alkene:
(a) 2-chloro-2-methylbutane
(b) 3-bromo-2,2,3-trimethylpentane
Answer:
(i) (CH₃)₃C–Cl (2-chloro-2-methylpropane), the most branched isomer, has the lowest boiling point, as it has the smallest surface area and weakest van der Waals forces.
(ii) (a) (CH₃)₂CCl–CH₂CH₃ gives (CH₃)₂C=CH–CH₃ (2-methylbut-2-ene, major) and CH₂=C(CH₃)–CH₂CH₃ (2-methylbut-1-ene, minor).
(b) (CH₃)₃C–CBr(CH₃)–CH₂CH₃ gives (CH₃)₃C–C(CH₃)=CH–CH₃ (3,4,4-trimethylpent-2-ene, major) and (CH₃)₃C–C(=CH₂)–CH₂CH₃ (2-ethyl-3,3-dimethylbut-1-ene, minor). C-2 has no hydrogen, so no alkene forms on that side.
The major products follow Saytzeff rule (more substituted alkene).

Teacher's Note:
1. First list every β-carbon that carries a hydrogen; a quaternary β-carbon (no H) cannot take part in elimination.

Question 35. Write the structure of the major product in each of the following:
(i) C₆H₅–CH(CH₃)₂ + Br₂ –[Δ]→
(ii) p-HO–C₆H₄–CH₂CH₂OH + HBr →
(iii) 1-chloro-2,4-dinitrobenzene + C₂H₅ONa –[ethanol, Δ]→
Answer:
(i) C₆H₅–C(Br)(CH₃)₂ (2-bromo-2-phenylpropane) – free radical substitution at the benzylic carbon.
(ii) p-HO–C₆H₄–CH₂CH₂Br – only the alcoholic –OH is replaced; the phenolic –OH does not react with HBr.
(iii) 2,4-(NO₂)₂C₆H₃–OC₂H₅ (1-ethoxy-2,4-dinitrobenzene) – the two nitro groups at ortho and para positions activate Cl for nucleophilic substitution.

Teacher's Note:
1. The phenolic C–O bond has partial double bond character, so phenols, like aryl halides, resist substitution of the group on the ring.

Question 36. Write the main products when:
(i) n-butyl chloride is treated with alcoholic KOH
(ii) 2,4,6-trinitrochlorobenzene is subjected to hydrolysis
(iii) Methyl chloride is treated with AgCN
Answer:
(i) CH₃CH₂CH₂CH₂Cl + KOH(alc) → CH₃CH₂CH=CH₂ (but-1-ene) + KCl + H₂O
(ii) 2,4,6-Trinitrophenol (picric acid) is formed; with three activating nitro groups the hydrolysis occurs even with warm water.
(iii) CH₃Cl + AgCN → CH₃NC (methyl isocyanide) + AgCl

Teacher's Note:
1. Compare: chlorobenzene needs NaOH at 623 K and 300 atm; each o/p nitro group makes the hydrolysis much easier.

Question 37. Explain why chlorination of n-butane in presence of light at 298 K gives a mixture of 72% of 2-chlorobutane and 28% of 1-chlorobutane.
Answer:
According to the question,
CH₃CH₂CH₂CH₃ (n-butane) →(Cl₂, 298 K, light) CH₃–CHCl–CH₂CH₃ (2-chlorobutane, 72%) + CH₃CH₂CH₂CH₂–Cl (1-chlorobutane, 28%)
The relative ratios of these two isomeric chlorobutanes can be easily calculated by knowing: (i) the number and type of hydrogens (i.e., 1°, 2° or 3°) to be substituted and (ii) their relative rates of substitution (i.e., 1 : 3.8 : 5.0 for Cl₂ at 298 K). Thus,
\( \frac{\text{1-Chlorobutane}}{\text{2-Chlorobutane}} = \frac{\text{No. of 1° H}}{\text{No. of 2° H}} \times \frac{\text{Reactivity of 1° H}}{\text{Reactivity of 2° H}} \)
\( = \frac{6}{4} \times \frac{1}{3.8} = \frac{6}{15.2} \) = 28% : 72%

Teacher's Note:
1. Product ratio = (number of H of each type) × (relative reactivity); here 6 × 1 = 6 and 4 × 3.8 = 15.2, i.e., 6/21.2 = 28%.

Question 38. Explain the following in one or two sentences
(i) Displacement of cyanide and amide ion is never observed in nucleophilic substitution reactions.
(ii) RCl is hydrolysed to ROH slowly but the reaction is rapid if a catalytic amount of KI is added to the reaction mixture.
Answer:
(i) HCN (pK_a 10) and NH₃ (pK_a about 38) are very weak acids. Therefore, their conjugate bases, i.e., CN⁻ ion and NH₂⁻ are very strong bases. Since strong bases are bad leaving groups, their displacement in nucleophilic substitution reactions is never observed.
(ii) Iodide ion is a powerful nucleophile and hence reacts rapidly with RCl to form RI.
KI → K⁺ + I⁻; I⁻ + R–Cl → R–I + Cl⁻
Further because I⁻ ion is a better leaving group than Cl⁻ ion, therefore, RI is more rapidly hydrolysed than RCl to form ROH,
HO⁻ + R–I → R–OH + I⁻
The I⁻ ion thus regenerated recycles in the above reaction thereby explaining its catalytic effect.

Teacher's Note:
1. Good leaving groups are weak bases (I⁻, Br⁻, TsO⁻); strong bases such as OH⁻, NH₂⁻ and CN⁻ are poor leaving groups.

Question 39. Write down the structure of the product of the following reactions:
(i) 3-Methyl-1-butene + HBr → ?
(ii) (E)-1-Phenylpropene, Ph(H)C=C(H)CH₃ + HBr → Product
(iii) 1-Methylcyclohexene + HBr →(Peroxide)
Answer:
(i) Addition of H⁺ to 3-methyl-1-butene first gives 2° carbocation (I) which being less stable rearranges to the more stable 3° carbocation (II). Nucleophilic attack by Br⁻ ion on this carbocation gives 2-bromo-2-methylbutane as the major product,
CH₃–CH(CH₃)–CH=CH₂ →(H⁺) CH₃–CH(CH₃)–C⁺H–CH₃ [2° carbocation (I), less stable] →(1,2-hydride shift) CH₃–C⁺(CH₃)–CH₂–CH₃ [3° carbocation (II), more stable] →(Br⁻) CH₃–CBr(CH₃)–CH₂CH₃ (2-bromo-2-methylbutane)
(ii) Electrophilic addition reaction occurs. In the first step, H⁺ adds to that carbon atom of the double bond which carries the CH₃ group since it produces carbocation (I) which is stabilized by +R-effect of the Ph group and +I-effect of the CH₂CH₃ group. Subsequent nucleophilic attack by Br⁻ on carbocation (I) gives 1-bromo-1-phenylpropane (II) as the final product.
Ph–CH=CH–CH₃ →(H⁺, slow) Ph–C⁺H–CH₂CH₃ (I) →(Br⁻) Ph–CHBr–CH₂CH₃ [1-bromo-1-phenylpropane (II)]
(iii) In presence of peroxides, addition of HBr to alkenes occurs anti to Markovnikov’s rule. Thus,
1-Methylcyclohexene →(HBr/Peroxide, anti-Markovnikov’s addition) 1-bromo-2-methylcyclohexane (Br on the ring carbon next to the one carrying CH₃)

Teacher's Note:
1. In ionic HBr addition always check whether the first carbocation can become more stable by a hydride or methyl shift; in peroxide addition no carbocation forms, so there is no rearrangement.

Question 40. Which reaction in each pair shown below will show the faster rate of disappearance of starting material?
(a) (i) (CH₃)₃CBr →(EtOH/H₂O)   (ii) (CH₃)₃CBr →(EtOH)
(b) (i) CH₃CH₂CH₂Br →(CN⁻)   (ii) CH₃CH₂CH₂I →(CN⁻)
(c) (i) C₆H₅CH₂Br →(H₂O)   (ii) C₆H₅Br →(H₂O)
(d) (i) (CH₃)₃CCH₂Cl →(CH₃S⁻)   (ii) (CH₃)₃CCH₂Cl →((CH₃)₂CHS⁻)
Answer:
(a) EtOH–H₂O being more polar than EtOH facilitates ionization of (CH₃)₃CBr and hence reaction (i) occurs faster than reaction (ii).
(b) I⁻ is a better leaving group than Br⁻ and hence reaction (ii) proceeds faster than reaction (i).
(c) In protic solvent H₂O, C₆H₅CH₂Br ionizes to form benzyl cation while C₆H₅Br does not ionize to form phenyl cation and hence reaction (i) occurs faster than reaction (ii).
C₆H₅CH₂Br →(H₂O) C₆H₅CH₂⁺ + Br⁻; C₆H₅Br ⇏(H₂O) C₆H₅⁺ + Br⁻ (does not occur)
(d) (CH₃)₂CHS⁻ being bigger in size than CH₃S⁻ causes steric hindrance and hence reaction (i) proceeds faster than reaction (ii).

Teacher's Note:
1. Decide first whether the substrate reacts by S_N1 (solvent polarity matters) or S_N2 (nucleophile size and leaving group matter).

Question 41. The following reactions involve reactions of an alkyl bromide with two possible reagents. Which reaction in each pair will show the faster rate of disappearance of the alkyl bromide?
(a) CH₃Br + (i) CH₃O⁻ (assume protic solvent) (ii) CH₃S⁻ (assume protic solvent) (iii) The reaction rates would be the same
(b) (CH₃)₃CBr + (i) EtOH in H₂O (ii) EtOH (iii) The reaction rates would be the same
(c) C₆H₅C(CH₃)₂Br + (i) CN⁻ (ii) HS⁻ (iii) The reaction rates would be the same
Answer:
(a) The rate of reaction (ii) will be higher than that of reaction (i) since CH₃S⁻ is a better nucleophile than CH₃O⁻.
(b) EtOH–H₂O being more polar than EtOH facilitates ionization of (CH₃)₃CBr and hence rate of reaction (i) is higher than that of reaction (ii).
(c) The starting material undergoes ionization instantaneously since it produces a highly stable carbocation (I).
C₆H₅C(CH₃)₂Br →(ionization) C₆H₅C⁺(CH₃)₂ [highly stable carbocation (I)] + Br⁻
Now we know that in S_N1 reactions, the rate of the reaction depends upon the ionization of the alkyl halide. Since in the present case, this step is fast because it produces a highly stable carbocation, therefore, the strength of the nucleophile has no effect on the overall rate of the reaction. Therefore, the rates of the two reactions would be the same.

Teacher's Note:
1. In S_N1 the nucleophile is not in the rate law, so changing the nucleophile does not change the rate; in S_N2 it does.

Long Answer Questions (5 Marks)

Question 42. (i) A primary alkyl halide (A), C₄H₉Br, reacted with hot alcoholic KOH to give compound (B). Compound (B) reacted with HBr to give (C), which is an isomer of (A). When (A) was reacted with sodium metal, it gave a compound (D), C₈H₁₈, which was different from the compound formed when n-butyl bromide was reacted with sodium. Give the structural formula of (A) and write equations of all the reactions.
(ii) Iodoform gives a precipitate with AgNO₃ on heating while CHCl₃ does not. Why?
Answer:
(i) A = (CH₃)₂CH–CH₂Br (1-bromo-2-methylpropane, a primary halide)
(CH₃)₂CH–CH₂Br + KOH(alc) –[Δ]→ (CH₃)₂C=CH₂ (B) + KBr + H₂O
(CH₃)₂C=CH₂ + HBr → (CH₃)₃C–Br (C, 2-bromo-2-methylpropane, an isomer of A)
2(CH₃)₂CH–CH₂Br + 2Na –[dry ether]→ (CH₃)₂CH–CH₂–CH₂–CH(CH₃)₂ (D, 2,5-dimethylhexane) + 2NaBr
n-Butyl bromide with sodium gives n-octane, CH₃(CH₂)₆CH₃, which is different from D.
(ii) The C–I bond in iodoform is much weaker than the C–Cl bond in chloroform. On heating, iodoform releases iodide ions which give a yellow precipitate of AgI with AgNO₃. The strong C–Cl bonds of CHCl₃ do not break, so no AgCl precipitate forms.

Teacher's Note:
1. A primary C₄H₉Br whose Wurtz product differs from n-octane must be branched – that leaves only isobutyl bromide.

Question 43. Give the preparation of the alkyl halide by the reaction of (i) HCl and (ii) PCl₅ on ethanol and give its reaction with :
(a) Aq.KOH,
(b) AgCN (Hr. Board 2011)
Answer:
Preparation of ethyl chloride :
(i) With HCl : Ethanol reacts with dry HCl gas in presence of anhydrous ZnCl₂ (Groove's process) to form ethyl chloride :
C₂H₅OH + HCl \( \xrightarrow{\text{Anhyd. ZnCl}_2,\ \Delta} \) C₂H₅Cl + H₂O
(ii) With PCl₅ : C₂H₅OH + PCl₅ → C₂H₅Cl + POCl₃ + HCl
Reactions of ethyl chloride :
(a) With aq. KOH : It undergoes hydrolysis (nucleophilic substitution) to form ethanol : C₂H₅Cl + KOH (aq) → C₂H₅OH + KCl
(b) With AgCN : It forms ethyl isocyanide as the main product, since AgCN is covalent and attack occurs through the N-atom : C₂H₅Cl + AgCN → C₂H₅–N⁺≡C⁻ + AgCl

Teacher's Note:
1. ZnCl₂ is a Lewis acid that helps the OH group leave; 1° alcohols need it, while 3° alcohols react with conc. HCl alone.