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Haloalkanes and Haloarenes: Class 12 Chemistry Practice Questions

30 original exam-pattern questions with full answers, matched to the current CBSE Class 12 paper design, including case-based questions. Attempt each question before opening the answer — or start a free 14-day trial ↓ for the full bank.

Q1Case-based4 marks

A forensic chemistry student is analysing two colourless liquid samples labelled P and Q. Sample P is 1-bromobutane and Sample Q is 2-bromobutane. Both samples are treated separately with aqueous NaOH under identical conditions. The student observes that the two samples undergo hydrolysis at different rates and give products with different stereochemical outcomes.

A forensic chemistry student is analysing two colourless liquid samples labelled P and Q. Sample P is 1-bromobutane and Sample Q is 2-bromobutane. Both samples are treated separately with aqueous NaOH under identical conditions. The student observes that the two samples undergo hydrolysis at different rates and give products with different stereochemical outcomes.

(a) Which sample, P or Q, undergoes hydrolysis faster? Give ONE reason for your answer. (2 marks)

(b) The hydrolysis of Sample Q with aqueous NaOH proceeds with Walden inversion. Identify the mechanism involved and justify why Walden inversion is NOT observed when Sample Q is instead heated with aqueous AgNO₃. (2 marks)

Show answer
MARKING SCHEME

(a) Sample P (1-bromobutane) undergoes hydrolysis FASTER. [½ mark for identifying P]

Reason: 1-Bromobutane is a primary alkyl halide. It undergoes hydrolysis by the SN2 mechanism, in which the nucleophile (OH⁻) attacks the carbon bearing the leaving group directly. The approach of OH⁻ to the less hindered primary carbon is sterically unhindered, so the reaction proceeds with a high rate. 2-Bromobutane is a secondary alkyl halide with greater steric hindrance around the C–Br carbon, which slows the backside attack by the nucleophile. [1½ marks for the correct reason: primary → less steric hindrance → faster SN2]

∴ P (1-bromobutane) is hydrolysed faster than Q (2-bromobutane).

(b) Mechanism involved in the hydrolysis of Sample Q (2-bromobutane) with aqueous NaOH: SN2 (bimolecular nucleophilic substitution). [½ mark]

In the SN2 mechanism, OH⁻ attacks the back of the C–Br bond in a single concerted step, inverting the configuration at the chiral carbon (Walden inversion) — the product has the opposite configuration to the reactant. [½ mark for the explanation of inversion]

When Sample Q is heated with aqueous AgNO₃, the reaction proceeds by the SN1 mechanism. [½ mark]

Ag⁺ assists the ionisation of the C–Br bond by precipitating AgBr↓, generating a planar carbocation intermediate. The nucleophile (H₂O) can attack this planar carbocation from either face with equal probability, giving a racemic mixture of products. Since attack occurs from both sides, there is no net inversion — Walden inversion is NOT observed. [½ mark for: planar carbocation → attack from both faces → racemisation, not inversion]
Q2Case-based4 marks

Priya is a chemistry student working in a laboratory. She has three unlabelled bottles, each containing one of the following compounds: 1-bromopropane (n-propyl bromide), 2-bromopropane (isopropyl bromide), and 2-bromo-2-methylpropane (tert-butyl bromide). She needs to identify each compound using their reactivity with aqueous AgNO₃ solution. She also knows that these three compounds undergo nucleophilic substitution by different mechanisms, which affects the rate of the reaction and the stereochemical outcome.

Priya is a chemistry student working in a laboratory. She has three unlabelled bottles, each containing one of the following compounds: 1-bromopropane (n-propyl bromide), 2-bromopropane (isopropyl bromide), and 2-bromo-2-methylpropane (tert-butyl bromide). She needs to identify each compound using their reactivity with aqueous AgNO₃ solution. She also recalls from her textbook that these three compounds undergo nucleophilic substitution by different mechanisms.

(a) Arrange the three compounds in decreasing order of their rate of reaction with aqueous AgNO₃ solution. Give one reason for the order you stated. (2 marks)

(b) Identify the mechanism (SN1 or SN2) by which 1-bromopropane reacts with aqueous NaOH, and state the stereochemical outcome of this reaction. (1 mark)

(c) Priya notices that when 2-bromo-2-methylpropane is treated with aqueous NaOH, the reaction rate is independent of the concentration of NaOH. Explain why this is so. (1 mark)

Show answer
(a) Decreasing order of rate of reaction with aqueous AgNO₃:

2-bromo-2-methylpropane > 2-bromopropane > 1-bromopropane

Reason: Aqueous AgNO₃ reacts via an SN1 pathway, which proceeds through carbocation formation. The rate depends on the stability of the carbocation intermediate. Due to the greater number of alkyl groups attached to the carbon bearing the leaving group, the carbocation stability increases as: 3° > 2° > 1°. Hence, 2-bromo-2-methylpropane (3° substrate) forms the most stable tertiary carbocation, reacts fastest, and gives an immediate precipitate of AgBr; 2-bromopropane (2°) reacts at an intermediate rate; and 1-bromopropane (1°) reacts slowest.

(Value points: correct order ½ + reason based on carbocation stability/class of substrate 1 + supporting conclusion ½ = 2 marks)

(b) 1-bromopropane undergoes nucleophilic substitution by the SN2 mechanism.

Stereochemical outcome: Inversion of configuration (Walden inversion) occurs at the carbon bearing the bromine atom, as the nucleophile (OH⁻) attacks from the back side, opposite to the leaving group (Br⁻).

(Value points: SN2 mechanism ½ + inversion of configuration / Walden inversion ½ = 1 mark)

(c) 2-bromo-2-methylpropane is a tertiary alkyl halide and reacts via the SN1 mechanism. In SN1, the rate-determining step is the slow, unimolecular ionisation of the substrate to form a tertiary carbocation:

(CH₃)₃CBr → (CH₃)₃C⁺ + Br⁻ (slow, rate-determining step)

Since NaOH (the nucleophile) does not participate in this slow step, the rate of the reaction depends only on the concentration of 2-bromo-2-methylpropane and is independent of the concentration of NaOH.

∴ Rate = k[(CH₃)₃CBr]

(Value point: SN1 mechanism / rate-determining step is unimolecular / NaOH not involved in slow step = 1 mark)
Q3Case-based4 marks

A student in a chemistry laboratory has three unlabelled bottles containing colourless liquids — n-propyl bromide (1-bromopropane), isopropyl bromide (2-bromopropane), and bromobenzene. The student uses chemical tests and reactions to identify each liquid and to study their behaviour.

A student in a chemistry laboratory is working with three unlabelled bottles, each containing a different colourless liquid: n-propyl bromide (1-bromopropane), isopropyl bromide (2-bromopropane), and bromobenzene. To identify each bottle, the student performs a series of chemical tests and reactions.

(a) The student adds each compound to a solution of AgNO₃ in ethanol. Only one compound gives an immediate white precipitate of AgBr. Identify the compound and explain why it reacts fastest. (2 marks)

(b) The student treats n-propyl bromide with alcoholic KOH. Write the balanced chemical equation for the reaction, name the type of reaction and state the major organic product. (1 mark)

(c) The student notes that bromobenzene does NOT undergo the same nucleophilic substitution as the alkyl bromides under ordinary conditions. Give ONE reason for its low reactivity. (1 mark)

Show answer
(a) The compound that gives an immediate white precipitate of AgBr is isopropyl bromide (2-bromopropane).

Reason: When alkyl halides react with AgNO₃ in ethanol, the reaction proceeds via the SN1 mechanism. The rate of SN1 reaction depends on the stability of the carbocation intermediate formed. 2-Bromopropane is a secondary alkyl halide; it forms a secondary carbocation which is more stable than the primary carbocation formed by 1-bromopropane. Therefore, 2-bromopropane reacts faster and gives an immediate precipitate of AgBr.

[Note: Bromobenzene does not react with AgNO₃ under ordinary conditions because the C–Br bond in haloarenes has partial double-bond character due to delocalisation of the lone pair of bromine into the benzene ring, making it very strong. n-Propyl bromide (primary) reacts slowly — a pale precipitate appears only on heating.]

∴ Compound giving immediate white precipitate = isopropyl bromide (2-bromopropane), due to greater stability of the secondary carbocation intermediate in the SN1 pathway.

(b) n-Propyl bromide (1-bromopropane) undergoes elimination (dehydrohalogenation) with alcoholic KOH:

CH₃CH₂CH₂Br + KOH (alcoholic) → CH₃CH=CH₂ + KBr + H₂O

Type of reaction: Elimination (E2 / dehydrohalogenation)
Major organic product: Propene (CH₃CH=CH₂)

(c) Bromobenzene is far less reactive towards nucleophilic substitution than alkyl bromides because the C–Br bond in bromobenzene acquires partial double-bond character. The lone pair of electrons on the bromine atom is delocalised into the benzene ring by resonance, making the C–Br bond shorter and stronger than in alkyl bromides. This prevents the nucleophile from attacking the carbon bearing bromine, so ordinary nucleophilic substitution does not occur under mild conditions.
Q4Case-based4 marks

A forensic chemist is analysing two unknown liquid samples, P and Q, recovered from a crime scene. Both samples have the molecular formula C₄H₉Br. Sample P reacts rapidly with AgNO₃ solution giving an immediate pale yellow precipitate, and when treated with ethanolic KOH it gives a single alkene product. Sample Q reacts slowly with AgNO₃, gives a precipitate only on warming, and when treated with ethanolic KOH it gives a mixture of two alkene products (Saytzeff product as major).

A forensic chemist is analysing two unknown liquid samples, P and Q, recovered from a crime scene. Both samples have the molecular formula C₄H₉Br. Sample P reacts rapidly with AgNO₃ solution giving an immediate pale yellow precipitate, and when treated with ethanolic KOH it gives a single alkene product. Sample Q reacts slowly with AgNO₃, gives a precipitate only on warming, and when treated with ethanolic KOH it gives a mixture of two alkene products (Saytzeff product as major).

(a) Identify samples P and Q, giving the IUPAC name of each. (2 marks)
(b) Explain why sample P gives an immediate precipitate with AgNO₃ while sample Q is slow to react. (1 mark)
(c) Name the reaction by which P and Q are converted to alkenes with ethanolic KOH, and state why Q gives two alkene products while P gives only one. (1 mark)

Show answer
(a) Identification of P and Q — 2 marks

Sample P undergoes rapid SN1 reaction with AgNO₃ (immediate pale yellow precipitate of AgBr) and gives only ONE alkene on elimination — both features indicate a primary alkyl halide with only one type of β-hydrogen, i.e., a straight-chain terminal halide.

∴ Sample P is 1-bromobutane (IUPAC: 1-bromobutane, CH₂BrCH₂CH₂CH₃).

Wait — a primary halide undergoes SN2 preferably and gives a slow reaction with AgNO₃. Correcting: rapid AgNO₃ precipitation indicates SN1 tendency, i.e., tertiary or secondary carbocation stability. P gives ONE alkene, which points to a tertiary halide where all β-H are equivalent.

∴ Sample P is 2-bromo-2-methylpropane (IUPAC: 2-bromo-2-methylpropane), a tertiary halide — carbocation formed is 3°, SN1 is fast, and only one alkene (2-methylpropene) is possible.

∴ Sample Q is 2-bromobutane (IUPAC: 2-bromobutane), a secondary halide — slower SN1/SN2, and two alkene products (but-1-ene and but-2-ene) are formed since β-H are available on two different carbon atoms.

(1 mark for correctly identifying P as 2-bromo-2-methylpropane with IUPAC name; 1 mark for correctly identifying Q as 2-bromobutane with IUPAC name)

(b) Reason for difference in rate of reaction with AgNO₃ — 1 mark

The reaction with AgNO₃ proceeds by the SN1 mechanism, which involves formation of a carbocation intermediate. Sample P (2-bromo-2-methylpropane) is a tertiary alkyl halide — it forms a stable 3° carbocation instantly, so AgBr precipitates immediately. Sample Q (2-bromobutane) is a secondary alkyl halide — it forms a less stable 2° carbocation, so the reaction is slower and precipitation occurs only on warming.

∴ Due to greater stability of the 3° carbocation formed from P compared to the 2° carbocation from Q, P reacts immediately while Q reacts slowly with AgNO₃. (1 mark)

(c) Name of reaction and reason for two alkene products from Q — 1 mark

The reaction is called Dehydrohalogenation (E2 / β-elimination), also known as the Saytzeff elimination.

Sample P (2-bromo-2-methylpropane) has only ONE type of β-carbon (all three methyl groups are equivalent), so only one alkene — 2-methylpropene — is possible.

Sample Q (2-bromobutane) has β-hydrogens on two DIFFERENT adjacent carbons (C-1 and C-3), so elimination can occur in two directions, giving but-1-ene (minor) and but-2-ene (major, Saytzeff product — more substituted, more stable alkene).

∴ Q gives two alkene products because its β-H atoms are present on two non-equivalent carbon atoms, allowing elimination in two directions. (1 mark)
Q5Case-based4 marks

A chemistry teacher demonstrates two reactions in the laboratory. In Reaction I, she treats 1-bromobutane with an aqueous solution of NaOH and observes that the reaction is fast, proceeds through a single step, and the product shows complete inversion of configuration. In Reaction II, she treats 2-bromo-2-methylpropane (tert-butyl bromide) with the same aqueous NaOH solution and observes that the reaction is relatively slower initially but the rate depends only on the concentration of the alkyl halide, not on NaOH. She also notes that the product is a racemic mixture.

Read the following passage and answer the questions that follow:

A chemistry teacher demonstrates two reactions in the laboratory. In Reaction I, she treats 1-bromobutane with an aqueous solution of NaOH and observes that the reaction is fast, proceeds through a single step, and the product shows complete inversion of configuration. In Reaction II, she treats 2-bromo-2-methylpropane (tert-butyl bromide) with the same aqueous NaOH solution and observes that the reaction is relatively slower initially but the rate depends only on the concentration of the alkyl halide, not on NaOH. She also notes that the product is a racemic mixture.

(a) Identify the type of mechanism (SN1 or SN2) operating in Reaction I and Reaction II. Give one reason for each. (2 marks)
(b) Why does Reaction I show complete inversion of configuration while Reaction II gives a racemic mixture? (1 mark)
(c) If the teacher replaces 1-bromobutane in Reaction I with chlorobenzene and uses the same aqueous NaOH under normal laboratory conditions, will the substitution reaction occur readily? Give one reason. (1 mark)

Show answer
(a) Reaction I — SN2 (Bimolecular Nucleophilic Substitution)
Because 1-bromobutane is a primary alkyl halide; primary substrates have less steric hindrance, so the nucleophile (OH⁻) can attack from the back in a single concerted step. The rate = k[CH₃CH₂CH₂CH₂Br][OH⁻], confirming bimolecular kinetics.

Reaction II — SN1 (Unimolecular Nucleophilic Substitution)
Because 2-bromo-2-methylpropane is a tertiary alkyl halide; the three methyl groups stabilise the intermediate carbocation (by hyperconjugation and inductive effect), making ionisation (C–Br bond breaking) the slow, rate-determining step. The rate = k[(CH₃)₃CBr], depending only on the concentration of the alkyl halide.

(b) In Reaction I (SN2), the nucleophile OH⁻ attacks the carbon from the side opposite to the leaving group (backside attack) in a single step. This results in complete inversion of configuration at the chiral centre (Walden inversion) — the spatial arrangement is turned inside-out like an umbrella in a storm.

In Reaction II (SN1), the C–Br bond breaks first to give a planar sp²-hybridised carbocation intermediate. The nucleophile OH⁻ can then attack from either face of the planar carbocation with equal probability, producing equal amounts of both enantiomers — a racemic mixture.

(c) No, the reaction will NOT occur readily under normal laboratory conditions.
Due to: The C–Cl bond in chlorobenzene acquires partial double-bond character through resonance (lone pair on Cl delocalises into the benzene ring), making the C–Cl bond shorter and stronger than in alkyl chlorides. Additionally, the sp²-hybridised ring carbon is more electronegative than an sp³ carbon, making it less susceptible to nucleophilic attack. The π-electron cloud above and below the ring also repels the incoming nucleophile. Hence, nucleophilic substitution of haloarenes requires drastic conditions (high temperature and pressure) and is not feasible under normal laboratory conditions.
Q6Case-based4 marks

A forensic chemist is analysing two colourless liquid samples recovered from a crime scene. Sample P is 2-bromobutane, Sample Q is 2-bromo-2-methylpropane, and Sample R is bromobenzene. The chemist subjects all three to nucleophilic substitution and elimination tests under controlled conditions.

A forensic chemist is analysing two colourless liquid samples recovered from a crime scene. Sample P is identified as 2-bromobutane, and Sample Q is identified as 2-bromo-2-methylpropane. Both samples are subjected to the following tests:

(i) Each sample is treated separately with aqueous KOH under identical conditions. The chemist observes that Sample Q reacts much faster than Sample P.

(ii) When Sample P is treated with alcoholic KOH, a mixture of two alkene products is obtained, of which one is the major product.

(iii) A third sample, Sample R (bromobenzene), is found at the scene. The chemist notes that Sample R is almost completely unreactive towards aqueous KOH under normal conditions.

Answer the following:
(a) Identify the mechanism by which Sample Q reacts with aqueous KOH. Give ONE reason, based on the structure of Sample Q, to justify why this mechanism operates preferentially. (2 marks)
(b) Name the major alkene formed when Sample P reacts with alcoholic KOH. State the rule that governs this outcome. (1 mark)
(c) Give ONE reason why Sample R (bromobenzene) is so much less reactive than Sample P towards nucleophilic substitution. (1 mark)

Show answer
(a) Sample Q (2-bromo-2-methylpropane) undergoes the Sᴺ1 mechanism (unimolecular nucleophilic substitution) with aqueous KOH.

Justification: 2-bromo-2-methylpropane is a tertiary alkyl halide. The central carbon bearing the bromine atom is surrounded by three bulky methyl groups, making backside attack by the nucleophile (OH⁻) sterically impossible. The reaction therefore proceeds through the formation of a stable tertiary carbocation intermediate in the slow, rate-determining step, followed by rapid attack of the nucleophile.

∴ The SN1 mechanism operates because the substrate forms a relatively stable tertiary carbocation intermediate, and steric hindrance prevents direct backside (SN2) attack.

(b) 2-bromobutane undergoes elimination with alcoholic KOH. The two possible alkene products are but-1-ene and but-2-ene. The major product is but-2-ene (the more substituted alkene).

This outcome is governed by Saytzeff’s Rule (Zaitsev’s Rule): In an elimination reaction, the major product is the more substituted (more stable) alkene, i.e., the alkene formed by removal of a hydrogen from the β-carbon that has fewer hydrogen atoms.

(c) Bromobenzene is far less reactive than 2-bromobutane towards nucleophilic substitution due to the partial double-bond character of the C–Br bond in bromobenzene.

In bromobenzene, the lone pair on the bromine atom is delocalised into the benzene π-system, giving the C–Br bond significant double-bond character (the bond is shorter and stronger than a typical C–Br single bond). As a result, the C–Br bond is very difficult to break, and the sp²-hybridised ring carbon is more electronegative and does not carry sufficient partial positive charge to attract a nucleophile. These factors together make nucleophilic substitution extremely difficult under normal conditions.
Q7Case-based4 marks

A chemist in a pharmaceutical laboratory is working with two substrates: (I) 1-bromopropane (a primary alkyl halide) and (II) 2-bromo-2-methylpropane (a tertiary alkyl halide). Both are treated separately with aqueous NaOH. The chemist observes that the two substrates react by different mechanisms and produce alcohols with different stereochemical outcomes.

Read the following passage and answer the sub-parts that follow:

A chemist in a pharmaceutical laboratory is working with two substrates: (I) 1-bromopropane (a primary alkyl halide) and (II) 2-bromo-2-methylpropane (a tertiary alkyl halide). Both are treated separately with aqueous NaOH (a nucleophilic substitution reaction). The chemist observes that the two substrates react by different mechanisms and produce alcohols with different stereochemical outcomes.

(a) Identify the mechanism by which each substrate (I) and (II) reacts with aqueous NaOH. Give one reason for each choice. [2]
(b) What is the stereochemical outcome of the reaction of substrate (I) with aqueous NaOH? Name the phenomenon. [1]
(c) The chemist replaces aqueous NaOH with alcoholic KOH for substrate (II). Name the type of reaction that now occurs and write the structural formula of the organic product formed. [1]

Show answer
(a) Substrate (I) 1-bromopropane reacts by the SN2 mechanism.
Reason: It is a primary alkyl halide — less sterically hindered — so the nucleophile (OH⁻) can attack the carbon from the back in a single concerted step.

Substrate (II) 2-bromo-2-methylpropane reacts by the SN1 mechanism.
Reason: It is a tertiary alkyl halide — the C–Br bond ionises to form a stable 3° carbocation intermediate, which is then attacked by the nucleophile.

(b) The reaction of substrate (I) proceeds with inversion of configuration at the carbon bearing the leaving group — the OH⁻ attacks from the side opposite to the departing Br⁻, pushing the three remaining groups to the other side.
This phenomenon is called Walden inversion (or inversion of configuration).

(c) With alcoholic KOH, substrate (II) undergoes an elimination reaction (E1/dehydrohalogenation).
The organic product formed is 2-methylpropene (isobutylene):

CH₂=C(CH₃)₂

(CH₃)₃CBr + alc. KOH → CH₂=C(CH₃)₂ + KBr + H₂O

∴ The product is 2-methylpropene.
Q8Case-based4 marks

A chemist is working with two alkyl halide compounds — Compound P (2-bromobutane) and Compound Q (bromomethane) — and carries out nucleophilic substitution reactions with aqueous KOH. She also investigates the stereochemical and product outcomes of these reactions.

A chemist in a pharmaceutical laboratory is working with two alkyl halide compounds — Compound P (2-bromobutane) and Compound Q (bromomethane). She needs to carry out nucleophilic substitution reactions with both compounds using aqueous KOH. Based on her observations and knowledge of reaction mechanisms, answer the following:

(a) Identify the mechanism (SN1 or SN2) most likely followed by each compound. Give one reason for your choice in each case. (2 marks)

(b) When optically pure (R)-2-bromobutane undergoes hydrolysis under SN2 conditions, what happens to the optical activity of the product? Name this stereochemical outcome. (1 mark)

(c) Compound Q (bromomethane) reacts with KCN. Write the IUPAC name of the organic product formed and state what type of isomerism is possible if a silver cyanide (AgCN) reagent were used instead. (1 mark)

Show answer
(a) Compound P (2-bromobutane) follows the SN1 mechanism.
Reason: 2-Bromobutane is a secondary alkyl halide; it can form a relatively stable secondary carbocation intermediate, and the reaction proceeds in a stepwise manner via ionisation.

Compound Q (bromomethane) follows the SN2 mechanism.
Reason: Bromomethane is a primary (methyl) alkyl halide with minimum steric hindrance, allowing the nucleophile to attack the carbon from the back in a single concerted step.

(b) When optically pure (R)-2-bromobutane undergoes SN2 hydrolysis, the nucleophile (OH⁻) attacks from the side opposite to the leaving group (Br⁻) — a backside attack. This results in complete inversion of configuration at the chiral centre, giving the (S)-2-butanol as the sole product.
∴ The product shows optical activity but with inverted configuration.
This stereochemical outcome is called Walden Inversion (inversion of configuration).

(c) When bromomethane reacts with KCN:
CH₃Br + KCN → CH₃CN + KBr
KCN is a carbon nucleophile — it attacks through the carbon of CN⁻, giving a nitrile (alkyl cyanide).
IUPAC name of product: Ethanenitrile (CH₃CN).

When AgCN is used instead, it reacts through the nitrogen atom (AgCN is a covalent compound — nitrogen acts as the nucleophile), giving an isocyanide (CH₃NC) — methyl isocyanide.
∴ The two products CH₃CN (ethanenitrile) and CH₃NC (methyl isocyanide) are functional group isomers (or isocyanide isomers — a type of functional isomerism).
Q9MCQ1 mark

Which of the following reactions proceeds via an SN2 mechanism?

Show answer
(B) Hydrolysis of CH₃Br with aqueous NaOH

Explanation: SN2 reactions occur by a single-step backside attack of the nucleophile on the substrate with simultaneous departure of the leaving group. This mechanism is favoured by primary alkyl halides because there is minimal steric hindrance at the carbon bearing the leaving group. CH₃Br (a primary halide) has no bulky substituents, so the hydroxide ion (OH⁻) can readily attack from the back, displacing Br⁻ in one concerted step. (CH₃)₃CBr is a tertiary halide and reacts via SN1 (carbocation intermediate) due to severe steric crowding, and ionisation in polar protic solvent further supports SN1 for the other options.
Q10MCQ1 mark

Which of the following is an example of an allyl halide?

Show answer
(B) CH₂=CH–CH₂–Cl

Explanation: In an allyl halide, the halogen atom is attached to a carbon that is adjacent to (next to) a carbon–carbon double bond (C=C), i.e. the halogen is on an sp³ carbon at the allylic position. CH₂=CH–CH₂–Cl fits this description. Option (A) CH₂=CH–Cl is a vinyl halide (halogen directly on the sp² carbon of the double bond). Option (C) C₆H₅–Cl is an aryl halide. Option (D) CH₃–CH₂–Cl is a simple alkyl halide with no double bond.
Q11MCQ1 mark

Which of the following compounds can undergo SN1 reaction most readily?

Show answer
(B) (CH₃)₃CCl

Explanation: SN1 reaction proceeds through the formation of a carbocation intermediate. The rate of SN1 depends on the stability of the carbocation formed. (CH₃)₃CCl is a tertiary alkyl halide; ionisation gives a tertiary carbocation, (CH₃)₃C⁺, which is the most stable due to hyperconjugation and inductive electron-donation from three methyl groups. The other options — CH₃Cl (methyl), CH₃CH₂Cl (primary) and (CH₃)₂CHCl (secondary) — form less stable carbocations and therefore undergo SN1 far less readily.
Q12MCQ1 mark

Which of the following compounds undergoes SN1 reaction most readily?

Show answer
(B) (CH₃)₃CBr

Explanation: SN1 reaction proceeds via a carbocation intermediate. The rate of SN1 depends on the stability of the carbocation formed. (CH₃)₃CBr is a tertiary alkyl bromide; it ionises to give a tertiary carbocation (CH₃)₃C⁺, which is the most stable due to hyperconjugation and inductive electron-donation by three methyl groups. Greater the stability of the carbocation, faster the SN1 reaction. Primary halides CH₃CH₂CH₂Br and CH₃CH₂Br give primary carbocations (highly unstable) and secondary halide (CH₃)₂CHBr gives a secondary carbocation — both less stable than the tertiary carbocation. ∴ (CH₃)₃CBr reacts most readily by the SN1 mechanism.
Q13MCQ1 mark

Which of the following alkyl halides undergoes SN2 reaction most readily?

Show answer
(D) CH₃Cl

Explanation: In SN2 reactions, the nucleophile attacks the carbon bearing the leaving group from the back side in a single concerted step. Steric hindrance at the carbon centre is the deciding factor — the less hindered the carbon, the faster the SN2 rate. CH₃Cl (methyl chloride) has no substituents on the carbon bearing the leaving group, making back-side attack completely unhindered. As the degree of substitution increases (1° → 2° → 3°), steric crowding increases progressively, slowing the SN2 rate. Hence the reactivity order for SN2 is: CH₃Cl > CH₃CH₂Cl > (CH₃)₂CHCl > (CH₃)₃CCl.
Q14Short Answer1 mark

Assertion (A) : The C–X bond length increases in the order C–F < C–Cl < C–Br < C–I in haloalkanes.
Reason (R) : As the size of the halogen atom increases down the group, the atomic radius increases, resulting in longer C–X bond length.

Show answer
Option (A)

Both A and R are TRUE, and R IS the correct explanation of A.

Explanation: Down Group 17, the atomic radius of the halogen increases in the order F < Cl < Br < I. A larger halogen atom forms a longer bond with carbon because the bonding electrons are held farther from both nuclei. Therefore, the C–X bond length increases in the order C–F < C–Cl < C–Br < C–I, and R correctly accounts for this trend stated in A.
Q15Short Answer2 marks

Identify the major organic products A and B in the following reactions:
(a) CH₃CH₂CH=CH₂ + HBr → A
(b) (CH₃)₃CH + Cl₂ → B (UV light)

Show answer
(a) CH₃CH₂CH=CH₂ + HBr → A

By Markovnikov's rule, the electrophile H⁺ adds to the carbon bearing more hydrogen atoms (C-1), and Br⁻ adds to the more substituted carbon (C-2).

∴ A = CH₃CH₂CHBrCH₃ (2-bromobutane)

(b) (CH₃)₃CH + Cl₂ → B (UV light)

Free-radical halogenation proceeds via a radical intermediate. The tertiary C–H bond is weaker than primary C–H bonds and gives the more stable tertiary radical, so substitution occurs preferentially at the tertiary carbon.

∴ B = (CH₃)₃CCl (2-chloro-2-methylpropane / tert-butyl chloride)
Q16Short Answer3 marks

Account for the following:
(a) tert-Butyl bromide undergoes hydrolysis faster than n-butyl bromide under identical conditions.
(b) Chlorobenzene is much less reactive towards nucleophilic substitution than chlorocyclohexane.
(c) p-Dichlorobenzene has a higher melting point than its ortho and meta isomers.

Show answer
(a) tert-Butyl bromide undergoes hydrolysis faster than n-butyl bromide.

Due to the formation of a stable 3° carbocation intermediate. tert-Butyl bromide undergoes SN1 hydrolysis via a tertiary carbocation [(CH₃)₃C]⁺, which is stabilised by the +I effect of three methyl groups. n-Butyl bromide would require formation of an unstable 1° carbocation, so it reacts by the much slower SN2 pathway. ∴ tert-Butyl bromide hydrolyses faster.

(b) Chlorobenzene is much less reactive towards nucleophilic substitution than chlorocyclohexane.

Due to the following reasons in chlorobenzene: (i) The C–Cl bond acquires partial double-bond character due to delocalisation of the lone pair on Cl into the benzene ring (resonance), making the bond stronger and shorter than a normal C–Cl bond. (ii) The carbon bearing Cl is sp² hybridised and therefore more electronegative, making the C–Cl bond harder to break. (iii) The π-electron cloud above and below the ring repels the incoming nucleophile. In chlorocyclohexane, the sp³ C–Cl bond has no such resonance stabilisation, so nucleophilic substitution proceeds readily.

(c) p-Dichlorobenzene has a higher melting point than the ortho and meta isomers.

Due to the highly symmetric structure of the para isomer, which allows it to pack more efficiently and closely into the crystal lattice. This symmetry leads to stronger intermolecular forces in the solid state and thus a higher lattice energy, resulting in a higher melting point compared to the less symmetric ortho and meta isomers.
Q17Short Answer3 marks

Give reasons for each of the following observations:
(a) Chlorobenzene is much less reactive towards nucleophilic substitution than chlorocyclohexane.
(b) p-Dichlorobenzene has a higher melting point than its ortho and meta isomers.
(c) The rate of SN1 reaction of alkyl halides follows the order: tertiary > secondary > primary.

Show answer
(a) Chlorobenzene is far less reactive than chlorocyclohexane towards nucleophilic substitution due to the following reasons:
(i) The C–Cl bond in chlorobenzene acquires partial double-bond character due to resonance — the lone pair on chlorine is delocalised into the benzene ring, making the C–Cl bond shorter and stronger than a normal C–Cl single bond.
(ii) The carbon bearing the chlorine is sp² hybridised and therefore more electronegative than the sp³ carbon in chlorocyclohexane, making it less susceptible to nucleophilic attack.
(iii) The π-electron cloud above and below the ring repels the incoming nucleophile.
Due to all these factors, nucleophilic substitution in chlorobenzene is extremely difficult.

(b) Among the three dichlorobenzene isomers, the p-isomer has a higher melting point because of its highly symmetrical structure. The symmetrical geometry of p-dichlorobenzene allows its molecules to pack more closely and efficiently in the crystal lattice, resulting in stronger intermolecular forces and a higher lattice energy. The ortho and meta isomers are less symmetrical and therefore pack less efficiently, giving them lower melting points.

(c) The rate of SN1 reaction depends on the ease of formation and stability of the carbocation intermediate. Tertiary carbocations are stabilised by three alkyl groups through hyperconjugation and inductive effect (+I effect), making them most stable and therefore forming fastest. Secondary carbocations are stabilised by two alkyl groups, and primary carbocations by only one alkyl group.
∴ Greater the number of alkyl groups on the carbon bearing the leaving group, the more stable the carbocation formed, and the faster the SN1 reaction.
Hence the order of SN1 reactivity is: tertiary > secondary > primary.
Q18Short Answer3 marks

A forensic chemist is analysing two unknown colourless liquids, P and Q, recovered from a crime scene. Both are known to be structural isomers of molecular formula C₄H₉Br. The following observations are recorded:

• Liquid P reacts with aqueous NaOH at room temperature to give a single optically active alcohol.
• Liquid Q reacts with the same reagent much faster than P and gives a racemic mixture of an alcohol.
• When both P and Q are treated separately with alcoholic KOH, P gives a single alkene product, whereas Q gives two alkene products, of which the major product follows Saytzeff's rule.

On the basis of these observations, answer the following:

(a) Identify liquids P and Q and write their IUPAC names. (2 marks)
(b) Name the mechanism by which Q reacts with aqueous NaOH. Give one reason why Q undergoes this mechanism faster than P. (1 mark)
(c) Write the IUPAC name of the major alkene formed when Q is treated with alcoholic KOH. (1 mark)

Show answer
(a) Identification of P and Q:

Liquid P reacts with aqueous NaOH to give a single optically active alcohol — this indicates P is a chiral, secondary or primary halide that undergoes substitution with inversion (Walden inversion) via SN2, giving a single enantiomeric product. However, P also gives only ONE alkene with alc. KOH, which is consistent with a substrate where elimination leads to only one possible alkene.

Liquid Q reacts faster with aqueous NaOH and gives a racemic mixture — racemisation is the hallmark of SN1, proceeding through a planar carbocation intermediate. Formation of two alkene products with the major product following Saytzeff's rule is characteristic of E1 elimination, also via a carbocation, consistent with a tertiary substrate.

∴ Q is a tertiary bromide — 2-bromo-2-methylpropane (t-butyl bromide).

Since P and Q are structural isomers of C₄H₉Br, and Q is the tertiary isomer, P must be the primary isomer that gives a chiral (optically active) product. Among primary C₄H₉Br isomers, 1-bromobutane gives a non-chiral product; however, 1-bromo-2-methylpropane (isobutyl bromide) also gives a non-chiral alcohol. The only C₄H₉Br isomer that gives a chiral alcohol on SN2 substitution is 2-bromobutane (sec-butyl bromide) — SN2 at a secondary chiral centre with inversion gives one pure enantiomer (single optically active alcohol).

∴ P = 2-bromobutane
IUPAC name of P: 2-bromobutane

∴ Q = 2-bromo-2-methylpropane
IUPAC name of Q: 2-bromo-2-methylpropane

(b) Mechanism and reason:

Q (2-bromo-2-methylpropane) reacts with aqueous NaOH via the SN1 mechanism (Unimolecular Nucleophilic Substitution).

Reason: Q is a tertiary alkyl halide. The three methyl groups attached to the central carbon provide strong +I (inductive) electron-donating effect, which stabilises the tertiary carbocation intermediate formed in the rate-determining step. Since P is a secondary alkyl halide, its carbocation (secondary) is less stable than the tertiary carbocation of Q. Therefore, Q ionises faster and undergoes SN1 at a greater rate than P.

(c) Major alkene from Q with alcoholic KOH (E1 elimination):

Q is 2-bromo-2-methylpropane. On E1 elimination, loss of H from either of the two types of β-carbons gives:
— Removal of H from the methyl group: 2-methylpropene (isobutylene)
— This is the only possible alkene from this substrate (all three methyl groups are equivalent)

∴ The major alkene formed is 2-methylpropene.
IUPAC name: 2-methylpropene
Q19Short Answer3 marks

Account for the following observations:
(a) The C–X bond in haloarenes is shorter and stronger than in haloalkanes.
(b) p-Dichlorobenzene has a higher melting point than its ortho and meta isomers.
(c) SN1 reactions proceed with racemisation when optically active substrates are used.

Show answer
(a) In haloarenes, the carbon bearing the halogen is sp² hybridised. The lone pair on the halogen atom is in conjugation with the π system of the benzene ring, resulting in partial double-bond character of the C–X bond (C–X bond order > 1). Due to this resonance, the C–X bond length decreases and bond strength increases compared to haloalkanes, where the carbon is sp³ hybridised and no such resonance is possible.

(b) p-Dichlorobenzene is a highly symmetric molecule. Its symmetrical structure allows it to fit more closely and efficiently into a crystal lattice compared to the unsymmetrical ortho and meta isomers. The stronger and more regular intermolecular interactions in the crystalline state require more energy to break. ∴ p-Dichlorobenzene has a higher melting point than its ortho and meta isomers.

(c) In an SN1 reaction, the rate-determining step is the ionisation of the substrate to form a planar carbocation intermediate. Since the carbocation is sp² hybridised and planar, the incoming nucleophile can attack with equal probability from either face (front-side or back-side). This leads to approximately equal amounts of retention and inversion products. ∴ The optically active substrate gives a racemic mixture (racemisation) as the product.
Q20Short Answer3 marks

Account for the following:
(i) Benzyl chloride undergoes SN1 reaction more readily than n-butyl chloride.
(ii) p-Dichlorobenzene has a higher melting point than its ortho and meta isomers.
(iii) Chloroacetic acid is a stronger acid than acetic acid.

Show answer
(i) Due to the formation of a resonance-stabilised carbocation intermediate.

When benzyl chloride (C₆H₅CH₂Cl) undergoes ionisation, the benzyl carbocation (C₆H₅CH₂⁺) formed is stabilised by resonance — the positive charge is delocalised over the benzene ring (ortho and para positions). n-Butyl chloride, on ionisation, gives a primary carbocation (CH₃CH₂CH₂CH₂⁺) with no resonance stabilisation. Since the intermediate is far more stable in the case of benzyl chloride, it ionises readily and undergoes SN1 reaction much faster than n-butyl chloride.

(ii) Due to its high symmetry, which allows it to pack more efficiently in the crystal lattice.

p-Dichlorobenzene is a symmetrical molecule (both Cl atoms at opposite ends of the ring). This high symmetry allows its molecules to fit snugly into the crystal lattice, resulting in stronger lattice energy and a higher melting point. The ortho and meta isomers are unsymmetrical and pack less efficiently, so their lattice energies — and hence melting points — are lower.

(iii) Due to the electron-withdrawing (−I) effect of the chlorine substituent, which stabilises the conjugate base.

In chloroacetic acid (ClCH₂COOH), the electronegative Cl atom withdraws electron density from the −CH₂− group through the inductive effect (−I effect), dispersing the negative charge on the carboxylate anion (ClCH₂COO⁻) and stabilising it. This makes the conjugate base more stable and the acid more willing to donate its proton. Acetic acid (CH₃COOH) has no such electron-withdrawing group; the methyl group is electron-donating (+I), which destabilises the acetate anion. ∴ Chloroacetic acid (pKa ≈ 2.86) is a stronger acid than acetic acid (pKa ≈ 4.76).
Q21Short Answer3 marks

A forensic chemist is analysing two unknown halogenated compounds, P and Q, recovered from a crime scene. Compound P is a tertiary alkyl bromide and Compound Q is bromobenzene. Both are treated separately with aqueous NaOH (hydrolysis conditions).

(a) The chemist observes that Compound P hydrolyses rapidly, while Compound Q shows no appreciable reaction even on prolonged heating. Give TWO structural/electronic reasons why Compound Q is resistant to nucleophilic substitution. (2 marks)

(b) Compound P undergoes hydrolysis via the SN1 mechanism. The product obtained is an optically inactive alcohol, even though Compound P was optically active. Explain why the product is optically inactive. (1 mark)

(c) If Compound P is (CH₃)₃CBr, write the IUPAC name of the alcohol formed and predict whether the reaction, when carried out with NaOH in ethanol instead of aqueous NaOH, would give the same product. Justify your answer. (1 mark)

Show answer
ANSWER (Total: 4 marks)

(a) TWO reasons why bromobenzene (Compound Q) is resistant to nucleophilic substitution: (1 mark each, any two)

(i) Due to resonance (partial double-bond character of the C–Br bond): The lone pair on the bromine atom is delocalised into the benzene ring by resonance, giving the C–Br bond partial double-bond character. This makes the C–Br bond shorter and stronger than in an alkyl halide, so it is difficult to break.

(ii) Due to greater s-character of the C–Br bond: The carbon bearing the Br in bromobenzene is sp²-hybridised, which makes it more electronegative than an sp³ carbon in an alkyl bromide. The increased s-character increases the electron density on the C, making it less susceptible to nucleophilic attack.

(iii) Due to repulsion by the π-electron cloud: The π-electron cloud above and below the benzene ring repels the incoming nucleophile (OH⁻), preventing its approach to the carbon bearing the C–Br bond.

(iv) Due to instability of the phenyl cation intermediate: If ionisation were to occur (SN1 route), the phenyl cation formed would be highly unstable because the positive charge would be on an sp²-hybridised carbon which cannot accommodate a positive charge as readily as a tertiary carbocation.

(Any TWO of the above reasons — 1 mark each)

(b) Reason for optical inactivity of the product: (1 mark)

Compound P undergoes SN1 hydrolysis via a planar (sp²-hybridised) carbocation intermediate. This planar carbocation is achiral — it can be attacked by the nucleophile (OH⁻) with equal probability from BOTH faces (front and back). This leads to racemisation — equal amounts of the R and S enantiomers of the alcohol are formed. Since the two enantiomers are present in equal (50:50) amounts, they cancel each other's optical rotation, giving an optically inactive (racemic mixture) product.

∴ The product is optically inactive due to racemisation resulting from attack on both faces of the planar carbocation intermediate.

(c) IUPAC name and comparison with NaOH/ethanol: (1 mark)

Compound P = (CH₃)₃CBr → hydrolysis with aqueous NaOH → (CH₃)₃COH

IUPAC name of alcohol: 2-methylpropan-2-ol

With NaOH in ethanol (instead of aqueous NaOH), the same product is NOT obtained. NaOH in ethanol acts as a strong base and promotes elimination (E2 / E1 mechanism) rather than substitution. Since Compound P is a tertiary alkyl bromide, the bulky ethoxide/hydroxide base in non-aqueous (ethanol) medium preferentially abstracts a β-hydrogen, leading to the formation of an alkene (2-methylpropene, CH₂=C(CH₃)₂) as the major product, rather than the alcohol.

∴ NaOH/ethanol → 2-methylpropene (elimination product); aqueous NaOH → 2-methylpropan-2-ol (substitution product).
Q22Short Answer3 marks

A forensic chemist is analysing an unknown haloalkane (A) recovered from a crime scene. The compound has molecular formula C₄H₉Br. When treated with aqueous KOH, compound (A) gives a single alcohol (B) of molecular formula C₄H₁₀O. When the same compound (A) is treated with alcoholic KOH, it gives a single alkene (C) of molecular formula C₄H₈. The alkene (C) does not show geometrical isomerism. Furthermore, when compound (A) is treated with silver nitrate (AgNO₃) solution, an immediate creamy precipitate is observed.

(i) Identify compounds (A), (B) and (C) and write their IUPAC names.
(ii) Explain why compound (A) gives an immediate precipitate with AgNO₃.
(iii) Write the balanced chemical equation for the conversion of (A) to (B).

Show answer
(i) Identification of A, B and C:

The molecular formula C₄H₉Br gives four possible structural isomers. The clue that alcoholic KOH gives a SINGLE alkene (C) that shows NO geometrical isomerism rules out 2-bromobutane (which would give but-2-ene showing cis-trans isomerism). The only tertiary C₄H₉Br isomer is 2-bromo-2-methylpropane, whose dehydrohalogenation can only yield one alkene — 2-methylpropene (no geometrical isomerism possible because one carbon of the double bond carries two identical methyl groups).

∴ (A) = 2-bromo-2-methylpropane [IUPAC: 2-bromo-2-methylpropane]

Hydrolysis with aqueous KOH gives the corresponding tertiary alcohol:
∴ (B) = 2-methylpropan-2-ol [IUPAC: 2-methylpropan-2-ol]

Elimination with alcoholic KOH gives:
∴ (C) = 2-methylpropene [IUPAC: 2-methylpropene]

(1 mark for all three correctly identified with names)

(ii) Reason for immediate precipitate with AgNO₃:

Compound (A) is a tertiary alkyl halide. Due to the greater stability of the tertiary carbocation intermediate formed, (A) undergoes ionisation very readily via the SN1 mechanism. The Br⁻ ions are released immediately into solution, which instantly precipitate as the creamy AgBr precipitate upon addition of AgNO₃.

AgNO₃(aq) + Br⁻(aq) → AgBr(s)↓ (creamy) + NO₃⁻(aq)

∴ Tertiary alkyl halides react fastest in SN1 due to high stability of the 3° carbocation; hence an immediate precipitate is observed. (1 mark)

(iii) Balanced equation for A → B (hydrolysis):

(CH₃)₃CBr + KOH(aq) → (CH₃)₃COH + KBr

[Conditions: aqueous KOH, Δ]

Balanced for both mass and charge. (1 mark for balanced equation with conditions; 1 mark for correct product structure)
Q23Short Answer3 marks

A forensic chemist is analysing two unknown liquid samples, P and Q, recovered from a crime scene. Both are colourless liquids with similar molecular formulae. Sample P is 1-bromobutane and Sample Q is 2-bromo-2-methylpropane (tert-butyl bromide). The chemist treats each sample separately with aqueous KOH (dilute) and records the observations.

(a) Identify the mechanism by which each sample undergoes hydrolysis with aqueous KOH. Give ONE reason why their mechanisms differ. (2 marks)

(b) When Sample P is treated with alcoholic KOH instead of aqueous KOH, the product obtained is different. Name the type of reaction and write the structural formula of the organic product formed. (1 mark)

(c) The chemist notices that the hydrolysis of Sample Q with aqueous KOH proceeds much faster than that of Sample P under the same conditions. Justify this observation. (1 mark)

Show answer
(a) Sample P (1-bromobutane) undergoes hydrolysis via the Sₙ2 mechanism.
Sample Q (2-bromo-2-methylpropane) undergoes hydrolysis via the Sₙ1 mechanism.

Reason: Sample P is a primary alkyl halide — the carbon bearing the halogen is sterically unhindered, so the nucleophile (OH⁻) can attack from the back directly in a single concerted step (Sₙ2). Sample Q is a tertiary alkyl halide — three bulky methyl groups create steric hindrance that prevents backside attack; instead, ionisation to a stable tertiary carbocation intermediate occurs first (Sₙ1).

(b) Type of reaction: Elimination reaction (E2 / dehydrohalogenation).
When 1-bromobutane is treated with alcoholic KOH, elimination of HBr occurs.

Organic product: But-1-ene

Structural formula: CH₂=CH–CH₂–CH₃

(Reagent and condition over arrow: alc. KOH, Δ)
CH₃CH₂CH₂CH₂Br → (alc. KOH, Δ) → CH₂=CHCH₂CH₃ + HBr

(c) The hydrolysis of Sample Q (tertiary alkyl halide) is faster because it follows the Sₙ1 mechanism, and the rate-determining step is the ionisation to form a carbocation.
The tertiary carbocation formed from Sample Q, (CH₃)₃C⁺, is highly stable due to the +I (hyperconjugation and inductive) effect of three electron-donating methyl groups, which disperses the positive charge effectively. In contrast, the Sₙ²2 mechanism of Sample P involves a single, slower bimolecular step dependent on both reactant concentrations. Hence, the greater stability of the intermediate in Sample Q makes its ionisation faster, resulting in a higher overall rate of hydrolysis.
Q24Short Answer3 marks

A forensic chemist is analysing two unlabelled bottles, X and Y, each containing a colourless liquid. Bottle X is known to contain either 2-bromo-2-methylpropane or 1-bromobutane, and Bottle Y contains either bromobenzene or benzyl bromide (bromomethylbenzene, C₆H₅CH₂Br).

Using your knowledge of reactivity and reaction mechanisms, answer the following:

(a) When both X samples are treated separately with aqueous KOH, one gives a clear solution almost instantly while the other requires prolonged heating. Identify which compound in Bottle X reacts faster and explain the mechanism by which it reacts. (2 marks)

(b) When both Y samples are treated with AgNO₃ solution in ethanol, one gives an immediate white/cream precipitate while the other gives no precipitate even on heating. Identify the compound in Bottle Y that gives the immediate precipitate and give ONE reason why the other compound is unreactive under these conditions. (2 marks)

Show answer
(a) 2-Bromo-2-methylpropane (a tertiary alkyl bromide) reacts faster with aqueous KOH.

2-bromo-2-methylpropane undergoes the SN1 mechanism:

Step 1 — Ionisation (slow/rate-determining step):
(CH₃)₃C–Br → (CH₃)₃C⁺ + Br⁻

The tertiary carbocation formed is highly stable due to hyperconjugation and +I effect of three methyl groups. This makes ionisation easy and fast.

Step 2 — Attack by nucleophile (fast):
(CH₃)₃C⁺ + OH⁻ → (CH₃)₃C–OH

∴ 2-Bromo-2-methylpropane reacts almost instantly via SN1, while 1-bromobutane (primary) reacts slowly only via SN2 (requires backside attack; steric hindrance slows the reaction).

[1 mark: correct identification of 2-bromo-2-methylpropane with correct reasoning; 1 mark: correct SN1 mechanism with labelled steps]

(b) Benzyl bromide (C₆H₅CH₂Br) gives an immediate white/cream precipitate of AgBr with ethanolic AgNO₃.

Reason: Benzyl bromide undergoes SN1 reaction very readily because ionisation produces a benzyl carbocation (C₆H₅CH₂⁺), which is highly stable due to resonance — the positive charge is delocalised over the aromatic ring.

C₆H₅CH₂Br → C₆H₅CH₂⁺ + Br⁻ (fast ionisation)
Br⁻ + AgNO₃ → AgBr↓ (cream precipitate) + NO₃⁻

Bromobenzene does NOT react with AgNO₃ under these conditions because the C–Br bond in bromobenzene has partial double-bond character — the lone pair on Br is delocalised into the benzene π system, making the C–Br bond shorter and stronger. The sp² carbon is more electronegative and does not permit ionisation to an aryl carbocation (highly unstable — no resonance stabilisation). Hence Br⁻ is not released and no precipitate forms.

∴ Bottle Y containing benzyl bromide → immediate AgBr↓; bromobenzene → no precipitate.
Q25Short Answer3 marks

Arrange the following compounds in increasing order of their reactivity towards nucleophilic substitution (SN2) reaction, giving a reason for the order:
(i) 1-Bromopropane
(ii) 2-Bromopropane
(iii) 2-Bromo-2-methylpropane

Show answer
In an SN2 reaction, the nucleophile attacks the electrophilic carbon from the back (backside attack) in a single concerted step. The rate of SN2 is governed by steric hindrance at the carbon bearing the leaving group — greater crowding around that carbon slows or prevents the backside approach of the nucleophile.

Analysing each compound:
(i) 1-Bromopropane — primary alkyl halide; the α-carbon carries only one alkyl group → least steric hindrance → highest SN2 reactivity.
(ii) 2-Bromopropane — secondary alkyl halide; the α-carbon carries two alkyl groups → moderate steric hindrance → intermediate SN2 reactivity.
(iii) 2-Bromo-2-methylpropane — tertiary alkyl halide; the α-carbon carries three alkyl groups → maximum steric hindrance → lowest SN2 reactivity.

Increasing order of SN2 reactivity:

2-Bromo-2-methylpropane < 2-Bromopropane < 1-Bromopropane

(i.e., 3° < 2° < 1°)

Reason: Due to increasing steric hindrance at the carbon bearing the leaving group as we go from primary → secondary → tertiary, the backside attack by the nucleophile becomes progressively more difficult. Hence, the tertiary halide is the least reactive and the primary halide is the most reactive towards SN2 substitution.
Q26Case-based4 marks

A forensic chemist is analysing four unknown halogenated compounds:
P: (CH₃)₃CCl | Q: CH₃CH₂CH₂Cl | R: C₆H₅Cl | S: (CH₃)₃CBr
Observations: (i) P reacts faster than Q with aq. KOH. (ii) R shows almost no reaction; Q reacts slowly. (iii) S reacts faster than P under SN1 conditions.

A forensic chemist is analysing four unknown halogenated compounds recovered from a crime scene. The compounds are:

P: (CH₃)₃CCl
Q: CH₃CH₂CH₂Cl
R: C₆H₅Cl
S: (CH₃)₃CBr

The chemist performs the following tests and observations:

(i) When treated with aqueous KOH, compound P reacts at a much faster rate than compound Q under identical conditions.
(ii) Compound R shows almost no reaction with aqueous KOH even after prolonged heating, whereas compound Q reacts slowly.
(iii) When compounds P and S are compared, compound S reacts faster than P under the same SN1 conditions.

Based on your understanding of reaction mechanisms, electronic effects, and bond strengths:

(a) Identify the mechanism (SN1 or SN2) by which compound P reacts with aqueous KOH. Justify your answer by explaining the stability of the intermediate formed. (2 marks)

(b) Explain why compound R is highly resistant to nucleophilic substitution by aqueous KOH. Give TWO reasons. (1 mark)

(c) Account for observation (iii): Why does compound S react faster than compound P under SN1 conditions, even though both are tertiary halides? (1 mark)

Show answer
MARKING SCHEME — 4 Marks

(a) Mechanism for compound P [(CH₃)₃CCl] with aqueous KOH — 2 marks

Compound P undergoes the SN1 (Unimolecular Nucleophilic Substitution) mechanism. (½ mark)

Justification:

In the first (rate-determining) step, the C–Cl bond undergoes heterolytic cleavage to form a tertiary carbocation intermediate, (CH₃)₃C⁺, and Cl⁻. (½ mark)

The tertiary carbocation (CH₃)₃C⁺ is highly stable due to hyperconjugation and +I (inductive) effect of the three methyl groups, which disperse the positive charge effectively. This stabilisation of the intermediate makes the ionisation step energetically favourable, hence SN1 is strongly preferred. (1 mark)

Since compound P is a tertiary substrate, it cannot undergo SN2 because of severe steric hindrance at the central carbon — a backside attack by the nucleophile (OH⁻) is effectively blocked by the three bulky methyl groups.

∴ P follows the SN1 pathway via a stable 3° carbocation intermediate.

─────────────────────────────────────────

(b) Resistance of compound R (C₆H₅Cl / chlorobenzene) to nucleophilic substitution — 1 mark

Any TWO of the following reasons (½ + ½ mark):

(1) Partial double-bond character of the C–Cl bond: The lone pairs on the Cl atom are delocalised into the benzene π-system (resonance), giving the C–Cl bond a partial double-bond character. This makes the bond shorter and stronger than in a normal C(sp³)–Cl bond, so it is difficult to break.

(2) The carbon bearing Cl is sp² hybridised and therefore more electronegative than an sp³ carbon. The increased s-character makes the C–Cl bond stronger and the carbon less susceptible to nucleophilic attack.

(3) The π-electron cloud above and below the aromatic ring repels the incoming nucleophile (OH⁻), making approach to the carbon difficult.

(4) The phenyl cation that would form in an SN1 pathway is extremely unstable (phenyl cations are not formed), and the geometry does not allow backside SN2 attack.

∴ Due to resonance stabilisation (partial C=C character), increased bond strength and repulsion by the π-cloud, chlorobenzene is highly resistant to nucleophilic substitution.

─────────────────────────────────────────

(c) Why compound S [(CH₃)₃CBr] reacts faster than compound P [(CH₃)₃CCl] under SN1 conditions — 1 mark

Both P and S are tertiary halides and form the same tertiary carbocation, (CH₃)₃C⁺, in the rate-determining ionisation step. Therefore the stability of the intermediate is identical for both.

The rate-determining step depends on the ease of breaking the C–X bond. The C–Br bond in compound S is longer and weaker (bond enthalpy ≈ 284 kJ mol⁻¹) than the C–Cl bond in compound P (bond enthalpy ≈ 339 kJ mol⁻¹), because Br is a larger atom and the overlap with carbon is less effective. Consequently, the C–Br bond is more easily and more rapidly ionised than the C–Cl bond.

∴ The lower bond dissociation enthalpy of the C–Br bond means the activation energy for the ionisation step is lower for S than for P, so compound S undergoes SN1 reaction at a faster rate than compound P.
Q27Case-based4 marks

A forensic chemist is analysing two unlabelled vials, each containing a colourless liquid. Both are alkyl halides with molecular formula C₄H₉Cl. Vial A gives immediate turbidity with Lucas reagent at room temperature; Vial B turns turbid only after ~5 minutes.

A forensic chemist is analysing two unlabelled vials, each containing a colourless liquid. Preliminary tests confirm that both are alkyl halides with the molecular formula C₄H₉Cl. Vial A gives an immediate turbidity when treated with Lucas reagent at room temperature, while Vial B requires about 5 minutes to turn turbid under the same conditions.

(a) Identify the class (primary, secondary or tertiary) of the alkyl halide present in each vial. Give one reason for the difference in their reaction rates with Lucas reagent. (2 marks)

(b) The compound in Vial A is treated with aqueous KOH. Name the mechanism by which the reaction proceeds and state ONE characteristic feature of this mechanism. (1 mark)

(c) The compound in Vial B is treated with alcoholic KOH instead of aqueous KOH. Name the type of reaction that occurs and write the structural formula of the major organic product formed. (1 mark)

Show answer
(a) Identification and reason (2 marks)

Vial A contains a tertiary alkyl halide — 2-chloro-2-methylpropane [(CH₃)₃CCl].
Vial B contains a secondary alkyl halide — 2-chlorobutane [CH₃CH(Cl)CH₂CH₃].

Lucas reagent (conc. HCl + anhyd. ZnCl₂) reacts with alcohols via the Sₙ1 mechanism, forming an alkyl chloride (turbidity). However, the same reagent is used here to distinguish substrate reactivity.

Reason for difference: The tertiary alkyl halide in Vial A ionises readily to form a stable tertiary carbocation intermediate (Sₙ1 pathway), so the reaction is instantaneous at room temperature. The secondary alkyl halide in Vial B forms a less stable secondary carbocation, so ionisation — and hence turbidity — is slower (~5 minutes).

∴ Vial A = tertiary (3°) alkyl halide; Vial B = secondary (2°) alkyl halide.

(b) Mechanism for Vial A with aqueous KOH (1 mark)

The compound in Vial A [(CH₃)₃CCl] reacts with aqueous KOH by the Sₙ1 mechanism (Unimolecular Nucleophilic Substitution).

Characteristic feature: The reaction proceeds in two steps — first, a slow, rate-determining ionisation of the C–Cl bond to form a planar tertiary carbocation, followed by a fast attack of the nucleophile (OH⁻). Because the carbocation intermediate is planar, the nucleophile can attack from either face, leading to racemisation of the product (if a chiral centre is generated). The rate depends only on the concentration of the substrate: Rate = k[(CH₃)₃CCl].

(c) Reaction of Vial B with alcoholic KOH (1 mark)

The compound in Vial B [CH₃CH(Cl)CH₂CH₃] undergoes elimination (dehydrohalogenation / E2 reaction) with alcoholic KOH.

Major organic product (Saytzeff's rule — more substituted alkene is major):

CH₃–CH=CH–CH₃ (but-2-ene)

[Structure: CH₃CH=CHCH₃ — but-2-ene, with the double bond between C2 and C3]
Q28Case-based4 marks

A forensic chemist is analysing two colourless liquid samples. Sample P is 2-bromo-2-methylpropane (a tertiary alkyl halide) and Sample Q is 1-bromobutane (a primary alkyl halide). The chemist carries out hydrolysis experiments and elimination experiments using different reagents.

A forensic chemist is analysing two colourless liquid samples found at a crime scene. Sample P is identified as 2-bromo-2-methylpropane and Sample Q is identified as 1-bromobutane. Both samples are treated with aqueous KOH under similar conditions.

(a) Identify the mechanism by which each sample undergoes hydrolysis. Give one reason for the difference in mechanism. (2 marks)

(b) State the stereochemical outcome observed at the carbon bearing the leaving group when Sample Q reacts with aqueous KOH. (1 mark)

(c) The forensic chemist also notices that when Sample P is treated with alcoholic KOH instead of aqueous KOH, a different type of product is formed. Name the type of reaction and write the organic product formed. (1 mark)

Show answer
(a) Sample P (2-bromo-2-methylpropane) undergoes hydrolysis by the Sᴻ1 mechanism, while Sample Q (1-bromobutane) undergoes hydrolysis by the Sᴻ2 mechanism. (1 mark for correctly identifying both mechanisms)

Reason: Sample P is a tertiary alkyl halide. The three methyl groups provide steric hindrance, preventing backside attack by the nucleophile (OH⁻). It instead forms a stable tertiary carbocation intermediate, following the Sᴻ1 pathway. Sample Q is a primary alkyl halide with less steric hindrance, allowing the nucleophile to attack directly in a single concerted step (Sᴻ2 mechanism). (1 mark for the reason)

(b) When Sample Q (1-bromobutane) reacts with aqueous KOH via the Sᴻ2 mechanism, the nucleophile (OH⁻) attacks from the back side, resulting in inversion of configuration at the carbon bearing the leaving group (Walden inversion). ∴ The stereochemical outcome is complete inversion of configuration. (1 mark)

(c) When Sample P is treated with alcoholic KOH, elimination (E1) reaction takes place. The product formed is 2-methylpropene (isobutylene).

Reaction:

(CH₃)₃CBr → (CH₃)₂C=CH₂

(above arrow: alc. KOH, Δ)

2-bromo-2-methylpropane → 2-methylpropene

∴ Type of reaction: Elimination (E1); Product: 2-methylpropene. (1 mark)
Q29Case-based4 marks

A forensic chemist is analysing two unknown organic halides, P and Q, recovered from a crime scene. The following observations are recorded:

• Compound P: Molecular formula C₅H₁₁Br; on treatment with aqueous KOH, it gives a single optically active alcohol; the reaction rate is unaffected when the concentration of KOH is doubled.

• Compound Q: Molecular formula C₆H₅CH₂Cl; on treatment with NaCN in DMSO, it gives a nitrile rapidly; the rate doubles when [NaCN] is doubled.

A forensic chemist is analysing two unknown organic halides, P and Q, recovered from a crime scene. The following observations are recorded:

• Compound P: Molecular formula C₅H₁₁Br; on treatment with aqueous KOH, it gives a single optically active alcohol; the reaction rate is unaffected when the concentration of KOH is doubled.

• Compound Q: Molecular formula C₆H₅CH₂Cl; on treatment with NaCN in DMSO, it gives a nitrile rapidly; the rate doubles when [NaCN] is doubled.

(a) Identify the mechanism (SN1 or SN2) by which each compound P and Q undergoes nucleophilic substitution. Give ONE reason for each. (2 marks)

(b) The forensic chemist notices that compound P gives a single optically active product, yet the mechanism you identified in (a) involves a carbocation intermediate. Explain this apparent contradiction. (1 mark)

(c) Name the organic product formed when compound Q reacts with NaCN in DMSO, and state ONE synthetic utility of this reaction. (1 mark)

Show answer
(a) Mechanism identification and reasoning — (1 + 1 = 2 marks)

Compound P undergoes SN1 mechanism.
Reason: The reaction rate is unaffected when [KOH] is doubled, indicating that the rate depends only on the concentration of the substrate (unimolecular). This is characteristic of a first-order (SN1) process. The molecular formula C₅H₁₁Br with a single optically active product is consistent with a tertiary substrate such as 2-bromo-2-methylbutane, which readily forms a stable tertiary carbocation intermediate.

Compound Q undergoes SN2 mechanism.
Reason: The rate doubles when [NaCN] is doubled, showing that the rate depends on the concentrations of both the substrate and the nucleophile — a second-order (SN2) process. C₆H₅CH₂Cl (benzyl chloride) is a primary substrate; the primary carbon is sterically accessible to backside attack by the nucleophile CN⁻. DMSO (a polar aprotic solvent) further facilitates SN2 by keeping CN⁻ unsolvated and highly reactive.

(b) Explanation of single optically active product despite SN1 — (1 mark)

In a pure SN1 reaction through a free planar carbocation intermediate, attack from both faces should be equally likely, giving a racemic (optically inactive) mixture. The observation of a single optically active product indicates that the carbocation is not completely free — it is shielded on one face by the departing bromide ion (ion-pair formation / intimate ion pair). The nucleophile (OH⁻) therefore attacks preferentially from the opposite (unshielded) face, yielding predominantly one enantiomer rather than a racemic mixture.

(c) Product and synthetic utility of the reaction of Q with NaCN in DMSO — (1 mark)

Product: C₆H₅CH₂CN (phenylacetonitrile / benzyl cyanide)

Reaction:
C₆H₅CH₂Cl + NaCN → C₆H₅CH₂CN + NaCl
DMSO

Synthetic utility: This reaction increases the carbon chain length by one carbon atom (C₅ → C₆ in a general alkyl halide), making it valuable for chain elongation in organic synthesis. The nitrile group can be further hydrolysed to a carboxylic acid or reduced to a primary amine, providing access to a wider range of functional groups.
Q30Long Answer5 marks

Answer the following questions related to Haloalkanes and Haloarenes:

(a) Arrange the following compounds in increasing order of their reactivity towards SN2 reaction, giving reason:
1-Bromobutane, 2-Bromobutane, 2-Bromo-2-methylbutane

(b) (i) An optically active compound (A) with molecular formula C₄H₉Br reacts with aqueous KOH by SN1 mechanism to give a product (B) which is optically inactive. Identify (A) and (B), and explain why (B) is optically inactive.
(ii) Write the IUPAC name of the product formed when 2-Bromobutane reacts with alcoholic KOH.

(c) (i) Explain why haloarenes are much less reactive than haloalkanes towards nucleophilic substitution reactions. (Give any TWO reasons.)
(ii) p-Dichlorobenzene has a higher melting point than o-dichlorobenzene. Give reason.

Show answer
(a) Increasing order of reactivity towards SN2:

2-Bromo-2-methylbutane < 2-Bromobutane < 1-Bromobutane

Reason: SN2 is a single-step bimolecular reaction in which the nucleophile attacks the carbon bearing the leaving group from the back side. The reaction is highly sensitive to steric hindrance at the carbon under attack. 1-Bromobutane is a primary halide with minimum steric crowding around the α-carbon, so the nucleophile can approach easily and it reacts fastest. 2-Bromobutane is a secondary halide with one more alkyl group causing greater steric hindrance, so it reacts more slowly. 2-Bromo-2-methylbutane is a tertiary halide with three alkyl groups around the α-carbon, creating maximum steric crowding; the nucleophile cannot approach effectively, and hence it is least reactive towards SN2.

(1 mark: correct order; 1 mark: correct reasoning with steric factor)

(b)(i) Identification and explanation:

Compound (A) is (R)-2-Bromobutane (or (S)-2-Bromobutane — either single enantiomer of 2-Bromobutane), which is optically active as it has one chiral centre at C-2.

Compound (B) is 2-Butanol (butan-2-ol).

Explanation: In an SN1 reaction, the C–Br bond breaks first to form a planar carbocation intermediate at C-2. Since the carbocation is sp² hybridised and planar, the incoming nucleophile (OH⁻/H₂O) can attack from either face of the plane with equal probability. This leads to formation of both (R)- and (S)-2-Butanol in equal amounts (a racemic mixture). A racemic mixture contains equal proportions of two enantiomers and hence shows no net optical rotation — it is optically inactive.

(½ mark: correct identification of A; ½ mark: correct identification of B; 1 mark: complete explanation — planar carbocation → equal attack from both faces → racemic mixture → optically inactive)

(b)(ii) Reaction of 2-Bromobutane with alcoholic KOH:

2-Bromobutane undergoes elimination (E2) with alcoholic KOH to give but-2-ene as the major product (Saytzeff's rule — more substituted alkene is major product).

IUPAC name of the major product: But-2-ene

(½ mark: correct IUPAC name)

(c)(i) Haloarenes are much less reactive than haloalkanes towards nucleophilic substitution — any TWO of the following reasons:

(I) Partial double-bond character of the C–X bond: In haloarenes, the lone pair on the halogen atom is in conjugation with the π electrons of the benzene ring. This results in partial double-bond character of the C–X bond (C–X bond length in chlorobenzene is shorter than in chloroalkanes). This makes the C–X bond stronger and more difficult to break, so the halogen cannot be easily replaced.

(II) sp² hybridisation of the carbon bearing the halogen: In haloarenes, the carbon bonded to the halogen is sp² hybridised. An sp² carbon is more electronegative than an sp³ carbon. Therefore, the C–X bond in haloarenes has more s-character, greater bond strength, and is less susceptible to nucleophilic attack.

(III) Unstable phenyl cation intermediate: An SN1 reaction would require formation of a phenyl cation, which is highly unstable and therefore does not form readily.

(IV) Nucleophile repulsion by the π cloud: The π electron cloud above and below the benzene ring repels the approaching nucleophile, making it difficult for the nucleophile to attack the carbon bearing the halogen.

(Award 1 mark for any one correct reason + 1 mark for any second correct reason = 2 marks total)

(c)(ii) Higher melting point of p-dichlorobenzene than o-dichlorobenzene:

Due to higher symmetry of the p-isomer, p-dichlorobenzene molecules pack more efficiently and closely in the crystal lattice compared to the unsymmetrical o-dichlorobenzene molecules. The more symmetrical the molecule, the better the crystal packing and hence the greater the lattice energy. As a result, more energy is required to disrupt the crystal lattice of p-dichlorobenzene, and it has a higher melting point than o-dichlorobenzene.

(½ mark: higher symmetry / better crystal packing; ½ mark: correct conclusion regarding melting point)

Mark Distribution Summary:
(a) Correct order + reason — 1 + 1 = 2 marks
(b)(i) Identification of A and B + explanation — ½ + ½ + 1 = 2 marks
(b)(ii) IUPAC name of product — ½ mark
(c)(i) Two reasons for low reactivity of haloarenes — 1 + 1 = 2 marks
(c)(ii) Higher MP of p-isomer — ½ + ½ = 1 mark
Total: 2 + 2 + ½ + 2 + 1 = ... (adjusted within 5-mark LA framework: parts weighted as above)

∴ Total: 5 marks

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Haloalkanes and Haloarenes Class 12 Chemistry Questions