CBSE Class 12 Chemistry Haloarenes Worksheet

Very Short Answer Questions

Question. Which halogen compound in each of the following pairs will react faster in S_N2 reaction :
(i) CH₃Br or CH₃I
(ii) (CH₃)₃CCl or CH₃Cl
Answer. (i) CH₃I will give faster S_N2 reaction.
(ii) CH₃Cl will give faster S_N2 reaction.

Question. How may methyl bromide be preferentially converted to methyl isocyanide? 
Answer. KCN is predominantly ionic and provides cyanide ions in solution
CH₃Br + KCN              →         CH₃C ≡ N + KBr
Methyl Methyl                             cyanide
bromide
AgCN is mainly covalent in nature and nitrogen is free to donate electron pair forming isocyanide as main product.
CH₃Br + AgCN          →          CH₃N ≡ C + AgBr
Methyl Methyl                            isocyanide
bromide

Question. What happens when ethyl chloride is treated with aqueous KOH?
Answer. When ethyl chloride is treated with aqueous KOH, ethanol is formed,
CH₃CH₂Cl + KOH_(aq) → CH₃CH₂OH + KCl

Question. Give reason :
n-Butyl bromide has higher boiling point than t-butyl bromide.
Answer. n-Butyl bromide, being a straight chain molecule have strong intermolecular forces whereas t-butyl bromide being a branched chain molecule have weaker intermolecular forces due to smaller surface area. Hence, boiling point of n-butyl bromide is higher than that of t-butyl bromide.

Question. Why are alkyl halides insoluble in water?
Answer. Alkyl halides are polar but are insoluble in water because energy required to break the intermolecular H – bond among water molecules is much higher than energy released by water halide interaction.

Question. Which would undergo S_N2 reaction faster in the following pair and why?
CH₃ CH₂ — Br and CH₃ — CH₂ — I
Answer. Since I^– is a better leaving group than Br–, thus, CH₃CH₂I undergoes S_N2 reaction faster than CH₃CH₂Br.

Short Answer Questions

Question. Suggest a possible reason for the following observations :
(i) The order of reactivity of haloalkanes is RI > RCl > RBr.
(ii) Neopentyl chloride (CH₃)₃CCH₂Cl does not follow S_N2 mechanism.
Answer. (i) Among the various halides with same alkyl group the order of reactivity is RI > RBr > RCl. Due to increasing bond strength of C I, C Br and C Cl the reactivity decreases.
(ii) Neopentyl chloride being a primary halide reacts slowly through S_N1 and the carbon carrying halogen is sterically more hindered. Hence it does not follow S_N2 mechanism.

Question. Which compound in the following couple will react faster in S_N2 displacement and why?
(i) 1-Bromopentane or 2-bromopentane
(ii) 1-Bromo-2-methylbutane or 2-bromo-2-methylbutane.
Answer. (i) 1-Bromopentane, as it is a primary alkyl halide.
(ii) 1-Bromo-2-methylbutane, as it is a primary alkyl halide.

1. Give IUPAC names of the following:-
(a) Iso-butylchloride
(b) tert-butylchloride
(c) neo-pentylbromide
(d) ethylidene chloride
(e) Ethylene dichloride
(f) ethylene dichloride
(g) bromoform
(h) vinylchloride
(i) allyl bromide
(j) methylene chloride
(k) benzyl chloride
Answer:
(a) 1-Chloro-2-methylpropane
(b) 2-Chloro-2-methylpropane
(c) 1-Bromo-2,2-dimethylpropane
(d) 1,1-Dichloroethane
(e) 1,2-Dichloroethane
(f) 1,2-Dichloroethane (repeated in source)
(g) Tribromomethane
(h) Chloroethene
(i) 3-Bromoprop-1-ene
(j) Dichloromethane
(k) Chlorophenylmethane (or (Chloromethyl)benzene)
In simple words: IUPAC names are systematically assigned based on the longest carbon chain and substituent positions.
Exam Tip: Always identify the principal functional group and number the carbon chain from the end that gives the lowest locants to substituents.

2. Give reasons for the following:-
(a) Fluorine is more electronegative than chlorine still dipole moment of C-F bond is less than that of C-Cl bond in methyl fluoride & methyl chloride.
Answer: Dipole moment is the product of charge \((q)\) and distance \((d)\) between the charges (\(\mu = q \times d\)). Although fluorine is much more electronegative than chlorine, the carbon-fluorine bond length is extremely short (\(139 \text{ pm}\)) compared to the carbon-chlorine bond length (\(178 \text{ pm}\)). The very small distance in the C-F bond outweighs the higher partial charge, resulting in a smaller dipole moment for methyl fluoride compared to methyl chloride.
In simple words: Dipole moment depends on both charge and bond length; the short bond length of C-F decreases its dipole moment.
Exam Tip: Remember that \(\mu = q \times d\). Do not rely solely on electronegativity differences when comparing bond dipoles.

2. (b) For the conversion of alcohols to alkyl halides, thionyl chloride is preferred over phosphorous halide or halogen acid.
Answer: Thionyl chloride (\(\text{SOCl}_2\)) is preferred because the by-products formed during the reaction are gases (\(\text{SO}_2\) and \(\text{HCl}\)). These gaseous by-products escape into the atmosphere, leaving behind the pure alkyl halide, which makes separation and purification unnecessary.
In simple words: Thionyl chloride leaves no liquid impurities because all by-products escape as gases.
Exam Tip: Mention the gaseous nature of \(\text{SO}_2\) and \(\text{HCl}\) to score full marks.

2. (c) Conversion of alcohols to alkyl iodide is done in absence of catalyst zinc chloride.
Answer: Hydroiodic acid (\(\text{HI}\)) is a strong reducing agent and is easily oxidized to iodine (\(\text{I}_2\)) by Lewis acid catalysts like \(\text{ZnCl}_2\) or by sulfuric acid. Therefore, \(\text{HI}\) is generated in situ using a non-oxidizing acid like concentrated \(\text{H}_3\text{PO}_4\) without needing \(\text{ZnCl}_2\).
In simple words: HI reduces or gets oxidized easily, so special conditions are used instead of standard Lewis acid catalysts.
Exam Tip: Highlight that \(\text{H}_3\text{PO}_4\) is used instead of \(\text{H}_2\text{SO}_4\) for preparing alkyl iodides.

2. (e) Free radical halogenation of alkanes is not a very good method for preparation of haloalkanes.
Answer: Free radical halogenation yields a complex mixture of mono-, di-, tri-, and poly-halogenated alkanes along with isomers. Because it is difficult to separate these structural isomers with similar boiling points, the yield of any single desired haloalkane is very poor.
In simple words: It gives a messy mixture of many different products that are hard to separate.
Exam Tip: Use the term "mixture of isomeric mono- and poly-haloalkanes" in your answer.

2. (f) Electrophilic substitution of benzene with iodine requires presence of oxidizing agents like \(\text{HIO}_3\).
Answer: Iodination of benzene is a reversible process. The hydrogen iodide (\(\text{HI}\)) formed as a by-product reduces iodobenzene back to benzene. An oxidizing agent like iodic acid (\(\text{HIO}_3\)) or nitric acid is added to oxidize \(\text{HI}\) to iodine (\(\text{I}_2\)), shifting the equilibrium in the forward direction.
In simple words: The oxidizing agent destroys the HI by-product so the reaction can move forward.
Exam Tip: State clearly that iodination is reversible and \(\text{HI}\) acts as a reducing agent.

2. (g) Electrophilic substitution of benzene with fluorine is not done for preparation of fluorobenzene.
Answer: Fluorine is extremely reactive and fluorination of benzene is a violently exothermic and explosive reaction, which cannot be controlled under normal laboratory conditions. Fluorobenzene is instead prepared indirectly via the Balz-Schiemann reaction.
In simple words: Fluorine reacts too violently with benzene to control safely.
Exam Tip: Mention the extreme exothermicity and explosive nature of direct fluorination.

2. (h) Finkelstein reaction is carried out in dry acetone medium.
Answer: Sodium halide (\(\text{NaCl}\) or \(\text{NaBr}\)) formed during the halogen exchange reaction is insoluble in dry acetone and precipitates out of solution. According to Le Chatelier's principle, precipitation of the byproduct shifts the equilibrium forward to favor the formation of alkyl iodide.
In simple words: Acetone precipitates out the salt byproduct, driving the reaction forward.
Exam Tip: Mention precipitation of \(\text{NaCl}/\text{NaBr}\) in dry acetone and Le Chatelier's principle.

2. (i) Rate of reaction between t-butyl bromide & hydroxide ion depends only on concentration of hydroxide ions. (Note: Statement corrected conceptually to reflect SN1 kinetics, or if assuming standard textbook phrasing: t-butyl bromide hydrolysis is SN1 and depends only on substrate concentration. Assuming the prompt implies standard SN1 behavior:)
Answer: Tertiary-butyl bromide undergoes hydrolysis via an \(\text{S}_\text{N}1\) mechanism involving the formation of a stable tertiary carbocation in the slow rate-determining step. Therefore, the rate depends only on the concentration of the alkyl halide and is independent of the concentration of hydroxide ions.
In simple words: The slow step only involves breaking the bond in t-butyl bromide, so nucleophile concentration does not affect the speed.
Exam Tip: Clearly state that \(\text{S}_\text{N}1\) reactions depend only on the concentration of the substrate.

2. (j) Allylic & benzylic halides show high reactivity towards SN1 reaction.
Answer: They react rapidly via \(\text{S}_\text{N}1\) because ionization yields allylic and benzylic carbocations, which are strongly stabilized by resonance, lowering the activation energy required for carbocation formation.
In simple words: The carbocations formed are very stable due to resonance, making them form easily.
Exam Tip: Draw or mention resonance stabilization of allylic and benzylic carbocations.

2. (k) Dehydrohalogenation of 2-bromopentane gives 2-pentene as the major product.
Answer: According to Zaitsev's rule, during dehydrohalogenation, elimination occurs to form the more substituted and more stable alkene. 2-pentene is a disubstituted alkene with greater hyperconjugative stability than 1-pentene (monosubstituted), making it the major product.
In simple words: Zaitsev's rule states that the more substituted alkene is more stable and forms as the main product.
Exam Tip: State Zaitsev's rule clearly: "In dehydrohalogenation, the preferred product is the alkene that has the greater number of alkyl groups attached to the doubly bonded carbon atoms."

2. (l) Optically active 2-bromobutane on treatment with aq KOH results in formation of optically inactive product.
Answer: Nucleophilic substitution of 2-bromobutane via \(\text{S}_\text{N}1\) mechanism proceeds with racemization (equal amounts of retention and inversion of configuration), while \(\text{S}_\text{N}2\) proceeds with complete inversion. In either case, starting with an optically active enantiomer yields a racemic mixture or inverted product, which is optically inactive due to external compensation (racemization).
In simple words: Equal amounts of both optical isomers are formed, canceling out optical rotation to form a racemic mixture.
Exam Tip: Mention the formation of a racemic mixture resulting in optical inactivity.

2. (m) Aryl halides are extremely less reactive towards nucleophilic substitution reactions.
Answer: Aryl halides are unreactive due to: (1) Resonance effect giving partial double bond character to the C-Cl bond; (2) Difference in hybridization (\(\text{sp}^2\) carbon in aryl halide vs \(\text{sp}^3\) in alkyl halide); (3) Instability of phenyl cation; and (4) Repulsion between the electron-rich ring and incoming nucleophile.
In simple words: Stronger carbon-halogen bonds, resonance, and electronic repulsion make aryl halides hard to attack.
Exam Tip: List at least two key reasons: resonance (partial double bond) and \(\text{sp}^2\) hybridization.

2. (n) Rate of electrophilic substitution reactions in haloarenes is less than that of benzene.
Answer: Halogens exhibit a strong -I (electron-withdrawing inductive) effect and a weaker +R (electron-donating resonance) effect. The net effect is deactivating, pulling electron density away from the benzene ring and making the ring less nucleophilic toward incoming electrophiles compared to unsubstituted benzene.
In simple words: Halogen atoms pull electron density away through inductive effect, deactivating the benzene ring.
Exam Tip: Contrast the strong inductive (-I) deactivating effect with the resonance effect.

2. (o) Chloroform is stored in dark bottles.
Answer: In the presence of sunlight and air, chloroform slowly oxidizes to form an extremely poisonous gas called phosgene (carbonyl chloride, \(\text{COCl}_2\)). Storing it in dark, airtight bottles completely filled to the brim cuts off light and air, preventing this dangerous oxidation.
In simple words: Sunlight and air turn chloroform into poisonous phosgene gas, so dark bottles block light.

Answer: The presence of an electron-withdrawing nitro group (\(-\text{NO}_2\)) at the para position withdraws electron density from the benzene ring through resonance (-R effect) and inductive effect (-I), stabilizing the carbanion intermediate formed during nucleophilic aromatic substitution.
In simple words: The nitro group pulls electron density away, making it easier for a nucleophile to attack the ring.
Exam Tip: Mention stabilization of the Meisenheimer intermediate by the electron-withdrawing nitro group.

2. (q) Reaction of (-) 2-Bromo octane with aq NaOH results in formation of (+) octane-2-ol.
Answer: This reaction proceeds via an \(\text{S}_\text{N}2\) mechanism, which involves complete inversion of configuration (Walden inversion). The incoming hydroxide ion attacks from the side opposite to the leaving bromide ion, transforming the levorotatory isomer into the dextrorotatory isomer.
In simple words: The backside attack in \(\text{S}_\text{N}2\) flips the spatial arrangement completely, changing the optical rotation sign.
Exam Tip: Mention Walden inversion and \(\text{S}_\text{N}2\) stereochemistry.

3. Identify A,B,C,D in each of the following parts:-
(a) \(\text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{PBr}_3} \text{A} \xrightarrow{\text{alc. KOH}} \text{B} \xrightarrow{\text{HBr}} \text{C} \xrightarrow{\text{NH}_3} \text{D}\)
(b) \(\text{CH}_3\text{CH}_2\text{CH}_2\text{I} \xrightarrow{\text{alc. KOH}} \text{A} \xrightarrow{\text{hydrolysis}} \text{B} \xrightarrow{\text{SOCl}_2} \text{C} \xrightarrow{\text{LiAlH}_4} \text{D}\)
(c) \(\text{CH}_3\text{CHBrCH}_3 \xrightarrow{\text{alc. KOH}} \text{A} \xrightarrow{\text{HBr/peroxide}} \text{B} \xrightarrow{\text{NaI/acetone}} \text{C} \xrightarrow{\text{Mg/ether}} \text{D}\)
Answer:
Part (a):
A = \(\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}\) (1-bromopropane)
B = \(\text{CH}_3\text{CH}=\text{CH}_2\) (propene)
C = \(\text{CH}_3\text{CHBrCH}_3\) (2-bromopropane - via Markovnikov addition)
D = \(\text{CH}_3\text{CH(NH}_2)\text{CH}_3\) (propan-2-amine)

Part (b):
A = \(\text{CH}_3\text{CH}=\text{CH}_2\) (propene)
B = \(\text{CH}_3\text{CH(OH)}\text{CH}_3\) (propan-2-ol - via acid-catalyzed hydration)
C = \(\text{CH}_3\text{CH(Cl)}\text{CH}_3\) (2-chloropropane)
D = \(\text{CH}_3\text{CH}_2\text{CH}_3\) (propane)

Part (c):
A = \(\text{CH}_3\text{CH}=\text{CH}_2\) (propene)
B = \(\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}\) (1-bromopropane - anti-Markovnikov addition in presence of peroxide)
C = \(\text{CH}_3\text{CH}_2\text{CH}_2\text{I}\) (1-iodopropane - Finkelstein reaction)
D = \(\text{CH}_3\text{CH}_2\text{CH}_2\text{MgI}\) (propylmagnesium iodide - Grignard reagent)
In simple words: Follow each step sequentially, applying substitution, elimination, and addition rules carefully.
Exam Tip: Double-check regioselectivity rules like Markovnikov's rule and peroxide effect (anti-Markovnikov) when adding HBr.

4. An alkyl halide A on reaction with Mg in dry ether followed by ethanol gave 2-methylbutane. Write structure of A & all its isomers.
Answer:
Reaction of alkyl halide A with Mg in dry ether forms a Grignard reagent (\text{R-MgX}), which upon reaction with ethanol (\text{CH}_3\text{CH}_2\text{OH}) undergoes protonolysis to yield an alkane where the Grignard carbon acquires a hydrogen atom. Since the product is 2-methylbutane (\(\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}_2-\text{CH}_3\)), alkyl halide A can be either 1-bromo-2-methylbutane or 2-bromo-2-methylbutane (or chloro derivatives). Taking 1-chloro-2-methylbutane or 1-bromo-2-methylbutane as A:
Structure of A: \(\text{CH}_3-\text{CH}_2-\text{CH}(\text{CH}_3)-\text{CH}_2\text{-Cl}\) (1-chloro-2-methylbutane)
Isomers of molecular formula \(\text{C}_5\text{H}_{11}\text{Cl}\):
1. \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl}\) (1-Chloropentane)
2. \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH(Cl)}\text{CH}_3\) (2-Chloropentane)
3. \(\text{CH}_3\text{CH}_2\text{CH(Cl)}\text{CH}_2\text{CH}_3\) (3-Chloropentane)
4. \(\text{CH}_3\text{CH}(\text{CH}_3)\text{CH}_2\text{CH}_2\text{Cl}\) (1-Chloro-3-methylbutane)
5. \(\text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)\text{CH}_2\text{Cl}\) (1-Chloro-2-methylbutane - A)
6. \(\text{CH}_3\text{CH}_2\text{C(Cl)}(\text{CH}_3)\text{CH}_3\) (2-Chloro-2-methylbutane)
7. \((\text{CH}_3)_2\text{CHCH(Cl)}\text{CH}_3\) (2-Chloro-3-methylbutane)
8. \((\text{CH}_3)_3\text{CCH}_2\text{Cl}\) (1-Chloro-2,2-dimethylpropane)
In simple words: Work backwards from 2-methylbutane to find A, then list all possible chain and position isomers with formula C5H11Cl.
Exam Tip: Systematically draw primary, secondary, and tertiary isomers to ensure none are missed.

5. Comp. A M.F. \(\text{C}_4\text{H}_8\) when treated with dil. \(\text{H}_2\text{SO}_4\) gives B. B with conc. \(\text{HCl} + \text{ZnCl}_2\) gives C. C + sodium ethoxide gives back A. Identify A, B, C. Give all reactions.
Answer:
- Compound A has molecular formula \(\text{C}_4\text{H}_8\) and is an alkene (butene).
- Treatment with dil. \(\text{H}_2\text{SO}_4\) hydrates it to alcohol B (Markovnikov addition).
- Alcohol B reacts rapidly with conc. \(\text{HCl} + \text{ZnCl}_2\) (Lucas reagent) at room temperature, indicating B is a tertiary alcohol (tert-butyl alcohol).
- Therefore, alkene A is 2-methylpropene (\(\text{CH}_3\text{C}(\text{CH}_3)=\text{CH}_2\)).
- Alcohol B is 2-methylpropan-2-ol (\((\text{CH}_3)_3\text{COH}\)).
- Alkyl halide C is 2-chloro-2-methylpropane (\((\text{CH}_3)_3\text{CCl}\)).
Reactions:
1. \(\text{CH}_3\text{C}(\text{CH}_3)=\text{CH}_2 + \text{H}_2\text{O} \xrightarrow{\text{dil. H}_2\text{SO}_4} (\text{CH}_3)_3\text{COH}\) (Compound B)
2. \((\text{CH}_3)_3\text{COH} + \text{HCl} \xrightarrow{\text{anhydrous ZnCl}_2} (\text{CH}_3)_3\text{CCl} + \text{H}_2\text{O}\) (Compound C)
3. \((\text{CH}_3)_3\text{CCl} + \text{C}_2\text{H}_5\text{ONa} \xrightarrow{\text{elimination}} \text{CH}_3\text{C}(\text{CH}_3)=\text{CH}_2 + \text{C}_2\text{H}_5\text{OH} + \text{NaCl}\) (Compound A)
In simple words: Lucas reagent test helps identify tertiary alcohol B, leading back to 2-methylpropene.
Exam Tip: Use the Lucas test reaction speed to quickly identify alcohol type (tertiary reacts immediately).

6. Give one chemical test to distinguish between chlorobenzene and benzyl chloride.
Answer:
Treatment with aqueous sodium hydroxide (\(\text{aq. NaOH}\)) followed by acidification with dilute \(\text{HNO}_3\) and addition of silver nitrate (\(\text{AgNO}_3\)):
- Benzyl chloride is an aliphatic-aromatic halide where the chlorine atom is bonded to an \(\text{sp}^3\) hybridized carbon. It undergoes easy nucleophilic substitution with aq. \(\text{NaOH}\) to form benzyl alcohol and chloride ions. Upon adding \(\text{AgNO}_3\), it gives a thick white precipitate of \(\text{AgCl}\).
- Chlorobenzene is an aryl halide with partial double bond character due to resonance. It does not hydrolyze under normal conditions and gives no precipitate with \(\text{AgNO}_3\).
In simple words: Benzyl chloride reacts with aq. NaOH followed by AgNO3 to give a white precipitate, while chlorobenzene does not react.
Exam Tip: Mention the formation of a white precipitate of AgCl with benzyl chloride and no reaction for chlorobenzene.