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Haloalkanes and Haloarenes – CBSE NCERT Study Resources

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12th

12th - Chemistry

Haloalkanes and Haloarenes

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Overview of Haloalkanes and Haloarenes

Haloalkanes and Haloarenes are hydrocarbons in which one or more hydrogen atoms are replaced by halogen atoms (F, Cl, Br, I). These compounds are classified based on the type of carbon atom to which the halogen is attached and the nature of the hydrocarbon skeleton.

Classification

Haloalkanes and Haloarenes are classified as:

  • Alkyl halides (Haloalkanes): Halogen atom attached to an alkyl group (sp³ hybridized carbon).
  • Aryl halides (Haloarenes): Halogen atom attached to an aryl group (sp² hybridized carbon).
  • Vinylic halides: Halogen attached to a carbon-carbon double bond.
  • Allylic halides: Halogen attached to a carbon adjacent to a double bond.

Nomenclature

IUPAC names for haloalkanes are derived by prefixing the name of the parent hydrocarbon with the halogen name (fluoro, chloro, bromo, iodo). For haloarenes, the prefix is added to the name of the aromatic ring.

Preparation Methods

  • From alcohols using halogen acids (HX) or phosphorus halides (PX₃, PX₅).
  • From hydrocarbons via free radical halogenation (for alkanes) or electrophilic substitution (for arenes).
  • From alkenes by addition of hydrogen halides (Markovnikov's rule applies).

Physical Properties

  • Boiling points increase with molecular weight and decrease with branching.
  • Haloalkanes are polar but insoluble in water due to inability to form hydrogen bonds.
  • Density increases with the size of the halogen atom.

Chemical Reactions

  • Nucleophilic substitution (SN1 and SN2): Replacement of halogen by nucleophiles like OH⁻, CN⁻, etc.
  • Elimination reactions: Formation of alkenes via dehydrohalogenation (β-elimination).
  • Reaction with metals: Formation of Grignard reagents (RMgX).
  • Reduction: Conversion to hydrocarbons using Zn/HCl or LiAlH₄.

Uses

  • Solvents (e.g., chloroform, carbon tetrachloride).
  • Refrigerants (e.g., Freons).
  • Pharmaceuticals and pesticides (e.g., DDT, BHC).

Environmental Impact

Some haloalkanes (e.g., CFCs) deplete the ozone layer, while others (e.g., DDT) are persistent organic pollutants.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:
Define haloalkanes.
Answer:
Definition: Haloalkanes are hydrocarbons with halogen atoms replacing hydrogen.
Question 2:
Name the IUPAC name of CH3CH2Br.
Answer:
Bromoethane
Question 3:
What is the common name of chlorobenzene?
Answer:
Phenyl chloride
Question 4:
Which reagent converts alcohols to haloalkanes?
Answer:
Thionyl chloride (SOCl2)
Question 5:
Give one example of vicinal dihalide.
Answer:
1,2-Dichloroethane (CH2ClCH2Cl)
Question 6:
What is the product of Wurtz reaction on CH3Br?
Answer:
Ethane (CH3CH3)
Question 7:
Which haloarene is used as a solvent?
Answer:
Chlorobenzene
Question 8:
What is the hybridization of carbon in chloromethane?
Answer:
sp3
Question 9:
Name the test to distinguish alkyl and aryl halides.
Answer:
Silver nitrate (AgNO3) test
Question 10:
Which haloalkane is used as a refrigerant?
Answer:
Chlorofluorocarbons (CFCs)
Question 11:
What is the major product of monochlorination of methane?
Answer:
Chloromethane (CH3Cl)
Question 12:
Why are aryl halides less reactive than alkyl halides?
Answer:

Due to resonance stabilization in aryl halides.

Question 13:
Give the structure of 1,3-dibromopropane.
Answer:
BrCH2CH2CH2Br
Question 14:
Which halogen shows the highest reactivity in nucleophilic substitution?
Answer:
Iodine (I)
Question 15:
What is the IUPAC name of CH3CH2CH2Cl?
Answer:

The IUPAC name is 1-Chloropropane.
It is a primary haloalkane with three carbon atoms and a chlorine atom attached to the first carbon.

Question 16:
Why are haloarenes less reactive towards nucleophilic substitution than haloalkanes?
Answer:

Haloarenes have partial double bond character due to resonance, making the C-X bond stronger.
In haloalkanes, the carbon is sp3 hybridized, allowing easier attack by nucleophiles.

Question 17:
What is the product formed when ethyl bromide reacts with alcoholic KOH?
Answer:

The product is ethene (CH2=CH2).
This is an elimination reaction where KOH removes HBr to form a double bond.

Question 18:
Name the reagent used to convert alkyl halides to alkanes.
Answer:

Zinc and hydrochloric acid (Zn/HCl) or Raney nickel can be used.
This is a reduction reaction where the halogen is replaced by hydrogen.

Question 19:
What is Finkelstein reaction? Give an example.
Answer:

It is the exchange of halogen in an alkyl halide using NaI in acetone.
Example: CH3Br + NaI → CH3I + NaBr.

Question 20:
Why is chlorobenzene less polar than cyclohexyl chloride?
Answer:

In chlorobenzene, the C-Cl bond has partial double bond character due to resonance, reducing polarity.
In cyclohexyl chloride, the bond is purely sigma, making it more polar.

Question 21:
What happens when iodoform is heated with silver powder?
Answer:

It undergoes dehalogenation to form acetylene (C2H2).
Reaction: 2CHI3 + 6Ag → C2H2 + 6AgI.

Question 22:
How does nucleophilic substitution in tertiary haloalkanes occur?
Answer:

It follows the SN1 mechanism:
1. Formation of a carbocation intermediate.
2. Attack by the nucleophile on the carbocation.

Question 23:
What is the role of pyridine in the Sandmeyer reaction?
Answer:

Pyridine acts as a base to neutralize the HBr formed during the reaction.
It helps in the formation of aryl diazonium salts.

Question 24:
Why is vinyl chloride less reactive than ethyl chloride?
Answer:

In vinyl chloride, the C-Cl bond has partial double bond character due to resonance.
In ethyl chloride, the carbon is sp3 hybridized, making it more reactive.

Question 25:
Give the structure of p-dichlorobenzene.
Answer:

The structure is:
Cl attached to para positions (1,4) of a benzene ring.
Formula: C6H4Cl2.

Question 26:
What is the major product when 2-bromobutane reacts with KOH (aqueous)?
Answer:

The product is 2-butanol (CH3CH(OH)CH2CH3).
This is a nucleophilic substitution reaction where OH replaces Br.

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:
Why are haloalkanes more reactive than haloarenes towards nucleophilic substitution reactions?
Answer:

Haloalkanes are more reactive because the carbon-halogen (C-X) bond is weaker due to less s-character in the hybrid orbital of carbon.
In haloarenes, the C-X bond acquires partial double bond character due to resonance, making it stronger and less reactive.

Question 2:
What is the Finkelstein reaction? Give an example.
Answer:

The Finkelstein reaction is the exchange of a halogen atom in an alkyl halide with another halogen using NaI in acetone.
Example: CH3Cl + NaI → CH3I + NaCl (Iodomethane is formed).

Question 3:
Explain why chlorobenzene is less reactive towards nucleophilic substitution than ethyl chloride.
Answer:

In chlorobenzene, the lone pair on chlorine delocalizes into the benzene ring via resonance, strengthening the C-Cl bond.
In ethyl chloride, no resonance occurs, making the C-Cl bond weaker and more reactive.

Question 4:
What happens when ethyl bromide is treated with alcoholic KOH? Write the reaction.
Answer:

Ethyl bromide undergoes elimination to form ethene.
CH3CH2Br + KOH (alcoholic) → CH2=CH2 + KBr + H2O

Question 5:
Why is iodoform stored in dark bottles?
Answer:

Iodoform decomposes in light to form iodine, which is volatile.
Dark bottles prevent photochemical decomposition.

Question 6:
Name the product formed when benzene diazonium chloride reacts with CuCN/KCN.
Answer:

Benzonitrile (C6H5CN) is formed via the Sandmeyer reaction.

Question 7:
What is the Swarts reaction? Give its general equation.
Answer:

The Swarts reaction converts alkyl chlorides/bromides to alkyl fluorides using AgF or SbF3.
General equation: R-Cl + AgF → R-F + AgCl

Question 8:
Why does p-dichlorobenzene have a higher melting point than its ortho and meta isomers?
Answer:

p-Dichlorobenzene has symmetrical packing in the crystal lattice, increasing intermolecular forces.
Ortho and meta isomers lack symmetry, reducing melting points.

Question 9:
How is chloroform prepared from ethanol? Write the reaction.
Answer:

Ethanol is oxidized to acetaldehyde, then chlorinated to form chloral, which undergoes hydrolysis.
CH3CH2OH → CH3CHO → CCl3CHO → CHCl3

Question 10:
What is the Wurtz reaction? Give its limitation.
Answer:

The Wurtz reaction couples two alkyl halides with Na in dry ether to form higher alkanes.
Limitation: It gives poor yields for unsymmetrical alkanes due to side products.

Question 11:
Why is vinyl chloride less reactive than ethyl chloride towards nucleophilic substitution?
Answer:

In vinyl chloride, the C-Cl bond has partial double bond character due to resonance with the π-electrons of the double bond.
In ethyl chloride, the bond is purely single and weaker.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:
Explain the mechanism of nucleophilic substitution reaction in haloalkanes with an example.
Answer:

The nucleophilic substitution reaction in haloalkanes occurs when a nucleophile replaces the halogen atom. For example, in the hydrolysis of chloromethane (CH3Cl) with aqueous KOH, the nucleophile OH- attacks the carbon atom bonded to chlorine.

Step 1: The nucleophile approaches the carbon atom from the opposite side of the leaving group (Cl).
Step 2: A transition state forms where the carbon is partially bonded to both OH and Cl.
Step 3: The C-Cl bond breaks, and the C-OH bond forms, yielding methanol (CH3OH) and Cl-.

Question 2:
Differentiate between SN1 and SN2 reactions with respect to their mechanism and stereochemistry.
Answer:

SN1 Reaction:
1. Unimolecular (depends only on the concentration of the substrate).
2. Forms a carbocation intermediate.
3. Results in racemization due to planar carbocation.

SN2 Reaction:
1. Bimolecular (depends on both substrate and nucleophile concentration).
2. Occurs in a single step with a transition state.
3. Leads to inversion of configuration (Walden inversion).

Question 3:
Why are aryl halides less reactive towards nucleophilic substitution compared to alkyl halides?
Answer:

Aryl halides are less reactive due to:
1. Resonance stabilization: The lone pair of the halogen delocalizes into the benzene ring, making the C-X bond stronger.
2. sp2 hybridization: The carbon in aryl halides is sp2-hybridized, which holds the halogen more tightly than sp3-hybridized carbon in alkyl halides.
3. Steric hindrance: The planar structure of benzene makes it difficult for nucleophiles to approach the carbon.

Question 4:
Describe the Finkelstein reaction with an example and its significance.
Answer:

The Finkelstein reaction involves the exchange of a halogen in an alkyl halide with another halogen using a metal halide salt.

Example: CH3Br + NaI (acetone) → CH3I + NaBr.

Significance:
1. Used to prepare iodoalkanes, which are more reactive than bromo/chloroalkanes.
2. The reaction is driven by the precipitation of the less soluble salt (NaBr in this case).
3. Works best in polar aprotic solvents like acetone.

Question 5:
Explain the Sandmeyer reaction with a suitable example.
Answer:

The Sandmeyer reaction converts aryl diazonium salts into aryl halides using copper(I) halides.

Example: C6H5N2+Cl- + CuCl → C6H5Cl + N2.

Mechanism:
1. The diazonium salt reacts with CuCl to form an intermediate.
2. Nitrogen gas is released, and the aryl radical combines with Cl to form chlorobenzene.
This reaction is useful for introducing halogens at specific positions in aromatic rings.

Question 6:
How does the presence of a nitro group at the ortho or para position increase the reactivity of aryl halides towards nucleophilic substitution?
Answer:

The nitro group (-NO2) is an electron-withdrawing group that stabilizes the negative charge on the intermediate in nucleophilic substitution.

1. At the ortho or para position, it withdraws electron density via resonance and inductive effect.
2. This stabilizes the Meisenheimer complex (intermediate) formed during the reaction.
3. Example: p-chloronitrobenzene undergoes substitution faster than chlorobenzene due to the nitro group's effect.

Question 7:
Explain the mechanism of nucleophilic substitution reaction in haloalkanes with an example.
Answer:

The nucleophilic substitution reaction in haloalkanes occurs when a nucleophile replaces the halogen atom. It follows two mechanisms: SN1 (unimolecular) and SN2 (bimolecular).

Example (SN2): Hydrolysis of chloromethane (CH3Cl) with NaOH.
Step 1: Nucleophile (OH-) attacks the carbon atom from the opposite side of the leaving group (Cl-).
Step 2: A transition state forms where the C-OH bond partially forms and the C-Cl bond partially breaks.
Step 3: Cl- leaves, forming methanol (CH3OH).

Key point: SN2 is a single-step process with inversion of configuration.

Question 8:
Why do haloarenes undergo electrophilic substitution reactions rather than nucleophilic substitution?
Answer:

Haloarenes undergo electrophilic substitution reactions due to the following reasons:
1. The C-X bond in haloarenes has partial double bond character due to resonance, making it stronger and less reactive towards nucleophiles.
2. The electron-withdrawing nature of the halogen decreases electron density on the benzene ring, making it more susceptible to attack by electrophiles.
3. Nucleophiles face steric hindrance from the bulky benzene ring and the halogen.

Example: Chlorobenzene undergoes nitration (electrophilic substitution) to form nitrochlorobenzene.

Question 9:
Differentiate between alkyl halides and aryl halides based on their reactivity towards nucleophilic substitution.
Answer:

Alkyl halides and aryl halides differ in reactivity as follows:

  • Alkyl halides: Highly reactive towards nucleophilic substitution due to the polar C-X bond and absence of resonance stabilization.
  • Aryl halides: Less reactive due to resonance stabilization of the C-X bond and sp2 hybridized carbon.

Reason: In aryl halides, the lone pair of the halogen delocalizes into the benzene ring, strengthening the C-X bond.

Question 10:
Explain the Finkelstein reaction with a suitable example and its significance.
Answer:

The Finkelstein reaction is a halogen exchange reaction where an alkyl chloride/bromide is converted to an alkyl iodide using NaI in acetone.
Example: CH3Br + NaI → CH3I + NaBr (ppt).
Significance:
1. Used to prepare iodoalkanes, which are more reactive in substitution reactions.
2. The reaction proceeds due to the precipitation of NaBr/NaCl, driving equilibrium forward (Le Chatelier’s principle).

Question 11:
How does the boiling point of haloalkanes vary with the size of the halogen atom? Explain with an example.
Answer:

The boiling point of haloalkanes increases with the size of the halogen atom due to increased van der Waals forces.
Example:
CH3F (bp: -78°C) < CH3Cl (bp: -24°C) < CH3Br (bp: 3.5°C) < CH3I (bp: 42°C).
Reason: Larger halogens have more electrons, leading to stronger intermolecular forces.

Question 12:
Describe the Sandmeyer reaction with a chemical equation and its application.
Answer:

The Sandmeyer reaction converts aryl diazonium salts to aryl halides using copper halides.
Equation: C6H5N2+Cl- + CuCl → C6H5Cl + N2↑.
Application: Used to synthesize chloro/bromo/iodoarenes, which are key intermediates in dyes and pharmaceuticals.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:
Explain the nucleophilic substitution reactions in haloalkanes with emphasis on SN1 and SN2 mechanisms. Provide examples.
Answer:
Theoretical Framework

Haloalkanes undergo nucleophilic substitution reactions where a nucleophile replaces the halogen atom. The two primary mechanisms are SN1 (unimolecular) and SN2 (bimolecular). SN1 involves a carbocation intermediate, while SN2 occurs in a single step with inversion of configuration.

Evidence Analysis
  • SN1 Mechanism: Observed in tertiary haloalkanes (e.g., (CH3)3C-Br) due to stable carbocation formation.
  • SN2 Mechanism: Favored in primary haloalkanes (e.g., CH3-Br) due to less steric hindrance.
Critical Evaluation

Our textbook shows that SN1 is faster in polar protic solvents, while SN2 dominates in polar aprotic solvents. Sterics and solvent effects critically influence the pathway.

Future Implications

Understanding these mechanisms aids in designing synthetic routes for pharmaceuticals, such as converting chloroethane to ethanol.

Question 2:
Compare the reactivity of haloalkanes and haloarenes towards nucleophilic substitution. Justify with electronic and steric factors.
Answer:
Theoretical Framework

Haloalkanes are more reactive than haloarenes in nucleophilic substitution due to differences in hybridization and resonance. The C-X bond in haloarenes has partial double-bond character.

Evidence Analysis
  • Electronic Factors: In haloarenes (e.g., chlorobenzene), the lone pair on halogen delocalizes into the ring, strengthening the C-Cl bond.
  • Steric Factors: Haloalkanes (e.g., ethyl chloride) lack resonance, making the C-Cl bond weaker and more accessible.
Critical Evaluation

Our textbook highlights that haloarenes require extreme conditions (e.g., high temperature/pressure) for substitution, whereas haloalkanes react readily.

Future Implications

This knowledge is vital for industrial processes like the Dow process, where chlorobenzene is converted to phenol.

Question 3:
Describe the preparation of chloroform from ethanol and acetone. Include chemical equations and mechanistic steps.
Answer:
Theoretical Framework

Chloroform (CHCl3) is prepared via the haloform reaction, where a methyl ketone (like acetone) or ethanol is treated with bleaching powder (CaOCl2) and water.

Evidence Analysis
  • From Ethanol: C2H5OH + Cl2 → CH3CHO → CHCl3 (via successive chlorination).
  • From Acetone: (CH3)2CO + 3Cl2 → CCl3-CO-CH3 + NaOH → CHCl3.
Critical Evaluation

Our textbook shows that the reaction proceeds through enolate formation and trihalogenation, followed by cleavage.

Future Implications

Chloroform’s anesthetic properties, though outdated, were historically significant, emphasizing the importance of this synthesis.

Question 4:
Analyze the environmental impact of polyhalogen compounds like DDT and BHC. Discuss their biodegradability and toxicity.
Answer:
Theoretical Framework

Polyhalogen compounds (e.g., DDT, BHC) are persistent organic pollutants (POPs) due to their stability and lipophilicity, leading to bioaccumulation.

Evidence Analysis
  • DDT: Causes eggshell thinning in birds; banned under Stockholm Convention.
  • BHC: Used as a pesticide but accumulates in fatty tissues, causing neurotoxicity.
Critical Evaluation

Our textbook highlights their slow biodegradability due to strong C-Cl bonds, posing long-term ecological risks.

Future Implications

Alternatives like neonicotinoids are being explored, but their environmental safety remains debated.

Question 5:
Explain the Sandmeyer reaction for synthesizing aryl halides from diazonium salts. Include mechanism and applications.
Answer:
Theoretical Framework

The Sandmeyer reaction converts diazonium salts (ArN2+) to aryl halides (Ar-X) using copper(I) halides (CuX) as catalysts.

Evidence Analysis
  • Mechanism: ArN2+ + CuX → Ar• + N2 + CuX2 → Ar-X (via radical intermediate).
  • Example: C6H5N2+Cl- + CuCl → C6H5-Cl (chlorobenzene).
Critical Evaluation

Our textbook shows this method is superior to direct halogenation for introducing halogens at specific positions.

Future Implications

Widely used in dye industries, e.g., synthesizing chlorobenzene for aniline production.

Question 6:
Explain the nucleophilic substitution reactions in haloalkanes with mechanisms. Compare SN1 and SN2 reactions.
Answer:
Theoretical Framework

We studied that nucleophilic substitution reactions involve the replacement of a halogen atom in haloalkanes by a nucleophile. Two mechanisms exist: SN1 (unimolecular) and SN2 (bimolecular).

Evidence Analysis
  • SN1: Two-step process forming a carbocation intermediate. Example: Tertiary butyl chloride hydrolysis.
  • SN2: One-step process with inversion of configuration. Example: Methyl bromide with NaOH.
Critical Evaluation

SN1 favors tertiary substrates due to stable carbocations, while SN2 prefers primary substrates due to less steric hindrance.

Future Implications

Understanding these mechanisms helps in designing synthetic routes for pharmaceuticals.

Question 7:
Discuss the preparation methods of haloarenes and their industrial significance.
Answer:
Theoretical Framework

Haloarenes are prepared via electrophilic substitution (e.g., chlorination) or Sandmeyer reaction (diazotization).

Evidence Analysis
  • Chlorobenzene is made by benzene chlorination using Cl2/FeCl3.
  • Sandmeyer reaction converts aryl diazonium salts to haloarenes (e.g., iodobenzene from aniline).
Critical Evaluation

These methods ensure high purity, crucial for pesticides like DDT.

Future Implications

Green chemistry approaches are emerging to reduce toxic byproducts.

Question 8:
Analyze the stability of aryl halides compared to alkyl halides in nucleophilic substitution.
Answer:
Theoretical Framework

Aryl halides resist nucleophilic substitution due to resonance stabilization and sp2 hybridization.

Evidence Analysis
  • Chlorobenzene’s C-Cl bond has partial double-bond character, unlike alkyl halides.
  • Example: NaOH doesn’t replace Cl in chlorobenzene but does in ethyl chloride.
Critical Evaluation

This stability makes aryl halides useful in high-temperature applications.

Future Implications

Catalysts like copper are being explored to activate aryl halides.

Question 9:
Describe the Wurtz-Fittig reaction and its limitations in synthesizing haloalkanes.
Answer:
Theoretical Framework

The Wurtz-Fittig reaction couples alkyl/aryl halides with sodium to form higher hydrocarbons.

Evidence Analysis
  • Example: Bromobenzene + ethyl bromide + Na → ethylbenzene.
  • Limitations: Poor yields with tertiary halides and formation of side products.
Critical Evaluation

Despite limitations, it’s valuable for asymmetric alkane synthesis.

Future Implications

Modern variants use palladium catalysts to improve efficiency.

Question 10:
Evaluate the environmental impact of polyhalogen compounds like CFCs and DDT.
Answer:
Theoretical Framework

Polyhalogen compounds like CFCs (ozone depletion) and DDT (bioaccumulation) have severe ecological effects.

Evidence Analysis
  • CFCs release chlorine radicals, destroying ozone (e.g., Antarctic ozone hole).
  • DDT persists in food chains, harming birds (e.g., eggshell thinning).
Critical Evaluation

Global bans (Montreal Protocol) show policy success but require alternatives.

Future Implications

Research focuses on biodegradable halogen substitutes.

Question 11:
Explain the mechanism of nucleophilic substitution reaction in haloalkanes with a suitable example. Discuss the factors affecting the rate of this reaction.
Answer:

The nucleophilic substitution reaction in haloalkanes occurs when a nucleophile replaces the halogen atom. The mechanism can proceed via SN1 (unimolecular) or SN2 (bimolecular) pathways, depending on the structure of the haloalkane and reaction conditions.

Example (SN2): Hydrolysis of chloromethane (CH3Cl) with aqueous NaOH:
CH3Cl + OH- → CH3OH + Cl-
The nucleophile (OH-) attacks the carbon from the opposite side of the leaving group (Cl-), resulting in inversion of configuration.

Factors affecting the rate:

  • Nature of substrate: Tertiary haloalkanes favor SN1 due to stability of carbocation, while primary favor SN2.
  • Nucleophile strength: Strong nucleophiles (e.g., OH-) favor SN2.
  • Solvent: Polar protic solvents favor SN1, while polar aprotic favor SN2.
  • Leaving group ability: Better leaving groups (e.g., I- > Br- > Cl-) increase reaction rate.

This reaction is fundamental in organic synthesis, enabling conversion of haloalkanes to alcohols, ethers, and other functional groups.

Question 12:
Compare the reactivity of haloalkanes and haloarenes towards nucleophilic substitution reactions. Justify your answer with reasons.
Answer:

Haloalkanes are more reactive than haloarenes towards nucleophilic substitution due to the following reasons:

1. Electronic Effects:
- In haloalkanes, the carbon-halogen bond is polar, making the carbon electrophilic.
- In haloarenes, the halogen's lone pair participates in resonance with the benzene ring, reducing electrophilicity.

2. Hybridization:
- Haloalkanes have sp3-hybridized carbon, which is more accessible to nucleophiles.
- Haloarenes have sp2-hybridized carbon, making the bond stronger and less reactive.

3. Stability of Intermediate:
- Haloalkanes form stable carbocations (in SN1) or transition states (in SN2).
- Haloarenes form unstable benzyne intermediates, requiring harsh conditions.

Example: Chlorobenzene does not undergo hydrolysis easily, while chloroethane reacts readily with NaOH.

Question 13:
Describe the preparation of chloroform from ethanol and its uses. Write the chemical equations involved.
Answer:

Preparation of chloroform (CHCl3) from ethanol:
1. Step 1: Ethanol is oxidized to acetaldehyde using K2Cr2O7/H+.
CH3CH2OH + [O] → CH3CHO + H2O
2. Step 2: Acetaldehyde reacts with chlorine in the presence of NaOH (haloform reaction).
CH3CHO + 3Cl2 + 4NaOH → CHCl3 + HCOONa + 3NaCl + 3H2O

Uses of chloroform:

  • As an anesthetic (now replaced due to toxicity).
  • Solvent for fats, alkaloids, and rubber.
  • Manufacture of pesticides like chloropicrin.
  • Preservation of anatomical specimens.

Note: Chloroform is stored in dark bottles to prevent oxidation to poisonous phosgene (COCl2).

Question 14:
Explain the Sandmeyer reaction with an example. How is it useful in the synthesis of aryl halides?
Answer:

The Sandmeyer reaction is a method to synthesize aryl halides by replacing the diazonium group (-N2+) with a halogen.

Example: Synthesis of chlorobenzene from benzene diazonium chloride.
1. Step 1: Diazotization of aniline.
C6H5NH2 + NaNO2 + 2HCl → C6H5N2+Cl- + NaCl + 2H2O
2. Step 2: Reaction with CuCl.
C6H5N2+Cl- + CuCl → C6H5Cl + N2 + Cu+

Utility in synthesis:

  • Provides a controlled method to introduce halogens at specific positions.
  • Works for Cl, Br, and CN (using CuCN).
  • Avoids direct halogenation, which may lead to mixtures.

Note: The reaction requires freshly prepared diazonium salts due to their instability.

Question 15:
Compare and contrast the chemical reactivity of haloalkanes and haloarenes towards nucleophilic substitution reactions. Justify your answer with suitable examples.
Answer:

Haloalkanes and haloarenes exhibit stark differences in reactivity towards nucleophilic substitution due to their structural and electronic properties.

Haloalkanes:

  • Undergo nucleophilic substitution readily via SN1 or SN2 mechanisms.
  • Example: CH3CH2Br + KOH → CH3CH2OH + KBr (SN2).
  • The sp3-hybridized carbon and weaker C-X bond facilitate substitution.

Haloarenes:

  • Resistant to nucleophilic substitution due to:
    • sp2-hybridization and resonance stabilization of C-X bond.
    • Partial double-bond character of C-X bond due to delocalization of lone pairs from halogen into the ring.
  • Substitution requires extreme conditions (e.g., high temperature/pressure or strong nucleophiles).
  • Example: Chlorobenzene reacts with NaOH only at 623K and 300 atm to form phenol.

Justification: The difference arises from the electronic effects (resonance in haloarenes) and hybridization (sp2 vs. sp3). Haloalkanes lack resonance stabilization, making their C-X bond more reactive.

Question 16:
Explain the mechanism of nucleophilic substitution reaction in haloalkanes with a suitable example. Discuss the factors affecting the rate of this reaction.
Answer:

The nucleophilic substitution reaction in haloalkanes occurs when a nucleophile replaces the halogen atom. There are two main mechanisms: SN1 (Unimolecular Nucleophilic Substitution) and SN2 (Bimolecular Nucleophilic Substitution).

Example: Hydrolysis of tert-butyl bromide (SN1) and methyl bromide (SN2).

Mechanism:
SN1:
1. Formation of a carbocation intermediate (slow step).
2. Nucleophile attacks the carbocation (fast step).
SN2:
1. Nucleophile attacks the carbon from the opposite side of the leaving group.
2. Simultaneous bond formation and breaking (single step).

Factors affecting the rate:

  • Nature of substrate: Tertiary haloalkanes favor SN1, while primary favor SN2.
  • Nucleophile strength: Strong nucleophiles favor SN2.
  • Solvent: Polar protic solvents favor SN1, while polar aprotic favor SN2.
  • Leaving group: Better leaving groups (e.g., I-) increase the rate.

This reaction is crucial in organic synthesis, such as converting haloalkanes to alcohols, ethers, or amines.

Question 17:
Explain the mechanism of nucleophilic substitution reaction in haloalkanes with a suitable example. Differentiate between SN1 and SN2 reactions.
Answer:

The nucleophilic substitution reaction in haloalkanes involves the replacement of a halogen atom by a nucleophile. There are two main mechanisms: SN1 (Unimolecular Nucleophilic Substitution) and SN2 (Bimolecular Nucleophilic Substitution).

Example: Hydrolysis of tert-butyl bromide (SN1) and methyl bromide (SN2) with aqueous NaOH.

Mechanism of SN1:
1. Step 1: The haloalkane undergoes ionization to form a carbocation and a halide ion.
2. Step 2: The nucleophile (OH-) attacks the carbocation to form the product.

Mechanism of SN2:
1. The nucleophile attacks the carbon atom from the opposite side of the leaving group (halogen).
2. A transition state is formed where the carbon is partially bonded to both the nucleophile and the leaving group.
3. The leaving group departs, and the product is formed.

Differences:

  • SN1 is a two-step process, while SN2 is a single-step process.
  • SN1 forms a carbocation intermediate, whereas SN2 forms a transition state.
  • SN1 is favored for tertiary haloalkanes, while SN2 is favored for primary haloalkanes.
  • SN1 shows racemization, while SN2 leads to inversion of configuration.

Question 18:
Explain the mechanism of nucleophilic substitution reaction in haloalkanes with the help of a suitable example. Discuss the factors affecting the rate of this reaction.
Answer:

The nucleophilic substitution reaction in haloalkanes involves the replacement of a halogen atom by a nucleophile. This reaction can proceed via two mechanisms: SN1 (Unimolecular Nucleophilic Substitution) and SN2 (Bimolecular Nucleophilic Substitution).


Example (SN2 Mechanism): Hydrolysis of chloromethane (CH3Cl) with aqueous NaOH to form methanol (CH3OH).
CH3Cl + NaOH → CH3OH + NaCl


Mechanism Steps (SN2):
1. The nucleophile (OH-) attacks the carbon atom bonded to the halogen from the opposite side of the leaving group (Cl-).
2. A transition state forms where the carbon is partially bonded to both OH- and Cl-.
3. The leaving group (Cl-) departs, and the product (CH3OH) is formed.


Factors Affecting the Rate:

  • Nature of the substrate: Primary haloalkanes favor SN2, while tertiary favor SN1.
  • Strength of the nucleophile: Strong nucleophiles (e.g., OH-) favor SN2.
  • Leaving group ability: Better leaving groups (e.g., I- > Br- > Cl-) increase the rate.
  • Solvent: Polar protic solvents favor SN1, while polar aprotic solvents favor SN2.


Additional Insight: Steric hindrance in tertiary haloalkanes slows down SN2 but accelerates SN1 due to carbocation stability.

Question 19:
Explain the mechanism of nucleophilic substitution reaction in haloalkanes with a suitable example. Differentiate between SN1 and SN2 reactions based on their kinetics, stereochemistry, and effect of solvent.
Answer:

The nucleophilic substitution reaction in haloalkanes involves the replacement of a halogen atom by a nucleophile. Two primary mechanisms are observed: SN1 (Unimolecular Nucleophilic Substitution) and SN2 (Bimolecular Nucleophilic Substitution).

Example: Hydrolysis of tert-butyl bromide (SN1) and methyl bromide (SN2) with aqueous NaOH.

  • SN1 Mechanism:
    1. Step 1: Slow ionization of haloalkane to form a carbocation.
    2. Step 2: Fast attack by nucleophile on carbocation.
    Kinetics: First-order (depends only on haloalkane concentration).
    Stereochemistry: Racemization due to planar carbocation.
    Solvent Effect: Polar protic solvents stabilize carbocation.
  • SN2 Mechanism:
    1. Single-step reaction where nucleophile attacks from the backside.
    Kinetics: Second-order (depends on both haloalkane and nucleophile concentration).
    Stereochemistry: Inversion of configuration (Walden inversion).
    Solvent Effect: Polar aprotic solvents favor SN2.

Key Difference: SN1 is favored for tertiary haloalkanes, while SN2 is favored for primary haloalkanes.

Question 20:
Discuss the environmental impact of haloalkanes and haloarenes, focusing on ozone layer depletion and bioaccumulation. Explain how the Montreal Protocol addressed these issues with examples.
Answer:

Haloalkanes and haloarenes, particularly chlorofluorocarbons (CFCs) and dichlorodiphenyltrichloroethane (DDT), have significant environmental consequences.

  • Ozone Layer Depletion:
    CFCs (e.g., Freon-12) release chlorine radicals in the stratosphere, which catalytically destroy ozone molecules.
    Reaction: Cl• + O3 → ClO• + O2 (cycle repeats).
    This leads to increased UV radiation exposure, causing skin cancer and ecosystem damage.
  • Bioaccumulation:
    DDT and other halogenated compounds persist in the environment, accumulating in fatty tissues of organisms.
    This causes biomagnification in food chains, leading to toxicity in higher trophic levels (e.g., birds).

Montreal Protocol (1987):
A global treaty to phase out ozone-depleting substances like CFCs.
Example: Replacement of CFCs with hydrofluorocarbons (HFCs), which have no chlorine and lower ozone depletion potential.

Outcome: Reduced CFC emissions and gradual ozone layer recovery, demonstrating effective international cooperation.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:
A student observed that chlorobenzene undergoes nucleophilic substitution slower than chloroethane. Explain this phenomenon with reference to hybridization and resonance effects.
Answer:
Case Deconstruction

Chlorobenzene has a sp² hybridized carbon bonded to chlorine, while chloroethane has sp³ hybridization. The C-Cl bond in chlorobenzene is shorter and stronger due to partial double bond character from resonance.

Theoretical Application
  • In chlorobenzene, lone pairs on chlorine delocalize into the benzene ring, making the C-Cl bond harder to break.
  • Chloroethane lacks resonance, so its C-Cl bond is weaker and more reactive.
Critical Evaluation

Our textbook shows that nucleophilic substitution in chlorobenzene requires harsh conditions, while chloroethane reacts readily under mild conditions.

Question 2:
Analyze why iodoalkanes are more reactive than chloroalkanes in SN2 reactions, considering bond dissociation energies and polarizability.
Answer:
Case Deconstruction

Iodoalkanes have weaker C-I bonds (234 kJ/mol) compared to C-Cl bonds (339 kJ/mol), making them easier to break.

Theoretical Application
  • Iodine's larger size increases polarizability, stabilizing the transition state in SN2 reactions.
  • Chlorine's higher electronegativity creates a stronger bond but reduces nucleophilic attack efficiency.
Critical Evaluation

We studied that while CH₃I reacts faster in SN2, CH₃Cl requires more energy. This aligns with bond energy data and transition state theory.

Question 3:
Compare the environmental impact of CFCs versus HCFCs as refrigerants, referencing ozone depletion potential and global warming potential.
Answer:
Case Deconstruction

CFCs like CCl₃F have high ozone depletion potential (ODP=1) due to chlorine release, while HCFCs like CHClF₂ have hydrogen atoms that reduce atmospheric lifetime.

Theoretical Application
PropertyCFCsHCFCs
ODP0.6-1.00.01-0.1
GWP4000-10000700-2000
Critical Evaluation

Our textbook shows HCFCs are transitional replacements, but still contribute to climate change, prompting research on hydrofluorocarbons.

Question 4:
Explain why aryl halides don't undergo SN1 or SN2 reactions easily, while allyl halides are highly reactive, using electronic and steric factors.
Answer:
Case Deconstruction

Aryl halides have sp² hybridization and resonance stabilization, while allyl halides form stabilized carbocations through p-π conjugation.

Theoretical Application
  • In aryl halides, backside attack is blocked by the ring, and carbocations are unstable.
  • Allyl carbocations are stabilized by delocalization across the π-system.
Critical Evaluation

We studied that CH₂=CH-CH₂Cl reacts 1000x faster than C₆H₅Cl in substitution, confirming the importance of intermediate stability.

Question 5:
A student observes that chlorobenzene undergoes nucleophilic substitution slower than chloroethane. Explain this phenomenon with reference to hybridization and resonance effects.
Answer:
Case Deconstruction

Chlorobenzene has a sp²-hybridized carbon bonded to chlorine, while chloroethane has sp³-hybridization. The C-Cl bond in chlorobenzene is shorter and stronger due to partial double bond character from resonance.

Theoretical Application
  • In chlorobenzene, the lone pair of chlorine delocalizes into the benzene ring, making the C-Cl bond harder to break.
  • Chloroethane lacks resonance, so the C-Cl bond is weaker and more reactive.
Critical Evaluation

Our textbook shows that nucleophilic substitution in aryl halides requires extreme conditions, unlike alkyl halides. This aligns with the observed reactivity difference.

Question 6:
While preparing 2-bromopropane from isopropyl alcohol, a student notices the formation of a white precipitate. Identify the precipitate and justify its formation using a chemical equation.
Answer:
Case Deconstruction

The white precipitate is sodium bromide (NaBr), formed due to the reaction between HBr (generated in situ) and NaOH (if used).

Theoretical Application
  • Isopropyl alcohol reacts with NaBr and H₂SO₄ to form 2-bromopropane and NaHSO₄.
  • Excess H₂SO₄ neutralizes NaOH, producing Na₂SO₄ and water.
Critical Evaluation

We studied that SN1 reactions favor tertiary alcohols, explaining the ease of 2-bromopropane formation. The precipitate confirms the presence of bromide ions.

Question 7:
Compare the boiling points of 1-chlorobutane, 1-bromobutane, and 1-iodobutane. Justify the trend using van der Waals forces.
Answer:
Case Deconstruction

The boiling points increase in the order: 1-chlorobutane (78°C) < 1-bromobutane (101°C) < 1-iodobutane (130°C).

Theoretical Application
  • Larger halogen atoms (I > Br > Cl) increase molecular size and van der Waals forces.
  • Greater electron cloud polarizability in iodoalkanes enhances intermolecular attraction.
Critical Evaluation

Our textbook shows that boiling points correlate with atomic mass and surface area. This trend is consistent across homologous series.

Question 8:
A Grignard reagent prepared from bromobenzene fails to react with carbon dioxide. Analyze two possible errors in the experimental procedure.
Answer:
Case Deconstruction

The Grignard reagent (phenylmagnesium bromide) may have decomposed or formed incorrectly due to procedural errors.

Theoretical Application
  • Exposure to moisture hydrolyzes the reagent to benzene.
  • Insufficient activation of magnesium or improper anhydrous conditions prevents formation.
Critical Evaluation

We studied that Grignard reagents are highly moisture-sensitive. The absence of reaction suggests compromised reagent integrity, as CO₂ normally forms benzoic acid.

Question 9:
A student observed that chloroform (CHCl3) stored in sunlight forms a poisonous gas. Identify the gas and explain the mechanism of its formation. How can this reaction be prevented?
Answer:
Case Deconstruction

Chloroform decomposes in sunlight to form phosgene (COCl2), a toxic gas. Our textbook shows this occurs via free radical oxidation.

Theoretical Application
  • Mechanism: CHCl3 + O2 → COCl2 + HCl
  • Prevention: Store in amber bottles with 1% ethanol to quench radicals.
Critical Evaluation

This highlights the instability of polyhalogen compounds. Similar decomposition occurs in carbon tetrachloride (CCl4).

Question 10:
In a lab, 2-bromopentane underwent substitution with KOH to form two products. Analyze the reaction mechanism and predict the major product with reasoning.
Answer:
Case Deconstruction

This is a nucleophilic substitution (SN2 vs SN1) scenario. 2-bromopentane is a secondary halide.

Theoretical Application
  • Major product: Pentan-2-ol (SN2 dominates due to strong nucleophile KOH)
  • Minor product: Pent-1-ene (E2 elimination)
Critical Evaluation

Our textbook shows secondary halides often show mixed mechanisms. Similar behavior is seen in 1-chlorobutane reactions.

Question 11:
Aryl halides are less reactive than alkyl halides in nucleophilic substitution. Compare their reactivity using resonance and hybridization concepts with two examples.
Answer:
Case Deconstruction

Aryl halides (e.g., chlorobenzene) have sp2 hybridized carbons vs sp3 in alkyl halides (e.g., ethyl chloride).

Theoretical Application
  • Resonance stabilizes C-Cl bond in aryl halides
  • Partial double bond character increases bond strength
Critical Evaluation

This explains why bromoethane undergoes SN2 easily while bromobenzene requires extreme conditions.

Question 12:
The boiling point of iodoethane (72°C) is higher than chloroethane (12°C). Correlate this trend with intermolecular forces and predict the order for fluoro/iodomethanes.
Answer:
Case Deconstruction

Boiling points increase with halogen size due to stronger van der Waals forces.

Theoretical Application
  • Order: CH3I > CH3Br > CH3Cl > CH3F
  • Reason: Increased electron cloud polarizability
Critical Evaluation

This trend matches our textbook data for dihalomethanes. Exception: HF shows H-bonding.

Question 13:

Rahul was performing a nucleophilic substitution reaction using 1-bromopropane with aqueous KOH. He observed the formation of propan-1-ol as the major product. However, when he used tert-butyl bromide under the same conditions, the reaction proceeded much faster but yielded a different product.

(a) Why did tert-butyl bromide react faster than 1-bromopropane?
(b) Identify the product formed in the case of tert-butyl bromide and explain the mechanism involved.

Answer:

(a) Tert-butyl bromide reacts faster than 1-bromopropane because it undergoes SN1 (unimolecular nucleophilic substitution) mechanism due to the formation of a stable tertiary carbocation. The steric hindrance and +I effect of the three alkyl groups stabilize the carbocation intermediate, accelerating the reaction. In contrast, 1-bromopropane follows the SN2 mechanism, which is slower due to steric crowding during the transition state.

(b) The product formed is tert-butyl alcohol ((CH3)3COH). The mechanism involves:

  • Step 1: Ionization of tert-butyl bromide to form a stable tertiary carbocation and bromide ion.
  • Step 2: Nucleophilic attack by OH− on the carbocation to form tert-butyl alcohol.

This is a classic example of SN1 reactivity, where the rate-determining step depends only on the substrate concentration.

Question 14:

Priya was studying the Wurtz-Fittig reaction and attempted to synthesize ethylbenzene by reacting bromoethane and bromobenzene with sodium in dry ether. However, she obtained a mixture of products.

(a) Write the balanced chemical equation for the formation of ethylbenzene.
(b) Explain why a mixture of products is formed and name the other possible products.

Answer:

(a) The balanced equation for the formation of ethylbenzene is:
C6H5Br + C2H5Br + 2Na → C6H5-C2H5 + 2NaBr

(b) A mixture of products is formed because the Wurtz-Fittig reaction involves coupling of two different alkyl/aryl halides, leading to cross-products and self-products. The other possible products are:

  • Biphenyl (C6H5-C6H5) from coupling of two bromobenzene molecules.
  • Butane (C4H10) from coupling of two bromoethane molecules.

This occurs due to the random collision of radicals during the reaction, making it less selective. The yield of ethylbenzene is often lower due to competitive side reactions.

Question 15:

Rahul performed a nucleophilic substitution reaction on 2-bromobutane with aqueous KOH and obtained two isomeric products. However, when he used alcoholic KOH, only one major product was formed.

(a) Identify the two products formed with aqueous KOH and explain why they are isomeric.
(b) Why does alcoholic KOH lead to only one major product? Write the reaction involved.

Answer:

(a) The two isomeric products formed with aqueous KOH are butan-2-ol and butan-1-ol.
These are structural isomers because they have the same molecular formula (C4H9OH) but different connectivity of atoms.

Explanation: Aqueous KOH favors SN2 mechanism, leading to substitution at both primary and secondary carbons, producing two alcohols.

(b) Alcoholic KOH causes elimination (dehydrohalogenation) via E2 mechanism, forming but-2-ene as the major product.

Reaction:
CH3-CH(Br)-CH2-CH3 + KOH (alc.) → CH3-CH=CH-CH3 + KBr + H2O

Reason: The strong base (OH–) in alcohol abstracts a β-hydrogen, leading to elimination rather than substitution.

Question 16:

Priya treated chlorobenzene with NaOH at 623K and 300 atm pressure, yielding Product A. However, when she used chloroethane under the same conditions, Product B was formed.

(a) Name Product A and B, and write their reactions.
(b) Explain why chlorobenzene requires harsh conditions for this reaction compared to chloroethane.

Answer:

(a) Product A: Phenol (C6H5OH)
Reaction: C6H5Cl + NaOH → C6H5OH + NaCl (Dow’s process)

Product B: Ethanol (C2H5OH)
Reaction: C2H5Cl + NaOH → C2H5OH + NaCl

(b) Reason:

  • Chlorobenzene has a C-Cl bond with partial double-bond character due to resonance, making it harder to break.
  • Chloroethane has a pure single C-Cl bond, which is more reactive toward nucleophilic substitution (SN2).

Thus, chlorobenzene requires high temperature and pressure to overcome resonance stabilization.

Question 17:

Rahul performed a nucleophilic substitution reaction between 2-bromopropane and aqueous KOH. He observed the formation of two products, one major and one minor.
(i) Identify the major and minor products formed in this reaction.
(ii) Explain the reason behind the formation of these products using the mechanism involved.

Answer:

(i) The major product is 2-propanol (CH3CH(OH)CH3), and the minor product is 1-propanol (CH3CH2CH2OH).

(ii) The reaction follows the SN2 mechanism for the minor product and the SN1 mechanism for the major product.
In SN1, the carbocation intermediate formed is stabilized by the +I effect of the two methyl groups, making 2-propanol the major product.
In SN2, the nucleophile attacks the less hindered primary carbon, forming 1-propanol as the minor product.

Question 18:

Priya was given two compounds: chlorobenzene and benzyl chloride. She was asked to distinguish between them using a chemical test.
(i) Name the test she can use.
(ii) Describe the observation and chemical reaction involved.

Answer:

(i) Priya can use the aqueous NaOH test followed by AgNO3 to distinguish between the two compounds.

(ii)

  • Benzyl chloride reacts with NaOH to form benzyl alcohol, which further reacts with AgNO3 to give a white precipitate of AgCl.
  • Chlorobenzene does not react with NaOH under normal conditions due to the resonance stabilization of the C-Cl bond, so no precipitate forms with AgNO3.
This difference in reactivity helps distinguish the two compounds.

Question 19:

Rahul was performing a nucleophilic substitution reaction using 1-bromopropane with aqueous KOH. However, he observed the formation of two different products. Explain the possible reaction mechanisms involved and justify the formation of two products with proper chemical equations.

Answer:

The reaction of 1-bromopropane with aqueous KOH can proceed via two mechanisms: SN2 and E2, leading to two different products.


SN2 Mechanism (Substitution):
KOH acts as a nucleophile, attacking the carbon bonded to bromine from the opposite side, resulting in the formation of 1-propanol.
CH3CH2CH2Br + KOH → CH3CH2CH2OH + KBr


E2 Mechanism (Elimination):
KOH acts as a base, abstracting a proton from the β-carbon, leading to the formation of propene.
CH3CH2CH2Br + KOH → CH3CH=CH2 + KBr + H2O


The competition between substitution and elimination depends on factors like the strength of the base and the nature of the alkyl halide.

Question 20:

Priya treated chlorobenzene with chlorine in the presence of anhydrous AlCl3. She observed the formation of a mixture of ortho and para dichlorobenzene. Explain the reaction mechanism and the reason behind the formation of these specific isomers.

Answer:

The reaction is an example of electrophilic aromatic substitution, where chlorobenzene undergoes chlorination in the presence of anhydrous AlCl3 (a Lewis acid catalyst).


Mechanism:
1. AlCl3 polarizes Cl2, generating Cl+ (electrophile).
2. The electrophile attacks the electron-rich benzene ring, forming a carbocation intermediate.
3. Loss of a proton restores aromaticity, yielding dichlorobenzene.


Isomer Formation:
The -Cl group is ortho/para directing due to its +M (mesomeric) effect, which stabilizes the intermediate carbocation at these positions. Hence, ortho and para isomers dominate, while the meta isomer is negligible.

Question 21:
A student performed a nucleophilic substitution reaction on 2-bromobutane with aqueous KOH and obtained two isomeric alcohols.

(a) Identify the two alcohols formed and write their structures.

(b) Explain the mechanism involved in this reaction.

Answer:

(a) The two isomeric alcohols formed are butan-2-ol and 2-methylpropan-2-ol.


Structures:
Butan-2-ol: CH3-CH(OH)-CH2-CH3
2-Methylpropan-2-ol: (CH3)3COH

(b) The reaction proceeds via SN2 and SN1 mechanisms. 2-Bromobutane undergoes SN2 substitution to form butan-2-ol, where the nucleophile (OH-) attacks the carbon from the opposite side of the leaving group (Br-). Simultaneously, a carbocation intermediate forms in SN1, leading to 2-methylpropan-2-ol due to rearrangement for stability.

Question 22:
An organic compound A (C6H5Cl) reacts with NaOH at 623K and high pressure to form compound B (C6H5OH).

(a) Name the reaction and write the chemical equation.

(b) Why is high pressure required in this reaction?

Answer:

(a) The reaction is called nucleophilic aromatic substitution (via benzyne mechanism).
Chemical equation:
C6H5Cl + NaOH → C6H5OH + NaCl


(b) High pressure is required because the reaction involves the formation of a reactive intermediate called benzyne, which is highly unstable. The high pressure ensures the reaction proceeds efficiently by overcoming the energy barrier for benzyne formation and subsequent nucleophilic attack by OH-.

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