Class 12 Chemistry Chapter 10: Haloalkanes and Haloarenes – NCERT Notes & Important Questions

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1. Classification

Haloalkanes and haloarenes are classified on the basis of the number of halogen atoms present and the nature of the carbon atom to which the halogen is bonded.

1.1 On the Basis of Number of Halogen Atoms

Depending upon whether one, two or more halogen atoms are present, organic halogen compounds are classified as monohaloalkanes, dihaloalkanes, or polyhaloalkanes.

C2H5X     CH2X–CH2X     CHX–CHX–CH2X
(Monohaloalkane)    (Dihaloalkane)    (Trihaloalkane)

Fig. 1.1: Classification based on number of halogen atoms.

1.2 Compounds Containing sp3 C–X Bond

In these compounds, the halogen atom is bonded to an sp3-hybridised carbon atom.

(A) Alkyl Halides or Haloalkanes (R–X)

These form a homologous series with general formula CnH2n+1X and are classified as primary (1°), secondary (2°), or tertiary (3°) depending upon the nature of the carbon atom bearing the halogen.

(B) Allylic Halides

In allylic halides, the halogen atom is bonded to an sp3-hybridised carbon atom adjacent to a C=C bond.

Example: CH2=CH–CH2–X (Allyl halide)

(C) Benzylic Halides

In benzylic halides, the halogen atom is bonded to an sp3-hybridised carbon atom adjacent to an aromatic ring.

Example: C6H5–CH2–X (Benzyl halide)

1.3 Compounds Containing sp2 C–X Bond

(A) Vinylic Halides

In vinylic halides, the halogen atom is bonded directly to an sp2-hybridised carbon atom of a C=C bond.

Example: CH2=CH–X (Vinyl halide)

(B) Aryl Halides

In aryl halides, the halogen atom is bonded directly to the sp2-hybridised carbon atom of an aromatic ring.

Example: C6H5–X (Halobenzene)

2. Nomenclature

Haloalkanes and haloarenes are named using both common and IUPAC systems. In the IUPAC system, they are named as halosubstituted hydrocarbons.

2.1 Common vs IUPAC Names

Structure Common Name IUPAC Name
CH3–CH2–CH(Cl)–CH3 sec-Butyl chloride 2-Chlorobutane
(CH3)3C–CH2–Br neo-Pentyl bromide 1-Bromo-2,2-dimethylpropane
(CH3)3C–Br tert-Butyl bromide 2-Bromo-2-methylpropane
CH2=CH–Cl Vinyl chloride Chloroethene
CH2=CH–CH2–Br Allyl bromide 3-Bromoprop-1-ene

2.2 Geminal and Vicinal Dihalides

  • Geminal dihalides: Both halogen atoms are attached to the same carbon atom.
    (Older term: Alkylidene halides)
  • Vicinal dihalides: Halogen atoms are attached to adjacent carbon atoms.
    (Older term: Alkylene dihalides)

3. Nature of C–X Bond

The carbon–halogen bond is polar due to higher electronegativity of halogens. Carbon carries a partial positive charge (δ⁺) while halogen carries a partial negative charge (δ⁻).

3.1 Periodic Trends

  • Bond length: C–F < C–Cl < C–Br < C–I
  • Bond enthalpy: C–F > C–Cl > C–Br > C–I
  • Dipole moment: Depends on both magnitude of charge separation and bond length.
Bond Bond Length (pm) Bond Enthalpy (kJ mol⁻¹)
C–F 139 452
C–Cl 178 351
C–Br 193 293
C–I 214 234

4. Methods of Preparation of Haloalkanes

Haloalkanes are commonly prepared from alcohols, hydrocarbons, and by halogen exchange reactions.

4.1 From Alcohols

The hydroxyl group (–OH) of an alcohol is replaced by a halogen atom using halogen acids (HX), phosphorus halides or thionyl chloride.

General Reactions:
1. R–OH + HCl $\xrightarrow{ZnCl_2}$ R–Cl + H2O (Lucas Reaction)
2. R–OH + PCl5 → R–Cl + POCl3 + HCl
3. 3R–OH + PCl3 → 3R–Cl + H3PO3
4. R–OH + SOCl2 → R–Cl + SO2↑ + HCl↑

4.2 From Hydrocarbons

(A) From Alkanes (Free Radical Halogenation)

Alkanes undergo free radical chlorination or bromination in the presence of ultraviolet light. The reaction produces a complex mixture of isomeric mono- and polyhaloalkanes, making separation of pure compounds difficult.

(B) From Alkenes

(i) Addition of Hydrogen Halides (HX):

Alkenes react with hydrogen halides to form alkyl halides according to Markovnikov’s Rule.

Markovnikov’s Rule:
The negative part of the addendum (halide ion) attaches to the carbon atom possessing a lesser number of hydrogen atoms.
CH3–CH=CH2 + HI → CH3–CH(I)–CH3 (Major product: 2-Iodopropane)

(ii) Addition of Halogens (X2):

Addition of bromine in CCl4 to alkenes produces vicinal dibromides. Decolourisation of reddish-brown bromine solution is used as a test for unsaturation.

4.3 Halogen Exchange Reactions (Named Reactions)

These reactions are specifically useful for the preparation of alkyl iodides and alkyl fluorides.

1. Finkelstein Reaction (For Alkyl Iodides):
R–X + NaI $\xrightarrow{Dry\ Acetone}$ R–I + NaX
(X = Cl, Br; NaCl/NaBr precipitates in dry acetone)
2. Swarts Reaction (For Alkyl Fluorides):
Heating alkyl chloride or bromide with metallic fluorides such as AgF, Hg2F2, CoF2 or SbF3.
CH3–Br + AgF → CH3–F + AgBr

5. Methods of Preparation of Haloarenes

5.1 From Hydrocarbons (Electrophilic Substitution)

Aryl chlorides and bromides are prepared by electrophilic substitution of arenes with chlorine or bromine in the presence of a Lewis acid catalyst (Fe or FeCl3) in the dark.

Toluene + Cl2 $\xrightarrow{Fe/FeCl_3,\ Dark}$ o-Chlorotoluene + p-Chlorotoluene

5.2 From Amines (Sandmeyer’s Reaction)

A primary aromatic amine dissolved in cold aqueous mineral acid (0–5 °C) reacts with sodium nitrite to form a diazonium salt. Treatment with cuprous chloride or cuprous bromide replaces the diazonium group by –Cl or –Br.

Step 1: Diazotisation
Ar–NH2 + NaNO2 + 2HX $\xrightarrow{273–278\ K}$ Ar–N2+X + 2H2O

Step 2: Sandmeyer Reaction
Ar–N2+X $\xrightarrow{CuCl/HCl}$ Ar–Cl + N2

6. Physical Properties

Haloalkanes are generally colourless when pure and possess a pleasant odour. However, alkyl bromides and iodides gradually develop colour on exposure to light due to photochemical decomposition with liberation of bromine or iodine.

6.1 Melting and Boiling Points

Due to the polar nature of the C–X bond and higher molecular mass, intermolecular forces of attraction (dipole–dipole and van der Waals forces) are stronger in haloalkanes than in the corresponding hydrocarbons.

  • Effect of Size of Halogen: For the same alkyl group, boiling points decrease in the order:
    R–I > R–Br > R–Cl > R–F
    Reason: As the size and mass of the halogen increase, van der Waals forces increase.
  • Effect of Branching: In isomeric haloalkanes, boiling points decrease with increase in branching.
    Reason: Branching reduces surface area, thereby decreasing intermolecular attractions.
    Order: Primary > Secondary > Tertiary

6.2 Solubility

Haloalkanes are very slightly soluble in water but are soluble in organic solvents.
Reason: The energy required to break hydrogen bonds in water is greater than the energy released when new attractions form between haloalkane and water molecules.


7. Chemical Reactions of Haloalkanes

The chemical reactions of haloalkanes are broadly classified into:
1. Nucleophilic substitution reactions
2. Elimination reactions
3. Reactions with metals

7.1 Nucleophilic Substitution Reactions

In these reactions, a nucleophile attacks the carbon atom of haloalkane which bears a partial positive charge. The halogen atom departs as a halide ion, acting as a leaving group.

General Reaction:
Nu⁻ + R–X → R–Nu + X⁻

Nucleophilic substitution reactions proceed by two mechanisms: SN2 and SN1.

(A) Substitution Nucleophilic Bimolecular (SN2)

This reaction follows second-order kinetics, i.e., the rate depends on the concentration of both the reactants.

  • Mechanism: Occurs in a single step with simultaneous bond breaking and bond formation.
  • Mode of Attack: Nucleophile attacks from the back side of the leaving group.
  • Stereochemistry: Leads to inversion of configuration (Walden inversion).
  • Reactivity Order: 1° > 2° > 3°
    Reason: Steric hindrance increases with substitution.

(B) Substitution Nucleophilic Unimolecular (SN1)

This reaction follows first-order kinetics and occurs in polar protic solvents such as water and alcohols.

  • Mechanism: Proceeds in two steps:
    Step 1: Formation of carbocation (slow, rate-determining step)
    Step 2: Nucleophilic attack (fast step)
  • Stereochemistry: Results in racemisation due to planar $sp^2$ carbocation. Racemisation occurs because the nucleophile can attack from both sides, leading to partial inversion and partial retention of configuration.
  • Reactivity Order: 3° > 2° > 1°
    Reason: Greater stability of carbocation.

7.2 Stereochemical Aspects of Nucleophilic Substitution

To understand the stereochemistry of SN1 and SN2 reactions, it is essential to understand basic stereochemical concepts such as optical activity, chirality, enantiomers, retention and inversion.

(A) Optical Activity and Chirality

Certain organic compounds rotate the plane of plane-polarised light when it is passed through their solution. Such compounds are called optically active.
• If rotation is towards the right (clockwise), the compound is dextrorotatory ($d$ or $+$).
• If rotation is towards the left (anticlockwise), the compound is laevorotatory ($l$ or $-$).

Chirality: A molecule which is non-superimposable on its mirror image is said to be chiral. A carbon atom attached to four different atoms or groups is called a chiral carbon or asymmetric carbon. Such a carbon atom is also known as a stereogenic centre.

(B) Enantiomers and Racemic Mixture

  • Enantiomers: Stereoisomers which are non-superimposable mirror images of each other. They have identical physical properties but rotate plane-polarised light in opposite directions.
  • Racemic Mixture: An equimolar mixture of two enantiomers. It shows no optical activity because the rotations cancel each other. The process of formation of a racemic mixture is called racemisation.

(C) Retention and Inversion of Configuration

  • Retention: The relative configuration of atoms around the chiral carbon remains unchanged during a reaction.
  • Inversion: The relative configuration of atoms around the chiral carbon is inverted. This occurs in SN2 reactions and is known as Walden inversion.

7.3 Elimination Reactions (Dehydrohalogenation)

When a haloalkane containing a β-hydrogen atom is heated with an alcoholic solution of potassium hydroxide (alc. KOH), there is elimination of a hydrogen atom from the β-carbon and a halide ion from the α-carbon, resulting in the formation of an alkene.

CH3–CH2–CH2–Br + KOH (alc.)  $\xrightarrow{\Delta}$  CH3–CH=CH2 + KBr + H2O

Zaitsev Rule

When a haloalkane can eliminate HX in more than one way, the preferred product is the more substituted alkene, i.e., the alkene having a greater number of alkyl groups attached to the double bonded carbon atoms.

Example: 2-Bromobutane
CH3–CH(Br)–CH2–CH3 $\xrightarrow{alc.\ KOH}$ Mixture of alkenes

But-2-ene (Major, ~81%)
But-1-ene (Minor, ~19%)

7.4 Reaction with Metals

(A) Reaction with Magnesium (Grignard Reagent)

Haloalkanes react with magnesium metal in dry ether to form alkyl magnesium halides, known as Grignard reagents.

R–X + Mg $\xrightarrow{Dry\ Ether}$ R–Mg–X

(B) Wurtz Reaction

When alkyl halides react with sodium metal in dry ether, a higher alkane containing double the number of carbon atoms is formed.

2R–X + 2Na $\xrightarrow{Dry\ Ether}$ R–R + 2NaX

7.5 Chemical Reactions of Haloarenes

(A) Nucleophilic Substitution (Low Reactivity)

Haloarenes are much less reactive towards nucleophilic substitution reactions than haloalkanes due to:

  1. Resonance effect: The C–X bond acquires partial double bond character, making it shorter and stronger.
  2. Hybridisation: The carbon atom is sp2 hybridised, leading to a stronger C–X bond.
  3. Instability of phenyl carbocation: SN1 mechanism is not feasible.

Nucleophilic substitution in haloarenes occurs only under drastic conditions or in the presence of strong electron-withdrawing groups (–NO2) at ortho/para positions.

Dow’s Process:
Chlorobenzene + NaOH $\xrightarrow{623\ K,\ 300\ atm}$ Sodium phenoxide $\xrightarrow{H^+}$ Phenol

(B) Electrophilic Substitution Reactions

Haloarenes undergo electrophilic substitution reactions such as nitration, sulphonation and halogenation. The halogen atom is ortho–para directing due to resonance effect but deactivating due to its –I effect.

Haloarenes generally do not undergo Friedel–Crafts reactions easily because the halogen forms a stable complex with Lewis acids like AlCl3, reducing the reactivity of the aromatic ring.

(C) Reaction with Metals

Wurtz–Fittig Reaction:
Ar–X + R–X + 2Na $\xrightarrow{Dry\ Ether}$ Ar–R + 2NaX

Fittig Reaction:
2Ar–X + 2Na $\xrightarrow{Dry\ Ether}$ Ar–Ar + 2NaX

8. Polyhalogen Compounds

Organic compounds containing more than one halogen atom are known as polyhalogen compounds. Many of these compounds are of industrial, medicinal, and agricultural importance.

8.1 Dichloromethane (Methylene Chloride)

  • Formula: CH2Cl2
  • Uses: Widely used as a solvent and paint remover.
  • Health Effects: Direct contact causes mild redness of the skin. High concentration in air causes dizziness and nausea.

8.2 Trichloromethane (Chloroform)

  • Formula: CHCl3
  • Uses: Solvent for fats and alkaloids. Earlier used as a general anaesthetic.
  • Storage Precaution: Chloroform is slowly oxidised by air in the presence of light to an extremely poisonous gas, carbonyl chloride (Phosgene).
2CHCl3 + O2 2COCl2 (Phosgene) + 2HCl

8.3 Triiodomethane (Iodoform)

  • Formula: CHI3
  • Properties: Yellow crystalline solid with an objectionable smell.
  • Use: Used as an antiseptic due to the liberation of free iodine (I2).

8.4 Tetrachloromethane (Carbon Tetrachloride)

  • Formula: CCl4
  • Uses: Manufacture of refrigerants and propellants. Earlier used as a fire extinguisher (Pyrene).
  • Environmental Impact: Causes depletion of the ozone layer and contributes to global warming.

8.5 Freons

  • The chlorofluorocarbon compounds of methane and ethane are collectively known as Freons.
  • Example: Freon-12 (CCl2F2).
  • Impact: They diffuse into the stratosphere and initiate radical chain reactions that disturb the natural ozone balance.

8.6 p,p’-Dichlorodiphenyltrichloroethane (DDT)

  • DDT was the first chlorinated organic insecticide.
  • Preparation: Heating chlorobenzene with chloral (CCl3CHO) in the presence of concentrated H2SO4.
  • Issue: It is non-biodegradable, fat-soluble and undergoes biomagnification. Hence, it is banned in many countries.

Fig 8.1: Chemical structure of DDT (p,p’-Dichlorodiphenyltrichloroethane).


9. Practice Problems (Board Level)

Q1. Why is sulphuric acid not used during the reaction of alcohols with KI?
Ans: Concentrated H2SO4 is a strong oxidising agent. It oxidises HI to iodine (I2), thus preventing the formation of alkyl iodide.
Q2. Which compound hydrolyses faster by SN1 mechanism: Benzyl chloride or 1-Chloro-1-phenylethane?
Ans: 1-Chloro-1-phenylethane hydrolyses faster because it forms a more stable secondary benzylic carbocation than the primary benzylic carbocation formed from benzyl chloride.
Q3. Convert Propene to 1-Propanol.
Ans:
Step 1: Propene + HBr (Peroxide) → 1-Bromopropane
Step 2: 1-Bromopropane + Aqueous KOH → 1-Propanol
Q4. Explain why the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
Ans:
• Due to resonance, the C–Cl bond in chlorobenzene acquires partial double bond character.
• The carbon atom bonded to chlorine in chlorobenzene is sp2 hybridised and holds the electron pair more tightly.

10. Frequently Asked Questions (FAQ)

Q1. What is the difference between enantiomers and diastereomers?

Enantiomers are non-superimposable mirror images of each other, whereas diastereomers are stereoisomers that are not mirror images.

Q2. Why are haloalkanes insoluble in water despite being polar?

Haloalkanes cannot form hydrogen bonds with water. The energy released on solvation is insufficient to break the strong hydrogen bonding in water.

Q3. Which reaction involves the formation of a carbocation?

The SN1 reaction proceeds through the formation of a carbocation intermediate. Reactivity depends on carbocation stability (3° > 2° > 1°).

🧪 Self Test: Haloalkanes & Haloarenes

Attempt each question first. Click on “View Answer” to check your understanding.

Q1. Why is sulphuric acid not used for preparing alkyl iodides from alcohols?

Answer: Concentrated H2SO4 is a strong oxidising agent. It oxidises HI to I2, preventing alkyl iodide formation. Hence, H3PO4 is used.

Q2. Classify the following: CH2=CH–CH2Cl

Answer: Allylic halide, because the halogen is attached to an sp3 carbon next to a C=C bond.

Q3. Arrange CH3Cl, CH3Br, CH3I in increasing boiling point.

Answer: CH3Cl < CH3Br < CH3I (boiling point increases with molecular mass).

Q4. Which mechanism leads to inversion of configuration?

Answer: SN2 mechanism, due to backside attack of the nucleophile (Walden inversion).

Q5. Why do tertiary alkyl halides not undergo SN2 reaction?

Answer: Due to severe steric hindrance around the carbon atom, backside attack by nucleophile is hindered.

Q6. Name the product formed when chlorobenzene reacts with NaOH at 623 K and 300 atm.

Answer: Phenol (Dow’s process).

Q7. Why are haloalkanes insoluble in water?

Answer: Haloalkanes cannot form hydrogen bonds with water, and the energy released is insufficient to break existing H-bonds in water.

Q8. Which reacts faster in SN1: Benzyl chloride or Chlorobenzene?

Answer: Benzyl chloride, because it forms a resonance-stabilised benzylic carbocation.

Q9. What is the role of dry acetone in Finkelstein reaction?

Answer: NaCl or NaBr precipitates in dry acetone, shifting equilibrium towards alkyl iodide formation.

Q10. Why is p-dichlorobenzene having higher melting point than o-isomer?

Answer: Due to symmetry, p-isomer packs better in crystal lattice, resulting in higher melting point.

Q11. Which reagent converts alkyl halide to alcohol?

Answer: Aqueous KOH (nucleophilic substitution).

Q12. Name the reaction: R–X + Mg → R–MgX

Answer: Formation of Grignard reagent.

Q13. Why are aryl halides less reactive towards nucleophilic substitution?

Answer: Due to resonance giving partial double bond character to C–X bond and sp2 hybridisation.

Q14. Which halide gives highest yield in Wurtz reaction?

Answer: Primary alkyl halides (symmetrical alkanes formed).

Q15. What type of product is formed in elimination using alc. KOH?

Answer: Alkene (β-elimination / dehydrohalogenation).

Q16. Why is chloroform stored in dark coloured bottles?

Answer: To prevent oxidation into poisonous phosgene gas in presence of light.

Q17. What is bio-magnification?

Answer: Accumulation and increase in concentration of non-biodegradable substances (like DDT) in higher trophic levels.

Q18. Which halogen shows maximum bond enthalpy with carbon?

Answer: Fluorine (C–F bond is strongest).

Q19. Identify the mechanism: Reaction proceeds via carbocation intermediate.

Answer: SN1 mechanism.

Q20. Why does chlorobenzene not give phenol on heating with aqueous NaOH under normal conditions?

Answer: Due to partial double bond character of C–Cl bond and instability of phenyl carbocation.

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