Class 11 Chemistry – Unit 17: Halogenoalkanes (FBISE)
This section provides complete, exam-oriented notes for Class 11 Chemistry Unit 17 – Halogenoalkanes strictly according to the Federal Board (FBISE) syllabus. The unit develops a strong understanding of alkyl halides and their chemical behavior.
Major topics include classification and nomenclature of halogenoalkanes, preparation methods, physical properties, nucleophilic substitution reactions (SN1 and SN2), elimination reactions, and important uses. Reaction mechanisms and comparison tables are highlighted for exam preparation.
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17.1 Types and Classification of Halogenoalkanes
Halogenoalkanes (also known as alkyl halides) are organic compounds where one or more hydrogen atoms in an alkane have been replaced by halogen atoms (Fluorine, Chlorine, Bromine, or Iodine).
Types based on Halogen Count
- Monohaloalkane: Contains one halogen atom (e.g., $CH_{3}X$).
- Dihaloalkane: Contains two halogen atoms (e.g., $CH_{2}X_{2}$).
- Trihaloalkane: Contains three halogen atoms (e.g., $CHX_{3}$).
Structural Classification
Alkyl halides are classified based on the nature of the carbon atom to which the halogen is attached:
| Class | Definition | Examples |
|---|---|---|
| Primary ($1^{\circ}$) | Halogen is attached to a carbon bonded to one or no other alkyl groups. | Methyl chloride ($CH_{3}Cl$), Ethyl chloride ($CH_{3}CH_{2}Cl$) |
| Secondary ($2^{\circ}$) | Halogen is attached to a carbon bonded to two other alkyl groups. | 2-Chloropropane (Isopropyl chloride), 2-Chlorobutane |
| Tertiary ($3^{\circ}$) | Halogen is attached to a carbon bonded to three other alkyl groups. | 2-Methyl-2-chloropropane (t-Butyl chloride) |
17.2 Physical Properties and Structure
- Bond Polarity: The $C-X$ bond is polar because halogens are more electronegative than carbon. This results in a partial positive charge ($\delta+$) on carbon and a partial negative charge ($\delta-$) on the halogen.
- Boiling/Melting Points: Alkyl halides have higher melting and boiling points compared to their parent alkanes due to these polar intermolecular forces.
- Hybridization: The carbon atom in the $R-X$ bond is $sp^{3}$ hybridized and forms a $\sigma$ (sigma) bond with the halogen.
17.3 Preparations of Alkyl Halides (From Alcohol)
Alcohols ($R-OH$) can be converted into alkyl halides ($R-X$) using various reagents:
- With $HCl$: Requires anhydrous $ZnCl_{2}$ as a catalyst.
Reaction: $CH_{3}CH_{2}OH + HCl \xrightarrow{ZnCl_{2}} CH_{3}CH_{2}Cl + H_{2}O$ - With $SOCl_{2}$ (Thionyl Chloride): Uses Pyridine as a solvent. This is often preferred as the by-products ($SO_{2}$ and $HCl$) are gases.
Reaction: $CH_{3}CH_{2}OH + SOCl_{2} \to CH_{3}CH_{2}Cl + HCl + SO_{2}$ - With Phosphorus Halides: Using $PBr_{3}$, $PCl_{3}/PCl_{5}$, or $PI_{3}$.
Review Questions & Answers
Q1: Why do alkyl halides have higher boiling points than alkanes of similar molecular mass?
A: This is due to the polarity of the $C-X$ bond. The electronegativity difference creates dipole-dipole interactions, which are stronger than the London dispersion forces found in non-polar alkanes, requiring more energy to break.
Q2: Define a secondary ($2^{\circ}$) alkyl halide.
A: A secondary alkyl halide is a compound where the halogen atom is attached to a secondary carbon—a carbon atom that is directly bonded to two other alkyl groups.
Q3: Which reagent is used alongside $SOCl_{2}$ to prepare an alkyl chloride from an alcohol?
A: Pyridine is used as a solvent in this reaction.
Q4: What is the hybridization of the carbon atom in a halogenoalkane?
A: The carbon atom is $sp^{3}$ hybridized.
Notes: Alkyl Halides (Halogenoalkanes)
1. Preparation of Alkyl Halides
- From Alkanes: Alkanes react with $Cl_{2}$ or $Br_{2}$ in the presence of diffused sunlight or UV light. This is a free radical substitution mechanism. It often produces a mixture of products (mono, di, or poly-substituted).
- From Alkenes:
- Hydrohalogenation: Addition of $HX$ (halogen acid) to an alkene. This follows Markovnikov's rule. Reactivity order: $HI > HBr > HCl$.
- Halogenation: Addition of $X_{2}$ in an inert solvent like $CCl_{4}$ to produce vicinal dihalides (halogens on adjacent carbons).
2. Reactivity Factors
The reactivity of alkyl halides is governed by two competing factors:
- Bond Polarity: Due to higher electronegativity of halogens, the $C-X$ bond is polar ($C^{\delta+} - X^{\delta-}$). Based on polarity alone, the reactivity order would be: $R-F > R-Cl > R-Br > R-I$.
- Bond Energy: This is the dominant factor. The $C-I$ bond is the weakest and easiest to break, while the $C-F$ bond is the strongest. Therefore, the actual reactivity order is:
$R-I > R-Br > R-Cl > R-F$
3. Nucleophilic Substitution ($S_{N}$) Reactions
A nucleophile ($Nu^{-}$) attacks the electrophilic carbon and replaces the leaving group ($X^{-}$).
- Substrate: The alkyl halide molecule being attacked.
- Nucleophile: Electron-rich species (e.g., $OH^{-}$, $CN^{-}$, $NH_{3}$).
- Leaving Group: The halogen atom that departs with the electron pair. $I^{-}$ is the best leaving group.
Common Substitution Reactions:
| Reagent | Product formed | General Equation |
|---|---|---|
| Aqueous $NaOH$ | Alcohol | $R-X + NaOH \rightarrow R-OH + NaX$ |
| Alcoholic $KCN$ | Nitrile | $R-X + KCN \rightarrow R-CN + KX$ |
| Excess Ammonia ($NH_{3}$) | Primary Amine | $R-X + NH_{3} \rightarrow R-NH_{2} + HX$ |
| Silver Nitrate ($AgNO_{3}$) | Nitroalkane | $R-X + AgNO_{3} \rightarrow R-NO_{2} + AgX_{(s)}$ |
4. Elimination Reactions ($\beta$-Elimination)
When treated with a strong base (like $NaOH$ in ethanol) and heated, alkyl halides lose a molecule of $HX$ to form an alkene. The hydrogen is removed from the $\beta$-carbon (the carbon adjacent to the one holding the halogen).
Questions and Answers
Q1: Why is the reaction of alkanes with halogens not preferred for preparing pure alkyl halides?A: The reaction proceeds via a free radical mechanism which is difficult to control. It results in a mixture of mono, di, and poly-halogenated derivatives rather than a single pure product.
Q2: How can you distinguish between $Cl^{-}$, $Br^{-}$, and $I^{-}$ ions using silver nitrate?A: By observing the color and solubility of the precipitate formed with $AgNO_{3}$:
- $Cl^{-}$: White precipitate, soluble in aqueous ammonia.
- $Br^{-}$: Cream-colored precipitate, partially soluble in aqueous ammonia.
- $I^{-}$: Yellow precipitate, insoluble in aqueous ammonia.
Q3: Complete the following reactions from Assessment Exercise 17.1:A:
- (a) $CH_{3}-CH=CH_{2} + HBr \rightarrow CH_{3}-CH(Br)-CH_{3}$ (2-bromopropane)
- (b) $CH_{3}-CH_{2}-Br + NaOH_{(aq)} \rightarrow CH_{3}-CH_{2}-OH + NaBr$ (Ethanol)
- (c) $C_{2}H_{5}-Cl + KCN \rightarrow C_{2}H_{5}-CN + KCl$ (Propanenitrile)
- (d) $C_{2}H_{5}-Cl + NH_{3} \rightarrow C_{2}H_{5}-NH_{2} + HCl$ (Ethylamine)
Q4: Why is $R-I$ more reactive than $R-F$ despite fluorine being more electronegative?A: Reactivity is primarily determined by bond energy rather than bond polarity. The $C-I$ bond is much longer and weaker (lower bond energy) than the $C-F$ bond, making it much easier to break during a reaction.