Haloalkanes & Haloarenes: Prep & Basics
Module 1 | CBSE Class 12 Chemistry | Organic Chemistry
1. Introduction & Importance
The replacement of hydrogen atom(s) in an aliphatic or aromatic hydrocarbon by halogen atom(s) results in the formation of alkyl halide (haloalkane) and aryl halide (haloarene), respectively.
- Haloalkanes contain halogen atom(s) attached to the sp3 hybridized carbon atom of an alkyl group.
- Haloarenes contain halogen atom(s) attached to sp2 hybridized carbon atom(s) of an aryl group.
- Chloramphenicol: A chlorine-containing antibiotic produced by soil microorganisms, highly effective for the treatment of typhoid fever.
- Thyroxine: An iodine-containing hormone produced by our body, deficiency of which causes goiter.
- Chloroquine: Used for the treatment of malaria.
- Halothane: Used as an anesthetic during surgery.
2. Classification of Halogen Derivatives
They can be classified based on the number of halogen atoms (mono, di, tri, polyhalogen) or based on the hybridization of the carbon atom to which the halogen is bonded.
2.1 Compounds containing sp3 C-X Bond (X = F, Cl, Br, I)
This class includes three major sub-types:
- Alkyl halides or Haloalkanes (R-X): The halogen atom is bonded to an alkyl group. They are further classified as primary (1°), secondary (2°), or tertiary (3°) depending on whether the halogen is attached to a primary, secondary, or tertiary carbon atom.
- Allylic halides: The halogen atom is bonded to an sp3-hybridized carbon atom adjacent to a carbon-carbon double bond (C=C).
Example: CH2=CH-CH2-X (Allyl halide). - Benzylic halides: The halogen atom is bonded to an sp3-hybridized carbon atom attached directly to an aromatic ring.
Example: C6H5-CH2-X (Benzyl halide).
2.2 Compounds containing sp2 C-X Bond
This class includes two major sub-types:
- Vinylic halides: The halogen atom is bonded directly to an sp2-hybridized carbon atom of a carbon-carbon double bond (C=C).
Example: CH2=CH-X (Vinyl halide). - Aryl halides (Haloarenes): The halogen atom is bonded directly to the sp2-hybridized carbon atom of an aromatic ring.
Example: C6H5-X (Halobenzene).
3. Nomenclature (Common & IUPAC)
In the IUPAC system, alkyl halides are named as haloalkanes. The longest carbon chain containing the halogen is chosen as the parent alkane.
- Geminal halides (gem-dihalides): Both halogen atoms are attached to the same carbon atom. Common name: Alkylidene halides (e.g., CH3CHCl2 is Ethylidene chloride).
- Vicinal halides (vic-dihalides): Halogen atoms are attached to adjacent carbon atoms. Common name: Alkylene dihalides (e.g., CH2Cl-CH2Cl is Ethylene dichloride).
| Structure | Common Name | IUPAC Name |
|---|---|---|
| CH3CH2CH(Cl)CH3 | sec-Butyl chloride | 2-Chlorobutane |
| (CH3)3C-CH2Br | neo-Pentyl bromide | 1-Bromo-2,2-dimethylpropane |
| CH2=CH-Cl | Vinyl chloride | Chloroethene |
| CH2=CH-CH2Br | Allyl bromide | 3-Bromopropene |
4. Nature of C-X Bond & Dipole Moments
Halogen atoms are more electronegative than carbon. As a result, the carbon-halogen bond is polarized. The carbon atom bears a partial positive charge (δ+) and the halogen atom bears a partial negative charge (δ-).
Bond Length and Enthalpy: As we go down group 17 in the periodic table (F to I), the size of the halogen atom increases. Therefore, the C-X bond length increases (C-F < C-Cl < C-Br < C-I) and the C-X bond enthalpy decreases, making C-I the most reactive (easiest to break).
Dipole moment (μ) depends on both charge difference (Q) and bond distance (r): μ = Q × r.
Even though Fluorine is the most electronegative element (maximum Q), the dipole moment of CH3Cl (1.86 D) is greater than CH3F (1.84 D).
Reason: The very small atomic size of fluorine makes the C-F bond distance (r) so small that the overall product (Q × r) is slightly less than that of the C-Cl bond.
Order of Dipole Moment: CH3Cl > CH3F > CH3Br > CH3I
5. Methods of Preparation of Haloalkanes
5.1 From Alcohols (R-OH)
Alcohols are the most widely used precursors for the synthesis of alkyl halides. The hydroxyl group can be replaced by halogen upon reaction with concentrated halogen acids, phosphorus halides, or thionyl chloride.
- With Halogen Acids (HX): Requires Lucas Reagent (conc. HCl + anhydrous ZnCl2) for primary and secondary alcohols. Tertiary alcohols react at room temperature without a catalyst.
R-OH + HX → R-X + H2O - With Phosphorus Halides:
3R-OH + PCl3 → 3R-Cl + H3PO3
R-OH + PCl5 → R-Cl + POCl3 + HCl - With Thionyl Chloride (Darzen's Procedure):
R-OH + SOCl2 → R-Cl + SO2(g) + HCl(g)
Darzen's method using SOCl2 is the preferred method for synthesizing alkyl chlorides because both by-products, SO2 and HCl, are escapable gases. Thus, pure alkyl halide is easily obtained without extensive purification.
5.2 From Hydrocarbons
A. From Alkanes by Free Radical Halogenation
Reaction of alkanes with Cl2 or Br2 in the presence of UV light (hν) or heat yields a complex mixture of isomeric mono- and polyhaloalkanes, making separation difficult. Hence, it is rarely used in laboratories, but common industrially.
CH3CH2CH2CH3 (n-Butane) + Cl2 (UV light) → CH3CH2CH2CH2Cl (1-Chlorobutane) + CH3CH2CH(Cl)CH3 (2-Chlorobutane - Major Product)
B. From Alkenes by Electrophilic Addition of HX
An alkene is converted to an alkyl halide by the addition of hydrogen halide (HCl, HBr, or HI).
CH3-CH=CH2 + HBr → CH3-CH(Br)-CH3 (2-Bromopropane - Major)
Anti-Markovnikov Addition (Peroxide Effect / Kharasch Effect): In the presence of organic peroxides, addition of HBr only takes place contrary to Markovnikov's rule due to a free radical mechanism.
CH3-CH=CH2 + HBr (Peroxide) → CH3-CH2-CH2Br (1-Bromopropane - Major)
5.3 Halogen Exchange (Name Reactions)
A. Finkelstein Reaction (For Alkyl Iodides)
Alkyl iodides are often prepared by the reaction of alkyl chlorides/bromides with NaI in dry acetone.
B. Swarts Reaction (For Alkyl Fluorides)
Direct fluorination is highly explosive. Alkyl fluorides are prepared by heating an alkyl chloride/bromide in the presence of a metallic fluoride like AgF, Hg2F2, CoF2, or SbF3.
6. Preparation of Haloarenes
Note: Haloarenes cannot be prepared from phenol reacting with HCl, because the C-O bond in phenol has partial double bond character due to resonance and is difficult to break.
A. Electrophilic Substitution
Aryl chlorides and bromides can be easily prepared by electrophilic substitution of arenes with chlorine and bromine respectively in the presence of Lewis acid catalysts like iron (Fe) or iron(III) chloride (FeCl3) in the dark.
Toluene + Cl2 (Fe/Dark) → o-Chlorotoluene + p-Chlorotoluene (Major)
B. Sandmeyer's Reaction (Highly Important)
When a primary aromatic amine (e.g., Aniline), dissolved or suspended in cold aqueous mineral acid (0-5°C), is treated with sodium nitrite (NaNO2), a diazonium salt is formed.
Mixing the freshly prepared diazonium salt with cuprous chloride (Cu2Cl2) or cuprous bromide (Cu2Br2) results in the replacement of the diazonium group by -Cl or -Br.
Note for Iodobenzene: For introducing iodine, cuprous halide is not required. Simply shaking the diazonium salt with Potassium Iodide (KI) is sufficient.
7. NCERT Solved Examples (Step-by-Step)
NCERT Example 10.3: Identify all the possible monochloro structural isomers expected to be formed on free radical monochlorination of (CH3)2CHCH2CH3.
The given compound is 2-methylbutane. Let's analyze the types of equivalent hydrogen atoms it possesses:
1. (CH3)2: Six equivalent primary hydrogens attached to C1 and the methyl branch on C2.
2. CH: One tertiary hydrogen attached to C2.
3. CH2: Two secondary hydrogens attached to C3.
4. CH3: Three primary hydrogens attached to C4.
Replacement of each type of hydrogen by chlorine yields four different structural isomers:
(i) (CH3)2CHCH2CH2Cl (1-Chloro-3-methylbutane)
(ii) (CH3)2CHCH(Cl)CH3 (2-Chloro-3-methylbutane)
(iii) (CH3)2C(Cl)CH2CH3 (2-Chloro-2-methylbutane)
(iv) ClCH2CH(CH3)CH2CH3 (1-Chloro-2-methylbutane)
NCERT Example 10.4: Write the products of the following reactions:
(i) CH3-CH=CH2 + HBr → ?
(ii) CH3-CH=CH2 + HBr + Peroxide → ?
(i) This is an electrophilic addition without peroxide. It follows Markovnikov's rule. The Bromine (negative part) attaches to the carbon with fewer hydrogens (the middle carbon).
Product: CH3-CH(Br)-CH3 (2-Bromopropane).
(ii) In the presence of peroxide, HBr undergoes free radical addition following the Anti-Markovnikov (Kharasch) effect. Bromine attaches to the terminal carbon with more hydrogens.
Product: CH3-CH2-CH2-Br (1-Bromopropane).
8. Previous Year Questions (PYQs) & Exhaustive Question Bank
Part A: Conceptual (1-2 Marks)
Q1. Out of CH3Cl and CH3F, which one has a higher dipole moment and why?
Q2. Why is sulphuric acid not used during the reaction of alcohols with KI to produce alkyl iodides?
2HI + H2SO4 → I2 + SO2 + 2H2O.
Therefore, the substitution of the -OH group by an Iodine atom does not take place. Instead, non-oxidizing acids like phosphoric acid (H3PO4) are used.
Part B: Assertion-Reason Type (1 Mark)
Q3. Assertion (A): Thionyl chloride (SOCl2) is preferred for the preparation of alkyl chlorides from alcohols.
Reason (R): The by-products of the reaction, SO2 and HCl, are gases and escape easily, leaving behind pure alkyl halide.
Part C: Application & Conversions (2-3 Marks)
Q4. Write the chemical equations for the following Name Reactions:
(a) Finkelstein Reaction
(b) Swarts Reaction
(a) Finkelstein Reaction (Synthesis of Alkyl Iodide):
CH3-CH2-Cl (Chloroethane) + NaI →(Dry Acetone, heat) CH3-CH2-I (Iodoethane) + NaCl↓
(b) Swarts Reaction (Synthesis of Alkyl Fluoride):
CH3-Br (Bromomethane) + AgF →(heat) CH3-F (Fluoromethane) + AgBr
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