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Haloalkanes and Haloarenes - Physical and Chemical Properties

Grade 12CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

The CXC-X bond is polar due to the electronegativity difference between Carbon and Halogen (XX). The dipole moment follows the order: CH3Cl>CH3F>CH3Br>CH3ICH_3Cl > CH_3F > CH_3Br > CH_3I.

Boiling Points: For the same alkyl group, the boiling point increases with the atomic mass of the halogen: RI>RBr>RCl>RFR-I > R-Br > R-Cl > R-F. For isomeric haloalkanes, boiling point decreases with branching because the surface area decreases, weakening Van der Waals forces.

Solubility: Although polar, haloalkanes are only slightly soluble in H2OH_2O because they cannot form hydrogen bonds with water molecules nor break the existing HH-bonds between water molecules.

SN2S_N2 Mechanism (Bimolecular Nucleophilic Substitution): Occurs in a single step via a pentacoordinate transition state. It involves the inversion of configuration (Walden Inversion). Reactivity order: CH3X>1>2>3CH_3X > 1^\circ > 2^\circ > 3^\circ.

SN1S_N1 Mechanism (Unimolecular Nucleophilic Substitution): Occurs in two steps involving a carbocation intermediate. It usually results in racemization. Reactivity order (based on carbocation stability): 3>2>1>CH3X3^\circ > 2^\circ > 1^\circ > CH_3X.

eta-Elimination (Dehydrohalogenation): When heated with alcoholic KOHKOH, haloalkanes form alkenes. According to Saytzeff's Rule, the highly substituted alkene (more alkyl groups attached to the double-bonded carbons) is the major product.

Reactivity of Haloarenes: They are extremely less reactive towards nucleophilic substitution due to resonance (partial double bond character of CXC-X), sp2sp^2 hybridization of carbon, and instability of the phenyl cation.

Electrophilic Substitution: In haloarenes, halogens are deactivating but orthoortho- and parapara-directing due to the +R+R effect (resonance) which increases electron density at oo and pp positions.

📐Formulae

RX+2Na+XRdry etherRR+2NaX (Wurtz Reaction)R-X + 2Na + X-R \xrightarrow{\text{dry ether}} R-R + 2NaX \text{ (Wurtz Reaction)}

RX+Mgdry etherRMgX (Grignard Reagent)R-X + Mg \xrightarrow{\text{dry ether}} R-Mg-X \text{ (Grignard Reagent)}

2ArX+2Nadry etherArAr+2NaX (Fittig Reaction)2Ar-X + 2Na \xrightarrow{\text{dry ether}} Ar-Ar + 2NaX \text{ (Fittig Reaction)}

RateSN2=k[RX][Nu]\text{Rate}_{S_N2} = k[R-X][Nu^-]

RateSN1=k[RX]\text{Rate}_{S_N1} = k[R-X]

ArX+2Na+XRdry etherArR+2NaX (Wurtz-Fittig Reaction)Ar-X + 2Na + X-R \xrightarrow{\text{dry ether}} Ar-R + 2NaX \text{ (Wurtz-Fittig Reaction)}

💡Examples

Problem 1:

Arrange the following in increasing order of boiling point: 11-Chloropropane, Isopropyl chloride, 11-Chlorobutane.

Solution:

Isopropyl chloride < 11-Chloropropane < 11-Chlorobutane

Explanation:

11-Chlorobutane has the highest molecular mass and surface area. Between 11-chloropropane and Isopropyl chloride (isomers), the branched Isopropyl chloride has a smaller surface area and thus lower Van der Waals forces.

Problem 2:

Which compound in the following pair will react faster in SN2S_N2 reaction with OHOH^-: CH3BrCH_3Br or CH3ICH_3I?

Solution:

CH3ICH_3I

Explanation:

In SN2S_N2 reactions, the leaving group ability is crucial. Since the CIC-I bond is weaker than the CBrC-Br bond due to the larger size of the Iodine atom, II^- is a better leaving group than BrBr^-.

Problem 3:

Predict the major product formed when 22-Bromopentane reacts with alcoholic KOHKOH.

Solution:

Pent2enePent-2-ene

Explanation:

According to Saytzeff's Rule, the elimination of HBrHBr occurs to prefer the more substituted alkene. Pent2enePent-2-ene (with two alkyl groups on the double bond) is more stable and thus the major product compared to Pent1enePent-1-ene.

Physical and Chemical Properties - Revision Notes & Key Formulas | CBSE Class 12 Chemistry