krit.club logo

Haloalkanes and Haloarenes - Physical and chemical properties

Grade 12ICSEChemistry

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

🔑Concepts

•

Physical Properties: Boiling points of alkyl halides follow the order R−I>R−Br>R−Cl>R−FR-I > R-Br > R-Cl > R-F due to the increase in van der Waalsvan\ der\ Waals forces with molecular mass. Among isomers, branching decreases the surface area, thus decreasing the boiling point.

•

Solubility: Haloalkanes are polar but are only sparingly soluble in water because they cannot form hydrogen bonds with water molecules. They are soluble in organic solvents.

•

SN1S_N1 Mechanism: A two-step nucleophilic substitution where the rate-determining step is the formation of a carbocation intermediate (Rate=k[RX]Rate = k[RX]). Reactivity order: 3∘>2∘>1∘3^\circ > 2^\circ > 1^\circ. It is favored by polar protic solvents.

•

SN2S_N2 Mechanism: A single-step process involving the simultaneous attack of the nucleophile and departure of the leaving group through a transition state (Rate=k[RX][Nu−]Rate = k[RX][Nu^-]). Reactivity order: CH3X>1∘>2∘>3∘CH_3X > 1^\circ > 2^\circ > 3^\circ due to steric hindrance.

•

Saytzeff's Rule: In dehydrohalogenation reactions, the major product is the alkene which has a greater number of alkyl groups attached to the doubly bonded carbon atoms (more substituted alkene).

•

Chemical Properties of Haloarenes: Haloarenes are less reactive towards nucleophilic substitution compared to haloalkanes due to resonance (partial double bond character of C−XC-X bond), sp2sp^2 hybridization of carbon, and instability of phenyl cation.

•

Electrophilic Substitution in Haloarenes: Halogens are deactivating but ortho- and para-directing due to +M+M effect (resonance) despite their −I-I effect.

•

Wurtz-Fittig and Fittig Reactions: Reaction of aryl halides with alkyl halides and sodium gives alkyl-substituted arenes (Wurtz-Fittig), while aryl halides reacting together give diphenyl (Fittig).

📐Formulae

R−X+KOH(aq)→R−OH+KXR-X + KOH(aq) \rightarrow R-OH + KX

R−X+2Na+X−R→dry etherR−R+2NaX (Wurtz Reaction)R-X + 2Na + X-R \xrightarrow{dry\ ether} R-R + 2NaX \text{ (Wurtz Reaction)}

R−X+Mg→dry etherR−Mg−X (Grignard Reagent)R-X + Mg \xrightarrow{dry\ ether} R-Mg-X \text{ (Grignard Reagent)}

CH3CH2CH(Br)CH3→alc.KOH,ΔCH3CH=CHCH3 (Major)+CH3CH2CH=CH2 (Minor)CH_3CH_2CH(Br)CH_3 \xrightarrow{alc. KOH, \Delta} CH_3CH=CHCH_3 \text{ (Major)} + CH_3CH_2CH=CH_2 \text{ (Minor)}

C6H5Cl+Cl2→FeCl3o−C6H4Cl2+p−C6H4Cl2C_6H_5Cl + Cl_2 \xrightarrow{FeCl_3} o-C_6H_4Cl_2 + p-C_6H_4Cl_2

💡Examples

Problem 1:

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

Solution:

CH3−CH=CH−CH2−CH3CH_3-CH=CH-CH_2-CH_3 (Pent−2−enePent-2-ene)

Explanation:

This is a β\beta-elimination reaction (dehydrohalogenation). According to Saytzeff's rule, the more substituted alkene is the major product. Pent−2−enePent-2-ene (disubstituted) is more stable and preferred over Pent−1−enePent-1-ene (monosubstituted).

Problem 2:

Which will undergo SN2S_N2 reaction faster: CH3CH2ClCH_3CH_2Cl or (CH3)3CCl(CH_3)_3CCl?

Solution:

CH3CH2ClCH_3CH_2Cl (Ethyl chloride)

Explanation:

SN2S_N2 reactions proceed via a transition state and are highly sensitive to steric hindrance. CH3CH2ClCH_3CH_2Cl is a primary halide with less steric crowding around the α\alpha-carbon compared to the tertiary halide (CH3)3CCl(CH_3)_3CCl, allowing the nucleophile to attack more easily.

Problem 3:

Why is Chlorobenzene less reactive than Methyl chloride towards aqueous NaOHNaOH?

Solution:

Due to resonance and hybridization.

Explanation:

In Chlorobenzene, the lone pair of ClCl enters into resonance with the benzene ring, giving the C−ClC-Cl bond partial double bond character. Also, the CC atom in C6H5ClC_6H_5Cl is sp2sp^2 hybridized (more electronegative), making the bond shorter and stronger than the C(sp3)−ClC(sp^3)-Cl bond in CH3ClCH_3Cl.