Section 1

⚗️ Introduction to Haloalkanes & Haloarenes

Why Study Haloalkanes?

From the anaesthetic that numbs pain during surgery (halothane: CF3CHClBr) to the antibiotic chloramphenicol that fights deadly infections — halogen-containing organic compounds are woven into medicine, industry, and everyday life. Understanding their chemistry unlocks an entire world of organic reactions.

What Are Haloalkanes?

A haloalkane (also called an alkyl halide) is an organic compound in which one or more hydrogen atoms of an alkane have been replaced by halogen atoms (F, Cl, Br, or I). The general formula is R—X, where R is an alkyl group and X is a halogen.

What Are Haloarenes?

A haloarene (also called an aryl halide) is a compound in which one or more halogen atoms are directly bonded to an aromatic ring. The general formula is Ar—X, where Ar represents an aryl group such as C6H5—.

Why Are These Compounds Important?

  • Solvents: Dichloromethane (CH2Cl2) is widely used in paint strippers and degreasers.
  • Anaesthetics: Halothane (CF3CHClBr) was one of the first safe inhaled anaesthetics.
  • Antibiotics: Chloramphenicol contains a —CHCl2 group and treats bacterial eye infections.
  • Refrigerants: Freon-12 (CCl2F2) was used in air conditioners before its ban due to ozone depletion.
  • Synthetic Intermediates: Alkyl halides serve as starting materials for making alcohols, ethers, amines, and countless other compounds.

The Nature of the C—X Bond

The carbon–halogen bond is polar because halogens are more electronegative than carbon. This makes the carbon atom electrophilic (electron-poor), which is precisely why nucleophiles attack it so readily.

In haloalkanes, the carbon bearing the halogen is sp3 hybridised — it has a tetrahedral geometry. In haloarenes (like chlorobenzene), the carbon bonded to the halogen is sp2 hybridised — it sits within a planar aromatic ring, which gives these compounds very different reactivity.
If the C—X bond is polar, why don't haloalkanes dissolve in water (which is also polar)? We'll explore this puzzle in Section 4!
Section 2

🏷️ Classification & IUPAC Nomenclature

Classification by Number of Halogen Atoms

  • Monohaloalkanes: Contain one halogen atom — e.g., CH3Cl (chloromethane)
  • Dihaloalkanes: Contain two halogen atoms — e.g., CH2Cl2 (dichloromethane)
  • Polyhaloalkanes: Contain three or more halogen atoms — e.g., CHCl3 (trichloromethane)

Classification by Type of Carbon Bearing the Halogen

  • Primary (1°): The halogen is on a carbon bonded to at most one other carbon — e.g., CH3CH2Cl
  • Secondary (2°): The halogen is on a carbon bonded to exactly two other carbons — e.g., CH3CHClCH3
  • Tertiary (3°): The halogen is on a carbon bonded to three other carbons — e.g., (CH3)3CCl

Special Halide Types

  • Allylic halide: Halogen on a carbon adjacent to a C═C double bond — e.g., CH2═CH—CH2Cl
  • Benzylic halide: Halogen on a carbon adjacent to an aromatic ring — e.g., C6H5CH2Cl
  • Vinylic halide: Halogen directly on a doubly-bonded carbon — e.g., CH2═CHCl
Don't confuse allylic with vinylic! An allylic halide has the halogen on the carbon next to the double bond, while a vinylic halide has it on the doubly-bonded carbon itself. Vinylic halides are far less reactive in nucleophilic substitution.

IUPAC Naming Rules

  1. Select the longest continuous carbon chain that includes the carbon bearing the halogen.
  2. Number the chain so that the halogen gets the lowest possible locant.
  3. Name the halogen as a prefix: fluoro-, chloro-, bromo-, iodo-.
  4. Use multiplying prefixes (di-, tri-, tetra-) for multiple identical halogens.

Nomenclature Reference Table

Structural FormulaIUPAC NameCommon Name
CH3CH2ClChloroethaneEthyl chloride
CH3CHBrCH32-BromopropaneIsopropyl bromide
(CH3)3CBr2-Bromo-2-methylpropanetert-Butyl bromide
CHCl3TrichloromethaneChloroform
CCl4TetrachloromethaneCarbon tetrachloride
C6H5ClChlorobenzene—
When two different halogens are present, list them in alphabetical order as prefixes. For example, CH2BrCl is bromochloromethane, not chlorobromomethane.
Section 3

🔧 Methods of Preparation

1. From Alcohols

Alcohols (R—OH) can be converted to haloalkanes by replacing the —OH group with a halogen atom using various reagents:

Using Hydrogen Halides
R—OH + HX → R—X + H2O
Reactivity order of HX: HI > HBr > HCl
Using Thionyl Chloride (Best Method)
R—OH + SOCl2 → R—Cl + SO2↑ + HCl↑
Both by-products are gases — pure product is obtained directly!
Using Phosphorus Halides
R—OH + PCl5 → R—Cl + POCl3 + HCl
3R—OH + PCl3 → 3R—Cl + H3PO3
Thionyl chloride (SOCl2) is often called the best reagent for preparing chloroalkanes from alcohols because both by-products (SO2 and HCl) escape as gases, leaving behind a pure product with no tedious purification needed.

2. From Alkenes

Addition of Hydrogen Halides (Markovnikov)
CH3CH═CH2 + HBr → CH3CHBrCH3 (major, 2-bromopropane)
Markovnikov's rule: H adds to C with more H atoms; halogen adds to C with fewer H atoms.
Addition of Halogens
CH2═CH2 + Br2 → CH2BrCH2Br (1,2-dibromoethane)

3. Halogenation of Alkanes (Free Radical)

Free Radical Substitution (requires UV light or heat)
R—H + X2 —UV→ R—X + HX
Mechanism: Initiation → Propagation → Termination (free radical chain)

4. Named Reactions for Halide Interchange

Sandmeyer Reaction
ArN2+Cl− + CuCl → ArCl + N2↑
ArN2+Cl− + CuBr → ArBr + N2↑
Converts diazonium salts to aryl halides using Cu2Cl2/Cu2Br2
Finkelstein Reaction
R—Cl + NaI —acetone→ R—I + NaCl↓
NaCl is insoluble in acetone and precipitates, pushing the equilibrium forward (Le Chatelier's principle).
Swarts Reaction
R—Cl + AgF → R—F + AgCl↓
Used to prepare fluoroalkanes — AgCl precipitates, driving the reaction forward.
Notice the clever trick in both the Finkelstein and Swarts reactions: a product precipitates out of solution, continuously pulling the equilibrium toward the product side. Le Chatelier's principle in action!
Section 4

⚙️ Physical Properties

Boiling Points

Boiling points of haloalkanes increase with:

  • Increasing molecular mass — more electrons → stronger London dispersion forces
  • Increasing chain length — larger surface area → greater intermolecular contact
  • For the same alkyl group: R—I > R—Br > R—Cl > R—F (due to increasing molecular mass and polarisability)

Density

Haloalkanes are generally denser than their parent hydrocarbons. Bromo and iodo alkanes with low carbon counts are even denser than water (density > 1 g/cm³).

Solubility

Despite the polar C—X bond, haloalkanes are insoluble in water. Why? To dissolve in water, a solute must be able to break the strong hydrogen-bond network among water molecules. Haloalkanes cannot form hydrogen bonds with water, so the energy cost of disrupting water's H-bond network outweighs the weak ion-dipole interactions that haloalkanes can offer.

Haloalkanes dissolve readily in non-polar organic solvents like benzene, ether, and chloroform because the intermolecular forces between haloalkane molecules and the solvent are comparable in strength — "like dissolves like."

Dipole Moments

A surprising trend: CH3Cl has a larger dipole moment than CH3F, even though fluorine is more electronegative. This is because the C—Cl bond is much longer than the C—F bond, and dipole moment depends on both charge separation and bond length (μ = q × d).

Bond Strength (Bond Dissociation Enthalpy)

BondBond Enthalpy (kJ/mol)Relative Strength
C—F~485Strongest
C—Cl~339↓
C—Br~285↓
C—I~213Weakest
Students often assume that C—F should be the most reactive bond because fluorine is the most electronegative. In reality, C—F is the strongest C—X bond and therefore the hardest to break. Reactivity in nucleophilic substitution is usually highest for C—I because the C—I bond is weakest.
Section 5

💥 Nucleophilic Substitution: SN1 & SN2

The Core Reaction of Haloalkane Chemistry

When a nucleophile (an electron-rich species like OH−, CN−, or NH3) approaches the electrophilic carbon of a haloalkane, it can replace the halogen — this is nucleophilic substitution. But how the replacement happens depends critically on the structure of the substrate, the nucleophile, and the solvent.

SN2: Bimolecular Nucleophilic Substitution

⚡ Mechanism — One Step

The nucleophile attacks the electrophilic carbon from the backside (opposite to the leaving group) in a single concerted step. The bond to the nucleophile forms at the same time as the bond to the leaving group breaks.

Key Features of SN2

  • Rate law: Rate = k[R—X][Nu−] — depends on both substrate and nucleophile concentrations
  • Stereochemistry: Walden inversion — complete inversion of configuration (like an umbrella flipping in the wind)
  • Substrate preference: CH3X > 1° > 2° ≫ 3° (steric hindrance blocks backside attack on bulky substrates)
  • Nucleophile: Strong nucleophiles (OH−, CN−, I−) are needed
  • Solvent: Polar aprotic solvents (like acetone, DMSO, DMF) — they don't solvate the nucleophile, keeping it "free" and reactive

SN1: Unimolecular Nucleophilic Substitution

⚡ Mechanism — Two Steps

Step 1 (slow, rate-determining): The C—X bond breaks heterolytically to form a carbocation and a halide ion.
Step 2 (fast): The nucleophile attacks the planar carbocation from either side.

Key Features of SN1

  • Rate law: Rate = k[R—X] — depends only on substrate concentration (first-order kinetics)
  • Stereochemistry: Racemisation — a mixture of both configurations (since the carbocation is flat and can be attacked from both faces)
  • Substrate preference: 3° > 2° > 1° (more substituted carbocations are more stable due to hyperconjugation and inductive effects)
  • Nucleophile: Weak nucleophiles (like H2O, ROH) are sufficient
  • Solvent: Polar protic solvents (like water, ethanol) — they stabilise the carbocation intermediate through solvation
Why does SN1 give racemisation rather than 100% retention? Because the carbocation intermediate is sp2 hybridised and planar. The incoming nucleophile can approach from the top or bottom face with equal probability — like flipping a coin!

SN1 vs SN2: Head-to-Head Comparison

FeatureSN1SN2
Full NameSubstitution, Nucleophilic, UnimolecularSubstitution, Nucleophilic, Bimolecular
Number of StepsTwo (via carbocation)One (concerted)
Rate LawRate = k[R—X]Rate = k[R—X][Nu−]
Substrate Order3° > 2° > 1°CH3X > 1° > 2° ≫ 3°
NucleophileWeak OK (e.g., H2O)Strong required (e.g., OH−)
SolventPolar proticPolar aprotic
StereochemistryRacemisationInversion (Walden)
IntermediateCarbocationTransition state only
RearrangementsPossible (carbocation can rearrange)Not possible
Exam shortcut: If the substrate is 3° or benzylic/allylic with a weak nucleophile → think SN1. If it's methyl or 1° with a strong nucleophile → think SN2. For 2° substrates, either mechanism is possible — look at the nucleophile and solvent to decide.
Section 6

🔄 Elimination Reactions

What Is Elimination?

Instead of replacing the halogen with a nucleophile (substitution), an elimination reaction removes HX from the haloalkane to form an alkene. The base abstracts a proton (H+) from a carbon adjacent to the one bearing the halogen, and the halide departs as a leaving group.

Dehydrohalogenation (General)
R—CH2—CHX—R' + alc. KOH → R—CH═CH—R' + KX + H2O
"alc. KOH" = KOH dissolved in ethanol (alcoholic potassium hydroxide)

E1 Mechanism (Unimolecular Elimination)

  • Step 1: C—X bond breaks to form a carbocation (same as SN1 first step)
  • Step 2: A base removes a β-hydrogen, and the electron pair forms a C═C double bond
  • Favoured by 3° substrates, weak bases, and polar protic solvents
  • Often competes with SN1

E2 Mechanism (Bimolecular Elimination)

  • One-step: The base abstracts a β-hydrogen at the same time as the leaving group departs
  • Rate = k[R—X][Base]
  • Requires a strong, bulky base (e.g., alc. KOH, t-BuOK)
  • Often competes with SN2

Saytzeff's Rule (Zaitsev's Rule)

When dehydrohalogenation can produce more than one alkene, the more substituted alkene is the major product. More substituted alkenes are more stable due to hyperconjugation.

Example: Saytzeff's Rule
CH3—CHBr—CH2—CH3 + alc. KOH →
  Major: CH3—CH═CH—CH3 (but-2-ene, more substituted)
  Minor: CH2═CH—CH2—CH3 (but-1-ene, less substituted)

Substitution vs Elimination: A Quick Guide

ReagentReaction TypeProduct
Aqueous KOH (aq. KOH)Nucleophilic substitutionAlcohol (R—OH)
Alcoholic KOH (alc. KOH)EliminationAlkene (C═C)
aq. KOH vs alc. KOH — this is one of the most commonly tested distinctions! Aqueous KOH provides OH− as a nucleophile (substitution), while alcoholic KOH provides ethoxide as a strong base (elimination). Missing this difference can cost you marks in exams.
Section 7

🌍 Haloarenes & Environmental Impact

Why Are Haloarenes Less Reactive Than Haloalkanes?

Chlorobenzene is far less reactive toward nucleophilic substitution compared to chloroethane. Two key reasons:

  1. Resonance effect: One of the lone pairs on the halogen delocalises into the aromatic ring, giving the C—X bond partial double-bond character. This makes it shorter and stronger than a typical C—X single bond, and harder to break.
  2. sp2 carbon: The carbon bearing the halogen in haloarenes is sp2 hybridised. This carbon holds its electrons more tightly (greater s-character), making it harder for a nucleophile to attack.
The C—Cl bond in chlorobenzene (1.69 Å) is shorter than the C—Cl bond in chloroethane (1.77 Å). This shortening is direct evidence of the partial double-bond character contributed by resonance.

Electrophilic Aromatic Substitution of Haloarenes

Halogens bonded to an aromatic ring are ortho, para-directing groups — they direct incoming electrophiles to the ortho and para positions. However, they are simultaneously deactivators (they slow down the reaction compared to benzene) because of their strong electron-withdrawing inductive effect.

This is a famous paradox: halogens are activators by resonance (+M effect, donating electrons) but deactivators by induction (−I effect, withdrawing electrons). The inductive effect wins overall, making halobenzenes less reactive than benzene, but the resonance effect decides the position of attack (ortho/para).

Nucleophilic Substitution in Haloarenes

Dow Process
C6H5Cl + NaOH —300°C, 200 atm→ C6H5OH + NaCl
Requires extreme conditions — evidence of how unreactive chlorobenzene is!

Environmental Impact of Organohalogen Compounds

DDT (Dichloro Diphenyl Trichloroethane)

DDT was once widely used as an insecticide. It is a persistent organic pollutant (POP) — it does not break down easily in the environment and bioaccumulates in the food chain, reaching dangerous concentrations in top predators. Its use is now heavily restricted worldwide.

CFCs (Chlorofluorocarbons)

CFCs like Freon-12 (CCl2F2) were used as refrigerants and aerosol propellants. In the upper atmosphere, UV radiation breaks the C—Cl bond, releasing chlorine radicals that catalytically destroy ozone (O3). A single Cl radical can destroy thousands of ozone molecules before it is deactivated.

Ozone Depletion Mechanism (Simplified)
Cl• + O3 → ClO• + O2
ClO• + O3 → Cl• + 2O2
Net: 2O3 → 3O2 (Cl• is regenerated — it acts as a catalyst)
The Montreal Protocol (1987) phased out CFCs worldwide and is considered one of the most successful international environmental agreements in history. The ozone layer is now slowly recovering!
Section 8

🧩 Practice Problems — Set 1

Test your understanding with these problems. Click "Reveal Answer" to check your reasoning.

PROBLEM 1

Give the IUPAC name of CH3CH2CH(Br)CH3.

2-Bromobutane. The longest chain has 4 carbons (butane). The bromine is on carbon-2. Hence the name is 2-bromobutane.
PROBLEM 2

Classify (CH3)3CBr as a primary, secondary, or tertiary halide.

Tertiary (3°). The carbon bonded to Br is also bonded to three other carbon atoms (three methyl groups), making it a tertiary alkyl halide.
PROBLEM 3

Which reacts faster in an SN2 reaction: CH3Br or (CH3)3CBr? Explain.

CH3Br reacts much faster. SN2 requires backside attack by the nucleophile. In (CH3)3CBr, three bulky methyl groups block the approach of the nucleophile (steric hindrance), making SN2 essentially impossible. CH3Br has no such steric barrier.
PROBLEM 4

What product forms when 2-bromopropane is treated with alcoholic KOH?

Propene (CH3CH═CH2). Alcoholic KOH acts as a strong base, causing elimination (dehydrohalogenation). HBr is removed to form an alkene.
PROBLEM 5

Haloalkanes are polar molecules. Why, then, are they insoluble in water?

Although haloalkanes have a polar C—X bond, they cannot form hydrogen bonds with water molecules. Dissolving in water requires breaking the extensive hydrogen-bond network of water, and the weak dipole-dipole interactions that haloalkanes can offer are insufficient to compensate for this energy cost.
PROBLEM 6

Show how ethanol can be converted to chloroethane using thionyl chloride (SOCl2).

C2H5OH + SOCl2 → C2H5Cl + SO2↑ + HCl↑
Thionyl chloride replaces the —OH group with —Cl. Both by-products (SO2 and HCl) are gases and escape, leaving behind pure chloroethane.
PROBLEM 7

What is Walden inversion? In which reaction mechanism does it occur?

Walden inversion is the complete inversion of spatial configuration (stereochemistry) at a chiral centre during a reaction. It occurs in the SN2 mechanism, where the nucleophile attacks from the backside, pushing the other groups to the opposite side — like an umbrella flipping inside out.
PROBLEM 8

Arrange the following bonds in decreasing order of bond dissociation enthalpy: C—F, C—Cl, C—Br, C—I.

C—F > C—Cl > C—Br > C—I. As we go down the halogen group, atomic size increases, the C—X bond becomes longer, and orbital overlap becomes less effective — resulting in weaker bonds.
PROBLEM 9

Explain why SN1 reactions typically give racemised products.

In SN1, the first step forms a planar carbocation (sp2 hybridised). This flat intermediate can be attacked by the nucleophile from either face with roughly equal probability, producing a near-equal mixture of both enantiomers — i.e., a racemic mixture.
PROBLEM 10

What is the Finkelstein reaction? Why is acetone used as the solvent?

Finkelstein reaction: R—Cl + NaI → R—I + NaCl. It converts a chloroalkane to an iodoalkane using sodium iodide. Acetone is used because NaCl is insoluble in acetone and precipitates out. This shifts the equilibrium toward the product side (Le Chatelier's principle), driving the reaction to completion.
Section 9

🧩 Practice Problems — Set 2 (Advanced)

These questions require deeper reasoning. Take your time before revealing the answer.

PROBLEM 1

Predict the major product when (CH3)2CHCH2Br is heated with alcoholic KOH.

2-Methylpropene (isobutylene), (CH3)2C═CH2. According to Saytzeff's rule, the more substituted alkene is the major product. Elimination removes HBr to give the double bond at the more substituted position.
PROBLEM 2

Chlorobenzene is far less reactive than chloroethane toward nucleophilic substitution. Provide two structural reasons for this difference.

Reason 1: Resonance stabilisation. A lone pair from Cl delocalises into the ring, giving the C—Cl bond partial double-bond character. This makes it shorter, stronger, and harder to break.
Reason 2: sp2 carbon. The ring carbon bonded to Cl is sp2 hybridised (more s-character), holding the bonding electrons more tightly and making nucleophilic attack more difficult.
PROBLEM 3

Will the reaction of (CH3)3CCl with water proceed by SN1 or SN2? Justify.

SN1. (CH3)3CCl is a tertiary halide — the three methyl groups create severe steric hindrance, ruling out backside attack (SN2). Additionally, the resulting tertiary carbocation is highly stable. Water is a weak nucleophile and a polar protic solvent — all conditions favour SN1.
PROBLEM 4

Concentrated H2SO4 cannot be used with KI to prepare alkyl iodides from alcohols. Why?

H2SO4 is a strong oxidising acid. It oxidises HI to I2 (HI is a strong reducing agent), consuming the HI before it can react with the alcohol. The net effect is that no alkyl iodide is produced. Instead, H3PO4 (a non-oxidising acid) is used with KI.
PROBLEM 5

Explain why para-dichlorobenzene has a higher melting point than its ortho- and meta-isomers.

The para-isomer has the most symmetrical molecular shape. Greater symmetry allows the molecules to pack more efficiently into a crystal lattice, resulting in stronger intermolecular forces in the solid state and therefore a higher melting point.
PROBLEM 6

Draw all structural isomers of C4H9Br and classify each as 1°, 2°, or 3°.

There are four structural isomers:
1. CH3CH2CH2CH2Br — 1-bromobutane (1°)
2. CH3CH2CHBrCH3 — 2-bromobutane (2°)
3. (CH3)2CHCH2Br — 1-bromo-2-methylpropane (1°)
4. (CH3)3CBr — 2-bromo-2-methylpropane (3°)
PROBLEM 7

What are Grignard reagents? Why must they be prepared under strictly anhydrous conditions?

Grignard reagents have the formula R—MgX (e.g., CH3MgBr). They are prepared by reacting an alkyl halide with magnesium metal in dry ether: R—X + Mg → R—MgX. They must be prepared in anhydrous (water-free) conditions because water immediately decomposes them: R—MgX + H2O → R—H + Mg(OH)X, destroying the reagent.
PROBLEM 8

In an SN2 reaction, which reacts faster: CH3I or CH3Cl? Why?

CH3I reacts faster. Both have the same substrate (methyl group), so steric factors are identical. The key difference is the leaving group ability: I− is a better leaving group than Cl− because the C—I bond is weaker (lower bond dissociation enthalpy) and I− is more stable (larger, more polarisable ion).
PROBLEM 9

What are CFCs? Why were they banned globally under the Montreal Protocol?

CFCs (Chlorofluorocarbons) are compounds containing carbon, chlorine, and fluorine (e.g., CCl2F2, Freon-12). They were used as refrigerants and aerosol propellants. In the stratosphere, UV radiation breaks C—Cl bonds, releasing Cl• radicals that catalytically destroy ozone molecules. Each Cl radical can destroy thousands of O3 molecules, causing dangerous thinning of the ozone layer.
PROBLEM 10

Predict which mechanism operates when neopentyl bromide ((CH3)3CCH2Br) reacts with OH−. Explain.

This is a tricky case! SN2 is extremely slow because three methyl groups on the adjacent carbon create massive steric hindrance for backside attack. SN1 is also unfavourable because it would generate a highly unstable primary carbocation. The most likely pathway is an E2 elimination (if conditions are forcing), or the reaction is simply very sluggish. This substrate is famously unreactive in nucleophilic substitution.
Section 10

🎯 Quick Quiz — 10 MCQs

Select the correct answer for each question. Your score will be shown at the end!

QUESTION 1

The IUPAC name of CHCl3 is:

AChloroform
BMethyl chloride
CTrichloromethane
DCarbon trichloride
QUESTION 2

SN2 reactions are favoured by:

ATertiary (3°) halides
BPrimary (1°) halides
CWeak nucleophiles
DPolar protic solvents
QUESTION 3

The rate-determining step in an SN1 reaction is:

ANucleophilic attack on the carbocation
BFormation of the carbocation (ionisation)
CFormation of the final product
DProton transfer to the solvent
QUESTION 4

Markovnikov addition of HBr to propene gives:

A1-Bromopropane
B2-Bromopropane
C3-Bromopropane
DPropyne
QUESTION 5

The correct order of C—X bond strength is:

AC–I > C–Br > C–Cl > C–F
BC–F > C–Cl > C–Br > C–I
CC–Cl > C–F > C–Br > C–I
DAll C—X bonds are equally strong
QUESTION 6

According to Saytzeff's rule, the major product of elimination is:

AThe less substituted alkene
BThe more substituted alkene
CA rearranged carbocation product
DNo reaction occurs
QUESTION 7

Which of the following is a Grignard reagent?

ACH3MgBr
BCH3Br
CMg(OH)2
DNaBr
QUESTION 8

Chlorobenzene is less reactive than chloroethane towards nucleophilic substitution primarily because of:

AGreater molecular mass of chlorobenzene
BResonance stabilisation of the C—Cl bond
CInductive effect of the ring
DSteric crowding by ring hydrogens
QUESTION 9

DDT stands for:

ADichloro Diphenyl Trichloroethane
BDimethyl Diphenyl Tetrachloride
CDi-trinitrotoluene
DDichlorodiphenyldichloroethane
QUESTION 10

Walden inversion is a characteristic feature of:

ASN1 reactions
BSN2 reactions
CE1 reactions
DE2 reactions

Your Score

0/10
Section 11

📋 Chapter Summary

🔑 Key Concepts at a Glance

  • Haloalkanes (R—X) contain a halogen bonded to an sp3 carbon; haloarenes (Ar—X) have a halogen bonded to an sp2 aromatic carbon.
  • The C—X bond is polar, making the carbon electrophilic and susceptible to nucleophilic attack.
  • Haloalkanes are classified as 1°, 2°, or 3° based on the number of carbons attached to the halogen-bearing carbon.

🧪 Preparation Methods

  • From alcohols: Using HX, SOCl2, PCl3, or PCl5. SOCl2 is the preferred reagent for chloroalkanes (gaseous by-products).
  • From alkenes: Addition of HX (Markovnikov's rule) or X2.
  • Free radical halogenation: R—H + X2 → R—X + HX (UV light needed).
  • Named reactions: Sandmeyer (ArN2+ → ArX), Finkelstein (R—Cl → R—I), Swarts (R—Cl → R—F).

⚙️ Physical Properties

  • Boiling points: increase with molecular mass and chain length.
  • Insoluble in water despite being polar (cannot break water's H-bond network).
  • Bond strength: C—F > C—Cl > C—Br > C—I (but reactivity is often the reverse!).
  • Dipole moment of CH3Cl > CH3F (longer bond length compensates for lower electronegativity).

💥 SN1 vs SN2 — The Big Picture

  • SN2: One step, backside attack, inversion. Favoured by 1°/methyl halides, strong nucleophiles, polar aprotic solvents.
  • SN1: Two steps via carbocation, racemisation. Favoured by 3° halides, weak nucleophiles, polar protic solvents.

🔄 Elimination Reactions

  • Dehydrohalogenation with alc. KOH gives alkenes.
  • Saytzeff's rule: The more substituted alkene is the major product.
  • Key distinction: aq. KOH → substitution (alcohol); alc. KOH → elimination (alkene).

🌍 Haloarenes & Environment

  • Haloarenes are less reactive than haloalkanes in nucleophilic substitution (resonance + sp2 effects).
  • Halogens on benzene are o,p-directors but deactivators.
  • CFCs deplete the ozone layer; DDT is a persistent organic pollutant that bioaccumulates.
  • The Montreal Protocol successfully phased out CFCs worldwide.

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