Class 12 Chemistry · NCERT chapter 10 · 7 topics · free to read
Classification & Nomenclature
Haloalkanes and haloarenes are classified based on the type of carbon bearing the halogen (1°, 2°, 3°, allylic, vinylic, benzylic, aryl). IUPAC naming treats halo groups as prefixes on the parent alkane chain. Understanding the classification determines reactivity in SN1, SN2, and elimination reactions.
R—X (X = F, Cl, Br, I)
Types of Organic Halides
Organic halides are classified by the hybridization and environment of the carbon bearing the halogen: alkyl (sp³ C), vinylic (sp² C=C), allylic (sp³ C next to C=C), benzylic (sp³ C next to benzene), and aryl (sp² C of benzene ring).
The classification of organic halides is entirely about where the halogen is attached. This one decision — which carbon holds the X — controls whether the compound undergoes SN1, SN2, elimination, or almost nothing at all.
Alkyl halide (haloalkane): X bonded to an sp^3 C of an alkyl group → R—X
Aryl halide (haloarene): X bonded directly to the sp^2 C of a benzene ring → Ar—X
Vinylic halide: X bonded to sp^2 C of a C=C → CH_2=CHX (extremely unreactive toward nucleophilic substitution)
Allylic halide: X on the sp^3 C adjacent to C=C → CH_2=CH—CH_2X (very reactive — SN1 due to resonance-stabilized carbocation)
Benzylic halide: X on the sp^3 C adjacent to benzene ring → C_6H_5CH_2X (reactive in both SN1 and SN2)
Alkyl halides (1°, 2°, 3°)
Vinylic halides (C=C—X)
Allylic halides (C=C—C—X)
Aryl halides (Ar—X)
Benzylic halides (Ar—C—X)
—
1°, 2°, 3° Classification
Alkyl halides are classified as primary, secondary, or tertiary based on the number of carbon atoms directly attached to the carbon bearing the halogen. This classification is the single biggest predictor of SN1 vs SN2 behavior.
Look at the carbon that holds the halogen. Count how many other carbons are bonded to it (not hydrogens — carbons).
Primary (1°): halogen-bearing C is attached to 1 other carbon → RCH_2X
Secondary (2°): halogen-bearing C is attached to 2 other carbons → R_2CHX
Tertiary (3°): halogen-bearing C is attached to 3 other carbons → R_3CX
Methyl halide ( CH_3X ) is a special case — 0 carbons attached to the halogen-bearing C
Quick recall
Classify: (CH_3)_3C—CH_2—Br
It's a primary (1°) halide. The carbon bearing Br is —CH_2— , which is attached to only one other carbon (the bulky tert-butyl group). The bulkiness of the neighboring group doesn't change the classification.
IUPAC Nomenclature
In IUPAC naming, the halogen is always treated as a prefix (fluoro-, chloro-, bromo-, iodo-) on the parent hydrocarbon chain. The parent chain is the longest chain containing the halogen substituent, numbered to give the halogen the lowest possible locant.
Unlike in common naming (where you say 'ethyl bromide'), IUPAC treats the halogen as a substituent prefix. The naming algorithm is the same as for any substituted alkane:
Step 1: Find the longest continuous carbon chain containing the C—X bond → this is your parent chain
Step 2: Number the chain to give the lowest locant to the halogen (or use lowest set of locants if multiple substituents)
Step 3: Name halogen as prefix: F → fluoro, Cl → chloro, Br → bromo, I → iodo
Step 4: Alphabetical order for multiple substituents (bromo before chloro before fluoro)
Step 5: Use multiplying prefixes (di, tri) for identical halogens
CH_3CHBrCH_2CH_3 arrow 2-Bromobutane
CH_2=CHCl arrow Chloroethene (vinyl chloride)
Common Names
In the common naming system, the alkyl/aryl group is named first, followed by the halide (fluoride, chloride, bromide, iodide). Many common names are still widely used in NEET questions alongside IUPAC names.
Common naming is simple: name the alkyl group, then say the halide. C_2H_5Br = ethyl bromide. (CH_3)_2CHCl = isopropyl chloride. This system breaks down for complex structures, which is why IUPAC exists.
CH_3Cl → methyl chloride (IUPAC: chloromethane)
CH_2Cl_2 → methylene chloride / dichloromethane
CHCl_3 → chloroform / trichloromethane
CCl_4 → carbon tetrachloride / tetrachloromethane
CH_2=CHCl → vinyl chloride / chloroethene
CH_2=CHCH_2Cl → allyl chloride / 3-chloropropene
Gem-dihalides (both halogens on same C) are called 'alkylidene dihalides'. Vic-dihalides (halogens on adjacent C's) are called 'alkylene dihalides'. These terms appear in NCERT and NEET MCQs.
Preparation Methods
Haloalkanes are prepared from alcohols (using HX, PBr₃, PCl₅, SOCl₂), from hydrocarbons (free radical halogenation, electrophilic addition to alkenes), and from halide exchange (Finkelstein, Swarts). Haloarenes are prepared by electrophilic aromatic substitution and the Sandmeyer reaction. Each method has specific advantages and limitations that NEET tests frequently.
R—OH + SOCl_2 Pyridine R—Cl + SO_2 + HCl
From Alcohols
Alcohols react with hydrogen halides (HX), phosphorus halides (PX₃, PX₅), or thionyl chloride (SOCl₂) to replace —OH with —X. The choice of reagent affects yield, side products, and stereochemistry.
Replacing —OH with —X is the bread-and-butter of haloalkane synthesis. Three families of reagents do this job, each with different trade-offs.
Reactivity of HX: HI > HBr > HCl (follows acid strength and nucleophilicity of X⁻)
Reactivity of alcohols: 3° > 2° > 1° (3° alcohols react fastest because they go through SN1 / carbocation)
Lucas test uses this reaction: ZnCl_2 + conc. HCl — 3° alcohol gives turbidity immediately, 2° in 5 min, 1° only on heating
3R—OH + PBr_3 arrow 3R—Br + H_3PO_3
PBr_3 is preferred for converting 1° and 2° alcohols to bromides. PCl_5 works but gives POCl_3 and HCl as by-products, making purification harder.
R—OH + PCl_5 arrow R—Cl + POCl_3 + HCl
R—OH + SOCl_2 Pyridine R—Cl + SO_2 + HCl
From Hydrocarbons
Alkanes undergo free radical halogenation (Cl₂/hv or Br₂/hv), alkenes undergo electrophilic addition of HX (Markovnikov) or HBr/peroxide (anti-Markovnikov), and arenes undergo electrophilic aromatic substitution.
Hydrocarbons can be directly halogenated, but the mechanism depends on whether the substrate is an alkane, alkene, or arene.
Free radical substitution (alkanes): R—H + X_2 h R—X + HX — works best with Cl_2 and Br_2
Anti-Markovnikov addition: R—CH=CH_2 + HBr peroxide R—CH_2—CH_2Br (only with HBr, not HCl or HI)
Electrophilic aromatic substitution (arenes): Ar—H + X_2 Lewis acid Ar—X + HX (catalyst: AlCl_3 or FeBr_3 )
Halide Exchange Reactions
Alkyl halides can be converted from one halide to another via halide exchange: Finkelstein reaction (NaI in acetone for alkyl iodides) and Swarts reaction (AgF or SbF₃ for alkyl fluorides).
Sometimes you need to swap one halogen for another. Two named reactions handle this elegantly, and both exploit solubility differences to drive the equilibrium forward.
NaI is soluble in acetone, but NaCl and NaBr are not. As NaCl/NaBr precipitate out, Le Chatelier's principle drives the reaction to completion. This is why Finkelstein works — it's a solubility trick.
R—X + AgF arrow R—F + AgX ( Swarts reaction )
AgF is soluble, but AgCl, AgBr, AgI are insoluble. The precipitation of AgX drives the equilibrium. Alternatively, Hg_2F_2 , CoF_2 , or SbF_3 can replace AgF.
Quick recall
Why does Finkelstein reaction use acetone as solvent?
Because NaI dissolves in acetone but NaCl/NaBr do not. The precipitation of NaCl/NaBr shifts the equilibrium forward, ensuring complete conversion to the alkyl iodide.
Sandmeyer Reaction
Aryl halides are best prepared from aryldiazonium salts (ArN₂⁺) using CuCl/HCl or CuBr/HBr (Sandmeyer) or NaBF₄ followed by heating (Balz-Schiemann for ArF). This route is critical because direct nucleophilic substitution on arenes is very difficult.
Since aryl halides can't be made by simple SN2 on benzene, the diazonium salt route is the go-to method. Start with an aniline ( ArNH_2 ), diazotize it, then replace N_2^+ with a halide.
For Ar—I : use KI directly (no copper catalyst needed)
For Ar—F : use ArN_2^+BF_4^- → heat → ArF + BF_3 + N_2 (Balz-Schiemann reaction)
Physical Properties
Haloalkanes are denser than hydrocarbons but lighter than water (except polyhalides). Boiling points increase with chain length and halogen size (I > Br > Cl > F for same alkyl group). They are insoluble in water but soluble in organic solvents. Dipole moments depend on electronegativity and bond length, creating a non-obvious trend.
Boiling point: R—I > R—Br > R—Cl > R—F
Boiling Point Trends
Boiling points of haloalkanes increase with increasing molecular mass (longer chain, heavier halogen) due to stronger van der Waals forces. For the same alkyl group, RI > RBr > RCl > RF.
Boiling points in haloalkanes are governed by van der Waals forces (London dispersion forces), which increase with molecular size and surface area. Two independent trends operate:
Same halogen, increasing chain length: BP increases ( CH_3Cl < C_2H_5Cl < C_3H_7Cl ) — more surface area = stronger London forces
Same alkyl group, different halogens: R—I > R—Br > R—Cl > R—F — heavier atoms are more polarizable
Branching decreases BP: iso-halide < n-halide for same formula — branching reduces surface area
Among isomeric haloalkanes: 1° > 2° > 3° for boiling point (less branching = more surface area)
Density & Solubility
Most monohalogenated alkanes are lighter than water, but di- and polyhalides (CH₂Cl₂, CHCl₃, CCl₄) are denser than water. Haloalkanes are insoluble in water (cannot break H-bonds) but miscible with organic solvents.
The density and solubility behavior of haloalkanes is a common source of NEET questions, especially for chloroform and carbon tetrachloride.
Density increases with: number of halogen atoms, atomic mass of halogen, decreasing chain length
CH₂Cl₂ (1.33 g/mL), CHCl₃ (1.48 g/mL), CCl₄ (1.59 g/mL) — all denser than water
Bromo and iodo compounds are generally denser than their chloro analogs
Insoluble in water: cannot form H-bonds strong enough to break the water H-bond network
Soluble in organic solvents: like dissolves like (non-polar/weakly polar with non-polar)
Dipole Moment
Dipole moment depends on both electronegativity difference AND bond length. CH₃F has a lower dipole moment than CH₃Cl because the very short C—F bond partially compensates for the high electronegativity of F. For polyhalides, vector addition determines the net dipole.
Dipole moment = charge separation distance. Two opposing factors compete:
= q d
Higher electronegativity of X → larger charge separation ( q ) → increases
Longer C—X bond → larger distance ( d ) → increases
Observed order: CH_3Cl > CH_3F > CH_3Br > CH_3I
CH₃Cl has the highest because it has the best balance of electronegativity and bond length
CH₃F: very high q but very short d → net is less than CH₃Cl
For polyhalides, use vector addition of individual C—X dipoles:
CCl_4 : four C—Cl dipoles cancel due to tetrahedral symmetry → = 0
CHCl_3 : three C—Cl dipoles don't fully cancel → 0
CH_2Cl_2 : two C—Cl dipoles at ~109° → 0 (and it's higher than CHCl₃)
Quick recall
Arrange in order of dipole moment: CH₃F, CH₃Cl, CH₃Br, CH₃I
CH_3Cl > CH_3F > CH_3Br > CH_3I . CH₃Cl wins because it has the optimal balance of electronegativity (charge) and bond length (distance).
SN1 & SN2 Reactions
Nucleophilic substitution at sp³ carbon proceeds by two mechanisms: SN2 (one step, backside attack, Walden inversion, favored by 1° substrates and strong nucleophiles) and SN1 (two steps, carbocation intermediate, racemization, favored by 3° substrates and weak nucleophiles). This is the highest-weightage topic in the chapter for NEET.
SN2 rate = k[R—X][Nu^-] SN1 rate = k[R—X]
SN2 Mechanism
In SN2, the nucleophile attacks the electrophilic carbon from the back side (opposite to the leaving group) in a single concerted step. The rate depends on both substrate and nucleophile concentrations. The stereochemical outcome is 100% inversion of configuration (Walden inversion).
SN2 = Substitution, Nucleophilic, 2 = bimolecular (both substrate and nucleophile are in the rate-determining step). Everything happens in one step — bond making and bond breaking occur simultaneously.
Mechanism: Nu⁻ attacks from the back side of C—X → transition state has 5 groups around C (trigonal bipyramidal) → X leaves from opposite side
Rate law: Rate = k[R—X][Nu^-] — second order overall
Stereochemistry: 100% inversion of configuration (Walden inversion) — like an umbrella flipping in the wind
Energy profile: Single transition state, no intermediate — one energy maximum
Polar aprotic solvents favor SN2: DMF, DMSO, acetone (don't solvate the nucleophile, leaving it free to attack)
SN1 Mechanism
In SN1, the leaving group departs first to form a planar carbocation, then the nucleophile attacks from either face. The rate depends only on substrate concentration. The stereochemical outcome is racemization (with slight excess of inversion product).
SN1 = Substitution, Nucleophilic, 1 = unimolecular (only the substrate is in the rate-determining step). It's a two-step process with a carbocation intermediate.
Step 1 (slow, RDS): R—X arrow R^+ + X^- — leaving group departs, forming a planar carbocation
Step 2 (fast): R^+ + Nu^- arrow R—Nu — nucleophile attacks the flat carbocation from either face
Rate law: Rate = k[R—X] — first order (nucleophile not in RDS)
Stereochemistry: Racemization — attack from both faces gives ~50% R + ~50% S (slight excess inversion in practice)
Polar protic solvents favor SN1: water, methanol, ethanol (stabilize the carbocation by solvation)
SN1 vs SN2
Choosing between SN1 and SN2 depends on four factors: substrate (1°/2°/3°), nucleophile strength, solvent polarity and type, and leaving group ability. This comparison is the most tested aspect of the chapter.
One step (concerted)
Two steps (carbocation intermediate)
Rate = k[RX][Nu⁻] (2nd order)
Rate = k[RX] (1st order)
Strong nucleophile required
Weak nucleophile sufficient
CH₃X > 1° > 2° ≫ 3°
3° > 2° > 1° > CH₃X
Walden inversion (100%)
Racemization (~50:50)
Polar aprotic solvents
Polar protic solvents
No rearrangement
Rearrangement possible (carbocation)
Backside attack
Attack from either face
Leaving group ability: I^- > Br^- > Cl^- > F^- — same for both SN1 and SN2
Better leaving group = weaker base = conjugate base of stronger acid
F^- is the worst leaving group (strongest C—F bond, smallest ion, poor at stabilizing negative charge in solution)
This is why C—F bonds are so hard to break — relevant to Teflon's chemical inertness
Quick recall
Predict mechanism: (CH₃)₃CBr + H₂O →?
SN1. Substrate is 3° (cannot do SN2 — too sterically hindered). Nucleophile is H₂O (weak). Solvent is polar protic. All three factors point to SN1. Product: (CH₃)₃COH with racemization.
Optical Activity in Substitution
SN2 converts an optically active substrate to the product with inverted configuration (optically active, opposite rotation). SN1 gives racemization (loss of optical activity). These stereochemical outcomes are testable predictions of the mechanisms.
If the substrate has a chiral center at the C—X carbon, the mechanism directly predicts what happens to optical activity:
SN2 → Inversion: (R)-substrate gives (S)-product (or vice versa). Optical activity is retained but sign flips — the product is optically active
SN1 → Racemization: Planar carbocation allows attack from both faces → equal mixture of (R) and (S) → optical activity is lost (or nearly lost)
Retention (same configuration preserved): does NOT happen in simple SN1 or SN2 — requires special mechanisms like neighboring group participation
Quick recall
An optically active alkyl halide undergoes hydrolysis. The product is optically inactive. What mechanism operated?
SN1. Racemization produces a 50:50 mixture of enantiomers → optically inactive (racemic mixture). SN2 would have given an optically active product (inverted configuration).
Elimination Reactions
Elimination reactions remove HX from haloalkanes to form alkenes. E2 is a one-step, bimolecular mechanism (strong base, anti-periplanar geometry). E1 is a two-step process via carbocation (same conditions as SN1). Saytzeff's rule predicts the more substituted alkene as the major product. The competition between substitution and elimination is a critical NEET concept.
R—CH_2—CHX—R' Strong base R—CH=CH—R' + HX
E2 Mechanism
E2 elimination is a one-step, bimolecular process where the base abstracts a β-hydrogen simultaneously as the leaving group departs. The H and X must be in anti-periplanar arrangement. Rate depends on both substrate and base.
E2 = Elimination, 2 = bimolecular. The base pulls off a -hydrogen while the leaving group departs — all in one concerted step. No intermediate is formed.
Rate law: Rate = k[R—X][Base] — second order
Geometry requirement: H and X must be anti-periplanar (180° dihedral angle) — this is a strict stereoelectronic requirement
Saytzeff rule: The more substituted alkene is the major product (more stable due to hyperconjugation)
E2 is concerted — no carbocation → no rearrangement possible
CH_3CH_2CHBrCH_3 alc. KOH CH_3CH=CHCH_3 _ major (Saytzeff) + CH_3CH_2CH=CH_2 _ minor (Hofmann)
E1 Mechanism
E1 elimination occurs in two steps: first the leaving group departs to form a carbocation (same as SN1), then a base removes a β-hydrogen to form the alkene. E1 always competes with SN1.
E1 shares the same first step as SN1 — formation of a carbocation. The difference is what happens next: in SN1, a nucleophile attacks the carbocation; in E1, a base removes a -hydrogen.
Step 1 (slow, RDS): R—X arrow R^+ + X^- — same as SN1
Step 2 (fast): Base removes a -H from the carbocation → alkene forms
Rate law: Rate = k[R—X] — first order (base not in RDS)
Saytzeff rule applies — more substituted alkene is major product
E1 and SN1 always compete — both go through the same carbocation
Elimination vs Substitution
SN1, SN2, E1, and E2 all compete. The outcome depends on substrate class (1°/2°/3°), nucleophile/base strength, temperature, and solvent. NEET tests this decision-making frequently.
Every time a haloalkane meets a nucleophile/base, four reactions compete: SN1, SN2, E1, E2. The "winner" depends on the conditions.
1° + strong base/nucleophile: SN2 dominates (if base is bulky, E2 can compete)
3° + any conditions: E2 dominates with strong base; E1 + SN1 mixture with weak base/solvolysis
2°: Gray zone — strong base → E2; strong nucleophile (not bulky) → SN2; weak nucleophile → SN1/E1 mixture
Higher temperature always favors elimination over substitution
Bulky base (t-BuOK) favors elimination over substitution — base can't reach the carbon but can reach the hydrogen
Low temperature
High temperature
Strong, small nucleophile (CN⁻, I⁻)
Strong, bulky base (t-BuOK)
1° substrate
3° substrate
Polar aprotic solvent (SN2)
Alcoholic solution (E2)
Weak nucleophile in protic solvent (SN1)
Heated weak base in protic solvent (E1)
Quick recall
What happens when (CH₃)₃CBr is treated with (a) aq. NaOH and (b) alc. KOH?
(a) SN1 → (CH₃)₃COH (3° can't do SN2, aqueous = protic solvent). (b) E2 → (CH₃)₂C=CH₂ (strong base in alcohol = elimination). This is a classic NEET question.
Reactions with Metals
Haloalkanes react with metals to form organometallic compounds or coupled products. Key reactions include Wurtz reaction (2RX + 2Na → R-R), Grignard reagent formation (RX + Mg → RMgX), Wurtz-Fittig reaction (ArX + RX + 2Na → Ar-R), and Frankland reaction (2RX + 2Zn → R-R + 2ZnX₂). These reactions are tools for carbon-carbon bond formation.
R—X + Mg dry ether R—MgX (Grignard reagent)
Wurtz Reaction
Two molecules of alkyl halide react with metallic sodium in dry ether to form a higher alkane with double the carbon count. The reaction works best with primary alkyl halides and always gives a symmetrical product when the same halide is used.
2R—X + 2Na dry ether R—R + 2NaX
Wurtz reaction couples two alkyl groups by removing the halogen with sodium. It's one of the oldest C—C bond-forming reactions (1855). The product always has an even number of carbons when the same halide is used.
Substrate: Works best with 1° alkyl halides (3° halides undergo elimination instead)
Reagent: Metallic sodium (Na) in dry ether
Product: Symmetrical alkane with 2× the carbon count: 2CH_3Br + 2Na arrow C_2H_6 + 2NaBr
Limitation: Mixed Wurtz reaction (two different RX) gives a mixture of three products — not useful for synthesis
Cannot make methane by Wurtz reaction (need two CH₃X, product is ethane)
Quick recall
Can you prepare methane using Wurtz reaction?
No. Wurtz reaction always couples two alkyl groups → minimum product is ethane ( CH_3—CH_3 ). Methane cannot be made by Wurtz reaction.
Grignard Reagent (RMgX)
Alkyl or aryl halides react with magnesium in dry ether to form Grignard reagents (RMgX), which are powerful nucleophiles and strong bases. They react with water, CO₂, aldehydes, ketones, epoxides, and esters to build larger molecules.
R—X + Mg dry ether R—MgX
Victor Grignard won the Nobel Prize (1912) for this reagent. The carbon bonded to Mg is highly nucleophilic (almost carbanion-like) because Mg is electropositive. This makes RMgX one of the most versatile reagents in organic chemistry.
Condition: Absolutely dry ether as solvent (ether coordinates with Mg, stabilizing the reagent)
Moisture kills it: RMgX + H_2O arrow R—H + Mg(OH)X — even trace moisture destroys the reagent
The C—Mg bond is highly polar: R^ - —Mg^ + — carbon acts as a nucleophile/base
Reactivity order for RX: R—I > R—Br > R—Cl (iodides react fastest with Mg)
Key reactions of Grignard reagents:
RMgX + H_2O arrow R—H (alkane — used to count active H atoms, Zerewitinoff test)
Organolithium: R—X + 2Li ether R—Li + LiX — even more reactive than Grignard
Quick recall
How would you prepare toluene (C₆H₅CH₃) from halides?
Wurtz-Fittig reaction: C₆H₅Br + CH₃Br + 2Na → C₆H₅CH₃ + 2NaBr (dry ether). This couples the aryl and alkyl groups selectively.
Haloarenes & Environmental Effects
Haloarenes (Ar—X) are much less reactive than haloalkanes toward nucleophilic substitution due to resonance stabilization and sp² C—X bond character. They undergo electrophilic aromatic substitution (ortho/para-directing). Environmental effects of organohalides include ozone depletion by CFCs, bioaccumulation of DDT, and persistence of polyhalogenated compounds in ecosystems.
C_6H_5—X + NaOH 300°C, 200 atm C_6H_5—OH + NaX
Why Haloarenes Resist Nucleophilic Substitution
The C—X bond in aryl halides has partial double bond character due to resonance between the lone pair on X and the aromatic ring. This makes the bond shorter, stronger, and resistant to nucleophilic attack. Additionally, the sp² carbon is more electronegative and holds onto X more tightly.
Chlorobenzene is notoriously unreactive toward NaOH under normal conditions, while chloroethane reacts easily. The explanation involves three reinforcing factors:
Resonance effect: Lone pair on X delocalizes into the aromatic ring → C—X bond acquires partial double bond character → harder to break
sp² hybridization: The ring carbon is sp^2 (more s-character → shorter, stronger bond) vs sp^3 in alkyl halides
Stability of phenyl cation: C_6H_5^+ is extremely unstable (positive charge on sp^2 C without resonance stabilization in the plane) → SN1 almost impossible
Steric shielding: The electron cloud of the aromatic ring physically blocks nucleophilic attack from above/below
Harsh conditions CAN force nucleophilic substitution on haloarenes:
Electron-withdrawing groups at ortho/para positions activate the ring for nucleophilic substitution (e.g., 2,4-dinitro-chlorobenzene reacts with mild nucleophiles)
Benzyne mechanism operates when NaNH₂ is used — proceeds through an unstable triple-bond-like intermediate
Electrophilic Aromatic Substitution
Despite being deactivating (electron-withdrawing by induction), halogens direct incoming electrophiles to ortho and para positions due to resonance donation of lone pairs. This creates a paradox: halogens slow down the reaction but control its regioselectivity.
The halogen substituent has two opposing electronic effects on the ring:
Inductive effect (-I): Halogen is electronegative → withdraws electron density from the ring → deactivates the ring (slower than benzene)
Resonance effect (+R/+M): Lone pair on halogen donates into the ring → increases electron density at ortho and para positions → directs there
Net result: deactivating (−I dominates over +R for overall rate) but ortho/para-directing (+R controls where the electrophile attacks)
Halogens are the ONLY substituents that are deactivating yet ortho/para-directing — a unique combination
Organohalogens have severe environmental impacts. CFCs destroy stratospheric ozone through free radical chain reactions. DDT bioaccumulates in food chains, causing ecological damage. These issues make organohalide environmental chemistry a regular NEET topic.
The very stability that makes haloalkanes useful (non-flammable, chemically inert, long shelf life) is also what makes them environmental nightmares — they persist in nature for decades.
CFCs (Chlorofluorocarbons): Freon-12 ( CCl_2F_2 ) was used as refrigerant and aerosol propellant
In the stratosphere, UV radiation breaks the C—Cl bond: CF_2Cl_2 UV CF_2Cl + Cl
One Cl radical destroys ~100,000 ozone molecules through a chain reaction before being deactivated
The chain: Cl + O_3 arrow ClO + O_2 then ClO + O arrow Cl + O_2 — Cl is regenerated
DDT (dichlorodiphenyltrichloroethane): First synthetic insecticide, won Paul Muller the Nobel Prize (1948)
DDT is non-biodegradable and bioaccumulates — concentration increases up the food chain
Caused eggshell thinning in birds (especially raptors) → population crashes → banned in most countries
Persistent organic pollutants (POPs): DDT, PCBs, dioxins — all organohalides that resist degradation
Carbon tetrachloride ( CCl_4 ): Depletes ozone, toxic to liver → banned as cleaning solvent
Iodoform ( CHI_3 ): Was used as antiseptic — the 'antiseptic' property was actually due to liberation of free iodine
Chloroform ( CHCl_3 ): Early anesthetic, now known to be hepatotoxic and carcinogenic
Teflon (PTFE): —(CF_2—CF_2)—_n — chemically inert, used in non-stick cookware. The strong C—F bond makes it nearly indestructible
Quick recall
Why are CFCs harmful to the ozone layer but HFCs are not?
CFCs contain C—Cl bonds that are broken by UV to release Cl radicals, which catalytically destroy O₃. HFCs (hydrofluorocarbons) contain only C—F and C—H bonds — the C—F bond is too strong to be broken by stratospheric UV, so no ozone-destroying radicals are released.