Haloalkanes and Haloarenes
Haloalkanes and Haloarenes for JEE Main & Advanced
Haloalkanes
Nomenclature, Preparation, and PropertiesTopic 1
Haloalkanes (Alkyl halides): Hydrocarbons in which one or more H atoms replaced by halogens (X = F, Cl, Br, I).
Classification:
| Type | Description | Example |
|---|---|---|
| Primary (1°) | -X on C bonded to one other C | $CH_3CH_2Cl$ (ethyl chloride) |
| Secondary (2°) | -X on C bonded to two other C's | $(CH_3)_2CHCl$ (isopropyl chloride) |
| Tertiary (3°) | -X on C bonded to three other C's | $(CH_3)_3CCl$ (tert-butyl chloride) |
| Allyl/benzyl | -X adjacent to C=C or aromatic ring | $CH_2=CHCH_2Cl$ (allyl chloride); $C_6H_5CH_2Cl$ (benzyl chloride) |
| Vinyl | -X on $sp^2$ C of C=C | $CH_2=CHCl$ (vinyl chloride) |
| Aryl | -X on aromatic ring | $C_6H_5Cl$ (chlorobenzene) |
Nomenclature (IUPAC): halo-X-...alkane.
- $CH_3CH_2Br$: bromoethane
- $(CH_3)_2CHCl$: 2-chloropropane
- $CH_3CHBrCH_2CH_3$: 2-bromobutane
Preparation:
1. From Alcohols:
- $R-OH + HX \to RX + H_2O$ (HX = HCl with $ZnCl_2$ as catalyst; HBr; HI)
- $3R-OH + PX_3 \to 3RX + H_3PO_3$
- $R-OH + PCl_5 \to RCl + POCl_3 + HCl$
- $R-OH + SOCl_2 \to RCl + SO_2 + HCl$ (Darzen's method — clean reaction, gases released easily)
Reactivity of HX: $HI > HBr > HCl$. Reactivity of ROH: $3° > 2° > 1°$ (carbocation stability).
Lucas Test: $ZnCl_2$ + conc. HCl distinguishes alcohols:
- $3°$ alcohol: cloudy immediately
- $2°$: cloudy in $\sim 5$ min
- $1°$: no cloudiness at room T (needs heating)
2. From Alkanes (Free Radical Halogenation): $CH_4 + Cl_2 \xrightarrow{h\nu} CH_3Cl + HCl$ (and further products). Mix of products; not selective in industry.
3. From Alkenes:
- HX addition (Markovnikov): $CH_2=CH_2 + HBr \to CH_3CH_2Br$
- Halogenation: $CH_2=CH_2 + Br_2 \to BrCH_2CH_2Br$ (1,2-dibromoethane)
4. Halogen Exchange:
- Finkelstein Reaction: $RCl + NaI \xrightarrow{\text{acetone}} RI + NaCl$ (NaCl insoluble in acetone, drives equilibrium)
- Swarts Reaction: $RCl + AgF \to RF + AgCl$ (or $Hg_2F_2, CoF_2$)
Physical Properties:
- C-X bond polar (X is more EN); molecules polar; have higher BP than corresponding alkanes
- BP order: $RF < RCl < RBr < RI$ (larger size → more London forces; polarizability ↑)
- BP among same X: $primary > secondary > tertiary$ (more compact = less surface area)
- Insoluble in water; soluble in organic solvents
- Density: $RF, RCl < $ water; $RBr, RI > $ water (denser due to heavy halogen)
Predict product of Lucas test on: (a) ethanol, (b) tert-butanol, (c) cyclohexanol.
Show solution
Lucas reagent = conc. HCl + ZnCl₂. Reaction proceeds via $S_N1$, so faster with more stable carbocation.
- (a) Ethanol ($1°$): no cloud at RT; heating needed.
- (b) tert-butanol ($3°$): cloudy immediately ($3°$ carbocation very stable).
- (c) Cyclohexanol ($2°$): cloud in $\sim$5-10 min.
Final Answer: Tertiary (b) reacts immediately; secondary takes time; primary very slow.
Predict products of $CH_3CH_2OH + SOCl_2 \to$.
Show solution
Darzen's method (preferred for clean reactions): $CH_3CH_2OH + SOCl_2 \to CH_3CH_2Cl + SO_2 \uparrow + HCl \uparrow$ Both byproducts are gases; easy purification.
Final Answer: Ethyl chloride + $SO_2$ + HCl (gases escape).
$(CH_3)_3CCl$ is:
Darzen's method uses:
Finkelstein reaction:
Lucas test distinguishes:
BP order of RCl vs RBr vs RI (same R):
SN1, SN2 Mechanisms and EliminationsTopic 2
Nucleophilic Substitution Reactions ($S_N$):
Nucleophile (Nu⁻) replaces leaving group (X⁻) at $sp^3$ C: $R-X + Nu^- \to R-Nu + X^-$
Two main mechanisms: $S_N1$ and $S_N2$.
SN2 Mechanism (Bimolecular):
- Concerted single-step: Nu attacks while X leaves
- Transition state: Nu and X partially bonded to same C
- Inversion of configuration (Walden inversion) — like umbrella turning inside out
Rate Law: Rate = $k[R-X][Nu^-]$ — second order; depends on both.
Reactivity: $1° > 2° > 3°$ (steric hindrance increases with branching → 3° hardest).
- Methyl > primary > secondary >> tertiary
Factors Favoring SN2:
- Strong nucleophile
- Polar aprotic solvent (acetone, DMSO, DMF)
- Less steric hindrance (1° preferred)
- Good leaving group ($I^- > Br^- > Cl^- > F^-$)
SN1 Mechanism (Unimolecular):
- Two-step:
- Slow: $R-X \to R^+ + X^-$ (ionization; rate-determining)
- Fast: $R^+ + Nu^- \to R-Nu$
- Rate Law: Rate = $k[R-X]$ — first order; independent of Nu concentration.
- Stereochemistry: Racemization (planar $R^+$ → attack from both faces equally).
Reactivity: $3° > 2° > 1°$ (more stable carbocation; opposite of $S_N2$).
- Allyl, benzyl > tertiary > secondary > primary > methyl (carbocation stability)
Factors Favoring SN1:
- Stable carbocation (3° alkyl)
- Polar protic solvent (water, alcohol — stabilizes ionic intermediates by H-bonding)
- Weak/moderate nucleophile
- Good leaving group
Summary Comparison:
| Property | SN1 | SN2 |
|---|---|---|
| Order | First | Second |
| Mechanism | Two-step (carbocation intermediate) | Concerted (one step) |
| Stereochemistry | Racemization | Inversion |
| Substrate | $3° > 2° > 1°$ | $1° > 2° > 3°$ |
| Nucleophile | Weak/moderate | Strong |
| Solvent | Polar protic | Polar aprotic |
| Rearrangement | Possible (via carbocation) | Not seen |
Elimination Reactions:
E1 Mechanism: Two-step. Carbocation intermediate, then loss of $\beta$-H by base.
- Reactivity: $3° > 2° > 1°$ (same as $S_N1$)
- Often competes with $S_N1$ in 3° alkyl halides
E2 Mechanism: Concerted. Base removes $\beta$-H as X⁻ leaves simultaneously.
- Reactivity: $3° > 2° > 1°$ (more substituted alkene more stable — Saytzeff)
- Requires anti-periplanar arrangement of H and X
- Rate = $k[R-X][B^-]$
SN vs E Competition:
| Conditions | Favors |
|---|---|
| Strong, bulky base (KOC(CH₃)₃, tBuOK) | E (over Nu) |
| Strong, small nucleophile (CN⁻, alkoxide RO⁻) | SN |
| Heat | E (high entropy product) |
| Cool conditions | SN |
| Aqueous KOH | SN1 (in alcohols) |
| Alcoholic KOH | E2 (eliminates HX) |
Hofmann's Rule: Less substituted alkene (more H-rich double bond) preferred when:
- Bulky base used (e.g., $t$-butoxide)
- Quaternary ammonium hydroxide (Hofmann degradation)
Predict mechanism and major product: (a) $CH_3CH_2Br + OH^-$ in water (b) $(CH_3)_3CBr + OH^-$ in water
Show solution
(a) Primary halide; OH⁻ strong Nu; water polar protic but also good Nu favors SN2. Mechanism: $S_N2$. Product: $CH_3CH_2OH$ (ethanol).
(b) Tertiary halide; OH⁻ present but $3°$ favors $S_N1$ (carbocation stable) or E. With weak base/strong Nu (OH⁻ in water): mostly $S_N1$ giving $(CH_3)_3COH$ (and some elimination giving $(CH_3)_2C=CH_2$).
Final Answer: (a) $S_N2$ → ethanol; (b) $S_N1$ → tert-butanol (with some isobutene).
Predict major product of $(CH_3)_2CHBr + KOC(CH_3)_3$ (potassium tert-butoxide) at heat.
Show solution
Secondary alkyl halide + bulky strong base → E2 elimination. Bulky base prefers H abstraction over substitution.
But here, only one alkene possible: $CH_3CH=CH_2 + (CH_3)_3COH + KBr$.
(Note: With Hofmann conditions for $2°$ halides with two different β-H sources, would give less substituted alkene preferentially.)
Final Answer: Propene (via E2).
$S_N2$ reaction shows:
Order of $S_N1$ reactivity (substrate):
$S_N1$ vs $S_N2$ in tertiary halide:
E2 needs:
Anti-periplanar requirement is for:
Haloarenes
Properties, Reactivity, and ReactionsTopic 1
Haloarenes (Aryl Halides): Halogen on aromatic ring carbon, e.g., chlorobenzene ($C_6H_5Cl$), bromobenzene.
Structure: $sp^2$ C-X bond. The lone pair on X overlaps with $\pi$ system of ring.
Resonance Structures: $X$ donates lone pair to ring → 4 resonance structures with double bond character in C-X.
Why aryl halides are unreactive toward $S_N$?
- C-X bond is shorter and stronger (due to partial double bond from resonance)
- Resonance stabilization of ground state makes ionization difficult
- $sp^2$ C is more electronegative than $sp^3$ C — holds halide tightly
- Direct $S_N1$/$S_N2$ on aryl C is sterically hindered and electronically unfavorable
Preparation of Haloarenes:
1. From Benzene (Electrophilic Aromatic Substitution): $C_6H_6 + Br_2 \xrightarrow{FeBr_3} C_6H_5Br + HBr$
- Lewis acid catalyst polarizes halogen
2. From Diazonium Salt (Sandmeyer Reaction): $C_6H_5N_2^+Cl^- + CuCl \to C_6H_5Cl + N_2 + CuCl$ (or CuBr → ArBr)
- For ArI: just KI; for ArF: Schiemann reaction with $HBF_4$ → $ArF + BF_3 + N_2$
Reactions of Aryl Halides:
1. Electrophilic Aromatic Substitution:
- $-X$ is o/p director, deactivating (due to +R donating but -I dominating)
- $C_6H_5Cl + HNO_3/H_2SO_4 \to o$- and $p$-nitrochlorobenzenes
2. Nucleophilic Substitution (Difficult):
- Requires high temperature and pressure
- $C_6H_5Cl + NaOH \xrightarrow{623K, 300\,atm} C_6H_5ONa \to C_6H_5OH$ (Dow's process for phenol from chlorobenzene)
- Activated by -NO₂, -CN groups at o/p positions (EW groups stabilize Meisenheimer intermediate)
Nucleophilic Aromatic Substitution Mechanism: Addition-elimination via Meisenheimer complex: $ArX + Nu^- \to [Ar(Nu)X]^- \to ArNu + X^-$ (intermediate has sp³ C).
With strong EW groups: $p$-nitrochlorobenzene + NaOH/heat → $p$-nitrophenol; $2,4$-dinitrochlorobenzene reacts at room T.
3. Benzyne Mechanism (in absence of activating groups, with very strong base like $NH_2^-$): $C_6H_5Cl + NaNH_2/NH_3(liq) \to$ benzyne intermediate → addition of Nu. Gives random position addition.
4. Wurtz-Fittig Reaction: $ArX + RX + 2Na \xrightarrow{ether} Ar-R + 2NaX$
- Aryl halide + alkyl halide + Na → alkyl aryl compound.
5. Fittig Reaction: $2ArX + 2Na \xrightarrow{ether} Ar-Ar + 2NaX$
- Two aryl halides + Na → biaryl.
Why is chlorobenzene less reactive than chloroethane toward NaOH?
Show solution
Chlorobenzene:
- C-Cl bond has partial double bond character (resonance) → shorter, stronger.
- $sp^2$ C is more EN than $sp^3$; C-Cl bond more polar, harder to break.
- Aryl C cannot undergo $S_N1/S_N2$ easily.
Chloroethane:
- $sp^3$ C, normal C-Cl single bond.
- Easy $S_N2$ with strong nucleophile like OH⁻.
Therefore chlorobenzene is much less reactive — requires harsh conditions (high T, P) or activating substituents.
Final Answer: Resonance, sp² C, and bond strength make chlorobenzene unreactive vs chloroethane.
Predict product: $2$-chloronitrobenzene + NaOH (heat).
Show solution
$-NO_2$ at ortho position activates C-Cl by stabilizing Meisenheimer intermediate (negative charge delocalized onto -NO₂). Therefore $S_N^{Ar}$ proceeds: $o$-nitrochlorobenzene + NaOH → $o$-nitrophenol (or sodium salt).
Final Answer: $o$-nitrophenol.
Aryl halides are:
$-Cl$ on benzene is:
Sandmeyer reaction:
Benzyne intermediate forms when:
$p$-nitrochlorobenzene + NaOH:
Polyhalogen Compounds and Important ApplicationsTopic 2
Polyhalogen Compounds: Have more than one halogen atom.
1. Dichloromethane ($CH_2Cl_2$):
- Solvent for paint removal; aerosol propellants (banned now in many countries)
- Slightly soluble in water
- Toxic; carcinogen
2. Chloroform ($CHCl_3$):
- Anesthetic (historical use; now replaced)
- Solvent for fats, alkaloids, etc.
- On exposure to light + air → poisonous phosgene ($COCl_2$): $2CHCl_3 + O_2 \to 2COCl_2 + 2HCl$
- Stored in dark amber bottles with small amount of ethanol (which reacts with phosgene)
3. Iodoform ($CHI_3$):
- Yellow crystalline solid; sweet smell
- Antiseptic (mild; due to release of free iodine)
- Iodoform test: for $CH_3CO-$ or $CH_3CH(OH)-$ groups: react with I₂/NaOH → yellow ppt of $CHI_3$ + carboxylate.
- Positive: acetone, ethanol, acetaldehyde, methyl ketones, secondary alcohols with CH₃-CH(OH)-
- Negative: methanol, formaldehyde
4. Carbon Tetrachloride ($CCl_4$):
- Solvent
- Fire extinguishers (Pyrene; now phased out due to toxicity)
- Depletes ozone layer (limited use now)
- Can cause liver damage; carcinogen
5. Freons (Chlorofluorocarbons, CFCs):
- $CCl_2F_2$ (Freon-12), $CCl_3F$ (Freon-11), etc.
- Refrigerants, propellants
- Banned (Montreal Protocol, 1987) due to ozone layer depletion
- Replaced by HFCs (hydrofluorocarbons): no chlorine, less harmful
Mechanism of Ozone Depletion: $CFCl_3 \xrightarrow{UV} CFCl_2^\bullet + Cl^\bullet$ $Cl^\bullet + O_3 \to ClO^\bullet + O_2$ $ClO^\bullet + O \to Cl^\bullet + O_2$ (Cl regenerated; chain reaction)
6. DDT ($p,p'$-Dichlorodiphenyltrichloroethane):
- $Cl_2C_6H_4-CCl_3-C_6H_4Cl$ structure
- First synthetic insecticide (1939, Müller, Nobel Prize 1948)
- Highly effective against malaria mosquitoes
- Banned in many countries (1970s-2000s) due to:
- Bio-accumulation in food chain
- Toxicity to wildlife (especially birds — eggshell thinning)
- Carcinogenic concerns
- Still used in some places for malaria control
Applications of Haloalkanes:
| Use | Example |
|---|---|
| Anesthetic | $CHCl_3$ (historical), $CH_2=CFCF_3$ (modern) |
| Solvent | $CHCl_3$, $CCl_4$, $CH_2Cl_2$ |
| Fire extinguisher | $CCl_4$ (Pyrene); $CF_3Br$ |
| Refrigerant | Freons (banned); HFCs |
| Aerosols | $CCl_4, CFCs$ (replaced) |
| Pesticides | DDT (largely banned), Lindane |
| Plastic | Polyvinyl chloride (PVC) from vinyl chloride |
| Pharmaceutical | Various |
Why is chloroform stored in dark bottles?
Show solution
On exposure to light and air, $CHCl_3$ slowly oxidizes to phosgene ($COCl_2$): $2CHCl_3 + O_2 \xrightarrow{light} 2COCl_2 + 2HCl$ Phosgene is highly toxic (used as chemical weapon in WWI). Storage in dark amber bottles prevents light-induced oxidation. Small amount of ethanol added to chloroform to convert any phosgene formed to harmless ethyl carbonate ($CO(OC_2H_5)_2$).
Final Answer: Prevents formation of toxic phosgene by light.
Identify if positive in iodoform test: (a) Acetaldehyde ($CH_3CHO$) (b) Methanol ($CH_3OH$) (c) Ethanol ($CH_3CH_2OH$) (d) Propan-2-ol ($CH_3CH(OH)CH_3$)
Show solution
Iodoform test requires $CH_3CO-$ or $CH_3CH(OH)-$ groups. (a) Has $CH_3CO-$ → positive (yellow $CHI_3$ ppt + HCOOH or HCOONa) (b) Has only $-OH$ on methyl C; no $CH_3CO-$ or $CH_3CH(OH)-$ → negative (c) Has $CH_3CH(OH)$ — methyl carbinol with H → positive (d) Has $CH_3CH(OH)CH_3$ (methyl carbinol) → positive
Final Answer: Positive: a, c, d; Negative: b (methanol).
$CCl_4$ is used as:
Freons deplete ozone via:
DDT is:
Phosgene from chloroform forms by:
Iodoform test is positive for:
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