JEE Main & Advanced

Haloalkanes and Haloarenes

Haloalkanes and Haloarenes for JEE Main & Advanced

1
Module 1

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:

TypeDescriptionExample
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)
Worked Examples
1

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.

2

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).

✎ Self-Check — 5 questions0 / 5
Q1.

$(CH_3)_3CCl$ is:

Q2.

Darzen's method uses:

Q3.

Finkelstein reaction:

Q4.

Lucas test distinguishes:

Q5.

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:
  1. Slow: $R-X \to R^+ + X^-$ (ionization; rate-determining)
  2. 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:

PropertySN1SN2
OrderFirstSecond
MechanismTwo-step (carbocation intermediate)Concerted (one step)
StereochemistryRacemizationInversion
Substrate$3° > 2° > 1°$$1° > 2° > 3°$
NucleophileWeak/moderateStrong
SolventPolar proticPolar aprotic
RearrangementPossible (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:

ConditionsFavors
Strong, bulky base (KOC(CH₃)₃, tBuOK)E (over Nu)
Strong, small nucleophile (CN⁻, alkoxide RO⁻)SN
HeatE (high entropy product)
Cool conditionsSN
Aqueous KOHSN1 (in alcohols)
Alcoholic KOHE2 (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)
Worked Examples
1

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).

2

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).

✎ Self-Check — 5 questions0 / 5
Q1.

$S_N2$ reaction shows:

Q2.

Order of $S_N1$ reactivity (substrate):

Q3.

$S_N1$ vs $S_N2$ in tertiary halide:

Q4.

E2 needs:

Q5.

Anti-periplanar requirement is for:

2
Module 2

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 $S_N$?

  1. C-X bond is shorter and stronger (due to partial double bond from resonance)
  2. Resonance stabilization of ground state makes ionization difficult
  3. $sp^2$ C is more electronegative than $sp^3$ C — holds halide tightly
  4. 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.
Worked Examples
1

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.

2

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.

✎ Self-Check — 5 questions0 / 5
Q1.

Aryl halides are:

Q2.

$-Cl$ on benzene is:

Q3.

Sandmeyer reaction:

Q4.

Benzyne intermediate forms when:

Q5.

$p$-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:

UseExample
Anesthetic$CHCl_3$ (historical), $CH_2=CFCF_3$ (modern)
Solvent$CHCl_3$, $CCl_4$, $CH_2Cl_2$
Fire extinguisher$CCl_4$ (Pyrene); $CF_3Br$
RefrigerantFreons (banned); HFCs
Aerosols$CCl_4, CFCs$ (replaced)
PesticidesDDT (largely banned), Lindane
PlasticPolyvinyl chloride (PVC) from vinyl chloride
PharmaceuticalVarious
Worked Examples
1

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.

2

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).

✎ Self-Check — 5 questions0 / 5
Q1.

$CCl_4$ is used as:

Q2.

Freons deplete ozone via:

Q3.

DDT is:

Q4.

Phosgene from chloroform forms by:

Q5.

Iodoform test is positive for:

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