Haloalkanes and Haloarenes

Haloalkanes and Haloarenes - Class 12 Chemistry

Class 12 Chemistry | Chapter 6

Haloalkanes and Haloarenes

Classification • Preparation • SN1 & SN2 Mechanisms • Optical Activity

1. Classification and Nature of C–X Bond

Replacement of hydrogen atom(s) in a hydrocarbon by halogen atom(s) results in formation of alkyl halide (haloalkane) and aryl halide (haloarene).

1.1 Classification Based on Carbon Hybridization

  • sp3 C–X Bond:
    • Alkyl halides (R–X): Halogen is bonded to an alkyl group. Can be primary (1°), secondary (2°), or tertiary (3°).
    • Allylic halides: Halogen is bonded to an sp3 hybridized carbon adjacent to a carbon-carbon double bond (allylic carbon). Example: CH2=CH–CH2X
    • Benzylic halides: Halogen is bonded to an sp3 hybridized carbon attached to an aromatic ring. Example: C6H5CH2X
  • sp2 C–X Bond:
    • Vinylic halides: Halogen is bonded directly to an sp2 hybridized carbon of a carbon-carbon double bond. Example: CH2=CHX
    • Aryl halides (Haloarenes): Halogen is bonded directly to the sp2 hybridized carbon of an aromatic ring. Example: C6H5X

1.2 Nature of C–X Bond

Halogens are more electronegative than carbon. Thus, the carbon-halogen bond is polarized: the carbon atom bears a partial positive charge (δ+) and the halogen atom bears a partial negative charge (δ−).

  • Bond length: Increases down the group (C–F < C–Cl < C–Br < C–I) as atomic size increases.
  • Bond enthalpy: Decreases down the group (C–F > C–Cl > C–Br > C–I) because longer bonds are weaker.
  • Dipole moment: Decreases in the order CH3Cl > CH3F > CH3Br > CH3I. (CH3Cl has a higher dipole moment than CH3F because the bond length d for C-Cl is much larger, compensating for the smaller charge difference q in μ = q × d).

2. Methods of Preparation of Haloalkanes

2.1 From Alcohols (Best Methods)

The hydroxyl group of an alcohol is replaced by halogen on reaction with concentrated halogen acids, phosphorus halides, or thionyl chloride.

1. R–OH + HX → (ZnCl2) → R–X + H2O (Lucas reagent is conc. HCl + ZnCl2)
2. 3R–OH + PCl3 → 3R–Cl + H3PO3
3. R–OH + PCl5 → R–Cl + POCl3 + HCl
💡 Darzens Procedure (Best method for R-Cl):
R–OH + SOCl2 → R–Cl + SO2(g) + HCl(g)
It is preferred because the by-products are gases and escape, leaving pure alkyl chloride.

2.2 From Hydrocarbons

  • Free Radical Halogenation: Cl2/UV light gives a complex mixture of isomeric mono- and polyhaloalkanes which is difficult to separate. (Not a good lab method).
  • Electrophilic Addition to Alkenes:
    CH2=CH–CH3 + HBr → CH3–CH(Br)–CH3 (Major, Markovnikov)

    Peroxide effect (Anti-Markovnikov): Occurs ONLY with HBr in presence of peroxide. Radical mechanism.

  • Addition of Halogens: Br2 in CCl4 adds to alkenes to form vicinal dibromides (Reddish-brown color of Br2 disappears – test for unsaturation).

2.3 Halogen Exchange Reactions

  • Finkelstein Reaction (for R-I):
    R–X + NaI → (Dry Acetone) → R–I + NaX↓ (X = Cl, Br). NaX precipitates out in acetone, driving the reaction forward (Le Chatelier's principle).
  • Swarts Reaction (for R-F):
    Heating alkyl chloride/bromide in presence of metallic fluorides like AgF, Hg2F2, CoF2 or SbF3.
    CH3Br + AgF → CH3F + AgBr

3. Preparation of Haloarenes

3.1 Electrophilic Substitution

Aryl chlorides and bromides can be prepared by electrophilic substitution of arenes with Cl2/Br2 in presence of Lewis acid catalysts (iron or iron(III) chloride).

Toluene + Cl2 → (Fe, dark) → o-Chlorotoluene + p-Chlorotoluene

Reaction with iodine requires an oxidizing agent (HNO3, HIO4) to oxidize the HI formed, as the reaction is reversible.

3.2 Sandmeyer's Reaction

When primary aromatic amine (aniline) is treated with NaNO2 + HCl at cold temp (0-5°C), a benzene diazonium salt is formed. Treating this with cuprous chloride (Cu2Cl2) or cuprous bromide (Cu2Br2) replaces the diazo group with -Cl or -Br.

1. Ar–NH2 + NaNO2 + 2HX → Ar–N2+X
2. Ar–N2+X + Cu2X2 → Ar–X + N2(g)

Note: Replacement by iodine does not require cuprous halide; just warming with KI is enough.

4. Chemical Reactions of Haloalkanes

4.1 Nucleophilic Substitution Reactions

The highly electronegative halogen atom is replaced by a nucleophile (Nu).

SN2 Mechanism (Substitution Nucleophilic Bimolecular)

  • Nucleophile approaches the carbon atom from the side opposite to the leaving group (halogen).
  • It is a one-step concerted mechanism. A transition state is formed where both incoming and outgoing groups are partially attached.
  • Inversion of configuration: If the reactant is optically active, the product has inverted stereochemistry (Walden inversion).
  • Reactivity order: CH3X > 1° > 2° > 3° (Due to Steric hindrance; bulky groups block the attack of nucleophile).

SN1 Mechanism (Substitution Nucleophilic Unimolecular)

  • It is a two-step process.
    Step 1: Slow cleavage of C-X bond to form a carbocation (Rate determining step).
    Step 2: Fast attack of nucleophile on the carbocation.
  • Racemization: The carbocation is planar, so the nucleophile can attack from either side, resulting in a 50:50 mixture of enantiomers (racemic mixture) if the chiral carbon is reacting.
  • Reactivity order: 3° > 2° > 1° > CH3X (Due to stability of the carbocation intermediate: 3° > 2° > 1°).
  • Note: Allylic and benzylic halides show high reactivity towards SN1 because the corresponding carbocations are resonance stabilized.
⚠️ Ambident Nucleophiles: Can attack through two different sites.
- Cyanide/Isocyanide: RX + KCN(ionic) → RCN (Alkyl cyanide, major). RX + AgCN(covalent) → RNC (Alkyl isocyanide, major).
- Nitrite/Nitro: RX + KNO2 → R-ONO (Alkyl nitrite). RX + AgNO2 → R-NO2 (Nitroalkane).

4.2 Elimination Reactions (β-Elimination)

When haloalkane is heated with alcoholic KOH, a hydrogen from the β-carbon and the halogen from the α-carbon are eliminated to form an alkene.

  • Zaitsev (Saytzeff) Rule: If more than one alkene can be formed, the more highly substituted alkene (alkene with more alkyl groups attached to the double bonded carbons) is the major product.
  • Example: 2-Bromobutane + alc. KOH → But-2-ene (Major, 81%) + But-1-ene (Minor, 19%).

4.3 Reaction with Metals

  • Grignard Reagent: RX + Mg → (dry ether) → RMgX. The C-Mg bond is highly polar. Grignard reagents are highly reactive and react with any source of proton (water, alcohols) to give hydrocarbons: RMgX + H2O → RH + Mg(OH)X.
  • Wurtz Reaction: 2RX + 2Na → (dry ether) → R-R + 2NaX. Used for preparing symmetrical alkanes with double the number of carbon atoms.

5. Chemical Reactions of Haloarenes

Haloarenes are extremely less reactive towards nucleophilic substitution compared to haloalkanes. Why?

  1. Resonance effect: Lone pairs on halogen are in conjugation with the benzene ring. The C-X bond gains partial double bond character, making it shorter and stronger.
  2. Difference in hybridization: The carbon atom in C-X of haloarenes is sp2 hybridized (more electronegative = holds electron pair more tightly), compared to sp3 in haloalkanes.
  3. Instability of phenyl cation: Phenyl cation formed in an SN1 mechanism is extremely unstable because the positive charge is on an sp2 carbon.

5.1 Replacement by Hydroxyl Group (Dow's Process)

Chlorobenzene can be converted to phenol but requires extreme conditions: aqueous NaOH at 623 K and 300 atm pressure. Presence of electron-withdrawing groups (-NO2) at ortho and para positions heavily increases reactivity.

5.2 Electrophilic Substitution Reactions

Halogen atom on benzene ring is ortho/para directing but deactivating. Reagents include Nitration (HNO3/H2SO4), Sulfonation (conc. H2SO4), Halogenation (Cl2/FeCl3), and Friedel-Crafts alkylation/acylation.

5.3 Reaction with Metals

  • Wurtz-Fittig Reaction: 1 Aryl halide + 1 Alkyl halide + 2Na → (dry ether) → Alkylbenzene.
  • Fittig Reaction: 2 Aryl halides + 2Na → (dry ether) → Diphenyl (biphenyl).

6. Stereochemistry Foundations

  • Chiral Carbon: A carbon atom attached to 4 different groups. The molecule lacks a plane of symmetry.
  • Enantiomers: Non-superimposable mirror images of each other. They rotate plane-polarized light in opposite directions.
  • Racemic Mixture: An equimolar mixture of two enantiomers. It is optically inactive due to external compensation. Racemization occurs in SN1 mechanism.
  • Walden Inversion: The turning inside out of an umbrella shape. Perfect inversion of configuration occurs in SN2 mechanism.

🎓 NEET Previous Year Questions

Q1. [NEET 2022] Which of the following is an example of allylic halide? (a) 1-Bromobut-2-ene (b) 4-Bromobut-1-ene (c) 3-Bromopropene (d) 1-Bromobenzene
Answer Allylic halide has the halogen attached to an sp³ carbon which is adjacent to a C=C bond. In CH₂=CH-CH₂Br (3-Bromopropene), Br is on C3, which is adjacent to the double bond. Correct is (c) 3-Bromopropene.
Q2. [NEET 2021] The major product formed in dehydrohalogenation reaction of 2-Bromopentane is Pent-2-ene. This product formation is based on:
Answer Zaitsev's Rule (Saytzeff Rule). The more highly substituted alkene is the major product in a β-elimination reaction.
Q3. [NEET 2020] Elimination reaction of 2-Bromo-pentane to form pent-2-ene is: (A) β-Elimination reaction (B) Follows Zaitsev rule (C) Dehydrohalogenation reaction (D) Dehydration reaction.
Answer It involves removal of H from β-carbon and Br from α-carbon (β-elimination, dehydrohalogenation) and yields the more substituted alkene (Zaitsev rule). So A, B, and C are correct.
Q4. [NEET 2019] For the following reactions: (a) CH₃CH₂CH₂Br + KOH(alc) → CH₃CH=CH₂ + KBr + H₂O (b) (CH₃)₃CBr + KOH(aq) → (CH₃)₃COH + KBr Which types are they?
Answer Alcoholic KOH causes Elimination (dehydrohalogenation) forming alkenes. Aqueous KOH causes Substitution forming alcohols. (a) is Elimination, (b) is Substitution.
Q5. [NEET 2018] Hydrocarbon (A) reacts with bromine by substitution to form an alkyl bromide which by Wurtz reaction is converted to gaseous hydrocarbon containing less than four carbon atoms. (A) is:
Answer Wurtz reaction doubles the carbons: 2 R-Br → R-R. The final gas has <4 carbons (i.e. ethane). So R must be CH₃-. The alkyl bromide is CH₃Br. The starting hydrocarbon (A) must be Methane (CH₄).

💡 Rapid Revision

  • SN1: Two steps → Carbocation intermediate → Racemization → 3° > 2° > 1°.
  • SN2: One step → Transition state → Inversion → Primary (1°) is fastest (least steric hindrance).
  • Aqueous KOH vs Alc. KOH: aq. KOH performs nucleophilic substitution (to give alcohol). alc. KOH performs elimination (to give alkene).
  • Grignard Reagents (RMgX) must be prepared under strictly anhydrous conditions, else they react with water to form alkanes.
  • Aryl halides are very unreactive toward nucleophiles due to resonance giving the C-X bond partial double-bond character.
Numericals & Solutions - Haloalkanes and Haloarenes - Class 12 Chemistry

CLASS 12 CHEMISTRY | NCERT SOLUTIONS

Chapter 6 — Haloalkanes and Haloarenes

22 Solved Questions — Mechanisms, Conversions & Reasoning

Note: Organic chemistry chapters focus on conversion chains, mechanism reasoning (SN1/SN2/E2), and predicting major/minor products (Markovnikov/Zaitsev rules) rather than mathematical numericals.

📝 SN1 vs SN2 and Reactivity (Q1 – Q7)

2 MarksQ1. Out of C₆H₅CH₂Cl and C₆H₅CHClC₆H₅, which is more easily hydrolysed by aqueous KOH? Why?
✓ Solution
Hydrolysis by aqueous KOH proceeds predominantly via SN1 mechanism in benzylic halides. SN1 reactivity depends on the stability of the intermediate carbocation.
1. C₆H₅CH₂Cl → forms 1° benzylic carbocation C₆H₅CH₂⁺.
2. C₆H₅CHClC₆H₅ → forms 2° benzylic carbocation (C₆H₅)₂CH⁺.
The secondary carbocation is highly stabilized by resonance with two phenyl rings. Thus, C₆H₅CHClC₆H₅ is hydrolysed more easily.
3 MarksQ2. Which compound in each of the following pairs will react faster in SN2 reaction with OH⁻? (i) CH₃Br or CH₃I (ii) (CH₃)₃CCl or CH₃Cl.
✓ Solution
(i) CH₃I: It will react faster than CH₃Br because iodine atom is larger in size than bromine, making the C–I bond weaker and easier to break. Iodide is a better leaving group.
(ii) CH₃Cl: In SN2 reactions, the nucleophile attacks from the backside. (CH₃)₃CCl (tertiary) has massive steric hindrance. CH₃Cl has negligible steric hindrance, so the incoming OH⁻ can attack easily.
2 MarksQ3. Predict the order of reactivity of the following compounds in SN1 and SN2 reactions: The four isomeric bromobutanes.
✓ Solution
The four isomers are: 1-bromobutane (1°), 1-bromo-2-methylpropane (1° branched), 2-bromobutane (2°), 2-bromo-2-methylpropane (3°).
SN1 Reactivity (based on carbocation stability: 3° > 2° > 1°):
3° > 2° > 1° branched > 1° unbranched.
SN2 Reactivity (based on steric hindrance: 1° > 2° > 3°):
1° unbranched > 1° branched > 2° > 3°.
2 MarksQ4. Haloalkanes react with KCN to form alkyl cyanides as main product while AgCN forms isocyanides as the chief product. Explain.
✓ Solution
KCN is predominantly ionic and provides cyanide ions (C≡N⁻) in solution. Although both C and N can donate electron pairs, attack takes place mainly through C atom since C–C bond is stronger than C–N bond. Thus, alkyl cyanides (R-CN) are formed.
AgCN is predominantly covalent. Only the nitrogen lone pair is available for attack. Therefore, attack takes place through N forming alkyl isocyanides (R-NC).
3 MarksQ5. Why is the dipole moment of chlorobenzene lower than that of cyclohexyl chloride?
✓ Solution
Dipole moment (μ) = charge (q) × distance (d).
1. The C atom in chlorobenzene is sp² hybridized (more electronegative), while in cyclohexyl chloride it is sp³ hybridized. Therefore, the C–Cl bond in chlorobenzene is less polar (smaller q).
2. Due to delocalization of lone pairs on Cl with the benzene ring (resonance), the C–Cl bond acquires partial double bond character. This decreases the bond length (smaller d).
Since both q and d are smaller in chlorobenzene, its overall dipole moment is lower.
2 MarksQ6. Out of ortho, meta, and para-dibromobenzenes, which one has highest melting point and why?
✓ Solution
The para-isomer has the highest melting point.
Reason: The p-isomer is more symmetrical than the ortho and meta isomers. Thus, it fits more closely in the crystal lattice, leading to stronger intermolecular forces of attraction, which requires more heat energy to break.
3 MarksQ7. Alkyl halides, though polar, are immiscible with water. Explain.
✓ Solution
To dissolve a solute in water, energy is required to overcome the attractions between the haloalkane molecules and to break the hydrogen bonds between water molecules.
When haloalkanes interact with water, new intermolecular attractions (dipole-dipole and van der Waals forces) form. However, these new forces are weaker than the original H-bonds in water. Since less energy is released in forming new interactions than is required to break the old ones, haloalkanes are sparingly soluble (essentially immiscible) in water.

💡 Reactions and Mechanisms (Q8 – Q14)

3 MarksQ8. Identify major and minor products when 2-bromo-pentane is heated with alcoholic KOH. What is the rule governing this?
✓ Solution
When 2-bromo-pentane is heated with alcoholic KOH, dehydrohalogenation (β-elimination) occurs.
Two β-carbons are available: C1 and C3.
Elimination from C1 → Pent-1-ene (Minor).
Elimination from C3 → Pent-2-ene (Major, ~81%).
Rule governing this is Zaitsev's (Saytzeff) Rule: In dehydrohalogenation reactions, the preferred product is that alkene which has the greater number of alkyl groups attached to the doubly bonded carbon atoms.
2 MarksQ9. Why are aryl halides extremely less reactive towards nucleophilic substitution reactions?
✓ Solution
1. Resonance effect: The lone pair on halogen is delocalized with the pi-electrons of the benzene ring, giving the C–X bond partial double bond character. It is very difficult to break.
2. Hybridization: The sp² hybridized carbon holds the electron pair of the C–X bond more tightly than sp³ carbon in alkyl halides.
3. Phenyl cation instability: If an SN1 mechanism were to occur, the phenyl cation formed is extremely unstable.
2 MarksQ10. Explain the preparation of alkyl halides by free radical halogenation of alkanes. Why is it not considered a good method for laboratory preparation?
✓ Solution
Free radical chlorination or bromination of alkanes gives a complex mixture of isomeric mono- and polyhaloalkanes.
CH₄ + Cl₂ → (UV) → CH₃Cl + CH₂Cl₂ + CHCl₃ + CCl₄
It is not a good lab method because it is very difficult to separate this mixture into pure compounds, resulting in a low yield of any single desired product.
3 MarksQ11. Write the mechanism for the reaction of n-butyl bromide with KCN.
✓ Solution
Reaction: CH₃CH₂CH₂CH₂Br + KCN(aq/ethanol) → CH₃CH₂CH₂CH₂CN + KBr.
Since n-butyl bromide is a primary (1°) halide, it will undergo nucleophilic substitution via SN2 mechanism.
Step: The nucleophile (C≡N⁻) attacks the primary carbon atom from the side opposite to the Br atom. A transition state forms where C-CN bond is forming and C-Br bond is breaking simultaneously. Finally, Br⁻ leaves, and inversion of configuration takes place.
2 MarksQ12. What are enantiomers? Give one example.
✓ Solution
Enantiomers are stereoisomers that are non-superimposable mirror images of each other. They have identical physical properties but rotate plane-polarized light in equal and opposite directions.
Example: (+)-Butan-2-ol and (-)-Butan-2-ol. The secondary carbon (C2) is chiral as it is attached to four different groups: -OH, -CH₃, -CH₂CH₃, and -H.
3 MarksQ13. How does the presence of electron withdrawing groups like -NO₂ on the benzene ring affect the nucleophilic substitution in haloarenes?
✓ Solution
The presence of an electron withdrawing group (EWG) like -NO₂ at ortho and para positions greatly increases the reactivity of haloarenes towards nucleophilic substitution.
Reason: The EWG stabilizes the intermediate carbanion (formed after nucleophile attack) through resonance. The negative charge can be delocalized onto the -NO₂ group. This stabilization lowers the activation energy.
Note: The effect is pronounced only when the -NO₂ group is at ortho or para positions, not meta.
2 MarksQ14. What are Grignard Reagents? How are they prepared?
✓ Solution
Grignard Reagents are alkyl magnesium halides (R-Mg-X). They are organometallic compounds where the carbon is covalently bonded to electropositive magnesium, making the carbon highly nucleophilic (carbanion character).
Preparation: They are prepared by reacting an alkyl/aryl halide with magnesium metal in dry ether.
R-X + Mg → (Dry Ether) → R-Mg-X

📈 Organic Conversions (Q15 – Q22)

3 MarksQ15. How will you bring about the following conversions? (i) Ethanol to but-1-yne (ii) Ethane to bromoethene.
✓ Solution
(i) Ethanol to but-1-yne:
1. CH₃CH₂OH + PCl₅ → CH₃CH₂Cl (Chloroethane)
2. HC≡CH + NaNH₂ → HC≡C⁻Na⁺ (Sodium acetylide)
3. CH₃CH₂Cl + HC≡C⁻Na⁺ → CH₃CH₂C≡CH (But-1-yne)
(ii) Ethane to bromoethene:
1. CH₃-CH₃ + Br₂ → (UV) → CH₃-CH₂Br (Bromoethane)
2. CH₃-CH₂Br + Alc. KOH → CH₂=CH₂ (Ethene)
3. CH₂=CH₂ + Br₂/CCl₄ → CH₂Br-CH₂Br (1,2-Dibromoethane)
4. CH₂Br-CH₂Br + Alc. KOH → CH₂=CHBr (Bromoethene - Vinyl bromide)
2 MarksQ16. Propene to 1-nitropropane
✓ Solution
We need anti-Markovnikov addition first to get terminal halide.
1. CH₃-CH=CH₂ + HBr → (Peroxide) → CH₃-CH₂-CH₂Br (1-Bromopropane)
2. CH₃-CH₂-CH₂Br + AgNO₂ → CH₃-CH₂-CH₂NO₂ (1-Nitropropane) + AgBr
Note: AgNO₂ is used, not KNO₂, to avoid farming nitrite (R-ONO) as major product.
2 MarksQ17. Toluene to benzyl alcohol
✓ Solution
1. C₆H₅-CH₃ + Cl₂ → (UV light / ht) → C₆H₅-CH₂Cl (Benzyl chloride - Free radical halogenation on side chain).
2. C₆H₅-CH₂Cl + Aq. KOH(heat) → C₆H₅-CH₂OH (Benzyl alcohol). Substitution.
2 MarksQ18. Propene to propyne
✓ Solution
1. CH₃-CH=CH₂ + Br₂/CCl₄ → CH₃-CHBr-CH₂Br (1,2-Dibromopropane).
2. CH₃-CHBr-CH₂Br + Alc. KOH → CH₃-C(Br)=CH₂ (2-Bromopropene).
3. CH₃-C(Br)=CH₂ + NaNH₂ (strong base) → CH₃-C≡CH (Propyne).
2 MarksQ19. Ethanol to ethyl fluoride
✓ Solution
Fluorides are made by Swarts reaction.
1. CH₃-CH₂OH + PCl₅ → CH₃-CH₂Cl (Chloroethane).
2. CH₃-CH₂Cl + AgF (or Hg₂F₂) → (Heat) → CH₃-CH₂F (Ethyl fluoride).
3 MarksQ20. Aniline to chlorobenzene
✓ Solution
This is Sandmeyer's reaction.
1. Diazotisation: C₆H₅-NH₂ + NaNO₂ + 2HCl → (0-5°C) → C₆H₅-N₂⁺Cl⁻ (Benzene diazonium chloride).
2. Sandmeyer: C₆H₅-N₂⁺Cl⁻ + Cu₂Cl₂/HCl → C₆H₅Cl (Chlorobenzene) + N₂.
3 MarksQ21. 2-Chlorobutane to 3,4-dimethylhexane
✓ Solution
Notice that the product has exactly double the number of carbon atoms (8 carbons) as the reactant (4 carbons). This is a classic Wurtz reaction.
2 CH₃-CHCl-CH₂-CH₃ + 2Na → (Dry ether) → CH₃-CH₂-CH(CH₃)-CH(CH₃)-CH₂-CH₃ + 2NaCl
The two 2-butyl radicals join at the C2 position to form 3,4-dimethylhexane.
2 MarksQ22. But-1-ene to n-butyl iodide
✓ Solution
We need terminal addition, then halogen exchange.
1. CH₃-CH₂-CH=CH₂ + HBr → (Peroxide) → CH₃-CH₂-CH₂-CH₂Br (1-Bromobutane - Anti-Markovnikov).
2. CH₃-CH₂-CH₂-CH₂Br + NaI → (Dry acetone) → CH₃-CH₂-CH₂-CH₂I (1-Iodobutane or n-butyl iodide). This is Finkelstein reaction.
✍ Score Guide — 22 Questions
All questions from NCERT Exercises covering Organometallics, SN1 vs SN2 reasoning, Zaitsev rule, and step-by-step interconversions.

High-Yield Facts & Formulas: Haloalkanes & Haloarenes

Alkyl Halide Bond
Haloalkanes have a polar Cδ+-Xδ- bond. Bond strength: C-F > C-Cl > C-Br > C-I.
Dipole Moment
Chloromethane (CH3Cl) has a higher dipole moment than fluoromethane (CH3F).
SN2 Mechanism
One step, bimolecular, transition state formed. Results in inversion of configuration.
SN1 Mechanism
Two steps, unimolecular, carbocation intermediate formed. Results in racemization.
Lucas Reagent
Anhydrous ZnCl2 + conc. HCl. Used to distinguish 1°, 2°, and 3° alcohols.
Finkelshtein Reaction
R-Cl/R-Br + NaI → R-I + NaCl (Acetone medium). Used for preparation of alkyl iodides.
Swarts Reaction
Reaction of alkyl halides with metallic fluorides (AgF, Hg2F2) to give alkyl fluorides.
Wurtz Reaction
2RX + 2Na (Dry Ether) → R-R + 2NaX. Used for symmetrical alkanes.
Grignard Reagent
R-Mg-X. Prepared by reacting RX with Mg in dry ether. Highly reactive with proton donors.
Sandmeyer Reaction
Conversion of diazonium salt to haloarene using Cu2Cl2/HCl or Cu2Br2/HBr.
Saytzeff Rule
In dehydrohalogenation, the more substituted alkene is the major product.
Chiral Center
A carbon atom bonded to four different groups.
Enantiomers
Non-superimposable mirror images that rotate plane-polarized light.
Racemic Mixture
Equimolar mixture of enantiomers, denoted by (dl) or (±).
Nucleophilicity
CN- > I- > OH- > CH3COO- > H2O.
Allylic Halides
Halogen bonded to sp3 carbon atom next to C=C bond. (Highly reactive in SN1).
Benzylic Halides
Halogen bonded to sp3 carbon atom attached to an aromatic ring.
Vinylic Halides
Halogen bonded directly to sp2 hybridized carbon of C=C bond. (Low reactivity).
Aryl Halides
Halogen bonded directly to aromatic ring. Less reactive towards nucleophilic substitution.
Phosgene formation
Chloroform (CHCl3) reacts with O2 in presence of light to form COCl2.
Iodoform test
Used to detect CH3CO- group or CH3CH(OH)- group. Forms yellow precipitate of CHI3.
Solubility
Haloalkanes are insoluble in water as they cannot form H-bonds with water molecules.
Density
Increases with increase in atomic mass of halogen and number of halogen atoms.
Wurtz-Fittig Reaction
Reaction between an aryl halide and an alkyl halide with sodium to give alkylarenes.
Fittig Reaction
Coupling of two aryl halides using sodium in dry ether to form diphenyl.
Electrophilic substitution in Haloarenes
Direction: Ortho and Para. Halogen is deactivating.
Boiling point trend
R-I > R-Br > R-Cl > R-F (due to Van der Waals forces).
Branching effect
Increased branching decreases the boiling point of haloalkanes.
Freons
Chlorofluorocarbons (CFCs). Freon-12 (CF2Cl2) is most common. Cause ozone depletion.
Gattermann Reaction
Modified Sandmeyer reaction using Cu powder instead of copper salts.
Reactivity of halides
Order for substitution: Tertiary > Secondary > Primary (for SN1).
Protic Solvents
Favor SN1 (e.g., Water, Alcohols).
Aprotic Solvents
Favor SN2 (e.g., DMSO, Acetone, DMF).
Dow's Process
Chlorobenzene to Phenol (High Temp/Pressure with NaOH).
Hunsdiecker Reaction
Silver salt of carboxylic acid + Br2 → Alkyl bromide + AgBr + CO2.
Leaving group ability
I- > Br- > Cl- > F-.
DDT toxicity
Persistent organic pollutant (POP). Bioaccumulates in food chain.
Haloform Reaction
Methyl ketones react with X2/NaOH to give haloform (CHX3).
Optical Activity
Ability of a substance to rotate plane polarized light.
Meso compound
A molecule with chiral centers but overall achiral due to internal plane of symmetry.
Halogenation of phenols
Occurs even in absence of Lewis acid due to high activation of ring by -OH.
C-X bond in Vinyl Halides
Shorter and stronger than C-X in alkyl halides due to resonance.
Nucleophile attacks electrophile
The base (nucleophile) attacks the electron deficient Carbon atom.
Ammonolysis
Reaction of alkyl halides with ammonia to give amines.
Reaction with AgCN
RX + AgCN → R-NC (Isocyanide) as major product.
Reaction with KCN
RX + KCN → R-CN (Cyanide) as major product.
Friedel-Crafts reaction
Alkyl halides react with aromatic compounds in AlCl3 to give alkylarenes.
Benzyne mechanism
Occurs in elimination-addition reactions of haloarenes with very strong bases.
Geminal vs Vicinal
Gem-dihalides: Two halogens on same carbon. Vic-dihalides: Halogens on adjacent carbons.
Carbon tetrachloride uses
Cleaning fluid, fire extinguisher (Pyrene), feedstock for CFCs.
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