**Haloalkanes** (alkyl halides) are organic compounds where one or more hydrogen atoms of an aliphatic hydrocarbon are replaced by halogen atoms. **Haloarenes** (aryl halides) are compounds where halogen atoms are directly bonded to aromatic carbon atoms.
**Compounds containing sp³ C—X bond:**
**Compounds containing sp² C—X bond:**
The distinction is critical because **vinylic and aryl halides do NOT undergo typical nucleophilic substitution reactions** due to strong C—X bonds and resonance stabilization.
---
1. **Identify the parent chain** containing the halogen atom
2. **Number the chain** to give the halogen atom the lowest number
3. **Name as halogenated hydrocarbons**: prefix the halogen name (fluoro-, chloro-, bromo-, iodo-) before the alkane name
4. **For multiple halogens**: use numerical prefixes (di-, tri-, tetra-, etc.) and list alphabetically
5. **For dihalogen compounds**:
| Structure | IUPAC Name | Classification |
|-----------|-----------|-----------------|
| CH₃CH₂CH₂CH₂Br | 1-Bromobutane | 1° alkyl halide |
| CH₃CHBrCH₂CH₃ | 2-Bromobutane | 2° alkyl halide |
| (CH₃)₃CBr | 2-Bromo-2-methylpropane | 3° alkyl halide |
| CH₂=CHCl | Chloroethene | Vinylic halide |
| CH₂=CHCH₂Br | 3-Bromopropene | Allylic halide |
| C₆H₅Cl | Chlorobenzene | Aryl halide |
| CH₂Cl₂ | Dichloromethane | Geminal dihalide |
| CHCl₃ | Trichloromethane | Geminal trihalide |
| CH₃CHClCH₂Cl | 1,2-Dichloropropane | Vicinal dihalide |
For benzene substitution, use **1,2- (ortho/o-), 1,3- (meta/m-), 1,4- (para/p-)** numbering in IUPAC (common names also acceptable):
---
**Carbon-halogen bonds are highly polarized** due to the high electronegativity of halogens. The carbon atom bears a **partial positive charge (δ+)** while the halogen bears a **partial negative charge (δ—)**.
| Bond | Bond Length (pm) | Bond Enthalpy (kJ/mol) | Dipole Moment (Debye) |
|------|------------------|----------------------|----------------------|
| C—F | 139 | 452 | 1.847 |
| C—Cl | 178 | 351 | 1.860 |
| C—Br | 193 | 293 | 1.830 |
| C—I | 214 | 234 | 1.636 |
---
**Alcohols are the most common and practical starting material** for preparing alkyl halides. The C—OH bond is replaced by C—X.
**Method 1: Reaction with hydrogen halides**
**Mechanism**:
1. Protonation of OH group (forms good leaving group)
2. Carbocation formation (SN1 mechanism for 2° and 3° alcohols)
3. Nucleophilic attack by halide
**Reactions**:
**Order of reactivity**: 3° > 2° > 1° (due to carbocation stability)
**Example**: (CH₃)₃C—OH + HCl (conc.) → (CH₃)₃C—Cl + H₂O
**Method 2: Reaction with phosphorus halides**
**Method 3: Reaction with thionyl chloride (preferred)**
**Why preferred**: Both gaseous byproducts (SO₂ and HCl) escape, driving reaction forward and leaving pure alkyl halide product. No catalyst needed; works for all primary, secondary, and tertiary alcohols.
**Important limitation**: These methods **cannot prepare aryl halides** because the C—O bond in phenols has **partial double bond character** (resonance stabilization) making it much stronger than a regular single bond and resistant to cleavage.
**Method 1: Free radical halogenation**
**Mechanism** (three steps):
1. **Initiation**: Cl₂ or Br₂ undergoes homolytic cleavage under UV light or heat → 2 X•
2. **Propagation**:
3. **Termination**: R• + R• → R—R (or other radical combinations)
**Reactions**:
**Problems**:
**Selectivity**: Tertiary C—H > Secondary C—H > Primary C—H (abstraction rates follow radical stability)
**Example**: Monochlorination of (CH₃)₂CHCH₂CH₃ yields four monochlorides:
**Method 1: Addition of hydrogen halides**
**Example**: CH₃CH=CH₂ + HCl → CH₃CHClCH₃ (2-chloropropane, major; CH₃CH₂CH₂Cl not formed)
**Method 2: Addition of halogens**
**Example**: CH₂=CH₂ + Br₂ → CH₂Br—CH₂Br (1,2-dibromoethane)
**Converting one alkyl halide to another by halogen substitution**:
**Finkelstein reaction** (alkyl chloride/bromide → alkyl iodide):
**Mechanism**: Nucleophilic substitution (SN2). NaCl and NaBr precipitate in dry acetone (low solubility), driving equilibrium forward via Le Chatelier's principle.
**Driving force**: Iodide is better nucleophile than chloride/bromide in polar aprotic solvents; precipitated NaCl/NaBr removes salts from solution.
**Swarts Reaction** (alkyl chloride/bromide → alkyl fluoride):
**Alternative agents**: CoF₂, Hg₂F₂ also used
**Why needed**: Direct fluorination of alcohols/hydrocarbons not feasible; fluorine too reactive
---
**Chlorination and bromination** of benzene and substituted benzenes:
**Mechanism**: Electrophilic aromatic substitution (SEAr):
1. Halogen activated by Lewis acid (FeCl₃ or FeBr₃) to form X⁺ species
2. Nucleophilic aromatic ring attacks electrophilic halogen
3. H⁺ eliminated to restore aromaticity
**Separation**: Ortho and meta isomers of dihalogenated benzenes have **very different melting points** (para isomers typically highest mp due to symmetry), allowing easy separation by fractional crystallization.
**Iodination**:
**Fluorination**: **NOT done by direct halogenation** because fluorine is too reactive and cannot be controlled; use Sandmeyer or other methods instead.
**General sequence**:
1. **Diazotization**: Ar—NH₂ + NaNO₂ + HCl (cold, <5°C) → [Ar—N≡N]⁺Cl⁻ (diazonium salt) + H₂O
2. **Halogenation**: [Ar—N≡N]⁺Cl⁻ + CuCl (or CuBr) → Ar—Cl (or Ar—Br) + N₂↑ + Cu⁺
**Mechanism**: Arene diazonium salts are highly reactive and unstable; copper(I) halides catalyze C—N bond cleavage with C—X bond formation.
**Examples**:
**Advantage**: Allows introduction of halogen into positions already bearing activating/deactivating groups via nucleophilic replacement of amine
**Key conditions**:
---
**General trends**:
**For same alkyl group, order of boiling points**: **R—I > R—Br > R—Cl > R—F**
**Effect of branching**: More branched isomers have **lower boiling points** due to decreased surface contact between molecules. Example:
**Dihalogenated benzenes**:
**Very low solubility** ("insoluble in water")
**Reason**:
**Solubility in organic solvents**: **High**
**Trend**: Density increases with:
1. Number of halogen atoms
2. Atomic mass of halogen (F < Cl < Br < I in density contribution)
3. Molecular mass increase
| Compound | Density (g/mL) |
|----------|----------------|
| n-C₃H₇Cl | 0.89 |
| n-C₃H₇Br | 1.335 |
| n-C₃H₇I | 1.747 |
| CH₂Cl₂ | 1.336 |
| CHCl₃ | 1.489 |
| CCl₄ | 1.595 |
---
**Definition**: Reaction where a nucleophile (Nu⁻: electron-rich species) replaces the halogen (leaving group, X) in an alkyl halide substrate.
**General equation**: R—X + Nu⁻ → R—Nu + X⁻
**Substrate requirement**: **sp³ hybridized C—X bond** (haloalkanes, allylic, benzylic halides)
**Nucleophiles involved** (Table 6.4):
**Ambident nucleophiles**: Possess two reactive sites
**Definition**: Bimolecular mechanism where rate depends on concentrations of both **nucleophile and substrate**.
**Rate law**: Rate = k[R—X][Nu⁻] (second order)
**Mechanism** (single-step):
**Example with CH₃Cl + OH⁻**:
```
HO⁻ + CH₃Cl → HO···C···Cl → HO—CH₃ + Cl⁻
(transition state)
```
**Stereochemistry**: **Complete inversion of configuration** (Walden inversion)
**Structural factors favoring SN2**:
1. **Primary alkyl halides (1°)**: MOST reactive
2. **Secondary alkyl halides (2°)**: Moderate reactivity
3. **Tertiary alkyl halides (3°)**: **LEAST reactive** in SN2
**Effect of leaving group**: **Better leaving groups favor SN2**
**Effect of solvent**: **Aprotic polar solvents strongly favor SN2**
**Effect of nucleophile strength**: Stronger nucleophiles (more negative, more polarizable) react faster in SN2
**Key characteristics summary**:
---
**Definition**: Reaction where H and X are removed from adjacent carbons, forming a C=C double bond.
**General equation**: R—CHX—CH₂—R' → R—CH=CH—R' + HX
**Substrate requirement**: Requires **β-hydrogen** (hydrogen on carbon adjacent to C—X)
**Competition**: In same substrate, SN2, SN1, E1, and E2 all possible; products determined by **mechanism which dominates**
**Definition**: Concerted bimolecular elimination where base attacks β-hydrogen as C—X bond breaks.
**Rate law**: Rate = k[R—X][Base] (second order)
**Mechanism** (single-step):
**Example**:
```
(CH₃)₂CHBr + KOH → (CH₃)₂C=CH₂ + KBr + H₂O
(3° halide → E2 predominates)
```
**Stereochemistry**: **Anti elimination** (stereoelectronic requirement)
**Zaitsev's rule**: When multiple alkenes possible, major product is alkene with **more substituted double bond** (thermodynamically more stable)
**Structural factors favoring E2**:
1. **Tertiary halides (3°)**: Most reactive (bulky, favors elimination over substitution)
2. **Secondary halides (2°)**: Moderate reactivity
3. **Primary halides (1°)**: **Cannot undergo E2** with bulky bases (no β-H abstraction; SN2 favored)
**Effect of base strength**: Stronger bases favor E2 over SN2
**Effect of solvent**: Aprotic solvents favor E2 (higher kinetic barrier to abstraction; base not solvated)
**Key characteristics**:
**Definition**: Two-step elimination where carbocation intermediate forms first (rate-determining), followed by base abstraction of β-hydrogen.
**Rate law**: Rate = k[R—X] (first order, depends only on halide concentration)
**Mechanism**:
1. **Slow step**: C—X bond breaks → carbocation intermediate
2. **Fast step**: Base abstracts β-hydrogen → alkene
**Driving force for step 1**: C—X cleavage driven by solvent polarity (polar solvent stabilizes carbocation)
**Conditions favoring E1**:
**Rearrangement in carbocation**: If more stable carbocation can form via **hydride or methyl shift**, both E1 and rearranged product possible
**Stereochemistry**: **No specific requirement** (two separate steps allow free rotation)
**Regioselectivity**: **Zaitsev product predominates** (most substituted alkene from most stable carbocation)
**Competition E1 vs SN1**: Both proceed via same carbocation intermediate; product ratio depends on nucleophilicity of solvent vs base concentration
**Key characteristics**:
| Mechanism | Order | Kinetics | Rate Law | Substrate | Base/Nucleophile | Solvent | Stereochemistry |
|-----------|-------|----------|----------|-----------|------------------|--------|-----------------|
| **SN2** | 1° > 2° > 3° | Bi | k[R-X][Nu] | Primary | Strong Nu | Aprotic | Inversion |
| **SN1** | 3° > 2° > 1° | Uni | k[R-X] | Tertiary | Weak Nu | Protic | Racemization |
| **E2** | 3° > 2°, 1° none | Bi | k[R-X][B] | Tertiary | Strong Base | Aprotic | Anti periplanar |
| **E1** | 3° > 2° > 1° | Uni | k[R-X] | Tertiary | Weak Base | Protic | No requirement |
---
**General synthesis**: Alkyl or aryl halides react with magnesium metal in dry ether solvent:
R—X + Mg (dry ether) → R—Mg—X (Grignard reagent)
**Examples**:
Q1. Which of the following is a secondary alkyl halide?
Answer: C — In option C, bromine is attached to a carbon bonded to two alkyl groups, making it secondary (2°).
Q2. The IUPAC name of CHCl₃ is:
Answer: B — CHCl₃ has three chlorine atoms on one carbon (CH); the IUPAC name is trichloromethane.
Q3. Which compound is a vic-dihalide?
Answer: C — Vic-dihalides have halogen atoms on adjacent carbons; option C has Cl on positions 2 and 3 of the chain.
Q4. The C-F bond is stronger than C-I bond because:
Answer: B — Bond strength increases with bond length decrease; fluorine's smaller size creates a shorter, stronger C-F bond.
Q5. How many structural isomers have the molecular formula C₄H₉Br?
Answer: C — Four isomers exist: 1-bromobutane (1°), 2-bromobutane (2°), 1-bromo-2-methylpropane (1°), and 2-bromo-2-methylpropane (3°).
Q6. Which of the following statements about allylic halides is correct? A: Halogen is bonded to sp³ carbon adjacent to C=C B: Halogen is bonded directly to sp² carbon of C=C Choose:
Answer: A — Allylic halides have halogen on sp³ carbon next to the double bond (A is correct); vinylic halides have halogen on the sp² carbon of C=C (B describes vinylic halides, not allylic).
Q7. The common name 'allyl bromide' corresponds to IUPAC name:
Answer: C — Allyl bromide is CH₂=CHCH₂Br; numbering from the double bond gives 3-bromopropene.
Q8. Which is NOT a correct statement about haloalkanes?
Answer: B — Benzylic halides have halogen bonded to sp³ carbon attached to the aromatic ring, not directly to the ring carbon (which would be aryl halide).
Q9. If a haloalkane contains 4 carbons in the main chain with Br at position 2 and a methyl group at position 3, the IUPAC name is: A: 2-Bromo-3-methylbutane B: 3-Bromo-2-methylbutane C: 2-Bromo-3-methylpentane D: Impossible structure Choose the correct option:
Answer: A — 4 carbons in main chain = butane; Br at position 2 and CH₃ at position 3 gives 2-bromo-3-methylbutane; numbering prioritizes the halogen.
Q10. Compared to alkyl halides, aryl halides are less reactive in nucleophilic substitution because: (i) sp² carbon is less polarized (ii) aryl carbocations are too unstable (iii) C-Ar bond is shorter and stronger Which statements are correct?
Answer: B — Aryl carbocations are highly unstable (ii is true); C-Ar bond is shorter due to sp² hybridisation (iii is true); sp² carbon is actually more polarized, not less (i is false).
What is the key difference between a primary and tertiary alkyl halide?
Primary (1°) has halogen on carbon bonded to 1 alkyl group; tertiary (3°) has halogen on carbon bonded to 3 alkyl groups.
Why is the C-F bond stronger than C-I bond?
Fluorine is smaller and more electronegative, forming a shorter and stronger bond with carbon.
Distinguish between gem-dihalide and vic-dihalide.
Gem-dihalide has both halogens on the same carbon atom; vic-dihalide has halogens on adjacent carbon atoms.
What is an allylic halide? Give one structural feature.
An allylic halide has halogen bonded to sp³ carbon adjacent to a C=C double bond.
Name the compound: CH3CH2CH(Br)CH3 using IUPAC rules.
2-Bromobutane.
Define benzylic halide with an example.
A benzylic halide has halogen bonded to sp³ carbon directly attached to an aromatic ring; example: C6H5CH2Cl (benzyl chloride).
Why does the carbon-halogen bond length increase from F to I?
Because halogen atomic size increases down the group (F < Cl < Br < I), increasing bond length.
What hybridisation must carbon have for a vinylic halide?
sp² hybridisation because the halogen is bonded to a carbon atom of a C=C double bond.
Convert common name 'tert-butyl bromide' to IUPAC name.
2-Bromo-2-methylpropane.
Why are aryl halides unreactive towards nucleophilic substitution compared to alkyl halides?
The C-Ar bond is shorter and stronger due to sp² hybridisation, and aryl carbocations are too unstable to form easily.
Define primary, secondary and tertiary alkyl halides. Give one structural example of each. [2 marks]
Define based on number of alkyl groups bonded to the carbon carrying halogen (1, 2, or 3 respectively). Use simple structures: CH₃CH₂CH₂Br (1°), CH₃CH(Br)CH₃ (2°), (CH₃)₃CBr (3°).
Write the IUPAC names of the following compounds and classify each as primary, secondary or tertiary halide: (a) CH₃CH₂CH(Br)CH₂CH₃ (b) (CH₃)₃CCl (c) CH₂=CHCH₂I [5 marks]
For (a): 5-carbon chain, Br at position 3 → 3-bromopentane (2°, check: carbon 3 has 2 alkyl groups). For (b): 4 carbons (2,2-dimethylpropane), Cl at quaternary carbon → 2-chloro-2-methylpropane (3°). For (c): 3-carbon chain with double bond from position 1, I at position 3 (allylic) → 3-iodopropene (1°). Show numbering priority: halogens get lowest numbers.
Explain why the C-F bond is much stronger than the C-I bond. Also explain why aryl halides are unreactive towards nucleophilic substitution compared to alkyl halides. Justify your answers with relevant structural and bonding arguments. [6 marks]
Part 1: Use periodic trends—fluorine is smallest, shortest C-F bond, strong overlap, lowest bond dissociation energy (recall Table 6.2 bond enthalpies). Iodine is largest, longest C-I bond, weak overlap, high bond dissociation energy. Show: bond strength ∝ 1/bond length. Part 2: Aryl halides have sp² C-X (shorter, stronger than sp³ C-X), aryl carbocations are unstable (no resonance stabilisation in this context), sp² carbon less electron-rich than sp³ (less susceptible to nucleophilic attack). Contrast with alkyl halides: sp³ C-X is longer/weaker, alkyl carbocations form readily (3° > 2° > 1°). Use examples: C₆H₅Cl vs CH₃CH₂Cl.
Practice with interactive flashcards, mind maps, upload your own chapters and get AI study kits instantly
Try StudyOS Free →