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Haloalkanes and Haloarenes

NCERT Class 12 · Chemistry Based on NCERT Class 12 Chemistry textbook · Free CBSE study kit

Chapter Notes

Classification of Haloalkanes and Haloarenes

**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.

Classification by Number of Halogen Atoms

  • **Monohalocompounds**: Single halogen atom (e.g., CH₃Cl)
  • **Dihalogen compounds**: Two halogen atoms (e.g., CH₂Cl₂)
  • **Polyhalogen compounds**: Three or more halogen atoms (e.g., CCl₄)
  • Classification by Hybridization and Position of Carbon-Halogen Bond

    **Compounds containing sp³ C—X bond:**

  • **Primary alkyl halides (1°)**: Halogen bonded to primary carbon with one other carbon substituent (e.g., CH₃CH₂Br). These are represented by the general formula CₙH₂ₙ₊₁X.
  • **Secondary alkyl halides (2°)**: Halogen bonded to carbon with two other carbon substituents (e.g., CH₃CHBrCH₃)
  • **Tertiary alkyl halides (3°)**: Halogen bonded to carbon with three other carbon substituents (e.g., (CH₃)₃CBr)
  • **Allylic halides**: Halogen bonded to sp³ carbon adjacent to C=C double bond (e.g., CH₂=CHCH₂Br, allyl bromide). The carbon bearing halogen is not part of the double bond.
  • **Benzylic halides**: Halogen bonded to sp³ carbon attached directly to aromatic ring (e.g., C₆H₅CH₂Br, benzyl bromide). The halogenated carbon is not aromatic.
  • **Compounds containing sp² C—X bond:**

  • **Vinylic halides**: Halogen bonded to sp² carbon of C=C double bond (e.g., CH₂=CHCl, vinyl chloride). The halogen is part of the double bond.
  • **Aryl halides**: Halogen directly bonded to sp² aromatic carbon (e.g., C₆H₅Cl, chlorobenzene)
  • 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.

    ---

    IUPAC Nomenclature of Haloalkanes and Haloarenes

    Rules for Haloalkanes

    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**:

  • **Geminal halides (gem-dihalides)**: Both halogens on same carbon → dihaloalkane
  • **Vicinal halides (vic-dihalides)**: Halogens on adjacent carbons → dihalalkane
  • Examples

    | 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 |

    Dihalogen Derivatives of Benzene

    For benzene substitution, use **1,2- (ortho/o-), 1,3- (meta/m-), 1,4- (para/p-)** numbering in IUPAC (common names also acceptable):

  • **1,2-dichlorobenzene** = o-dichlorobenzene
  • **1,3-dichlorobenzene** = m-dichlorobenzene
  • **1,4-dichlorobenzene** = p-dichlorobenzene
  • ---

    Nature of C—X Bond

    Bond Characteristics

    **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 Properties Table

    | 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 |

    Trends

  • **Bond length increases** (F < Cl < Br < I) down the halogen group as atomic size increases
  • **Bond enthalpy decreases** (F > Cl > Br > I) down the group, making C—I bonds weakest and most easily cleaved
  • **Dipole moment** shows relative constancy across the series; bonds are all significantly polar
  • **Reactivity in substitution**: C—I most reactive, C—F least reactive (inverse of bond strength)
  • ---

    Preparation of Haloalkanes

    6.4.1: From Alcohols

    **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**:

  • **With HCl**: R—OH + HCl → R—Cl + H₂O (catalyst ZnCl₂ required for 1° and 2° alcohols; 3° alcohols react at room temperature)
  • **With HBr**: R—OH + HBr (48%) → R—Br + H₂O (requires heating; no catalyst needed for 3° alcohols)
  • **With HI**: R—OH + HI → R—I + H₂O (best results with dilute acid; HI is most reactive)
  • **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**

  • **PBr₃**: 3 R—OH + PBr₃ → 3 R—Br + H₃PO₃ (used for large-scale preparations; generates PBr₃ in situ using red phosphorus + Br₂)
  • **PI₃**: 3 R—OH + PI₃ → 3 R—I + H₃PO₃ (PI₃ prepared in situ using red phosphorus + I₂)
  • **Method 3: Reaction with thionyl chloride (preferred)**

  • **SOCl₂**: R—OH + SOCl₂ → R—Cl + SO₂↑ + HCl↑
  • **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.

    6.4.2: From Hydrocarbons

    **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**:

  • R—H + X• → R• + HX
  • R• + X₂ → R—X + X•
  • 3. **Termination**: R• + R• → R—R (or other radical combinations)

    **Reactions**:

  • **Chlorination**: CH₄ + Cl₂ → CH₃Cl + HCl (and further chlorination products CH₂Cl₂, CHCl₃, CCl₄)
  • **Bromination**: C₂H₆ + Br₂ → C₂H₅Br + HBr (UV light or heat required)
  • **Problems**:

  • **Mixture of products**: Both mono- and polychloro/polybromo isomers formed
  • **Low yield of single product**: Example, monochlorination of butane gives 4 isomeric monochloroproducts
  • **Poor selectivity**: Cannot easily isolate pure compounds
  • **Selectivity**: Tertiary C—H > Secondary C—H > Primary C—H (abstraction rates follow radical stability)

    **Example**: Monochlorination of (CH₃)₂CHCH₂CH₃ yields four monochlorides:

  • (CH₃)₂CCl—CH₂CH₃ (from tertiary C—H)
  • (CH₃)₂CH—CHCl—CH₃ (from secondary C—H)
  • (CH₃)₂CH—CH₂—CH₂Cl (from primary C—H on CH₂)
  • CH₃—CHCl—CH₂—CH₃ (from primary C—H on CH₂)
  • 6.4.3: From Alkenes

    **Method 1: Addition of hydrogen halides**

  • **Mechanism**: Follows **Markovnikov's rule** (halogen adds to carbon of C=C with more hydrogen atoms)
  • **Reaction**: R—CH=CH₂ + HX → R—CHX—CH₃
  • **Example**: CH₃CH=CH₂ + HCl → CH₃CHClCH₃ (2-chloropropane, major; CH₃CH₂CH₂Cl not formed)

    **Method 2: Addition of halogens**

  • **Reaction**: R—CH=CH—R + X₂ → R—CHX—CHX—R (forms **vic-dihalides**)
  • **Product**: Colorless liquid; used to detect C=C (Br₂ in CCl₄ decolorization test)
  • **Example**: CH₂=CH₂ + Br₂ → CH₂Br—CH₂Br (1,2-dibromoethane)

    6.4.4: Halogen Exchange (Finkelstein Reaction)

    **Converting one alkyl halide to another by halogen substitution**:

    **Finkelstein reaction** (alkyl chloride/bromide → alkyl iodide):

  • R—Cl + NaI (dry acetone) → R—I + NaCl↓
  • R—Br + NaI (dry acetone) → R—I + NaBr↓
  • **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):

  • R—Cl + AgF (heating) → R—F + AgCl↓
  • R—Cl + SbF₃ (heating) → R—F + SbCl₃
  • **Alternative agents**: CoF₂, Hg₂F₂ also used

    **Why needed**: Direct fluorination of alcohols/hydrocarbons not feasible; fluorine too reactive

    ---

    Preparation of Haloarenes

    Method 1: Electrophilic Aromatic Substitution

    **Chlorination and bromination** of benzene and substituted benzenes:

  • **Chlorination**: C₆H₆ + Cl₂ (FeCl₃ catalyst) → C₆H₅Cl + HCl
  • **Bromination**: C₆H₆ + Br₂ (FeBr₃ catalyst) → C₆H₅Br + HBr
  • **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**:

  • C₆H₆ + I₂ (HNO₃ or HIO₄ oxidant) → C₆H₅I + HI
  • **Requires oxidizing agent** because I₂ addition is normally reversible; oxidant converts generated HI to I₂, driving equilibrium forward
  • **Fluorination**: **NOT done by direct halogenation** because fluorine is too reactive and cannot be controlled; use Sandmeyer or other methods instead.

    Method 2: Sandmeyer Reaction (From Primary Aromatic Amines)

    **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**:

  • **Chlorination**: C₆H₅—NH₂ + NaNO₂/HCl → [C₆H₅—N≡N]⁺Cl⁻ + CuCl → C₆H₅—Cl
  • **Bromination**: C₆H₅—NH₂ + NaNO₂/HCl → [C₆H₅—N≡N]⁺Cl⁻ + CuBr → C₆H₅—Br
  • **Iodination**: [C₆H₅—N≡N]⁺Cl⁻ + KI → C₆H₅—I (no copper needed; simply mix with KI)
  • **Advantage**: Allows introduction of halogen into positions already bearing activating/deactivating groups via nucleophilic replacement of amine

    **Key conditions**:

  • Diazotization must be done **at cold temperature** (0–5°C) to prevent decomposition
  • Diazonium salt solution should be **freshly prepared** (unstable)
  • Reaction mixture should be **cold** during halogenation to prevent side reactions
  • ---

    Physical Properties of Haloalkanes and Haloarenes

    Melting and Boiling Points

    **General trends**:

  • **Higher than parent hydrocarbons**: Due to polarity of C—X bond and stronger intermolecular forces (dipole-dipole interactions + van der Waals forces)
  • **Increase with molecular mass**: More electrons → stronger London forces
  • **State at room temperature**:
  • Gases: CH₃Cl, CH₃Br, etc.
  • Liquids: Most C₂–C₆ haloalkanes
  • Solids: Longer-chain and polyhalogenated compounds
  • **For same alkyl group, order of boiling points**: **R—I > R—Br > R—Cl > R—F**

  • Despite F forming strongest C—X bond, fluorine is smallest; I is largest with most van der Waals interactions
  • F compounds have lowest boiling point due to minimal van der Waals forces (small size)
  • **Effect of branching**: More branched isomers have **lower boiling points** due to decreased surface contact between molecules. Example:

  • n-Butyl chloride (CH₃CH₂CH₂CH₂Cl): bp 78°C
  • Isopropyl chloride ((CH₃)₂CHCl): bp 35°C
  • tert-Butyl chloride ((CH₃)₃CCl): bp 51°C
  • **Dihalogenated benzenes**:

  • Boiling points of ortho, meta, para isomers very similar
  • **Melting points vary significantly**: Para isomers highest (symmetrical, pack better in crystal lattice)
  • Example: o-dichlorobenzene (mp 17°C), m-dichlorobenzene (mp –25°C), p-dichlorobenzene (mp 53°C)
  • Solubility in Water

    **Very low solubility** ("insoluble in water")

    **Reason**:

  • Breaking hydrogen bonds between water molecules requires energy
  • Weak dipole-dipole attractions between haloalkane (C—X polar) and water cannot compensate
  • Unfavorable enthalpy and entropy change
  • **Solubility in organic solvents**: **High**

  • Intermolecular forces (van der Waals, dipole-dipole) between haloalkane and nonpolar/weakly polar solvents similar in strength
  • New attractions formed ≈ original attractions broken → favorable ΔH ≈ 0
  • Density

  • **Bromine and iodine derivatives**: Denser than water (sink in water)
  • **Chlorine derivatives**: Generally less dense than water
  • **Polychlorinated compounds**: Density increases with Cl number (CCl₄ = 1.595 g/mL)
  • **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 |

    Color and Smell

  • **Color**: Colorless when pure; develop color (Br₂ orange, I₂ brown) on exposure to light due to photodecomposition
  • **Smell**: Many volatile haloalkanes have characteristic sweet smell (e.g., chloroform, tetrachloromethane); some toxic by inhalation
  • ---

    Chemical Reactions of Haloalkanes: Nucleophilic Substitution

    Overview of Nucleophilic Substitution

    **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)

  • **Vinylic halides (C=C—X) and aryl halides (Ar—X) do NOT undergo nucleophilic substitution** (strong C—X bond; resonance stabilization prevents carbocation/carbanion character)
  • **Nucleophiles involved** (Table 6.4):

  • OH⁻ → alcohols
  • OR⁻ → ethers
  • I⁻ → alkyl iodides
  • NH₃ → primary amines (RNH₂)
  • RNH₂ → secondary amines (RNHR')
  • R₂NH → tertiary amines (RNR'R")
  • CN⁻ → nitriles (RCN)
  • NC⁻ → isonitriles (RNC)
  • NO₂⁻ → nitroalkanes (RNO₂)
  • ONO⁻ → alkyl nitrites (RONO)
  • RCOO⁻ → esters (RCOOR')
  • H⁻ → hydrocarbons (RH)
  • **Ambident nucleophiles**: Possess two reactive sites

  • **Cyanide (CN⁻)**: [C≡N⁻ ↔ C⁻=N] attacks through C (nitrile, favored) or N (isonitrile)
  • **Nitrite (NO₂⁻)**: [—O—N=O ↔ —O=N⁺—O⁻] attacks through O (alkyl nitrite, kinetic) or N (nitroalkane, thermodynamic)
  • SN2 Mechanism (Substitution Nucleophilic Bimolecular)

    **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):

  • Nucleophile attacks from **rear** (opposite side to leaving group), backside attack
  • C—X bond breaks as C—Nu bond forms simultaneously (concerted)
  • Transition state shows **partial bonds**: Nu···C···X
  • **Example with CH₃Cl + OH⁻**:

    ```

    HO⁻ + CH₃Cl → HO···C···Cl → HO—CH₃ + Cl⁻

    (transition state)

    ```

    **Stereochemistry**: **Complete inversion of configuration** (Walden inversion)

  • If substrate is optically active (stereocenter), product has **opposite stereochemistry**
  • Example: (R)-2-Iodobutane + OH⁻ → (S)-2-butanol (inverted)
  • **Structural factors favoring SN2**:

    1. **Primary alkyl halides (1°)**: MOST reactive

  • Least steric hindrance around C—X bond
  • Unhindered approach for nucleophile
  • Example: CH₃CH₂Br → fastest SN2
  • 2. **Secondary alkyl halides (2°)**: Moderate reactivity

  • Intermediate steric hindrance
  • Example: (CH₃)₂CHBr → intermediate rate
  • 3. **Tertiary alkyl halides (3°)**: **LEAST reactive** in SN2

  • Extreme steric hindrance (three bulky groups on C—X carbon)
  • Nucleophile cannot approach from rear
  • Example: (CH₃)₃CBr → extremely slow SN2
  • **Effect of leaving group**: **Better leaving groups favor SN2**

  • Order: I⁻ > Br⁻ > Cl⁻ > F⁻ > OH⁻
  • Larger, more polarizable halides are better leaving groups (stabilize negative charge in product)
  • **Effect of solvent**: **Aprotic polar solvents strongly favor SN2**

  • Aprotic polar (DMSO, acetone, DMF): Nucleophile not solvated by H-bonding; increased reactivity
  • Protic polar (H₂O, alcohols): Nucleophile solvated; decreased reactivity
  • Nonpolar (hexane, benzene): Very low reactivity (no solvation of ions)
  • **Effect of nucleophile strength**: Stronger nucleophiles (more negative, more polarizable) react faster in SN2

  • Order in aprotic solvent: I⁻ > Br⁻ > Cl⁻ > F⁻
  • Order in protic solvent: F⁻ > Cl⁻ > Br⁻ > I⁻ (reversed due to solvation effects)
  • **Key characteristics summary**:

  • Bimolecular kinetics
  • Single-step mechanism
  • Backside attack / inversion of stereochemistry
  • Favored by: 1° > 2° > 3° (primary most reactive)
  • Favored by: strong nucleophile, aprotic solvent, good leaving group
  • ---

    Chemical Reactions of Haloalkanes: Elimination Reactions

    Overview

    **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)

  • Primary halides with no β-H: undergo only substitution
  • Secondary and tertiary halides: both elimination and substitution possible
  • **Competition**: In same substrate, SN2, SN1, E1, and E2 all possible; products determined by **mechanism which dominates**

    E2 Mechanism (Elimination Bimolecular)

    **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):

  • Base abstracts β-H while C—X bond breaks
  • **Anti periplanar geometry required**: H and X on opposite sides of C—C bond (dihedral angle ≈ 180°)
  • Concerted transition state with partial C—H and C—X bond breaking
  • **Example**:

    ```

    (CH₃)₂CHBr + KOH → (CH₃)₂C=CH₂ + KBr + H₂O

    (3° halide → E2 predominates)

    ```

    **Stereochemistry**: **Anti elimination** (stereoelectronic requirement)

  • H and X must be **anti periplanar** (opposite sides on Newman projection)
  • In flexible molecules, can rotate to achieve anti geometry
  • Zaitsev product predominates: double bond forms with substitution on carbon having more alkyl groups
  • **Zaitsev's rule**: When multiple alkenes possible, major product is alkene with **more substituted double bond** (thermodynamically more stable)

  • Example: (CH₃)₂CHCHBrCH₃ + base → (CH₃)₂C=CHCH₃ + CH₃CH=C(CH₃)₂ (both possible; first is Zaitsev/major product)
  • Hofmann product (less substituted alkene) is minor
  • **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

  • Strong bases (KOH, KOtBu, NaH): E2 predominates
  • Weak bases (Br⁻, I⁻): SN2 favored
  • Example: (CH₃)₃CBr + KOH → (CH₃)₂C=CH₂ (E2); (CH₃)₃CBr + Br⁻ → no reaction (3° halide resists SN2)
  • **Effect of solvent**: Aprotic solvents favor E2 (higher kinetic barrier to abstraction; base not solvated)

    **Key characteristics**:

  • Bimolecular kinetics
  • Single-step, concerted mechanism
  • Anti periplanar H and X geometry
  • Zaitsev rule: more substituted alkene product
  • Favored by: strong base, high temperature, 3° halide
  • E1 Mechanism (Elimination Unimolecular)

    **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

  • (CH₃)₃CBr → (CH₃)₃C⁺ + Br⁻
  • 2. **Fast step**: Base abstracts β-hydrogen → alkene

  • (CH₃)₃C⁺ + B⁻ → (CH₃)₂C=CH₂ + BH
  • **Driving force for step 1**: C—X cleavage driven by solvent polarity (polar solvent stabilizes carbocation)

    **Conditions favoring E1**:

  • 3° halides (most stable carbocation)
  • Polar protic solvents (H₂O, alcohols; solvate carbocation)
  • Mild heating (provides activation energy)
  • Weak or no base (no competing SN2)
  • Elevated temperature
  • **Rearrangement in carbocation**: If more stable carbocation can form via **hydride or methyl shift**, both E1 and rearranged product possible

  • Example: 1-bromo-1-methylethylcyclopropane → rearranged E1 product (carbocation ring opens)
  • **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**:

  • Unimolecular kinetics
  • Two-step mechanism with carbocation intermediate
  • Rearrangement possible
  • Favored by: 3° halide, polar protic solvent, heat, weak base
  • Zaitsev rule: more substituted alkene
  • Summary: SN1, SN2, E1, E2 Mechanistic Competition

    | 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 |

    ---

    Grignard Reagents (Organomagnesium Compounds)

    Preparation

    **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**:

  • CH₃—Br + Mg → CH₃—Mg—Br (methylmagnesium bromide)
  • C₂H₅—I + Mg → C₂H₅—Mg—I (ethylmagnesium iodide)
  • C₆H₅—Br + Mg → C₆H
  • MCQs — 10 Questions with Answers

    Q1. Which of the following is a secondary alkyl halide?

    • A. CH₃CH₂CH₂Br
    • B. (CH₃)₃CBr
    • C. CH₃CH(Br)CH₂CH₃ ✓
    • D. C₆H₅CH₂Cl

    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:

    • A. Chloroform
    • B. Trichloromethane ✓
    • C. Trichloroethane
    • D. Methyl trichloride

    Answer: B — CHCl₃ has three chlorine atoms on one carbon (CH); the IUPAC name is trichloromethane.

    Q3. Which compound is a vic-dihalide?

    • A. CH₂ClCHCl₂
    • B. CHCl₂CHCl₂
    • C. CH₃CHClCHClCH₃ ✓
    • D. CCl₂(CH₃)₂

    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:

    • A. Fluorine has higher atomic number
    • B. Fluorine is smaller, resulting in shorter and stronger bond ✓
    • C. Iodine is radioactive
    • D. Fluorine forms ionic bonds with carbon

    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?

    • A. 2
    • B. 3
    • C. 4 ✓
    • D. 5

    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:

    • A. Only A is true ✓
    • B. Only B is true
    • C. Both A and B are true
    • D. Neither A nor B is true

    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:

    • A. 1-Bromopropane
    • B. 2-Bromopropene
    • C. 3-Bromopropene ✓
    • D. 1-Bromopropene

    Answer: C — Allyl bromide is CH₂=CHCH₂Br; numbering from the double bond gives 3-bromopropene.

    Q8. Which is NOT a correct statement about haloalkanes?

    • A. Primary alkyl halides have halogen on carbon bonded to one alkyl group
    • B. Benzylic halides have halogen directly bonded to aromatic ring carbon ✓
    • C. Bond length increases as C-F < C-Cl < C-Br < C-I
    • D. Haloalkanes are used as solvents and synthetic intermediates

    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:

    • A. 2-Bromo-3-methylbutane ✓
    • B. 3-Bromo-2-methylbutane
    • C. 2-Bromo-3-methylpentane
    • D. Impossible structure

    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?

    • A. Only (i) and (ii)
    • B. Only (ii) and (iii) ✓
    • C. Only (i) and (iii)
    • D. All three 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).

    Flashcards

    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.

    Important Board Questions

    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.

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