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8. Haloalkanes (Alkyl halides)

Haloalkanes (Alkyl Halides) — Chapter 8 Notes
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Learning Outcomes

By the end of the chapter, students should be able to:

  1. Describe the nomenclature, isomerism, and classification of monohaloalkanes.
  2. Explain the preparation of monohaloalkanes from alkanes, alkenes, and alcohols.
  3. State physical properties of haloalkanes.
  4. Describe chemical properties of haloalkanes: substitution reactions, SN1 and SN2 reactions (basic concept only).
  5. Explain the formation of alcohol, nitrile, amine, ether, thioether, carbylamines, nitrite, and nitroalkane using haloalkanes.
  6. Describe elimination reactions (dehydrohalogenation — Saytzeff’s rule), reduction reactions, and the Wurtz reaction.
  7. Show the preparation of trichloromethane from ethanol and propanone.
  8. Explain the chemical properties of trichloromethane: oxidation, reduction, action on silver powder, conc. nitric acid, propanone, and aq. alkali.

Introduction

Did you know that some of the most useful compounds in medicine, agriculture, and industry come from a simple idea: replacing hydrogen atoms in an alkane with halogen atoms? These compounds are called haloalkanes (also known as alkyl halides). They form the backbone of many drugs, refrigerants, solvents, and pesticides.

🔑 What is a Haloalkane?

A haloalkane is an organic compound in which one or more hydrogen atoms of an alkane are replaced by halogen atoms (F, Cl, Br, or I).

General formula: R–X

Where R = alkyl group (e.g. –CH₃, –C₂H₅ …) and X = halogen atom (–F, –Cl, –Br, –I)

Nature of the C–X Bond

It is a polar covalent bond: Carbon is δ⁺, Halogen is δ⁻.

Since halogens are more electronegative than carbon, the shared electron pair is pulled towards the halogen atom. As a result, carbon becomes slightly positive (δ⁺) and the halogen becomes slightly negative (δ⁻).

Classification

A. Based on the Number of Halogen Atoms

  • Monohaloalkanes → CH₃Cl
  • Dihaloalkanes → CH₂Cl₂
  • Trihaloalkanes → CHCl₃
  • Polyhaloalkanes → CCl₄

B. Based on the Nature of the Carbon Atom

TypeDescriptionExample
Primary (1°)Carbon attached to 1 carbonCH₃CH₂Cl
Secondary (2°)Attached to 2 carbonsCH₃CHClCH₃
Tertiary (3°)Attached to 3 carbons(CH₃)₃CCl

A carbon bonded with only one or no other carbon is called a primary carbon. The carbon bonded with two next carbon atoms is called secondary, and carbons bonded with 3 and 4 next carbon atoms are respectively called tertiary and quaternary carbon atoms.

IUPAC and Common Nomenclature

IUPAC Rules

  1. Select the longest carbon chain.
  2. Number the chain to give the halogen the lowest position.
  3. Use prefixes: chloro-, bromo-, iodo-.

CH₃CH₂CHCH₃ (Br on C-2)

  • Longest chain = 4 carbons, word root = but
  • C–C single bonds; primary suffix = ane
  • Numbering the prefix bromo = 2-bromo
  • Name = 2-bromobutane

Common Names

Named as alkyl + halide. Example: CH₃Br → methyl bromide (IUPAC: bromomethane)

BromoalkaneIUPAC nameCommon nameDegree
CH₃BrBromomethaneMethyl bromide
CH₃CH₂BrBromoethaneEthyl bromide
CH₃CH₂CH₂Br1-bromopropanen-propyl bromide
CH₃CHBrCH₃2-bromopropaneiso-propyl bromide
CH₃CH₂CH₂CH₂Br1-bromobutanen-butyl bromide
CH₃CH₂CHBrCH₃2-bromobutanesec-butyl bromide
(CH₃)₂CHCH₂Br1-bromo-2-methylpropaneiso-butyl bromide
(CH₃)₃CBr2-bromo-2-methylpropanetert-butyl bromide

🧠 Observe the molecular formula (C₄H₉Br) of the last four compounds in the table above — same formula, different structures.

Isomerism

  • Different compounds having the same molecular formula are called isomers.
  • Structural isomers are compounds that differ in structure and have the same molecular formula.
  • Two types of structural isomerism are observed in monohaloalkanes:

Chain Isomerism (skeletal isomerism)

Chain isomers are compounds with the same molecular formula differing only in the main carbon chain or skeleton. Exhibited by haloalkanes having 4 or more carbon atoms.

e.g., 2-iodobutane and 2-iodo-2-methylpropane

Positional Isomerism

Position isomers are compounds with the same molecular formula differing only in the position of the halo group. Exhibited by haloalkanes having 3 or more carbon atoms.

e.g., 1-chloropropane and 2-chloropropane

🧩 Quick Quiz

Welcome to your 8 Haloalkanes - Introduction quizzes

General Methods of Preparation of Monohaloalkanes

Haloalkanes can be prepared from several organic compounds. The three methods included in the NEB syllabus are shown below.

1. From Alkanes (by direct halogenation)

Alkanes can be halogenated easily in the presence of sunlight (UV light). These reactions are free radical substitution reactions. Sunlight (UV light) provides the energy needed to break the X–X bond, producing highly reactive free radicals that initiate the reaction.

CH₃CH₂HEthane
+
Cl–Cl(limited)
UV light
CH₃CH₂Cl + HClChloroethane

If chlorine is supplied in excess and the reaction is allowed to continue, all the hydrogen atoms may gradually be replaced by chlorine atoms:

CH₄Methane
Cl₂/hν−HCl
CH₃ClChloromethane
Cl₂/hν−HCl
CH₂Cl₂Dichloromethane
Cl₂/hν−HCl
CHCl₃Trichloromethane
Cl₂/hν−HCl
CCl₄Tetrachloromethane

2. From Alkene (by hydrohalogenation)

Alkenes undergo addition with hydrohalic acids (HCl, HBr, and HI), resulting in haloalkanes.

H₂C=CH₂Ethene
+
HI
CH₃CH₂IIodoethane
▸ Markovnikov’s Rule

When an unsymmetrical reagent adds to an unsymmetrical alkene, the positive part of the reagent (e.g. H⁺) is attached to the carbon atom with more hydrogen atoms, and the negative part gets bonded to the carbon atom with a smaller number of hydrogen atoms.

CH₃–CH=CH₂Propene
+
HCl
CH₃–CH₂–CH₂Cl1-chloropropane (minor)
CH₃–CHCl–CH₃2-chloropropane (major)

Why? It operates because the reaction follows the pathway that forms the more stable carbocation intermediate.

💡 Memory TipHydrogen goes to the carbon that already has more hydrogens.
▸ Peroxide Effect (Anti-Markovnikov’s Rule)

When HBr is added to unsymmetrical alkenes in the presence of organic peroxide, the positive part (H) gets bonded with the carbon having a smaller number of hydrogens, and the negative part (Br) gets bonded with the carbon having a higher number of hydrogens.

CH₃CH₂CH=CH₂But-1-ene
+ HBrorganic peroxide (ROOR)
CH₃CH₂CHBrCH₂H1-bromobutane

3. From Alcohols

a. Reaction with halogen acids

  • Alcohols, when treated with halogen acids (HCl, HBr, HI), have the –OH group substituted with the –X group (halide).
  • The reaction of alcohols with hydrochloric acid is slow; it therefore requires the presence of ZnCl₂ as a catalyst.
CH₃OHMethanol
+
HCl
ZnCl₂
CH₃Cl + H₂OChloromethane

b. Reaction with PCl₅ / PX₃ / SOCl₂

Alcohols also react with PCl₅, PCl₃, and SOCl₂, resulting in chloroalkanes.

CH₃CH₂OHEthanol
+
PCl₅
CH₃CH₂ClChloroethane
+
POCl₃ + HCl
3 CH₃CH(OH)CH₃Propan-2-ol
+ PCl₃
3 CH₃CHClCH₃2-chloropropane
+
H₃PO₃
📝 Note In the case of PI₃ and PBr₃, being unstable, P₄ and I₂ or Br₂ are heated to produce the PI₃ or PBr₃ in situ.
📝 Note SOCl₂ is a preferred reagent as the side products SO₂ and HCl escape as gases.
CH₃CH₂CH₂OHPropan-1-ol
+ SOCl₂
CH₃CH₂CH₂Cl1-chloropropane
+
SO₂↑ + HCl↑

Preparations Summary

Starting compoundReagentReaction type
AlkaneCl₂ / Br₂ + UV lightSubstitution
AlkeneHCl / HBr / HIAddition
AlcoholHX, PCl₅, SOCl₂Substitution
🧩 Quick Quiz

Welcome to your 8. Haloalkanes- Preparation- MCQs

Physical Properties

  • Lower members are gases; higher members are liquids with a pleasant smell; even higher members are waxy solids.
  • Boiling points increase with molecular mass in the homologous series and among different halogen atoms: RI > RBr > RCl
  • Insoluble in water; soluble in organic solvents like benzene, ether, and acetone.

🧩 Quick Quiz

Welcome to your 8. Haloalkanes- Preparation- MCQs

Chemical Properties

[A] Nucleophilic Substitution Reactions (SN1 / SN2)

The most common type of reaction that haloalkanes undergo — the halogen is replaced by a nucleophile.

R–X
+
Y⁻
R–Y
+
X⁻

where Y⁻ is a nucleophile like OH⁻, NH₂⁻, CN⁻, etc. Nucleophilic substitution reactions are of two types:

  • SN1 — substitution nucleophilic unimolecular
  • SN2 — substitution nucleophilic bimolecular

I) SN1 Reaction

The rate of reaction depends upon the concentration of only one of the reactants — the haloalkane substrate. Hence called unimolecular.

Rate ∝ [Haloalkane]   →   Rate law: Rate = k[R–X]

It is a two-step process:

Step I — Haloalkane undergoes heterolytic fission to form a carbocation and a halide ion. It’s slow, hence the rate-determining step (rds).

R₃C–XHaloalkane
slow, r.d.s.
R₃C⁺Carbocation
+
X⁻

Step II — The carbocation, being highly reactive, reacts with nucleophiles such as OH⁻ to form the product in a fast step.

R₃C⁺
+
OH⁻
fast
R₃C–OHAlcohol

Order of reactivity of 1°, 2°, 3°, and methyl haloalkanes: R₃C–X > R₂CHX > RCH₂X > CH₃–X i.e. 3° > 2° > 1° > methyl

📝 Note SN1 is favoured by tertiary haloalkanes because the tertiary carbocation (3°) is the most stable (stabilised by three electron-donating alkyl groups).
Stability of carbocations: 3° > 2° > 1° > CH₃⁺

II) SN2 Reaction

This reaction occurs in a single step. The rate depends on the concentration of both the substrate (haloalkane) and the reagent (nucleophile) — a bimolecular reaction.

Rate law: Rate = k[R–X][Nu⁻]

In the transition state, the nucleophile and the leaving group (halogen) have partial bonds:

Nu⁻
+
CH₃–X
fast
[Nu⋯C⋯X]transition state
Nu–CH₃
+
X⁻
📝 Note SN2 is favoured by primary haloalkanes — there is less steric hindrance, so the nucleophile can approach easily.

Primary halogenoalkanes tend to react via the SN2 mechanism; tertiary halogenoalkanes via the SN1 mechanism; and secondary halogenoalkanes by a mixture of the two, depending on structure.

Quick Comparison

FeatureSN1SN2
StepsTwo (stepwise)One (concerted)
IntermediateCarbocationNone (transition state only)
Rate lawk[R–X]k[R–X][Nu⁻]
Favoured byTertiary (3°) substratesPrimary (1°) substrates
StereochemistryRacemisation (mixture of products)Inversion of configuration (Walden inversion)
▶ An Animated Explanation of Nucleophilic Substitution
🧩 Quick Quiz

Welcome to your 8 Haloalkanes -SN1,SN2 MCQs

1. Formation of Alcohols

Reaction with aqueous caustic alkali (substitution by –OH group). When treated with aq. NaOH or KOH, haloalkanes give alcohols.

CH₃BrBromomethane
aq. NaOH−NaBr
CH₃OHMethanol
CH₃CH₂ClChloroethane
aq. KOH−KCl
CH₃CH₂OHEthanol

2. Formation of Ether

Reaction with sodium alkoxide — substitution by –OR (Williamson’s etherification). The alkoxide ion (OR⁻) acts as a nucleophile and substitutes the halogen atom from the alkyl halide.

CH₃CHClCH₃2-chloropropane
+
NaOCH₃Sodium methoxide
−NaCl
CH₃CH(OCH₃)CH₃2-methoxypropane
▸ Importance of Williamson’s Etherification
  • It is an SN2 reaction of an alkoxide ion with a primary alkyl halide.
  • Helps to prove the structure of ethers.
  • Suitable for preparing a wide variety of symmetrical and unsymmetrical ethers.
🧩 Quick Quiz

Welcome to your 8. Haloalkanes- nucleophilic substitution MCQs I

3. Formation of Nitrile

Action of alcoholic potassium cyanide (substitution by –CN).

CH₃CH₂CH₂Br1-bromopropane (n-propyl bromide)
alc. KCN
CH₃CH₂CH₂CNButanenitrile (n-propyl cyanide)
+
KBr
📝 NoteThe chain length is increased by one carbon. CN⁻ is an ambident nucleophile — it can bond through C or through N.

4. Formation of Carbylamine (Isocyanide)

Reaction with alcoholic AgCN.

R–XAlkyl halide
alc. AgCN
R–NCAlkyl isocyanide
+
AgX
CH₃CH₂ClChloroethane
CH₃CH₂NCEthyl isocyanide
+
AgCl

Occurs due to the ambident nature of CN⁻.

5. Formation of Nitrites

Action of alcoholic potassium nitrite.

R–X
+
KNO₂
R–O–N=OAlkyl nitrite
+
KX

Reaction occurs via the oxygen of NO₂⁻ (ambident nucleophile).

6. Formation of Nitro Compound

Action of alcoholic silver nitrite.

R–X
+
AgNO₂
R–NO₂Nitroalkane
+
AgX
📝 NoteReaction occurs via the nitrogen of NO₂⁻. AgNO₂ drives the reaction through N, giving a nitroalkane instead of a nitrite ester.

7. Formation of Amines

Reaction with ammonia (Hoffmann’s ammonolysis) — forms different degrees of amines. When an alkyl halide is heated with an aqueous or alcoholic solution of ammonia, a 1° amine is formed. Excess of alkyl halide gives a mixture of primary, secondary, and tertiary amines and quaternary ammonium salts.

R–X
+
H–NH₂
−HX
R–NH₂1° amine
+ R–X−HX
R–NH–R2° amine
+ R–X−HX
R–N(R)–R3° amine
+ R–X
[R₄N]⁺X⁻Quaternary ammonium salt
📝 NoteExcess ammonia favours the primary amine. Excess alkyl halide gives a mixture of all degrees plus a quaternary ammonium salt.

8. Formation of Thioether

Action of thioalcohol.

R–X
+
Na–S–R’
R–S–R’Alkyl alkyl sulphide (thioether)
+
NaX
R–X
+
Na₂S
R–S–RDialkyl sulphide (thioether)
+
NaX
🧩 Quick Quiz

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[B] Elimination Reaction

Dehydrohalogenation (reaction with alc. KOH). Haloalkanes undergo elimination with alcoholic KOH or NaOH, giving alkenes.

CH₃CHBrCH₃2-bromopropane
+ alc. KOH−KBr, −H₂O
CH₃CH=CH₂Propene
▸ Saytzeff’s Rule

When two different alkenes are possible from an elimination reaction, the alkene with the highest number of side chains (alkyl groups) is formed as the major product — the more substituted alkene is the major product.

CH₃–CH₂–CHBr–CH₃  →  CH₃–CH=CH–CH₃  (but-2-ene, major)
                         →  CH₃–CH₂–CH=CH₂  (but-1-ene, minor)

But-2-ene is more substituted → major product (Saytzeff’s rule).

Competition: Substitution vs Elimination

  • Aq. KOH / NaOH → substitution
  • Alc. KOH / NaOH → elimination

🧩 Quick Quiz

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[C] Reduction

Haloalkanes are reduced to alkanes with lithium aluminium hydride.

R–XHaloalkane
+
2[H]
LiAlH₄
R–HAlkane
+
HX
📌 Exam Note Isocyanides (carbyl amines) also undergo reduction, giving 2° amines. This reaction is not particularly mentioned in the NEB syllabus but is asked frequently in NEB exams.
R–NC
+
[H]
LiAlH₄
R–NH–CH₃

or,

R–NC
+
H₂
Ni
R–NH–CH₃

[D] Reaction with Metals — Wurtz Reaction

Haloalkanes react with sodium metal in the presence of dry ether, forming an alkane with a double number of carbon atoms.

R–X + 2Na + X–RAlkyl halide
dry ether, Δ
R–RAlkane
+
2NaX
⚠ Limitation If two different haloalkanes are used, three different alkanes are produced (a mixture), making the reaction less useful for synthesis.
🧩 Quick Quiz

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Preparation of Chloroform

Chloroform is prepared in the laboratory by distilling ethanol (ethyl alcohol) or propanone (acetone) with an aqueous paste of bleaching powder. In the reaction, aqueous bleaching powder acts as an oxidising, chlorinating, and hydrolysing agent.

First, bleaching powder reacts with water:

CaOCl₂
+
H₂O
Ca(OH)₂ + Cl₂

Preparation from Ethyl Alcohol

Step 1 — Oxidation of ethyl alcohol into acetaldehyde:

CH₃CH₂OHEthanol
+
Cl₂
CH₃CHOEthanal (acetaldehyde)
+
2HCl

Step 2 — Chlorination of acetaldehyde into trichloroacetaldehyde (chloral):

CH₃CHOEthanal
+
3Cl₂
Cl₃C–CHO2,2,2-trichloroethanal (chloral)
+
3HCl

Step 3 — Hydrolysis of chloral into chloroform:

2 Cl₃C–CHOChloral
+
Ca(OH)₂
2 CHCl₃Chloroform
+
(HCOO)₂CaCalcium methanoate

Preparation from Propanone

Involves only 2 steps.

Step 1 — Chlorination of propanone into 1,1,1-trichloropropanone:

CH₃COCH₃Propanone (acetone)
+
3Cl₂
Cl₃C–CO–CH₃1,1,1-trichloropropanone
+
3HCl

Step 2 — Hydrolysis of 1,1,1-trichloropropanone into chloroform:

2 Cl₃C–CO–CH₃
+
Ca(OH)₂
2 CHCl₃Chloroform
+
(CH₃COO)₂CaCalcium ethanoate
🧩 Quick Quiz

Welcome to your 8. Haloalkanes- Chloroform Preparation MCQs

Properties of Trichloromethane

  • It is a sweet-smelling colourless liquid.
  • It is heavier than water with sp. gr. 1.485.
  • It boils at 61 °C and freezes at −63 °C.
  • It dissolves nonpolar compounds like fat, oil, and wax.
  • If inhaled in a small amount, it causes temporary unconsciousness.

1) Oxidation — Reaction with Air

When exposed to light and air, chloroform undergoes slow oxidation to give highly poisonous phosgene gas, which can cause death in higher amounts. This is one of the reasons for discarding chloroform as an anaesthetic.

CHCl₃Chloroform
+
½O₂(air)
sunlight
COCl₂Carbonyl chloride (phosgene)
+
HCl

To prevent the formation of carbonyl chloride, the following precautions should be taken:

  • Store in a dark bottle to cut off the light.
  • Fill the chloroform up to the stopper to exclude air.
  • 1% ethanol is also added, which reacts with any phosgene gas formed and changes it into a nontoxic compound — diethyl carbonate.
2 C₂H₅OHEthanol
+
COCl₂Carbonyl chloride
(C₂H₅O)₂CODiethyl carbonate (non-toxic)
+
2HCl
✏ Worked Example

Q1. Why does chloroform not give a white precipitate with aqueous silver nitrate? (1 mark)

Solution: Chloroform and other chloroalkanes contain covalently bonded chlorine, which does not ionise easily in an aqueous solution. Therefore, they do not give a white precipitate with aqueous silver nitrate.

Unlike inorganic chlorides like NaCl, HCl, etc., which ionise into Cl⁻ (chloride ion), which reacts with AgNO₃, giving a white precipitate of AgCl:

AgNO₃
+
Cl⁻
AgCl↓
+
NO₃⁻

(But actually, chloroalkanes do give a white precipitate very slowly — over about an hour — due to hydrolysis.)

2) Reduction

Chloroform can be reduced to dichloromethane and methane, respectively, by Zn/HCl in the presence of ethanol and Zn dust in water.

CHCl₃Trichloromethane
+
2[H]
Zn/HClC₂H₅OH
CH₂Cl₂Dichloromethane
+
HCl
CHCl₃Chloroform
+
6[H]
Zn/H₂O
CH₄Methane
+
3HCl

3) Reaction with Silver Powder

Chloroform and iodoform both, when heated with silver powder, give ethyne (acetylene).

2 CHCl₃Chloroform
+
6Ag
Δ−6AgCl
HC≡CHEthyne (acetylene)
2 CHI₃Iodoform
+
6Ag
Δ−6AgI
HC≡CHEthyne (acetylene)
✏ Worked Example

Q2. How would you obtain ethylene from trichloromethane?

2 CHCl₃Chloroform
+
6Ag
Δ−6AgCl
HC≡CHEthyne
+ H₂Pd/BaSO₄
CH₂=CH₂Ethene (ethylene)

4) Reaction with Aqueous Caustic Alkali (Hydrolysis)

Chloroform reacts with aqueous NaOH or KOH, forming a tri-alcohol — a typical substitution reaction like that of monohaloalkanes. But the triol, being very unstable, undergoes decomposition, losing a molecule of water and resulting in methanoic acid.

HCCl₃Trichloromethane
+ 3NaOH(aq)−3NaCl
HC(OH)₃Unstable triol
−H₂O
HCOOHMethanoic acid (formic acid)

5) Reaction with Concentrated HNO₃

Chloroform reacts with conc. nitric acid, resulting in chloropicrin, which is used as a broad-spectrum antimicrobial, fungicide, herbicide, insecticide, and a component of tear gas.

Cl₃C–HChloroform
+
HO–NO₂Nitric acid
Cl₃C–NO₂Chloropicrin
+
H₂O

6) Reaction with Propanone (Acetone)

Chloroform condenses with propanone (acetone) in the presence of KOH to give chloretone, which is used as a hypnotic drug (sleep-inducing).

H₃C–CO–CH₃Propanone (δ⁻ O)
+
H–CCl₃Chloroform (δ⁺H)
KOH
(CH₃)₂C(OH)CCl₃Chloretone — 1,1,1-trichloro-2-methylpropan-2-ol
🧩 Quick Quiz

Welcome to your 8. Haloalkanes- Chloroform propertiesMCQs

📋

Chapter Summary

TopicKey points
Preparation(1) Alkane + X₂/hν  (2) Alkene + HX (Markovnikov)  (3) Alcohol + HX / PX₃ / SOCl₂
SN1Two steps; carbocation intermediate; rate = k[R–X]; favoured by 3° R–X
SN2One step; no intermediate; rate = k[R–X][Nu⁻]; favoured by 1° R–X
Substitution productsAlcohols, ethers, nitriles, isocyanides, nitrites, nitroalkanes, amines, thioethers
EliminationAlc. KOH → alkene; major product = more substituted alkene (Saytzeff)
ReductionLiAlH₄ / dry ether → alkane
Wurtz reaction2R–X + 2Na → R–R; doubles the carbon chain
Chloroform prepEthanol or propanone + bleaching powder + water (distillation)
Chloroform oxidationAir + light → phosgene (toxic); prevent by dark bottle + 1% ethanol
Chapter 8 · Haloalkanes (Alkyl Halides) — notes sourced from subarnam.com.np
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