Haloalkanes and Haloarenes
Complete Revision Guide for CBSE Board Exams
1. Introduction & Classification
The replacement of hydrogen atom(s) in an aliphatic or aromatic hydrocarbon by halogen atom(s) results in the formation of alkyl halide (haloalkane) and aryl halide (haloarene), respectively. In haloalkanes, the halogen atom is attached to an $sp^3$ hybridized carbon atom, while in haloarenes, it is attached to an $sp^2$ hybridized carbon atom of an aromatic ring.
Crucial Classifications based on $sp^3$ C-X bond
- Alkyl Halides ($R-X$): The halogen is bonded to an alkyl group. Classified as $1^\circ$, $2^\circ$, or $3^\circ$ based on whether the carbon bearing the halogen is bonded to 1, 2, or 3 other carbon atoms.
- Allylic Halides: The halogen is bonded to an $sp^3$ hybridized carbon atom next to a carbon-carbon double bond ($C=C$). Structure: $C=C-C-X$.
- Benzylic Halides: The halogen atom is bonded to an $sp^3$ hybridized carbon atom next to an aromatic ring. Structure: $Ph-CH_2-X$.
Based on $sp^2$ C-X bond
- Vinylic Halides: The halogen is bonded directly to an $sp^2$ hybridized carbon atom of a carbon-carbon double bond. Structure: $C=C-X$.
- Aryl Halides (Haloarenes): The halogen atom is bonded directly to the $sp^2$ hybridized carbon atom of an aromatic ring. Structure: $Ph-X$.
2. Master Cheat Sheet: Name Reactions
A. Halogen Exchange Reactions (Preparation)
- 1. Finkelstein Reaction (For Alkyl Iodides):
Alkyl chlorides or bromides are treated with NaI in dry acetone. NaCl or NaNaBr precipitates out, driving the reaction forward (Le Chatelier's principle).
$$R-X + NaI \xrightarrow{\text{Dry Acetone}} R-I + NaX \downarrow$$ - 2. Swarts Reaction (For Alkyl Fluorides):
Heating an alkyl chloride/bromide with metallic fluorides like $AgF$, $Hg_2F_2$, $CoF_2$ or $SbF_3$.
$$R-Br + AgF \xrightarrow{\Delta} R-F + AgBr$$
B. Coupling Reactions (For forming C-C bonds)
- 1. Wurtz Reaction: Alkyl halides react with sodium in dry ether to form alkanes containing double the number of carbon atoms.
$$2R-X + 2Na \xrightarrow{\text{Dry Ether}} R-R + 2NaX$$
- 2. Wurtz-Fittig Reaction: A mixture of an alkyl halide and an aryl halide gives an alkylarene.
$$Ph-X + 2Na + X-R \xrightarrow{\text{Dry Ether}} Ph-R + 2NaX$$
- 3. Fittig Reaction: Aryl halides react with sodium in dry ether to form diaryls (diphenyl).
$$2Ph-X + 2Na \xrightarrow{\text{Dry Ether}} Ph-Ph (\text{Biphenyl}) + 2NaX$$
C. Haloarenes Preparation
- 1. Sandmeyer Reaction: Primary aromatic amine is converted to diazonium salt, which is treated with cuprous halides ($Cu_2Cl_2$ or $Cu_2Br_2$).
$$Ph-N_2^+X^- + Cu_2Cl_2/HCl \rightarrow Ph-Cl + N_2 \uparrow$$
- 2. Gattermann Reaction: Similar to Sandmeyer, but uses copper powder instead of cuprous halide.
$$Ph-N_2^+X^- \xrightarrow{Cu \text{ powder}/HCl} Ph-Cl + N_2 \uparrow$$
3. Exhaustive Theoretical Concepts
3.1 Nucleophilic Substitution Reactions (SN1 and SN2)
In this reaction, a nucleophile ($Nu^-$) replaces the halogen atom (leaving group) from the haloalkane. This is the most crucial part of this chapter.
| SN2 (Substitution Nucleophilic Bimolecular) | SN1 (Substitution Nucleophilic Unimolecular) |
|---|---|
| Mechanism: Single-step process. Nucleophile attacks from the back, and the leaving group leaves simultaneously through a transition state. | Mechanism: Two-step process. Step 1 (Slow): Formation of carbocation. Step 2 (Fast): Attack of nucleophile. |
| Rate Law: Rate $\propto [R-X][Nu^-]$ (Bimolecular) | Rate Law: Rate $\propto [R-X]$ (Unimolecular) |
| Reactivity Order: $1^\circ > 2^\circ > 3^\circ$. Reason: Steric hindrance. Bulky alkyl groups block the backside attack. |
Reactivity Order: $3^\circ > 2^\circ > 1^\circ$. Reason: Stability of carbocation. $3^\circ$ carbocation is most stable due to hyperconjugation and +I effect. |
| Stereochemistry: Complete Inversion of configuration (Walden Inversion) occurs, like an umbrella turning inside out in a strong wind. | Stereochemistry: Racemization occurs. The planar carbocation allows the nucleophile to attack from both faces, forming a racemic mixture (50% retention, 50% inversion). |
CBSE Favorite: Resonance stabilized Carbocations
Allylic ($CH_2=CH-CH_2-X$) and Benzylic ($Ph-CH_2-X$) halides show high reactivity towards SN1 reactions. Why? Because the carbocation formed after the halide leaves is highly stabilized by resonance.
3.2 Stereochemical Aspects of Nucleophilic Substitution
- Chiral Carbon (Stereocenter): A carbon atom bonded to four entirely different groups. It lacks any plane of symmetry.
- Chirality: The property of an object (or molecule) being non-superimposable on its mirror image (like your left and right hands). Such molecules are optically active.
- Enantiomers: Stereoisomers that are non-superimposable mirror images of each other. They have identical physical properties but rotate plane-polarized light in opposite directions.
- Racemic Mixture: A mixture containing two enantiomers in equal proportions (50:50). It will have zero optical rotation because the rotation caused by one enantiomer is exactly canceled by the other. The process of converting an enantiomer into a racemic mixture is called racemisation.
3.3 Elimination Reactions ($\beta$-Elimination)
When a haloalkane with a $\beta$-hydrogen atom is heated with alcoholic KOH, elimination of a hydrogen atom from the $\beta$-carbon and a halogen atom from the $\alpha$-carbon occurs, forming an alkene. This is called dehydrohalogenation.
Saytzeff (Zaitsev) Rule: If there is a possibility of forming more than one alkene, the major product is the highly substituted alkene (the one with the greater number of alkyl groups attached to the doubly bonded carbon atoms).
Exam Tip: Aqueous KOH causes Substitution (forming alcohols), Alcoholic KOH causes Elimination (forming alkenes).
3.4 Why are Haloarenes Unreactive towards Nucleophilic Substitution?
Aryl halides (like chlorobenzene) are extremely less reactive towards nucleophilic substitution compared to alkyl halides. This is a very common board exam question. Here are the four reasons:
- Resonance Effect: The lone pair of electrons on the halogen atom is in conjugation with the pi electrons of the benzene ring. This gives the C-X bond a partial double bond character, making it shorter, stronger, and much harder to break.
- Hybridization of Carbon: The carbon attached to halogen in haloarenes is $sp^2$ hybridized (more s-character, highly electronegative), holding the electron pair of the C-X bond tightly. In haloalkanes, it is $sp^3$ hybridized.
- Instability of Phenyl Cation: In SN1 mechanism, a carbocation must be formed. The phenyl cation formed by cleavage of the C-X bond in haloarenes is highly unstable (cannot be stabilized by resonance).
- Electronic Repulsions: The nucleophile (which is electron-rich) suffers electrostatic repulsion when it approaches the electron-rich arene cloud.
3.5 Electrophilic Substitution in Haloarenes
Haloarenes undergo the usual electrophilic substitution reactions of the benzene ring such as halogenation, nitration, sulphonation and Friedel-Crafts reactions.
Directing Influence: Halogen atom is ortho-para directing due to the +R (Resonance) effect which increases electron density at ortho and para positions.
Reactivity: However, halogens are overall deactivating groups because their strong -I (Inductive) effect withdraws electrons from the ring, making the ring less reactive than pure benzene towards electrophilic substitution.
4. Most Important CBSE Board Questions
Curated based on frequency in past papers and latest CBSE patterns.
Section A: 1 Mark Questions (MCQs & A-R)
Q1. Which of the following compounds is most reactive towards SN2 displacement? 1 Mark
- 1-Bromopentane
- 2-Bromopentane
- 2-Bromo-2-methylbutane
- Chlorobenzene
Explanation: For SN2 reactions, reactivity order is $1^\circ > 2^\circ > 3^\circ$ due to steric hindrance. 1-Bromopentane is a primary ($1^\circ$) halide and has the least steric hindrance around the carbon bearing the halogen.
Q2. Alkyl halides are immiscible in water though they are polar. Why? 1 Mark
To dissolve in water, energy is required to overcome the attractions between the haloalkane molecules and break the hydrogen bonds between water molecules. However, haloalkanes cannot form hydrogen bonds with water. The new intermolecular attractions (dipole-dipole) between water and haloalkane are weaker than the original hydrogen bonds in water. Hence, less energy is released, and they remain immiscible.
Assertion-Reasoning
Q3. Assertion (A): $KCN$ reacts with alkyl halide to give alkyl cyanide as the major product, while $AgCN$ forms isocyanide as the major product.
Reason (R): $KCN$ is ionic while $AgCN$ is covalent in nature. 1 Mark
Explanation: KCN is predominantly ionic, providing free cyanide ions ($C \equiv N^-$) in solution. Attack mainly occurs through Carbon (since C-C bond is stronger than C-N bond), giving cyanides. AgCN is predominantly covalent; the nitrogen lone pair is available for nucleophilic attack, leading to isocyanides ($R-N \equiv C$).
Section B: 2 Mark Questions (Very Short Answer)
Q4. Give reasons for the following:
(i) Chloroform ($CHCl_3$) is stored in dark colored bottles completely filled to the brim.
(ii) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride. 2 Marks
Equation: $2CHCl_3 + O_2 \xrightarrow{\text{light}} 2COCl_2 + 2HCl$
(ii) In chlorobenzene, the C-Cl bond is $sp^2$ hybridized (more electronegative), making the bond less polar. Also, the +R effect of the benzene ring opposes the -I effect of Chlorine, reducing the net dipole moment. In cyclohexyl chloride, the carbon is $sp^3$ hybridized, and there is no opposing resonance, leading to a higher dipole moment.
Q5. Which compound in the following pairs will react faster in SN1 reaction with OH⁻ and why?
(i) $CH_3-CH_2-Br$ or $CH_3-CH(Br)-CH_3$
(ii) $C_6H_5-CH_2-Cl$ or $C_6H_5-Cl$ 2 Marks
(ii) $C_6H_5-CH_2-Cl$ (Benzyl chloride) will react faster. It forms a benzyl carbocation ($C_6H_5-CH_2^+$) which is highly stabilized by resonance with the benzene ring. Chlorobenzene forms a phenyl cation which is highly unstable.
Section C: 3 Mark Questions (Short Answer)
Q6. How will you bring about the following conversions?
(i) Ethanol to but-1-yne
(ii) Propene to propan-1-ol
(iii) Benzene to biphenyl 3 Marks
Step 1: Convert ethanol to bromoethane using $SOCl_2$ or $PBr_3$.
$CH_3CH_2OH \xrightarrow{PBr_3} CH_3CH_2Br$
Step 2: Treat with sodium acetylide ($NaC \equiv CH$) (Nucleophilic substitution).
$CH_3CH_2Br + HC \equiv C^-Na^+ \rightarrow CH_3CH_2C \equiv CH$ (But-1-yne)
(ii) Propene to propan-1-ol:
Step 1: Anti-Markovnikov addition of HBr (using peroxide).
$CH_3-CH=CH_2 + HBr \xrightarrow{\text{Peroxide}} CH_3-CH_2-CH_2-Br$ (1-Bromopropane)
Step 2: Nucleophilic substitution with aqueous KOH.
$CH_3-CH_2-CH_2-Br \xrightarrow{\text{aq. KOH}} CH_3-CH_2-CH_2-OH$ (Propan-1-ol)
(iii) Benzene to biphenyl:
Step 1: Halogenation of benzene to form chlorobenzene.
$C_6H_6 + Cl_2 \xrightarrow{FeCl_3} C_6H_5-Cl$
Step 2: Fittig reaction using Sodium in dry ether.
$2C_6H_5-Cl + 2Na \xrightarrow{\text{Dry Ether}} C_6H_5-C_6H_5$ (Biphenyl) $+ 2NaCl$
Q7. Distinguish chemically between the following pairs of compounds:
(i) Chlorobenzene and Benzyl chloride
(ii) Chloroform and Carbon tetrachloride 3 Marks
Silver Nitrate ($AgNO_3$) Test: Boil both compounds separately with aqueous NaOH, then acidify with dilute $HNO_3$ and add $AgNO_3$ solution.
- Benzyl chloride will give a white precipitate of AgCl because it undergoes nucleophilic substitution easily.
- Chlorobenzene will NOT give a white precipitate because its C-Cl bond has partial double bond character and does not break easily to give $Cl^-$ ions.
(ii) Chloroform vs Carbon tetrachloride:
Carbylamine (Isocyanide) Test: Heat both compounds with a primary amine (like aniline) and alcoholic KOH.
- Chloroform ($CHCl_3$) will produce an extremely foul/offensive smelling gas (phenyl isocyanide).
- Carbon tetrachloride ($CCl_4$) does not give this test.
Section D: 4 Mark Case-Based Questions
Q8. Read the passage given below and answer the following questions:
(i) Out of Chloromethane and 2-Chloro-2-methylpropane, which is more reactive towards SN2 reaction and why? (1 Mark)
(ii) What is the stereochemical outcome of an SN1 reaction on an optically active alkyl halide? (1 Mark)
(iii) (a) Explain why Allyl chloride is highly reactive towards SN1 reaction. (2 Marks)
OR
(iii) (b) Write the mechanism of the reaction of n-butyl bromide with aqueous KOH. (2 Marks)
(ii) The stereochemical outcome of an SN1 reaction is Racemization. The planar carbocation intermediate allows the nucleophile to attack with equal probability from both the front and the back, resulting in a 50:50 mixture of retention and inversion of configuration (racemic mixture).
(iii) (a) (Internal Choice 1): Allyl chloride ($CH_2=CH-CH_2-Cl$) undergoes SN1 easily because the resulting intermediate is an allyl carbocation ($CH_2=CH-CH_2^+$). This carbocation is highly stabilized by resonance (the positive charge is delocalized over the adjacent double bond), making the first, slow step of SN1 energetically favorable.
(iii) (b) (Internal Choice 2): n-butyl bromide ($1^\circ$ halide) reacts with aq. KOH via SN2 mechanism. It is a one-step concerted mechanism.
Mechanism: The $OH^-$ nucleophile attacks the $\alpha$-carbon from the side opposite to the $Br$ atom. A transition state forms where C-OH bond is partially forming and C-Br bond is partially breaking simultaneously. Finally, $Br^-$ leaves, causing an inversion of configuration to form n-butyl alcohol.
$HO^- + CH_3CH_2CH_2CH_2-Br \xrightarrow{\text{Transition State}} CH_3CH_2CH_2CH_2-OH + Br^-$
Section E: 5 Mark Long Answer Questions
Q9. An organic compound 'A' having molecular formula $C_4H_9Br$ on treatment with alcoholic KOH gives compound 'B' ($C_4H_8$). Ozonolysis of 'B' yields only ethanal ($CH_3CHO$) as the product. Compound 'A' reacts with Sodium metal to give a symmetric hydrocarbon 'C'.
Identify A, B, and C and write all the corresponding chemical reactions involved. 5 Marks
Ethanal + Ethanal $\Rightarrow CH_3-CH=CH-CH_3$.
Therefore, Compound 'B' is But-2-ene.
2. Identify 'A': Compound 'A' ($C_4H_9Br$) undergoes elimination with alcoholic KOH (dehydrohalogenation) to give But-2-ene. According to Saytzeff's rule, But-2-ene is the major product if the starting material is a secondary halide. Therefore, the Bromine must be on the 2nd carbon.
Therefore, Compound 'A' is 2-Bromobutane ($CH_3-CH(Br)-CH_2-CH_3$).
3. Identify 'C': Compound 'A' reacts with Sodium metal. This is the Wurtz Reaction, which joins two alkyl groups together.
2 molecules of 2-Bromobutane react with 2Na to give 3,4-dimethylhexane.
Therefore, Compound 'C' is 3,4-dimethylhexane.
Chemical Reactions:
Reaction 1 (Elimination):
$$CH_3-CH(Br)-CH_2-CH_3 \xrightarrow{\text{alc. KOH, } \Delta} CH_3-CH=CH-CH_3 \text{ (B: But-2-ene)} + KBr + H_2O$$
Reaction 2 (Ozonolysis of B):
$$CH_3-CH=CH-CH_3 \xrightarrow{1. O_3 \quad 2. Zn/H_2O} 2 CH_3CHO \text{ (Ethanal)}$$
Reaction 3 (Wurtz reaction forming C):
$$2 CH_3-CH(Br)-CH_2-CH_3 + 2Na \xrightarrow{\text{Dry Ether}} CH_3-CH_2-CH(CH_3)-CH(CH_3)-CH_2-CH_3 \text{ (C)} + 2NaBr$$
Mastering 'Haloalkanes and Haloarenes'
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The Reagent Trap (Aqueous vs Alcoholic): This is the #1 mistake students make. Aqueous KOH/NaOH leads to Nucleophilic Substitution (forming Alcohols). Alcoholic KOH/NaOH leads to Elimination (forming Alkenes). Never confuse the two!
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Ambidentate Nucleophiles: KCN yields Cyanides (major), AgCN yields Isocyanides (major). Similarly, $KNO_2$ yields alkyl nitrites (R-O-N=O), while $AgNO_2$ yields nitroalkanes ($R-NO_2$). Memorize these ionic vs covalent differences.
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Markovnikov vs Anti-Markovnikov: When adding HX to an alkene, Markovnikov says H goes to the carbon with more H's. Anti-Markovnikov (Peroxide effect / Kharasch effect) ONLY works with HBr in the presence of Peroxide. It does not work with HCl or HI.
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Step Up Conversions: If a conversion requires increasing the carbon chain length by one carbon, use KCN (to form cyanide) followed by reduction. If it requires doubling the chain, use the Wurtz reaction.
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