Haloalkanes and Haloarenes — Class 12 Chemistry NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 12 Chemistry chapter "Haloalkanes and Haloarenes" — 9 important questions with detailed answers for CBSE board exam preparation.
TL;DR: Free step-by-step NCERT solutions for Class 12 Chemistry chapter "Haloalkanes and Haloarenes" — 9 important questions with detailed answers for CBSE b…
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NCERT Solutions for Class 12 Chemistry — Haloalkanes and Haloarenes. Step-by-step answers to all 9 textbook questions from this chapter, written for CBSE board preparation and free to use.
Haloalkanes and Haloarenes — All 9 Questions Solved
Q1. What is the difference between haloalkanes and haloarenes? Give one example of each.
Haloalkanes are compounds where halogen atoms are bonded to sp3 hybridized carbon atoms of alkane chains. Example: CH3CH2Br (bromoethane).
Haloarenes are compounds where halogen atoms are bonded to sp2 hybridized carbon atoms of benzene rings. Example: C6H5Cl (chlorobenzene).
Key difference: In haloalkanes, the C-X bond can undergo nucleophilic substitution easily. In haloarenes, the C-X bond is stronger due to partial double bond character from resonance, making nucleophilic substitution difficult and requiring drastic conditions or a catalyst.
Q2. Explain the mechanism of nucleophilic substitution in alkyl halides. How does it differ from that in aryl halides?
In alkyl halides (haloalkanes):
SN2 mechanism: Nucleophile attacks from the back side, opposite to the leaving group. This is a one-step process with inversion of configuration. Occurs with primary haloalkanes.
SN1 mechanism: Forms carbocation intermediate, then nucleophile attacks. Two-step process with racemization. Occurs with tertiary haloalkanes.
In aryl halides (haloarenes):
Nucleophilic aromatic substitution is extremely slow under normal SN2/SN1 conditions because:
- The C-X bond has partial double bond character (sp2 carbon)
- Formation of aryl carbocation is highly unfavorable
- Benzene ring is very stable
Substitution in haloarenes requires:
- High temperature and pressure (like Dow's process: 300°C, phenoxide formation)
- Presence of electron-withdrawing groups ortho/para to halogen
- Or nucleophilic aromatic substitution (SNAr) mechanism with electron-withdrawing groups activating the ring
Q3. Write the IUPAC name of CH3-CHBr-CH2-CH3.
IUPAC name: 2-bromobutane
Naming steps:
1. Main chain has 4 carbons = butane
2. Bromine (halogen) is the functional group at position 2
3. Numbering starts from the end nearest to the substituent to give the lowest number
4. Position 2 gives '2', position 3 would give '3', so position 2 is correct
5. Common name: sec-butyl bromide or secondary butyl bromide
Q4. Explain the effect of the halogen atom on the reactivity of the benzene ring in haloarenes. Why is chlorobenzene less reactive than benzene toward electrophilic substitution?
In haloarenes (e.g., chlorobenzene C6H5Cl):
The halogen atom attached to the benzene ring is deactivating for electrophilic aromatic substitution despite being electron-donating by the inductive effect.
Reason:
- Halogens are electronegative (inductive effect: withdraw electron density)
- Through resonance, halogens can donate electrons to the ring: lone pairs on halogen can form π bonds with the benzene ring
- The resonance donation is weak because halogens are highly electronegative
- Net effect: Inductive withdrawal dominates over resonance donation
- This reduces electron density on the benzene ring, making it less susceptible to electrophilic attack
Result: Chlorobenzene is LESS reactive than benzene in electrophilic aromatic substitution (e.g., nitration, sulfonation, Friedel-Crafts reactions).
Direction of substitution: The halogen is ortho-para directing (weak) because the lone pair resonance effect, though weak, still operates at ortho and para positions.
Q5. Describe the Wurtz reaction. Write the equation for the reaction of CH3Br with Na in dry ether.
Wurtz reaction: Two alkyl halides react in the presence of sodium metal in dry ether to form an alkane with the union of the two alkyl groups.
General equation: 2R-X + 2Na → R-R + 2NaX
For CH3Br:
2 CH3Br + 2 Na → CH3-CH3 + 2 NaBr
(bromoethane gives ethane)
Mechanism:
1. R-X + Na → R-Na + NaX (formation of alkyl sodium, a very strong nucleophile)
2. R-Na + R-X → R-R + NaX (nucleophilic attack by R-Na on the other R-X)
Limitations:
When using two different alkyl halides (R-X and R'-X), a mixture of products forms (R-R, R'-R', R-R') because the reaction is not selective. Example: CH3Br + C2H5Br gives CH3-CH3 + C2H5-C2H5 + CH3-C2H5.
Useful only when a symmetric product is desired (using same halide twice) or when the mixed product can be separated.
Q6. What is the Finkelstein reaction? Explain with an example how it is used in the lab.
Finkelstein reaction: An alkyl halide is converted to an alkyl iodide by reaction with sodium iodide (NaI) in dry acetone solvent.
General equation: R-X + NaI → R-I + NaX
Example: CH3Br + NaI → CH3I + NaBr
(acetone solvent)
Mechanism:
This is an SN2 reaction. Iodide ion (I-) is an excellent nucleophile and attacks the alkyl halide, displacing the halide as a leaving group.
Why acetone is used:
- NaI is soluble in acetone
- NaBr and NaCl (products) are insoluble in acetone and precipitate out, shifting equilibrium forward
- Acetone is polar enough to support the reaction
Selectivity:
- Iodine is the best leaving group (least basic)
- Chlorine and bromine are better leaving groups than iodine
- This reaction shows that for SN2, leaving group ability is: I- > Br- > Cl- > F-
Practical use: Synthesis of alkyl iodides from more stable alkyl chlorides or bromides, as alkyl iodides are more reactive in subsequent reactions and direct halogenation to form alkyl iodides is difficult.
Q7. What happens when 1-bromobutane is treated with alcoholic KOH? Write the equation and name the product.
Reaction: 1-bromobutane + alcoholic KOH → alkene
Equation: CH3CH2CH2CH2Br + KOH(alc.) → CH3CH2CH=CH2 + KBr + H2O
Product name: but-1-ene (or 1-butene) or butene
Mechanism: This is an elimination reaction (E2 mechanism)
1. Alcoholic KOH provides OH- as a strong base
2. The base abstracts a hydrogen atom from the carbon β to the halogen (next to the carbon bearing Br)
3. Simultaneous breaking of C-Br bond forms a C=C double bond
4. The double bond forms between C1 and C2 (terminal alkene)
Why elimination, not substitution?
- Alcoholic KOH is a strong base, favoring elimination
- Aqueous KOH would favor nucleophilic substitution (SN2) to form the alcohol
Regioselectivity (Zaitsev's rule): When the β-carbon has more than one hydrogen, the major product is the alkene with the more substituted double bond. In this case, only one product forms as there is only one β-carbon.
Q8. Compare the physical properties of haloalkanes with hydrocarbons of similar molar mass. Why are haloalkanes denser and have higher boiling points?
Comparison: For similar molar mass, haloalkanes have higher boiling points and densities than hydrocarbons.
Examples:
- CH3CH2CH3 (propane, M = 44): b.p. = -42°C, density = 0.58 g/mL
- CH3CH2Br (bromoethane, M = 109): b.p. = 38°C, density = 1.46 g/mL
Reasons:
1. Polarizability: Halogen atoms are large and highly polarizable. This increases temporary dipole-induced dipole interactions (London dispersion forces).
2. Molecular mass: Haloalkanes contain heavy halogen atoms (Br, I, Cl), significantly increasing molar mass, which increases intermolecular forces.
3. Dipole-dipole interactions: C-X bonds are polar. If many halogen atoms are present, the cumulative dipole-dipole attractions are significant.
4. Density: The presence of heavy halogen atoms increases density. Haloalkanes are generally denser than water and hydrocarbons.
Effect of halogen type on b.p.:
For alkyl halides: R-I > R-Br > R-Cl > R-F (due to increasing polarizability from I to F, despite opposite electronegativity trend)
Example: For methyl halides:
- CH3I: b.p. = 42°C
- CH3Br: b.p. = 4°C
- CH3Cl: b.p. = -24°C
- CH3F: b.p. = -78°C
Q9. Explain why chlorobenzene is immiscible with water but soluble in non-polar solvents. What does this tell us about the polar nature of the C-Cl bond in haloarenes?
Observation: Chlorobenzene (C6H5Cl) is immiscible with water but highly soluble in non-polar solvents like benzene, CCl4, and organic solvents.
Explanation:
The C-Cl bond in chlorobenzene is polar (Cl is more electronegative than C), but this polarity is not strong enough to overcome the hydrophobic nature of the benzene ring.
Reasons for water immiscibility:
1. The benzene ring is very hydrophobic (non-polar). The aromatic character and π electrons are hydrophobic.
2. Although the C-Cl bond is polar, the localized dipole is not sufficient to make the entire molecule polar enough to dissolve in water.
3. The benzene ring dominates the molecular properties, making the compound overall non-polar in character.
4. Formation of hydrogen bonds with water is not favorable.
Solubility in non-polar solvents:
- Non-polar solvents (benzene, CCl4, petroleum ether) can dissolve chlorobenzene through London dispersion forces and slight dipole-induced dipole interactions.
This tells us:
1. The C-Cl bond polarity in haloarenes is overridden by the hydrophobic benzene ring.
2. The overall molecule is essentially non-polar despite the presence of an electronegative halogen.
3. In haloarenes, the halogen's polarizing effect is localized to the C-X bond but does not significantly alter the overall molecular polarity due to the aromatic ring's dominance.
Contrast: In haloalkanes like CH3Br, which are somewhat more polar overall, they show slight miscibility or partial solubility in water due to the weaker hydrophobic effect of the alkyl chain compared to the benzene ring.
Haloalkanes and Haloarenes — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. What is the IUPAC name for CH3-CH(Cl)-CH3?
- A. 2-Chloropropane
- B. Isopropyl chloride
- C. 1-Chloropropane
- D. Propyl chloride
Answer: A. 2-Chloropropane — The longest carbon chain contains three carbons, making it a propane. The chlorine atom is attached to the second carbon, hence the IUPAC name is 2-Chloropropane.
MCQ 2. Which of the following compounds will undergo SN2 reaction most readily?
- A. 2-Bromo-2-methylpropane
- B. 1-Bromopropane
- C. 2-Bromopropane
- D. Bromobenzene
Answer: B. 1-Bromopropane — SN2 reactions are favored by less sterically hindered alkyl halides. Primary alkyl halides (like 1-bromopropane) are most reactive, followed by secondary, then tertiary. Aryl halides are generally unreactive towards SN2.
MCQ 3. Which statement is true regarding the SN1 reaction of (S)-3-bromo-3-methylhexane?
- A. The product formed will be a racemic mixture.
- B. The product formed will have retention of configuration.
- C. The product formed will have inversion of configuration.
- D. The reaction will not proceed via SN1 mechanism.
Answer: A. The product formed will be a racemic mixture. — SN1 reactions proceed via a planar carbocation intermediate. The nucleophile can attack from either face of this planar intermediate, leading to a racemic mixture of enantiomers if the starting material is chiral and the product retains chirality.
MCQ 4. Which reagent is typically used for the conversion of an alcohol to an alkyl chloride, often proceeding with retention of configuration under specific conditions (SNi mechanism)?
- A. HCl in presence of ZnCl2 (Lucas reagent)
- B. SOCl2 (Thionyl chloride)
- C. PCl5
- D. PCl3
Answer: B. SOCl2 (Thionyl chloride) — Thionyl chloride (SOCl2) can react with alcohols via an SNi (internal nucleophilic substitution) mechanism, especially in the absence of pyridine, leading to alkyl chlorides with retention of configuration.
MCQ 5. What is the name of the reaction that converts alkyl halides into alkyl fluorides using metallic fluorides like AgF, Hg2F2, CoF2, or SbF3?
- A. Finkelstein reaction
- B. Swarts reaction
- C. Wurtz reaction
- D. Sandmeyer reaction
Answer: B. Swarts reaction — The Swarts reaction is a specific halogen exchange reaction used for the preparation of alkyl fluorides from alkyl chlorides or bromides by heating them with metallic fluorides.
MCQ 6. Consider the reaction of 2-bromopropane with potassium tert-butoxide. What is the major product?
- A. Propene
- B. Propan-2-ol
- C. 2-tert-Butoxypropane
- D. 2-Methyl-2-propanol
Answer: A. Propene — Potassium tert-butoxide is a strong, bulky base and a poor nucleophile. It preferentially favors elimination (E2) over substitution, especially with secondary alkyl halides, leading to the formation of propene.
MCQ 7. Why are haloarenes less reactive towards nucleophilic substitution reactions compared to haloalkanes?
- A. Due to the resonance effect, which provides partial double bond character to the C-X bond.
- B. Due to the sp3 hybridization of the carbon atom attached to the halogen.
- C. Due to the bulky nature of the aryl group.
- D. Due to the high electronegativity of the halogen atom.
Answer: A. Due to the resonance effect, which provides partial double bond character to the C-X bond. — In haloarenes, the lone pair electrons on the halogen atom are delocalized into the benzene ring through resonance, imparting partial double bond character to the C-X bond, making it stronger and less susceptible to nucleophilic attack. Also, the carbon atom attached to the halogen is sp2 hybridized, making it more electronegative and the bond shorter.
MCQ 8. What happens when a Grignard reagent (RMgX) is treated with water?
- A. An alkene is formed.
- B. An alkane is formed.
- C. An alcohol is formed.
- D. A ketone is formed.
Answer: B. An alkane is formed. — Grignard reagents are strong bases and react readily with any protic source (like water, alcohols, acids) to form an alkane (RH) corresponding to the alkyl group of the Grignard reagent.
MCQ 9. Starting from propene, how can 1-bromopropane be selectively synthesized?
- A. Treat propene with HBr.
- B. Treat propene with Br2/CCl4 followed by KOH (alcoholic).
- C. Treat propene with HBr in the presence of peroxide.
- D. Treat propene with NBS.
Answer: C. Treat propene with HBr in the presence of peroxide. — The addition of HBr to unsymmetrical alkenes like propene in the presence of peroxide follows the anti-Markovnikov's rule, leading to the formation of 1-bromopropane.
MCQ 10. Which of the following compounds is primarily responsible for the depletion of the ozone layer?
- A. Chloroform
- B. Iodoform
- C. Freons (Chlorofluorocarbons)
- D. DDT
Answer: C. Freons (Chlorofluorocarbons) — Chlorofluorocarbons (CFCs), commonly known as Freons, are highly stable compounds that reach the stratosphere and release chlorine radicals under UV radiation, which then catalyze the destruction of the ozone layer.
MCQ 11. Which of the following reactions is used to prepare symmetrical alkanes by coupling two molecules of alkyl halides with sodium metal in dry ether?
- A. Fittig reaction
- B. Wurtz-Fittig reaction
- C. Wurtz reaction
- D. Sandmeyer reaction
Answer: C. Wurtz reaction — The Wurtz reaction involves treating two molecules of the same alkyl halide with sodium metal in dry ether to form a symmetrical alkane with an even number of carbon atoms.
MCQ 12. Arrange the following in increasing order of reactivity towards SN1 reaction: (i) 2-Bromo-2-methylpropane (ii) 1-Bromobutane (iii) 2-Bromobutane (iv) Bromobenzene
- A. (iv) < (ii) < (iii) < (i)
- B. (ii) < (iii) < (i) < (iv)
- C. (i) < (iii) < (ii) < (iv)
- D. (iii) < (ii) < (i) < (iv)
Answer: A. (iv) < (ii) < (iii) < (i) — SN1 reactivity depends on the stability of the carbocation intermediate formed. Tertiary carbocations are most stable, followed by secondary, then primary. Aryl halides are unreactive towards SN1. Thus, the order is Bromobenzene (aryl) < 1-Bromobutane (primary) < 2-Bromobutane (secondary) < 2-Bromo-2-methylpropane (tertiary).
MCQ 13. The industrial preparation of phenol from chlorobenzene involves heating chlorobenzene with NaOH at high temperature and pressure. This reaction is known as:
- A. Friedel-Crafts reaction
- B. Kolbe's reaction
- C. Dow's process
- D. Reimer-Tiemann reaction
Answer: C. Dow's process — Dow's process is an industrial method for synthesizing phenol from chlorobenzene by subjecting it to harsh conditions (high temperature and pressure) with aqueous NaOH, followed by acidification. It is a harsh nucleophilic aromatic substitution reaction.
MCQ 14. Dehydrohalogenation of 2-bromopentane with alcoholic KOH results in the formation of:
- A. Pent-1-ene as major product
- B. Pent-2-ene as major product
- C. Both pent-1-ene and pent-2-ene in equal amounts
- D. Pentane
Answer: B. Pent-2-ene as major product — According to Saytzeff's rule, in dehydrohalogenation reactions, the preferred alkene formed is the one that is more substituted (has more alkyl groups attached to the doubly bonded carbon atoms). Pent-2-ene is more substituted and thus the major product over pent-1-ene.
MCQ 15. When (R)-2-iodobutane reacts with sodium cyanide (NaCN) in ethanol, what is the configuration of the product?
- A. (R)-2-cyanobutane
- B. (S)-2-cyanobutane
- C. A racemic mixture of (R) and (S)-2-cyanobutane
- D. A product with no chiral center
Answer: B. (S)-2-cyanobutane — This is an SN2 reaction because NaCN is a strong nucleophile and 2-iodobutane is a secondary alkyl halide. SN2 reactions proceed with complete inversion of configuration (Walden inversion). Therefore, (R)-2-iodobutane will yield (S)-2-cyanobutane.
MCQ 16. What is the common name for trichloromethane?
- A. Methane chloride
- B. Chloroform
- C. Carbon tetrachloride
- D. Methyl chloride
Answer: B. Chloroform — Trichloromethane (CHCl3) is commonly known as chloroform, which was historically used as an anesthetic and is still a useful solvent.
MCQ 17. Chlorobenzene undergoes nitration to give:
- A. Only o-chloronitrobenzene
- B. Only p-chloronitrobenzene
- C. A mixture of o-chloronitrobenzene and p-chloronitrobenzene
- D. m-chloronitrobenzene
Answer: C. A mixture of o-chloronitrobenzene and p-chloronitrobenzene — Halogens are deactivating groups in electrophilic aromatic substitution reactions, but they are ortho-para directing due to the resonance effect which stabilizes the intermediate at these positions, overriding the inductive effect.
MCQ 18. Which reagent can be used to distinguish between chlorobenzene and cyclohexyl chloride?
- A. Aqueous AgNO3 solution
- B. Aqueous NaOH solution at room temperature
- C. Conc. H2SO4
- D. Ammoniacal cuprous chloride
Answer: A. Aqueous AgNO3 solution — Cyclohexyl chloride is an alkyl halide that can readily undergo SN1 reaction with aqueous AgNO3 to form a white precipitate of AgCl. Chlorobenzene, being an aryl halide, is much less reactive towards nucleophilic substitution and will not form a precipitate under these mild conditions.
MCQ 19. The addition of HBr to an unsymmetrical alkene in the presence of peroxide follows which rule?
- A. Markovnikov's rule
- B. Saytzeff's rule
- C. Hofmann's rule
- D. Anti-Markovnikov's rule (Peroxide effect)
Answer: D. Anti-Markovnikov's rule (Peroxide effect) — The addition of HBr to unsymmetrical alkenes in the presence of peroxide proceeds via a free radical mechanism, leading to the anti-Markovnikov product, where the bromine adds to the less substituted carbon atom.
MCQ 20. What is the primary purpose of the Finkelstein reaction?
- A. Preparation of alkyl fluorides
- B. Preparation of alkyl chlorides
- C. Preparation of alkyl iodides
- D. Preparation of alkyl bromides
Answer: C. Preparation of alkyl iodides — The Finkelstein reaction is a halogen exchange reaction specifically used for the preparation of alkyl iodides from alkyl chlorides or bromides by reacting them with NaI in dry acetone.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.