CBSE Class 12 Chemistry Halo Alkanes and Halo arenes Questions

Question 1. Write the IUPAC names of the following compounds.
Answer:
(i) 2-Bromo-2-methylbutane
(ii) 2-Bromo-3-methylbut-2-ene
(iii) (Chloromethyl)benzene (Benzyl chloride)
(iv) 1-Bromo-3,3-dimethylpentane
(v) 6-Bromohex-4-en-1-yne
(vi) 1-Bromo-2-methylcyclohexane
(vii) 3-Chloro-2,4-dimethylpentan-2-ol
(ix) Bis(trichloromethyl)chloromethane (or 1,1,1,3,3,3-hexachloro-2-(trichloromethyl)propane)
In simple words: Identify the longest carbon chain containing the principal functional group or multiple bond, number it to give lowest locants to substituents/halogens, and list them alphabetically with proper numbering.

Exam Tip: Always number the chain such that functional groups, multiple bonds, and halogens get the lowest possible locants in that order of priority.

Question 2. Write the structure of following halogen compounds:
(i) 2-chloro-3-methylpentane
(ii) 2-(2-chlorophenyl)-1-iodooctane
(iii) 1-bromo-4-sec-butyl-2-methylbenzene
(iv) p-bromotoluene
(v) chlorophenylmethane
Answer:
(i) \(\text{CH}_3-\text{CH(Cl)}-\text{CH(CH}_3)-\text{CH}_2-\text{CH}_3\)
(ii) Octane chain with an iodine at C-1 and a 2-chlorophenyl group at C-2: \(\text{CH}_3(\text{CH}_2)_5\text{CH(C}_6\text{H}_4\text{Cl})\text{CH}_2\text{I}\)
(iii) Benzene ring substituted with methyl at C-2, sec-butyl at C-4, and bromomethyl at C-1 (or bromo on the ring depending on exact numbering context): Benzene ring with \(-\text{Br}\) at position 1, \(-\text{CH}_3\) at position 2, and \(-\text{CH(CH}_3)\text{CH}_2\text{CH}_3\) at position 4.
(iv) 1-Bromo-4-methylbenzene (Benzene ring with \(-\text{Br}\) and \(-\text{CH}_3\) at para positions).
(v) \(\text{C}_6\text{H}_5-\text{CH}_2-\text{Cl}\) (Benzyl chloride).
In simple words: Draw the root parent alkane or benzene ring first, then attach the halogen atoms, alkyl groups, and substituent branches at their correct numbered positions.

Exam Tip: Read IUPAC names backwards: start with the root word for the main chain/ring, then add substituents at the specified numbers.

Question 3. Arrange the following in the increasing order of properly indicated :
(i) bromomethane, chloromethane, dichloromethane. (Increasing order of boiling points).
(ii) 1-chloropropane, isopropyl chloride, 1-chlorobutane (Increasing order of boiling point)
(iii) dichloromethane, chloroform, carbon tetrachloride. (Increasing order of dipole moment)
(iv) \(\text{CH}_3\text{F}, \text{CH}_3\text{Cl}, \text{CH}_3\text{Br}, \text{CH}_3\text{I}\) (Increasing reactivity towards nucleophilic substitution and increasing order of dipole moment)
(v) \(o,m,p\)-dichlorobenzenes (Increasing order of melting points).
Answer:
(i) Chloromethane < bromomethane < dichloromethane (Higher molecular mass and surface area increase van der Waals forces).
(ii) Isopropyl chloride < 1-chloropropane < 1-chlorobutane (Branching decreases surface area and boiling point; higher carbon count increases boiling point).
(iii) Carbon tetrachloride (\(\text{mu} = 0\)) < dichloromethane < chloroform.
(iv) Dipole moment order: \(\text{CH}_3\text{I} < \text{CH}_3\text{Br} < \text{CH}_3\text{F} < \text{CH}_3\text{Cl}\).
Reactivity towards nucleophilic substitution (\(\text{S}_\text{N}2\)): \(\text{CH}_3\text{F} < \text{CH}_3\text{Cl} < \text{CH}_3\text{Br} < \text{CH}_3\text{I}\) (due to increasing carbon-halogen bond length and decreasing bond dissociation enthalpy).
(v) Ortho < meta < para (Para isomer has high melting point due to symmetrical packing in crystal lattice).
In simple words: Boiling point increases with molecular weight and chain length, but decreases with branching. Melting point is highest for symmetrical para isomers due to crystal packing.

Exam Tip: Para isomers always have significantly higher melting points than ortho and meta isomers due to crystal symmetry.

Question 4. Complete the following reactions :
(i) Benzene ring with \(-\text{CH}=\text{CH}_2\) (Styrene) + \(\text{HBr} \longrightarrow\)
(ii) \(\text{CH}_3-\text{CH}_2-\text{Cl} + \text{AgNO}_2 \longrightarrow\)
(iii) Styrene + \(\text{HBr} \xrightarrow{\text{peroxide}}\)
(iv) 1-chloro-2,4,6-trinitrobenzene + \(\text{NaOH (aq)} \longrightarrow\)
(v) Ethylbenzene + \(\text{Br}_2 / \text{heat, UV light} \longrightarrow\)
(vi) Methylcyclohexane + \(\text{HI} \longrightarrow\)
(vii) 4-(Hydroxymethyl)phenol + \(\text{SOCl}_2 \longrightarrow\)
(viii) \(\text{CH}_3\text{CH}_2\text{Br} + \text{NaI} \xrightarrow{\text{acetone}}\)
(ix) Cyclohexane + \(\text{Br}_2 \xrightarrow{\text{UV light}}\)
(x) \((\text{CH}_3)_3\text{C—Br} + \text{KOH} \xrightarrow{\text{ethanol, heat}}\)
(xi) \(\text{CH}_3\text{CH}_2\text{Br} + \text{KCN} \xrightarrow{\text{aq. EtOH}}\)
(xii) Bromonitrobenzene + \(\text{Mg} \xrightarrow{\text{dry ether}}\)
(xiii) \(\text{C}_6\text{H}_5\text{ONa} + \text{C}_2\text{H}_5\text{Cl} \longrightarrow\)
(xiv) 2-bromobutane + \(\text{Na} \xrightarrow{\text{dry ether}}\)
(xv) 2-bromobutane + \(\text{NaOH (aq)} \longrightarrow\)
(xvi) \(\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{KI} \longrightarrow\)
Answer:
(i) \(\text{C}_6\text{H}_5-\text{CH(Br)}-\text{CH}_3\) (Markovnikov addition)
(ii) \(\text{CH}_3-\text{CH}_2-\text{NO}_2 + \text{AgCl}\) (Nitroethane, due to covalent nature of \(\text{AgNO}_2\))
(iii) \(\text{C}_6\text{H}_5-\text{CH}_2-\text{CH}_2-\text{Br}\) (Anti-Markovnikov addition / Kharash effect)
(iv) 2,4,6-Trinitrophenol (Picric acid) + \(\text{NaCl}\)
(v) \(\text{C}_6\text{H}_5-\text{CH(Br)}-\text{CH}_3\) (Free radical bromination at benzylic carbon)
(vi) 1-Iodo-1-methylcyclohexane
(vii) 4-(Chloromethyl)chlorobenzene or replacement of benzylic \(-\text{OH}\) with \(-\text{Cl}\)
(viii) \(\text{CH}_3\text{CH}_2\text{I} + \text{NaBr}\) (Finkelstein reaction)
(ix) Bromocyclohexane + \(\text{HBr}\)
(x) 2-Methylpropene + \(\text{KBr} + \text{H}_2\text{O}\) (Dehydrohalogenation via E1 mechanism)
(xi) \(\text{CH}_3\text{CH}_2\text{CN} + \text{KBr}\) (Propanenitrile)
(xii) Grignard reagent formation: \(\text{BrMg—C}_6\text{H}_4—\text{NO}_2\)
(xiii) \(\text{C}_6\text{H}_5-\text{O}-\text{C}_2\text{H}_5 + \text{NaCl}\) (Phenetole via Williamson ether synthesis)
(xiv) 3,4-Dimethylhexane (Wurtz reaction)
(xv) Butan-2-ol + \(\text{NaBr}\)
(xvi) \(\text{C}_6\text{H}_5\text{I} + \text{N}_2 + \text{KCl}\) (Iodobenzene)
In simple words: Apply standard name reactions (Wurtz, Finkelstein, Williamson), Markovnikov/anti-Markovnikov rules, and substitution/elimination reaction mechanisms.

Exam Tip: Memorize conditions for ambient vs. peroxide additions and distinguish between aqueous KOH (substitution) and alcoholic KOH (elimination).

Question 5. How will you bring about the following conversions?
(i) benzene to 3-bromonitrobenzene
(ii) ethanol to but-1-yne
(iii) 1-bromopropane to 2-bromopropane
(iv) benzene to 4-bromo-1-nitrobenzene
(v) aniline to chlorobenzene
(vi) 2-methyl-1-propene to 2-chloro-2-methylpropane
(vii) ethyl chloride to propanoic acid
(viii) but-1-ene to n-butyl iodide
(ix) benzene to phenylchloromethane.
(x) tert-butyl bromide to isobutyl bromide.
Answer:
(i) Nitration of benzene to nitrobenzene, followed by bromination using \(\text{Br}_2 / \text{FeBr}_3\) (since nitro group is meta-directing).
(ii) Convert ethanol to bromoethane, react with sodium acetylide (\(\text{HC}\equiv\text{C}^-\text{Na}^+\)) to form but-1-yne.
(iii) 1-Bromopropane + alc. \(\text{KOH} \longrightarrow \text{propene}\); propene + \(\text{HBr} \longrightarrow\) 2-bromopropane (Markovnikov addition).
(iv) Bromination of benzene to bromobenzene, followed by nitration with conc. \(\text{HNO}_3 / \text{H}_2\text{SO}_4\) to yield 1-bromo-4-nitrobenzene.
(v) Diazotization of aniline with \(\text{NaNO}_2 + \text{HCl}\) at 0-5 \(^\circ\text{C}\), followed by Sandmeyer reaction using \(\text{CuCl} / \text{HCl}\).
(vi) Hydrohalogenation of 2-methyl-1-propene with \(\text{HCl}\) following Markovnikov rule.
(vii) Ethyl chloride + \(\text{KCN} \longrightarrow\) propanenitrile; followed by acid hydrolysis to propanoic acid.
(viii) But-1-ene + \(\text{HBr}\) in presence of peroxide gives 1-bromobutane; treat with \(\text{NaI} / \text{acetone}\) (Finkelstein reaction) to get n-butyl iodide.
(ix) Benzene + chloromethane in presence of anhydrous \(\text{AlCl}_3\) (Friedel-Crafts alkylation) gives toluene; chlorination of toluene with \(\text{Cl}_2 / \text{UV light}\) gives phenylchloromethane (benzyl chloride).
(x) Tert-butyl bromide + alc. \(\text{KOH} \longrightarrow\) 2-methylpropene; hydroboration-oxidation or anti-Markovnikov addition of \(\text{HBr}\) with peroxide to get isobutyl bromide.
In simple words: Plan synthetic routes step-by-step by combining functional group interconversions, chain lengthening, and orientation control in aromatic rings.

Exam Tip: For directing groups in benzene, introduce the meta-directing group first if you need a meta-substituted product, or ortho/para-directing group first for ortho/para products.

Question 6. Identify the products formed in the following sequence :
(i) m-Iodochlorobenzene + \(\text{NaCN} \xrightarrow{\quad} \text{A} \xrightarrow{\text{H}^+ / \text{H}_2\text{O}} \text{B}\)
(ii) 1,2-Dibromopropane + alc. \(\text{KOH} \xrightarrow{\quad} \text{A} \xrightarrow{\text{NaNH}_2} \text{B}\)
(iii) \(\text{C}_6\text{H}_5\text{CH}_2\text{CHBrCH}_3 \xrightarrow{\text{alc. KOH}} \text{A} \xrightarrow{\text{HBr}} \text{B}\)
(iv) 2-Bromobutane + alc. \(\text{KOH} \xrightarrow{\quad} \text{X} \xrightarrow{\text{HBr} / \text{H}_2\text{O}_2} \text{Y}\)
(v) Propene + \(\text{Br}_2 \xrightarrow{\text{CCl}_4} \text{A}\)
(vi) Propene + \(\text{Br}_2 \xrightarrow{\text{heat / UV light}} \text{B}\)
(vii) 3-bromocyclohexene + \(\text{A} / \text{B} \longrightarrow \text{Y}\)
(viii) \(\text{CH}_3\text{Br} \xrightarrow{\text{KCN}} \text{A} \xrightarrow{\text{H}_3\text{O}^+} \text{B} \xrightarrow[\text{ether}]{\text{LiAlH}_4}\)
Answer:
(i) A = m-Chlorobenzonitrile, B = m-Chlorobenzoic acid
(ii) A = Allyl bromide or bromoalkene / propyne intermediate, B = Propyne
(iii) A = \(\text{C}_6\text{H}_5\text{CH}=\text{CHCH}_3\) (major alkene), B = \(\text{C}_6\text{H}_5\text{CH(Br)}—\text{CH}_2\text{CH}_3\) or isomer depending on addition.
(iv) X = But-2-ene (major), Y = 1-bromobutane (via anti-Markovnikov addition with peroxide)
(v) A = 1,2-Dibromopropane
(vi) B = 3-Bromoprop-1-ene (Allylic bromination)
(vii) Y = Product of substitution or addition on cyclohexene system.
(viii) A = Acetonitrile (\(\text{CH}_3\text{CN}\)), B = Acetamide (\(\text{CH}_3\text{CONH}_2\)), Final product = Propan-1-amine (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2\))
In simple words: Follow each chemical reagent step in sequence to determine how functional groups change from starting material to final product.

Exam Tip: Pay close attention to reaction conditions such as catalysts, light, heat, and peroxides, as they completely alter the reaction pathway (substitution vs. addition vs. elimination).

Question 7. Explain the following reactions with suitable example :
(i) Finkelstein reaction.
(ii) Swarts reaction.
(iii) Wurtz reaction.
(iv) Wurtz-Fitting reaction.
(v) Friedel-Crafts alkylation reaction.
(vi) Friedel-Crafts acylation reaction.
(vii) Sandmeyer reaction.
Answer:
(i) Finkelstein reaction: Alkyl chlorides or bromides are treated with sodium iodide in dry acetone to form alkyl iodides. Example: \(\text{CH}_3\text{Cl} + \text{NaI} \xrightarrow{\text{acetone}} \text{CH}_3\text{I} + \text{NaCl} \downarrow\).
(ii) Swarts reaction: Alkyl chlorides/bromides are heated with metal fluorides like \(\text{AgF}, \text{Hg}_2\text{F}_2, \text{CoF}_2\) to form alkyl fluorides. Example: \(\text{CH}_3\text{Br} + \text{AgF} \longrightarrow \text{CH}_3\text{F} + \text{AgBr} \downarrow\).
(iii) Wurtz reaction: Alkyl halides react with sodium metal in dry ether to give higher alkanes with double the carbon atoms. Example: \(2\text{CH}_3\text{Br} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{CH}_3-\text{CH}_3 + 2\text{NaBr}\).
(iv) Wurtz-Fitting reaction: A mixture of an aryl halide and alkyl halide is treated with sodium in dry ether to give an alkylarenes. Example: \(\text{C}_6\text{H}_5\text{Cl} + 2\text{Na} + \text{CH}_3\text{Cl} \xrightarrow{\text{dry ether}} \text{C}_6\text{H}_5\text{CH}_3 + 2\text{NaCl}\).
(v) Friedel-Crafts alkylation: Benzene reacts with an alkyl halide in the presence of anhydrous \(\text{AlCl}_3\) to form alkylbenzene. Example: Benzene + \(\text{CH}_3\text{Cl} \xrightarrow{\text{anhyd. AlCl}_3} \text{Toluene} + \text{HCl}\).
(vi) Friedel-Crafts acylation: Benzene reacts with an acyl chloride or acid anhydride in the presence of anhydrous \(\text{AlCl}_3\) to form an acylbenzene (ketone). Example: Benzene + \(\text{CH}_3\text{COCl} \xrightarrow{\text{anhyd. AlCl}_3} \text{Acetophenone} + \text{HCl}\).
(vii) Sandmeyer reaction: Primary aromatic amines treated with nitrous acid form diazonium salts, which react with cuprous halides (\(\text{CuCl / HCl}\) or \(\text{CuBr / HBr}\)) to form aryl halides. Example: \(\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- \xrightarrow{\text{CuCl} / \text{HCl}} \text{C}_6\text{H}_5\text{Cl} + \text{N}_2\).
In simple words: These are named organic reactions used to build carbon chains, introduce specific halogens, or functionalize benzene rings.

Exam Tip: Always mention the specific catalysts (dry ether, anhydrous \(\text{AlCl}_3\), acetone) required for these named reactions.

Question 8. Write the major products and name the rule responsible for the formation of the product.
(i) \(\text{CH}_3-\text{CH}_2-\text{CH(Br)}-\text{CH}_3 \xrightarrow{\text{KOH, EtOH}}\)
(ii) \(\text{CH}_3-\text{CH}_2-\text{CH}=\text{CH}_2 + \text{HBr} \xrightarrow{\text{organic peroxide}}\)
Answer:
(i) Major product: But-2-ene (\(\text{CH}_3-\text{CH}=\text{CH}-\text{CH}_3\)). Rule responsible: Saytzeff's Rule (elimination yields the more substituted, more stable alkene).
(ii) Major product: 1-Bromobutane (\(\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2\text{Br}\)). Rule responsible: Anti-Markovnikov's Rule or Peroxide Effect (Kharash Effect).
In simple words: Saytzeff's rule governs elimination to form the most substituted double bond, while peroxide effect governs anti-Markovnikov addition of HBr across alkenes.

Exam Tip: Saytzeff rule applies to elimination reactions (\(\text{E}2 / \text{E}1\)), while Markonikov/Anti-Markovnikov rules apply to electrophilic addition across double bonds.

Question 9. Write the difference between
(i) enantiomers and diastereomers
(ii) retention and inversion of configuration.
(iii) electrophilic and nucleophilic substitution reactions.
Answer:
(i) Enantiomers: Non-superimposable mirror images of each other; have identical physical properties except rotation of plane-polarized light. Diastereomers: Stereoisomers that are not mirror images of each other; have different physical properties (melting point, boiling point, solubility).
(ii) Retention of configuration: Preservation of spatial arrangement of bonds around a chiral center during a chemical reaction. Inversion of configuration: Spatial inversion or "umbrella flipping" of the chiral center configuration (Walden inversion), typical of \(\text{S}_\text{N}2\) mechanisms.
(iii) Electrophilic substitution: Substitution in which an electrophile (electron-deficient species) attacks an electron-rich aromatic ring (characteristic of haloarenes). Nucleophilic substitution: Substitution in which a nucleophile attacks an electron-deficient carbon bonded to a leaving group (characteristic of haloalkanes).
In simple words: Enantiomers are exact mirror images, diastereomers are not. Retention keeps the same spatial layout, inversion flips it inside out.

Exam Tip: Clearly state that enantiomers differ only in optical rotation, whereas diastereomers have entirely different physical and chemical constants.

Question 10. Give a chemical test to distinguish between the following pairs of compounds:
(i) chlorobenzene and cyclohexylchloride.
(ii) vinyl chloride and ethyl chloride.
(iii) n-propyl bromide and isopropyl bromide.
Answer:
(i) Chlorobenzene and cyclohexylchloride: Treat both with aqueous \(\text{NaOH}\) followed by acidification with \(\text{HNO}_3\) and addition of \(\text{AgNO}_3\). Cyclohexyl chloride gives a white precipitate of \(\text{AgCl}\) (due to reactive aliphatic C-Cl bond), whereas chlorobenzene does not react under normal conditions due to partial double bond character of C-Cl bond.
(ii) Vinyl chloride and ethyl chloride: Treat with alcoholic silver nitrate (\(\text{AgNO}_3\)). Ethyl chloride gives an immediate white precipitate of \(\text{AgCl}\), while vinyl chloride does not react due to strong carbon-chlorine bond stability from resonance.
(iii) n-Propyl bromide and isopropyl bromide: Treat with aqueous sodium hydroxide followed by acidification and addition of Lucas reagent or oxidation test / silver nitrate test. Alternatively, treat with alcoholic KOH followed by ozonolysis or silver nitrate test where primary and secondary alkyl halides show different rates of precipitate formation with \(\text{AgNO}_3\).
In simple words: Use aqueous silver nitrate tests or reactivity differences to show that aliphatic or reactive alkyl halides precipitate AgCl quickly, whereas aryl and vinyl halides do not.

Exam Tip: Always mention the observable visual change, such as formation of a white precipitate (\(\text{AgCl}\)) or colored precipitate.

Question 11. Give mechanism of the following reactions :
(i) \((\text{CH}_3)_3\text{C—Cl} + \text{OH}^- \longrightarrow (\text{CH}_3)_3\text{C—OH}\)
(ii) \(\text{CH}_3-\text{Cl} + \text{OH}^- \longrightarrow \text{CH}_3-\text{OH}\)
(iii) \(\text{CH}_3-\text{CH(CH}_3)\text{—Cl} + \text{OH}^- \xrightarrow{\text{ethanol}} \text{CH}_3-\text{CH}=\text{CH}_2\)
(iv) Benzene + \(\text{Cl}_2 \xrightarrow[\text{dark}]{\text{FeCl}_3} \text{Chlorobenzene}\)
(v) 1-chloro-4-nitrobenzene + \(\text{NaOH} \xrightarrow[\text{pressure}]{\text{high temp.}} \text{4-Nitrophenol}\)
Answer:
(i) \(\text{S}_\text{N}1\) Mechanism:
Step 1: Slow ionization of tert-butyl chloride to form a stable tertiary carbocation: \((\text{CH}_3)_3\text{C—Cl} \rightleftharpoons (\text{CH}_3)_3\text{C}^+ + \text{Cl}^-\).
Step 2: Fast attack of nucleophile \(\text{OH}^-\) on the planar carbocation to yield tert-butyl alcohol.
(ii) \(\text{S}_\text{N}2\) Mechanism:
Concerted single-step bimolecular process where the nucleophile \(\text{OH}^-\) attacks from the back while the chloride ion leaves simultaneously, passing through an unstable pentavalent transition state with inversion of configuration.
(iii) \(\text{E}2\) Mechanism:
Bimolecular elimination where base \(\text{OH}^-\) abstracts a proton from the beta-carbon simultaneously with the departure of chloride ion in a single concerted transition state.
(iv) Electrophilic aromatic substitution:
Step 1: Generation of electrophilic chloronium ion (\(\text{Cl}^+\)) using Lewis acid catalyst \(\text{FeCl}_3\): \(\text{Cl}_2 + \text{FeCl}_3 \longrightarrow \text{Cl}^+ + \text{FeCl}_4^-\).
Step 2: Attack of \(\text{Cl}^+\) on benzene ring to form resonance-stabilized carbocation complex (arenium ion).
Step 3: Loss of proton from the sp3 carbon to restore aromaticity.
(v) Nucleophilic aromatic substitution (\(\text{S}_\text{N}\text{Ar}\)):
The strongly electron-withdrawing nitro group at para position stabilizes the negatively charged carbanion intermediate (Meisenheimer complex) formed by the attack of \(\text{OH}^-\), followed by elimination of chloride ion.
In simple words: Write step-by-step reaction pathways showing bond cleavage, intermediate formation (carbocations or transition states), and final product generation.

Exam Tip: Clearly distinguish between unimolecular two-step mechanisms (\(\text{S}_\text{N}1\)) and concerted single-step mechanisms (\(\text{S}_\text{N}2\)).

Question 12. Which compound in each of the following pairs will react faster in \(\text{S}_\text{N}2\) reaction with \(\text{OH}^-\) and why?
(i) \(\text{CH}_3\text{Br}\) or \(\text{CH}_3\text{I}\)
(ii) \((\text{CH}_3)_3\text{CCl}\) or \(\text{CH}_3\text{Cl}\)
Answer:
(i) \(\text{CH}_3\text{I}\) reacts faster because iodide is a much better leaving group than bromide due to the weaker carbon-iodine bond.
(ii) \(\text{CH}_3\text{Cl}\) reacts faster because primary alkyl halides experience minimal steric hindrance compared to bulky tertiary alkyl halides like \((\text{CH}_3)_3\text{CCl}\).
In simple words: \(\text{S}_\text{N}2\) reactions favor unhindered substrates and better leaving groups.

Exam Tip: Leaving group ability order for \(\text{S}_\text{N}2\) is \(\text{I}^- > \text{Br}^- > \text{Cl}^- > \text{F}^-\).

Question 13. In the following pairs which halogen compound undergoes faster \(\text{S}_\text{N}1\) reaction?
(i) 1-Chlorobutane vs 2-chlorobutane (or corresponding cyclic structures)
(ii) Chlorocyclohexane vs 1-chlorohexane
(iii) \((\text{CH}_3)_3\text{C—Cl}\) and \(\text{C}_6\text{H}_5\text{CH}_2\text{Cl}\)
(iv) \(\text{C}_6\text{H}_5\text{CH}_2\text{Cl}\) and \(\text{C}_6\text{H}_5\text{C(Cl)C}_6\text{H}_5\)
(v) \(\text{CH}_2=\text{CH—Cl}\) and \(\text{CH}_2=\text{CH—CH}_2\text{Cl}\)
Answer:
(i) Secondary alkyl halide reacts faster than primary alkyl halide via \(\text{S}_\text{N}1\) due to greater stability of secondary carbocations.
(ii) Secondary cyclic halide reacts faster than primary straight-chain halide.
(iii) \((\text{CH}_3)_3\text{C—Cl}\) and \(\text{C}_6\text{H}_5\text{CH}_2\text{Cl}\): Both form very stable carbocations (tert-butyl carbocation is stabilized by +I effect of three methyl groups; benzyl carbocation is stabilized by resonance with the benzene ring). Benzyl chloride reacts extremely fast due to resonance stabilization.
(iv) \(\text{C}_6\text{H}_5\text{C(Cl)C}_6\text{H}_5\) reacts faster than \(\text{C}_6\text{H}_5\text{CH}_2\text{Cl}\) because it forms a diphenylmethyl carbocation stabilized by resonance across two benzene rings.
(v) \(\text{CH}_2=\text{CH—CH}_2\text{Cl}\) reacts much faster than \(\text{CH}_2=\text{CH—Cl}\) because allyl chloride forms a resonance-stabilized allyl carbocation, whereas vinyl chloride has partial double bond character making ionization extremely difficult.
In simple words: \(\text{S}_\text{N}1\) depends entirely on carbocation stability: Tertiary > Secondary > Primary, with allylic and benzylic halides being exceptionally reactive.

Exam Tip: Always evaluate carbocation stability (\(3^\circ > 2^\circ > 1^\circ\) + resonance) to determine \(\text{S}_\text{N}1\) reactivity.

Question 14. Give reasons for the following :
(i) The bond length of C—Cl bond is larger in haloalkanes than that in haloarenes.
(ii) Although alkyl halides are polar in nature but are not soluble in water.
(iii) tert-butyl bromide has lower boiling point than n-butyl bromide.
(iv) Haloalkanes react with KCN to form alkyl cyanide as main product while with AgCN alkyl isocyanide is the main product.
(v) Sulphuric acid is not used in the reaction of alcohol with KI.
(vi) Thionyl chloride is the preferred reagent for converting ethanol to chloroethane.
(vii) Haloalkanes undergo nucleophilic substitution reaction easily but haloarenes do not undergo nucleophilic substitution under ordinary conditions.
(viii) Chlorobenzene on reaction with fuming sulphuric acid gives ortho and para chlorosulphonic acids.
(ix) 2,4-Dinitrochlorobenzene is much more reactive than chlorobenzene towards hydrolysis reaction with NaOH.
(x) Grignard reagent should be prepared under anhydrous conditions.
(xi) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(xii) Neopentyl bromide undergoes nucleophilic substitution reactions very slowly.
(xiii) Vinyl chloride is unreactive in nucleophilic substitution reaction.
(xiv) An optically inactive product is obtained after the hydrolysis of optically active 2-bromobutane.
(xv) Methyl iodide is hydrolysed at faster rate than methyl chloride.
Answer:
(i) In haloalkanes, carbon is \(\text{sp}^3\) hybridized, whereas in haloarenes, carbon is \(\text{sp}^2\) hybridized with partial double bond character due to resonance, shortening the bond length.
(ii) Alkyl halides cannot form hydrogen bonds with water molecules and cannot break strong hydrogen-bonding networks of water.
(iii) Branching in tert-butyl bromide makes the molecule spherical, decreasing surface area and van der Waals intermolecular forces compared to straight-chain n-butyl bromide.
(iv) KCN is predominantly ionic and furnishes cyanide ions (\(\text{C}\equiv\text{N}^-\)), where carbon attacks to form stable C-C bonded nitriles. AgCN is covalent, so nitrogen is more available for nucleophilic attack, yielding isocyanides.
(v) Conc. \(\text{H}_2\text{SO}_4\) oxidizes KI to iodine (\(\text{I}_2\)) instead of forming alkyl iodide.
(vi) Thionyl chloride (\(\text{SOCl}_2\)) is preferred because by-products formed are gases (\(\text{SO}_2\) and \(\text{HCl}\)), escaping easily and leaving pure alkyl chloride.
(vii) Due to resonance, partial double bond character, \(\text{sp}^2\) hybridized carbon, and repulsion of incoming nucleophile by electron-rich benzene ring.
(viii) Sulphonation is an electrophilic substitution reaction where the chlorobenzene ring directs the incoming sulphonic acid group to ortho and para positions.
(ix) Presence of two electron-withdrawing nitro groups at ortho and para positions strongly depletes electron density on the benzene ring, facilitating nucleophilic attack by hydroxide ions.
(x) Grignard reagents react violently with moisture/water to form alkanes, destroying the reagent.
(xi) In chlorobenzene, the electron-withdrawing inductive effect of chlorine opposes the +M electron-releasing resonance effect, reducing net dipole moment compared to aliphatic cyclohexyl chloride.
(xii) Neopentyl bromide experiences extreme steric hindrance around the reaction center, impeding \(\text{S}_\text{N}2\) attack.
(xiii) Partial double bond character of C—Cl bond due to resonance keeps vinyl chloride stable against nucleophilic substitution.
(xiv) Hydrolysis proceeds via \(\text{S}_\text{N}1\) mechanism through a planar carbocation, which is attacked equally from both sides, producing a racemic mixture (equimolar d and l forms) which is optically inactive.
(xv) Carbon-iodine bond has lower bond dissociation enthalpy than carbon-chlorine bond, making methyl iodide split and react much faster.
In simple words: These reasons involve fundamental chemical principles like resonance, hybridization, steric hindrance, leaving group ability, and hydrogen bonding.

Exam Tip: For haloarene stability and sluggishness in substitution, always mention "resonance" and "partial double bond character".

Question 15. Write the different products and their number formed by the monochlorination of following compounds :
(i) \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_3\)
(ii) \((\text{CH}_3)_2\text{CHCH}_2\text{CH}_3\)
(iii) \((\text{CH}_3)_2\text{CHCH(CH}_3)_2\)
Answer:
(i) Two products: 1-chlorobutane and 2-chlorobutane.
(ii) Four products: 1-chloro-2-methylbutane, 2-chloro-2-methylbutane, 2-chloro-3-methylbutane, and 1-chloro-3-methylbutane.
(iii) Three products: 1-chloro-2,3-dimethylbutane, 2-chloro-2,3-dimethylbutane, and substitution on methyl groups (total 3 distinct structural isomers).
In simple words: Identify all chemically non-equivalent hydrogen atoms in each hydrocarbon structure and replace each type with a chlorine atom.

Exam Tip: Count non-equivalent sets of hydrogens by checking symmetry in the carbon skeleton.

Question 16. (a) When 3-methylbutan-2-ol is treated with HBr, give the mechanism for this reaction.
(b) In the reaction of 3-methylbutan-2-ol with HBr, identify the major and minor products among the possible isomers.
Answer:
(a) Mechanism:
Step 1: Protonation of hydroxyl group of 3-methylbutan-2-ol by \(\text{H}^+\) to form an oxonium ion.
Step 2: Departure of water molecule to form a secondary carbocation: \((\text{CH}_3)_2\text{CH—CH}^+—\text{CH}_3\).
Step 3: 1,2-Hydride shift from C-3 to C-2 to form a more stable tertiary carbocation: \((\text{CH}_3)_2\text{C}^+—\text{CH}_2\text{CH}_3\).
Step 4: Attack of bromide ion \(\text{Br}^-\) on the tertiary carbocation to yield 2-bromo-2-methylbutane as major product.
(b) Major product: 2-bromo-2-methylbutane (formed via rearrangement to tertiary carbocation). Minor product: 2-bromo-3-methylbutane.
In simple words: Carbocations undergo rearrangement via hydride or methide shifts if it leads to a more stable tertiary carbocation.

Exam Tip: Always check for possible carbocation rearrangements (1,2-hydride or 1,2-methyl shifts) in reactions proceeding through carbocation intermediates.

Question 17. Give one use of each of following :
(i) Freon-12
(ii) DDT
(iii) Carbon tetrachloride
(iv) Iodoform
Answer:
(i) Freon-12 (Dichlorodifluoromethane): Used as a refrigerant in refrigerators and air conditioners.
(ii) DDT (Dichlorodiphenyltrichloroethane): Used as a powerful insecticide to control mosquitoes (malaria) and lice (typhus).
(iii) Carbon tetrachloride (\text{CCl}_4): Used as a solvent for oils, fats, and grease, and previously in fire extinguishers (Pyrene).
(iv) Iodoform (\text{CHI}_3): Used as an antiseptic for wounds due to the liberation of free iodine.
In simple words: These are historically important organohalogen compounds used in refrigeration, pest control, antisepsis, and industrial cleaning.

Exam Tip: Remember environmental concerns (such as ozone depletion by Freons and bioaccumulation of DDT) alongside their uses.

Question 18. An optically active compound having molecular formula \(\text{C}_7\text{H}_{15}\text{Br}\) reacts with aqueous KOH to give \(\text{C}_7\text{H}_{15}\text{OH}\), which is optically inactive. Give mechanism for the reaction.
Answer:
The reaction proceeds via an \(\text{S}_\text{N}1\) mechanism.
Step 1: Slow ionization of chiral alkyl halide to form a planar carbocation intermediate.
Step 2: Nucleophilic attack by \(\text{OH}^-\) from both upper and lower faces of the planar carbocation with equal probability, forming a 50:50 mixture of enantiomers (a racemic mixture). Since a racemic mixture is optically inactive due to external compensation, the resulting alcohol is optically inactive.
In simple words: Optically active reactant forms a flat carbocation intermediate, which gets attacked from both sides equally, producing a racemic mixture that does not rotate plane-polarized light.

Exam Tip: Whenever a question mentions an optically active reactant turning into an optically inactive product during substitution, think \(\text{S}_\text{N}1\) and racemic mixture formation.

Question 19. An organic compound \(\text{C}_8\text{H}_9\text{Br}\) has three isomers A, B and C. A is optically active. Both A and B gave the white precipitate when warmed with alcoholic \(\text{AgNO}_3\) solution in alkaline medium. Benzoic acid, terephthalic acid and p-bromobenzoic acid were obtained on oxidation of A, B and C respectively. Identify A, B and C.
Answer:
A = \(\text{C}_6\text{H}_5-\text{CH(Br)}-\text{CH}_3\) (1-bromo-1-phenylethane; optically active, gives benzoic acid on oxidation, gives white ppt with \(\text{AgNO}_3\) because bromine is on benzylic position).
B = \(p-\text{Br—C}_6\text{H}_4—\text{CH}_2\text{CH}_3\) (1-(4-bromophenyl)ethane; gives terephthalic acid on oxidation, gives white ppt with \(\text{AgNO}_3\) due to active benzylic/side-chain bromine).
C = \(p-\text{CH}_3\text{C}_6\text{H}_4\text{CH}_2\text{Br}\) or 4-bromotoluene derivative yielding p-bromobenzoic acid on oxidation.
In simple words: Analyze the oxidation products and reactions with silver nitrate to locate where the bromine atom and alkyl side-chains are attached on the benzene ring.

Exam Tip: Oxidation of any alkyl side chain on a benzene ring with \(\text{KMnO}_4\) always converts it into a carboxylic acid (\(-\text{COOH}\)) group attached directly to the ring.

Question 20. An alkyl halide X having molecular formula \(\text{C}_6\text{H}_{13}\text{Cl}\) on treatment with potassium tert-butoxide gives two isomeric alkenes Y and Z, but alkene Y is symmetrical. Both alkenes on hydrogenation give 2,3-dimethylbutane. Identify X, Y and Z.
Answer:
X = 2-Chloro-2,3-dimethylbutane (\((\text{CH}_3)_2\text{CH—C(Cl)(CH}_3)_2\))
Y = 2,3-Dimethylbut-2-ene (\((\text{CH}_3)_2\text{C}=\text{C(CH}_3)_2\) - symmetrical alkene)
Z = 2,3-Dimethylbut-1-ene (\(\text{CH}_2=\text{C(CH}_3)\text{CH(CH}_3)_2\))
In simple words: Work backwards from the hydrogenation product (2,3-dimethylbutane) to determine the carbon skeleton, then place the chlorine atom to allow elimination into the two specified alkenes.

Exam Tip: Symmetrical alkenes have identical groups on both carbons of the double bond (like tetramethylethylene).

Question 21. An organic compound (A) having molecular formula \(\text{C}_3\text{H}_7\text{Cl}\) on reaction with alcoholic solution of KCN gives compound B. The compound B on hydrolysis with dilute HCl gives compound C. C on reduction with \(\text{H}_2 / \text{Ni}\) gives 1-aminobutane. Identify A, B and C.
Answer:
A = 1-Chloropropane (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{Cl}\))
B = Butanenitrile (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CN}\))
C = Butanamide (\(\text{CH}_3\text{CH}_2\text{CH}_2\text{CONH}_2\))
In simple words: Trace the carbon count: 1-aminobutane has 4 carbons, so nitrile B has 4 carbons, meaning alkyl halide A must be 3-carbon 1-chloropropane.

Exam Tip: Reduction of nitriles gives primary amines with one extra carbon atom compared to the starting alkyl halide.

Question 22. Identify A, B, C, D, E, R and R´ in the following sequence of reactions :
(a) Bromocyclohexane + Mg \(\xrightarrow{\text{dry ether}}\) A \(\xrightarrow{\text{H}_2\text{O}}\) B
(b) R—Br + Mg \(\xrightarrow{\text{dry ether}}\) C \(\xrightarrow{\text{D}_2\text{O}}\) \( \text{CH}_3-\text{CH(D)}-\text{CH}_3\
(c) Alkene + Na / dry ether / \(\text{R}'\text{X}\) / Mg / \(\text{H}_2\text{O}\) \(\longrightarrow\) E
Answer:
(a) A = Cyclohexylmagnesium bromide (Grignard reagent), B = Cyclohexane.
(b) R—Br = 2-Bromopropane, C = Isopropylmagnesium bromide, D = Magnesium halide derivative, E = Corresponding hydrocarbon product.
In simple words: Grignard reagents react with water or heavy water (\(\text{D}_2\text{O}\)) to form deuterated or normal alkanes by replacing the halogen with hydrogen/deuterium.

Exam Tip: Grignard reagents + \(\text{D}_2\text{O}\) is a standard method for introducing deuterium into specific positions of organic molecules.

Question 23. Which nomenclature is not according to IUPAC system.
(i) \(\text{Br—CH}_2-\text{CH}=\text{CH}_2\); 1-bromoprop-2-ene
(ii) 4-bromo-2,4-dimethylhexane
(iii) 2-methyl-3-phenylpentane
(iv) 5-oxohexanoic acid
Answer:
(i) Correct IUPAC name is 3-bromoprop-1-ene (numbering must start from the principal functional group / double bond, giving it position 1).
(ii) Correct numbering should give lower locants, e.g., 3-bromo-2,5-dimethylhexane.
(iii) Check chain length and locants according to substituent priority.
(iv) Correct according to priority of carboxylic acid group.
In simple words: Check whether numbering follows the lowest locant rule giving priority to principal functional groups and multiple bonds.

Exam Tip: Multiple bonds (\(\text{alkenes / alkynes}\)) take precedence over halogen substituents when numbering carbon chains.