CBSE Class 12 Chemistry Halo Alkanes and Halo arenes Assignment

Haloalkanes and Haloarenes

LEVEL-1 QUESTIONS

Question 1. Out of \(\text{PCl}_5\) and \(\text{SOCl}_2\) which one is better reagent for conversion of alcohol to alkyl chloride. Why?
Answer: \(\text{SOCl}_2\) is a better reagent because all the byproducts except the alkyl halide are in the gaseous state and escape easily, leaving behind pure alkyl chloride.
\(\text{ROH} + \text{SOCl}_2 \rightarrow \text{R-Cl} + \text{SO}_2 \uparrow + \text{HCl} \uparrow\)
In simple words: Thionyl chloride produces gases as byproducts, so you get pure product without messy separation.

Exam Tip: Mention the gaseous nature of \(\text{SO}_2\) and \(\text{HCl}\) to score full marks in this frequently asked question.

Question 2. Why conc. \(\text{HI}\) is not preferred in the preparation of alkyl iodide from alcohol?
Answer: Concentrated \(\text{HI\) is a strong reducing agent. A part of the alkyl iodide formed is reduced back to alkane, hence it is not preferred for preparing alkyl iodides directly.
\(\text{R-OH} + \text{HI} \rightarrow \text{RI} + \text{H}_2\text{O}\)
\(\text{RI} + \text{HI} \rightarrow \text{RH} + \text{I}_2\)
In simple words: Concentrated hydroiodic acid reduces the desired alkyl iodide into an alkane.

Exam Tip: Always write both equations (formation and reduction) to justify why \(\text{HI}\) is not used.

Question 3. Haloalkanes undergo nucleophilic substitution. Why?
Answer: The C-X bond in haloalkanes is polar in nature due to the high electronegativity of the halogen atom. This creates a partial positive charge on the carbon atom, making it electron-deficient and vulnerable to attack by nucleophiles.
In simple words: The halogen pulls electrons away from carbon, making carbon positive and inviting negative nucleophiles to attack.

Exam Tip: Use terms like electronegativity difference, polarity, and electron-deficient carbon.

Question 4. Haloarenes are less reactive than haloalkanes towards nucleophilic substitution reaction. Why?
Answer: Haloarenes are less reactive than haloalkanes towards nucleophilic substitution due to:
1. Resonance effect: The C-X bond acquires partial double bond character due to resonance, making bond cleavage difficult.
2. Hybridization of carbon: The carbon atom attached to halogen in haloarenes is \(sp^2\) hybridized (more electronegative with smaller bond length), whereas in haloalkanes it is \(sp^3\) hybridized.
3. Instability of phenyl cation: The phenyl cation formed by self-ionization is not stabilized by resonance.
In simple words: Resonance gives the carbon-halogen bond extra strength (double bond character), making it hard to break.

Question 5. How will you convert: (i) Benzyl chloride to benzyl alcohol (ii) Methyl magnesium bromide to 2-methylpropan-2-ol
Answer:
(i) Benzyl chloride to benzyl alcohol:
Treat benzyl chloride with aqueous \(\text{NaOH}\) followed by acidification.
\(\text{C}_6\text{H}_5\text{CH}_2\text{Cl} + \text{NaOH (aq)} \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{OH} + \text{NaCl}\)
(ii) Methyl magnesium bromide to 2-methylpropan-2-ol:
React methyl magnesium bromide with acetone (\(\text{CH}_3\text{COCH}_3\)), followed by acid hydrolysis.
\(\text{CH}_3\text{MgBr} + \text{CH}_3\text{COCH}_3 \rightarrow \text{Adduct} \xrightarrow{\text{H}_3\text{O}^+} (\text{CH}_3)_3\text{COH} + \text{Mg(OH)Br}\)
In simple words: Nucleophilic addition of Grignard reagent to a ketone yields a tertiary alcohol after hydrolysis.

Exam Tip: Always show the intermediate addition product (adduct) before writing the final hydrolysis step.

Question 6. How will you convert ethylmagnesium chloride to propan-1-ol?
Answer:
React ethylmagnesium chloride with formaldehyde (\(\text{HCHO}\)), followed by acid hydrolysis.
\(\text{C}_2\text{H}_5\text{MgCl} + \text{HCHO} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{OMgCl}\)
\(\text{CH}_3\text{CH}_2\text{CH}_2\text{OMgCl} + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} + \text{Mg(OH)Cl}\)
In simple words: Grignard reagent reacting with formaldehyde increases the carbon chain by one carbon atom to form a primary alcohol.

Exam Tip: Remember: Formaldehyde gives 1 degrees alcohol, other aldehydes give 2 degrees alcohols, and ketones give 3 degrees alcohols with Grignard reagents.

Question 7. Write the structure of 4-bromo-2,3-dimethylpentan-1-ol.
Answer:
Structure:
\(\text{CH}_3-\text{CH(Br)}-\text{CH(CH}_3)-\text{CH(CH}_3)-\text{CH}_2-\text{OH}\)
In simple words: Start with a 5-carbon chain, put an alcohol group at carbon 1, methyl groups at carbons 2 and 3, and a bromine atom at carbon 4.

Exam Tip: Number the longest carbon chain starting from the principal functional group (\(-\text{OH}\)).

Question 8. Complete the reaction: \(\text{CH}_3-\text{CH}_2-\text{CH}=\text{CH}_2 + \text{HBr} \xrightarrow{\text{peroxide}}\) ?
Answer:
\(\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_2\text{Br}\) (1-bromobutane)
In simple words: In the presence of peroxide, addition of HBr to unsymmetrical alkenes follows anti-Markovnikov rule.

Exam Tip: Peroxide effect is applicable only to HBr, not to HCl or HI.

Question 9. Explain why alkyl halides, though polar, are immiscible with water?
Answer: Alkyl halides are polar and held together by dipole-dipole interactions. Water molecules are strongly held together by intermolecular hydrogen bonding. The new force of attraction between alkyl halide and water molecules is weaker than the breaking of existing hydrogen bonds in water and dipole-dipole bonds in alkyl halides. Hence, they are immiscible.
In simple words: They cannot form hydrogen bonds with water molecules.

Question 10. Which one of the given secondary and primary halogen compounds undergoes \(\text{S}_\text{N}1\) substitution reaction faster and why?
Answer: The secondary halide undergoes \(\text{S}_\text{N}1\) substitution faster than the primary halide because it forms a secondary carbocation intermediate, which is more stable than a primary carbocation due to greater hyperconjugation and inductive effect.
In simple words: More stable carbocation means faster reaction.

Exam Tip: Stability order for \(\text{S}_\text{N}1\): \(3^\circ > 2^\circ > 1^\circ\).

Question 11. Convert: (a) Chlorobenzene to p-nitrophenol (b) Aniline to phenylisocyanide
Answer:
(a) Chlorobenzene to p-nitrophenol:
1. Nitration: Treat chlorobenzene with conc. \(\text{HNO}_3\) and conc. \(\text{H}_2\text{SO}_4\) to get 1-chloro-4-nitrobenzene.
2. Nucleophilic substitution: Treat with 15% \(\text{NaOH}\) at 433 K followed by acidification (dil. \(\text{HCl}\)) to yield p-nitrophenol.
(b) Aniline to phenylisocyanide:
Treat aniline with chloroform (\(\text{CHCl}_3\)) and alcoholic potassium hydroxide (\(\text{KOH}\)) upon warming (Carbylamine reaction).
\(\text{C}_6\text{H}_5\text{NH}_2 + \text{CHCl}_3 + 3\text{KOH (alc.)} \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{NC} + 3\text{KCl} + 3\text{H}_2\text{O}\)
In simple words: Carbylamine test turns primary amines into foul-smelling isocyanides.

Exam Tip: Carbylamine reaction is a specific test for primary amines.

Question 12. Write the name reactions: (a) Finkelstein reaction (b) Dow's process
Answer:
(a) Finkelstein reaction: Preparation of alkyl iodides by heating alkyl chloride or bromide with sodium iodide in dry acetone.
\(\text{R-X} + \text{NaI} \xrightarrow{\text{dry acetone}} \text{R-I} + \text{NaX} \quad (\text{X} = \text{Cl, Br})\)
(b) Dow's process: Chlorobenzene is fused with NaOH at 623 K and 300 atm pressure to form sodium phenoxide, which on acidification gives phenol.
\(\text{C}_6\text{H}_5\text{Cl} + 2\text{NaOH} \xrightarrow{623\text{ K, 300 atm}} \text{C}_6\text{H}_5\text{ONa} + \text{NaCl} + \text{H}_2\text{O}\)
\(\text{C}_6\text{H}_5\text{ONa} + \text{HCl} \rightarrow \text{C}_6\text{H}_5\text{OH} + \text{NaCl}\)
In simple words: Finkelstein swaps halogens for iodine using acetone, while Dow's process turns chlorobenzene into phenol under harsh conditions.

Exam Tip: Mention temperature and pressure conditions for Dow's process.

Question 13. Distinguish between chlorocyclohexane and chlorobenzene.
Answer: Chlorocyclohexane is an aliphatic alkyl halide and readily gives a white precipitate of \(\text{AgCl}\) when treated with alcoholic silver nitrate (\(\text{AgNO}_3\)). Chlorobenzene is an aryl halide in which the C-Cl bond has partial double bond character and is unreactive; hence it does not give a white precipitate with \(\text{AgNO}_3\).
In simple words: Aliphatic halides react with silver nitrate to give white precipitate, aromatic halides do not.

Exam Tip: Use \(\text{AgNO}_3\) test to distinguish between alkyl and aryl halides.

Question 14. Arrange each set of compounds in the order of increasing boiling points: (i) Bromomethane, Bromoform, Chloromethane, Dibromomethane (ii) 1-chloropropane, Isopropyl chloride, 1-chlorobutane
Answer:
(i) Chloromethane < Bromomethane < Dibromomethane < Bromoform (Boiling point increases with molecular mass due to increased van der Waals forces).
(ii) Isopropyl chloride < 1-chloropropane < 1-chlorobutane (Branching decreases surface area, thereby lowering the boiling point).
In simple words: Heavier molecules and straight-chain molecules have higher boiling points.

Question 15. (i) Why are haloarenes less reactive than haloalkanes towards nucleophilic substitution reactions? (ii) Haloalkane react with KCN to form alkyl cyanides as main product while AgCN forms isocyanides as the chief product. Explain.
Answer:
(i) Due to resonance and \(sp^2\) hybridization of carbon in the C-X bond of haloarenes.
(ii) KCN is predominantly ionic and furnishes cyanide ions (\(\text{CN}^-\)) in solution. Since the C-C bond is more stable than the C-N bond, attack occurs mainly through the carbon atom, yielding alkyl cyanides. On the other hand, AgCN is covalent; nitrogen has a lone pair of electrons readily available for attack, forming isocyanides.
In simple words: Ionic cyanide attacks through carbon, covalent cyanide attacks through nitrogen.

Exam Tip: Mention ionic nature of KCN versus covalent nature of AgCN.

Question 16. Explain the following name reactions: (i) Swarts reaction (ii) Wurtz reaction (iii) Sandmeyer's reaction
Answer:
(i) Swarts reaction: Alkyl fluorides are prepared by heating alkyl chloride or bromide in the presence of metallic fluorides like \(\text{AgF}, \text{Hg}_2\text{F}_2, \text{CoF}_2\).
\(\text{CH}_3\text{Br} + \text{AgF} \rightarrow \text{CH}_3\text{F} + \text{AgBr}\)
(ii) Wurtz reaction: Alkyl halides react with sodium metal in dry ether to give higher alkanes containing twice the number of carbon atoms.
\(2\text{R-X} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{R-R} + 2\text{NaX}\)
(iii) Sandmeyer's reaction: Diazonium salts react with cuprous halides (\(\text{Cu}_2\text{Cl}_2 / \text{HCl}\) or \(\text{Cu}_2\text{Br}_2 / \text{HBr}\)) to replace the diazonium group with a halogen atom.
In simple words: Swarts makes fluorides, Wurtz doubles the carbon chain, and Sandmeyer converts diazonium salts to aryl halides.

Exam Tip: Clearly write the catalysts and reagents used in each name reaction.

Question 17. Draw the structure of major monohalo product in each of the following reactions: (i) Cyclohexanol + \(\text{SOCl}_2\) (ii) 1-phenylpropene + HBr in presence of peroxide
Answer:
(i) Chlorocyclohexane + \(\text{SO}_2 + \text{HCl}\)
(ii) Addition follows anti-Markovnikov rule in the presence of peroxide to yield 1-bromo-3-phenylpropane.
In simple words: Alcohol becomes chloride with thionyl chloride, and peroxide adds bromine to the less substituted carbon.

Question 18. (a) Write the formula of main product formed in the following chemical reactions: (i) \((\text{CH}_3)_2\text{CH-Cl} + \text{Na}\) in dry ether (ii) \(\text{CH}_3\text{Br} + \text{AgF}\) (iii) \(\text{CH}_3\text{CH}_2\text{Br} + \text{NaI}\) in dry acetone (b) Predict the order of reactivity of the four isomeric bromobutanes in \(\text{S}_\text{N}1\) and \(\text{S}_\text{N}2\) reactions.
Answer:
(a)(i) \((\text{CH}_3)_2\text{CH-CH(CH}_3)_2 + 2\text{NaCl}\) (2,3-dimethylbutane)
(ii) \(\text{CH}_3\text{F} + \text{AgBr}\)
(iii) \(\text{CH}_3\text{CH}_2\text{I} + \text{NaBr}\)
(b) \(\text{S}_\text{N}1\) order: \(1^\circ \text{ (primary)} < \text{isobutyl bromide} < \text{secondary} < \text{tertiary}\)
\(\text{S}_\text{N}2\) order: \(\text{tertiary} < \text{secondary} < \text{isobutyl bromide} < 1^\circ \text{ (primary)}\)
In simple words: Tertiary halides favor \(\text{S}_\text{N}1\), while primary halides favor \(\text{S}_\text{N}2\).

Exam Tip: Steric hindrance dictates \(\text{S}_\text{N}2\) reactivity, whereas carbocation stability dictates \(\text{S}_\text{N}1\) reactivity.

LEVEL-2 QUESTIONS

Question 1. What happens when: (i) n-butyl chloride is treated with alcoholic KOH (ii) bromobenzene is treated with Mg in the presence of dry ether (iii) chlorobenzene is subjected to hydrolysis (iv) ethyl chloride is treated with aqueous KOH (v) methyl bromide is treated with sodium in the presence of dry ether?
Answer:
(i) Undergoes dehydrohalogenation to form but-1-ene: \(\text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2\)
(ii) Forms phenylmagnesium bromide (Grignard reagent): \(\text{C}_6\text{H}_5\text{MgBr}\)
(iii) Does not undergo hydrolysis under normal conditions; requires drastic conditions (\(623\text{ K}, 300\text{ atm}\)) to form phenol.
(iv) Undergoes nucleophilic substitution to form ethanol: \(\text{CH}_3\text{CH}_2\text{OH}\)
(v) Undergoes Wurtz reaction to form ethane: \(\text{CH}_3-\text{CH}_3\)
In simple words: Alcoholic KOH eliminates, aqueous KOH substitutes, and sodium in dry ether couples alkyl groups.

Question 2. Primary alkyl halide \(\text{C}_4\text{H}_9\text{Br}\) (a) reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), \(\text{C}_8\text{H}_{18}\) which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write equations for all reactions.
Answer:
Compound (a) is isobutyl bromide (\((\text{CH}_3)_2\text{CHCH}_2\text{Br}\)).
Reactions:
1. \((\text{CH}_3)_2\text{CHCH}_2\text{Br} \xrightarrow{\text{alc. KOH}} (\text{CH}_3)_2\text{C}=\text{CH}_2\) (Compound b: 2-methylpropene)
2. \((\text{CH}_3)_2\text{C}=\text{CH}_2 + \text{HBr} \rightarrow (\text{CH}_3)_3\text{C-Br}\) (Compound c: 2-bromo-2-methylpropane)
3. \(2(\text{CH}_3)_2\text{CHCH}_2\text{Br} + 2\text{Na} \rightarrow (\text{CH}_3)_2\text{CHCH}_2\text{CH}_2\text{CH}(\text{CH}_3)_2 + 2\text{NaBr}\) (Compound d: 2,5-dimethylhexane)
In simple words: Trace out the reactions step by step to deduce that the branched chain isomer is isobutyl bromide.

Exam Tip: Structure determination questions require working backward and forward using given molecular formulas.

Question 3. (a) How can the following conversions be carried out? (i) Propene to propan-1-ol (ii) 1-Bromopropane to 2-bromopropane (b) Explain why (i) dipole moment of chlorobenzene is lower than that of cyclohexyl chloride (ii) alkyl halides are immiscible with water (iii) Grignard reagents should be prepared under anhydrous conditions?
Answer:
(a)(i) Propene to propan-1-ol: Treat propene with HBr in presence of peroxide to get 1-bromopropane, then treat with aqueous \(\text{KOH}\).
(ii) 1-Bromopropane to 2-bromopropane: Treat with alcoholic KOH to get propene, then add HBr via Markovnikov addition.
(b)(i) In chlorobenzene, \(sp^2\) hybridized carbon is more electronegative and the benzene ring exerts an electron-withdrawing \(-R\) effect, reducing polarity.
(ii) Alkyl halides cannot form hydrogen bonds with water molecules.
(iii) Grignard reagents react violently with moisture to form alkanes.
In simple words: Dehydrohalogenation followed by addition helps change positions of functional groups.

Exam Tip: Always mention dry conditions for Grignard reagent preparation to prevent protonolysis.

Question 4. (a) Draw the structures of major monohalo products in each of the following reactions: (i) 1-ethyl-4-nitrobenzene + \(\text{Br}_2\) (heat/UV light) (ii) 1-methylcyclohexene + HI (iii) Cyclohexene + \(\text{Br}_2\) (heat/UV) (b) Predict the order of reactivity of the four isomeric bromobutanes in \(\text{S}_\text{N}1\) and \(\text{S}_\text{N}2\) reactions.
Answer:
(a)(i) Free radical bromination at the benzylic carbon gives 4-(1-bromoethyl)nitrobenzene.
(ii) Ionic addition across the double bond yields 1-iodo-1-methylcyclohexane.
(iii) Allylic bromination gives 3-bromocyclohexene.
(b) \(\text{S}_\text{N}1\): Primary < Isobutyl < Secondary < Tertiary
\(\text{S}_\text{N}2\): Tertiary < Secondary < Isobutyl < Primary
In simple words: Benzylic and allylic positions are highly reactive toward free radical halogenation.

Exam Tip: Differentiate between electrophilic addition, nucleophilic substitution, and free-radical substitution conditions.

LEVEL-3 QUESTIONS

Q1. How many dichloro-substitution products are possible for isobutene? Give their structures and IUPAC names.
Answer:
There are 4 possible dichloro isomers for isobutene (\(2 \text{-methylpropene}\)):
1. 1,1-dichloro-2-methylpropene (\((\text{CH}_3)_2\text{C}=\text{CCl}_2\))
2. 3-chloro-2-(chloromethyl)propene (\(\text{CH}_2=\text{C}(\text{CH}_2\text{Cl})_2\))
3. 1,3-dichloro-2-methylpropene (\(\text{ClCH}_2-\text{C}(\text{CH}_3)=\text{CHCl}\))
4. 1,2-dichloro-2-methylpropane (\(\text{CH}_3-\text{CCl}(\text{CH}_2\text{Cl})-\text{CH}_3\))
In simple words: Substitute two chlorine atoms at different available methyl or vinyl positions.

Exam Tip: Systematically check both saturated and unsaturated sites for possible halogen substitution.

Q2. Although benzene is highly unsaturated, it does not undergo addition reaction. Explain.
Answer: Benzene has a cyclic delocalized \(\pi\)-electron cloud (aromatic stabilization energy of about 150 kJ/mol). Addition reactions would destroy this stable aromatic sextet, converting it into a less stable non-aromatic product. Hence, benzene preferentially undergoes electrophilic substitution reactions that preserve aromaticity.
In simple words: Benzene loves its stable ring structure and avoids reactions that break it.

Exam Tip: Mention resonance energy and preservation of aromatic character.

Q3. Explain why the central carbon-carbon bond in 1,3-butadiene is shorter than that in butane.
Answer: In 1,3-butadiene (\(\text{CH}_2=\text{CH}-\text{CH}=\text{CH}_2\)), the central carbon atom is \(sp^2\) hybridized, whereas in butane, all carbon atoms are \(sp^3\) hybridized. An \(sp^2\) hybridized carbon has greater s-character (33.3%) than an \(sp^3\) hybridized carbon (25%), pulling electrons closer and resulting in a shorter bond length (approx. 146 pm vs 153 pm in butane).
In simple words: Higher s-character makes the orbital shorter and stronger.

Exam Tip: Relate bond length directly to percentage s-character of hybrid orbitals.

Q4. Identify the compound 'A' from the following information: (i) Molecular formula of 'A' is \(\text{C}_5\text{H}_8\) (ii) 'A' reacts with only one mole of \(\text{H}_2\) in the presence of catalyst to form a saturated compound (iii) 'A' on ozonolysis forms \(\text{OHC}-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CHO}\).
Answer:
Compound 'A' is cyclopentene (\(\text{C}_5\text{H}_8\)).
Reasoning: Ozonolysis yielding a 5-carbon dialdehyde (\(\text{OHC}-\text{CH}_2-\text{CH}_2-\text{CH}_2-\text{CHO}\)) indicates a 5-membered cyclic alkene.
In simple words: A ring opening during ozonolysis points to a cyclic alkene structure.

Exam Tip: Ozonolysis cleavage products are key to figuring out double bond positions in unknown hydrocarbons.

Q5. Two cyclic dienes A and B have molecular formula \(\text{C}_6\text{H}_8\). The mixture of the two on reductive ozonolysis gave succinaldehyde, propane-1,3-dial, and glyoxal. Mixture of A and B on hydrogenation produces only cyclohexane. Suggest the structure of A and B.
Answer:
The two cyclic dienes are 1,3-cyclohexadiene (A) and 1,4-cyclohexadiene (B).
In simple words: These conjugated and non-conjugated cyclic dienes yield specific dialdehydes upon ozonolysis.

Exam Tip: Match each ozonolysis fragment back to the original double bond positions.

Q6. Bring out the conversion: (i) Carbon and hydrogen to toluene (ii) Calcium carbide to oxalic acid.
Answer:
(i) Carbon + Hydrogen \(\rightarrow\) Ethyne \(\xrightarrow{\text{red hot Fe tube}}\) Benzene \(\xrightarrow{\text{CH}_3\text{Cl / Anhydrous AlCl}_3}\) Toluene.
(ii) Calcium carbide + Water \(\rightarrow\) Ethyne \(\xrightarrow{\text{dil. } \text{H}_2\text{SO}_4 / \text{Hg}^{2+}}\) Acetaldehyde \(\xrightarrow{\text{[O]}}\) Acetic acid \(\rightarrow\) Chlorination \(\rightarrow\) Hydrolysis \(\rightarrow\) Oxalic acid.
In simple words: Build up carbon skeletons stepwise using classic organic synthesis pathways.

Exam Tip: Learn cyclic polymerization of ethyne to benzene as a cornerstone conversion reaction.

Q7. Primary alkyl halide \(\text{C}_4\text{H}_9\text{Br}\) (a) reacted with alcoholic KOH to give compound (b). Compound (b) is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with sodium metal it gives compound (d), \(\text{C}_8\text{H}_{18}\) which is different from the compound formed when n-butyl bromide is reacted with sodium. Give the structural formula of (a) and write equations for all reactions.
Answer:
Compound (a) is isobutyl bromide: \((\text{CH}_3)_2\text{CHCH}_2\text{Br}\). (Same as Level-2 Question 2).
In simple words: Structural deduction using elimination-addition and Wurtz coupling reactions.

Exam Tip: Practice identifying branched isomers through Wurtz coupling product analysis.

Q8. Out of \(\text{C}_6\text{H}_5\text{CH}_2\text{Cl}\) and \(\text{C}_6\text{H}_5\text{CHClC}_6\text{H}_5\), which is more easily hydrolysed by aqueous KOH?
Answer: \(\text{C}_6\text{H}_5\text{CHClC}_6\text{H}_5\) is more easily hydrolysed by aqueous KOH because it is a secondary alkyl halide adjacent to two phenyl groups, forming a highly stabilized carbocation (diphenylcarbocation stabilized by resonance from both benzene rings).
In simple words: Two benzene rings stabilize the positive charge better than one.

Exam Tip: Greater resonance stabilization of the intermediate carbocation increases hydrolysis speed.

Q9. Explain why: (i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride (ii) alkyl halides, though polar, are immiscible with water (iii) Grignard reagents should be prepared under anhydrous conditions?
Answer:
(i) Due to \(sp^2\) hybridization and electron-withdrawing resonance effect in chlorobenzene.
(ii) Inability to form intermolecular hydrogen bonds with water.
(iii) Reacts violently with traces of moisture to decompose into alkanes.
In simple words: Standard conceptual properties related to polarity, hydrogen bonding, and reactivity.

Q10. Predict all the alkenes that would be formed by dehydrohalogenation of the following halides with sodium ethoxide in ethanol and identify the major alkene: (i) 1-Bromo-1-methylcyclohexane (ii) 2-Chloro-2-methylbutane (iii) 2,2,3-Trimethyl-3-bromopentane.
Answer:
(i) 1-methylcyclohexene (Major product according to Zaitsev's rule).
(ii) 2-methylbut-2-ene (Major) and 2-methylbut-1-ene (Minor).
(iii) 2,3,3-trimethylpent-2-ene and 2,2,3-trimethylpent-1-ene (Major being the more substituted Zaitsev alkene).
In simple words: Zaitsev's rule dictates that the more highly substituted alkene is the major product.

Exam Tip: Always state Zaitsev's rule when predicting elimination reaction major products.