CBSE Class 12 Chemistry Haloalkanes and Haloarenes MCQs Set 09

Question 1. Which one of the following has the highest dipole moment?
(a) \( \text{CH}_3\text{F} \)
(b) \( \text{CH}_3\text{Cl} \)
(c) \( \text{CH}_3\text{I} \)
(d) \( \text{CH}_3\text{Br} \)
Answer: (b) \( \text{CH}_3\text{Cl} \)
In simple words: Even though fluorine has a stronger pull on electrons, its bond with carbon is very short. This makes the overall dipole moment of methyl chloride slightly larger than methyl fluoride.

Exam Tip: Remember that dipole moment is charge multiplied by distance, so bond length plays a crucial role in deciding the order between fluoride and chloride.

Question 2. What should be the correct IUPAC name for diethylbromomethane?
(a) 1-bromo-1, 1-diethylmethane
(b) 3-bromopentane
(c) 1-bromo-1-ethylpropane
(d) 1-bromopentane
Answer: (b) 3-bromopentane
In simple words: Diethylbromomethane consists of a central carbon bonded to a bromine and two ethyl groups, which forms a five-carbon chain with bromine at the third carbon.

Exam Tip: Always find the longest continuous carbon chain first to name any branched alkyl halide.

Question 3. The IUPAC name of benzene hexachloride is
(a) 1, 2, 3, 4, 5, 6-hexachlorocyclohexane
(b) 1, 1, 1, 6, 6, 6-hexachlorocyclohexane
(c) 1, 6-phenyl-1, 6-chlorohexane
(d) 1, 1-phenyl-6, 6-chlorohexane
Answer: (a) 1, 2, 3, 4, 5, 6-hexachlorocyclohexane
In simple words: This compound consists of a ring with six carbons where each carbon is bonded to one chlorine atom, forming a saturated cyclohexane ring.

Exam Tip: Do not get confused by the word 'benzene' in the common name, as the actual molecule is a saturated cyclohexane ring.

Question 4. Match the Column I with Column II and choose the correct option using the codes given below.

Codes
(a) A-(i), B-(ii), C-(iii)
(b) A-(ii), B-(i), C-(iii)
(c) A-(ii), B-(iii), C-(i)
(d) A-(i), B-(iii), C-(ii)
Answer: (c) A-(ii), B-(iii), C-(i)
In simple words: Methylene chloride is dichloromethane, vinyl chloride is chloroethene, and allyl bromide is 3-bromopropene.
Exam Tip: Practice common names alongside IUPAC names for famous chlorinated solvents and alkenes.

Question 5. Which of the following reagent produces pure alkyl halides when heated with alcohols?
(a) \( \text{PCl}_5 \)
(b) \( \text{PCl}_3 \)
(c) \( \text{SOCl}_2 \)
(d) Dry \( \text{HCl} \)
Answer: (c) \( \text{SOCl}_2 \)
In simple words: When thionyl chloride reacts with an alcohol, the side products formed are sulfur dioxide and hydrochloric acid, which are both gases and fly away easily, leaving behind the pure alkyl halide.

Exam Tip: Thionyl chloride is the most preferred reagent because it avoids any tedious purification steps.

Question 6. Toluene reacts with a halogen in the presence of iron (III) chloride giving ortho and para halo compounds. The reaction is
(a) electrophilic elimination reaction
(b) electrophilic substitution reaction
(c) free radical addition reaction
(d) nucleophilic substitution reaction
Answer: (b) electrophilic substitution reaction
In simple words: The iron (III) chloride helper acts as a Lewis acid to produce a positive halogen ion (electrophile) that attacks the electron-rich benzene ring.

Exam Tip: Recall that reactions on the aromatic ring itself in the presence of a Lewis acid catalyst are always electrophilic substitutions.

Question 7. The reaction of toluene with \( \text{Cl}_2 \) in presence of \( \text{FeCl}_3 \) gives 'X', while the reaction of toluene with \( \text{Cl}_2 \) in presence of light gives 'Y'. Thus, 'X' and 'Y' are
(a) X = benzyl chloride, Y = o- and p-chlorotoluene
(b) X = m-chlorotoluene, Y = p-chlorotoluene
(c) X = o- and p-chlorotoluene, Y = trichloromethylbenzene
(d) X = benzyl chloride, Y = m-chlorotoluene
Answer: (c) X = o- and p-chlorotoluene, Y = trichloromethylbenzene
In simple words: Using a Lewis acid like iron (III) chloride places chlorine atoms on the ortho and para positions of the ring, whereas using light and heat replaces all side-chain hydrogens with chlorine.

Exam Tip: Remember that light and heat promote side-chain free radical substitution, while Lewis acids promote ring electrophilic substitution.

Question 9. Which reagent is required for one step conversion of benzene diazonium chloride to bromobenzene?
(a) \( \text{PBr}_3 \)
(b) \( \text{HBr} \)
(c) \( \text{Cu}_2\text{Br}_2 \)
(d) \( \text{Br}_2 \)
Answer: (c) \( \text{Cu}_2\text{Br}_2 \)
In simple words: Copper(I) bromide is used in Sandmeyer's reaction to substitute the diazonium group on the benzene ring with a bromine atom.

Exam Tip: Sandmeyer's reaction uses cuprous halides (\( \text{Cu}_2\text{Cl}_2 \) or \( \text{Cu}_2\text{Br}_2 \)) to prepare haloarenes from diazonium salts.

Question 10. Arrange the following compounds in increasing order of their boiling points:
(i) \( \text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}_2\text{Br} \)
(ii) \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} \)
(iii) \( (\text{CH}_3)_3\text{C}-\text{Br} \)
The correct order is
(a) (ii) < (i) < (iii)
(b) (i) < (ii) < (iii)
(c) (iii) < (i) < (ii)
(d) (iii) < (ii) < (i)
Answer: (c) (iii) < (i) < (ii)
In simple words: More branching in a molecule reduces its surface area and weakens its van der Waals forces, leading to a lower boiling point.

Exam Tip: For isomeric alkyl halides, the boiling point order is always tertiary < secondary < primary.

Question 12. Which one of the following undergoes nucleophilic substitution exclusively by \( \text{S}_{\text{N}}1 \) mechanism?
(a) Ethyl chloride
(b) Isopropyl chloride
(c) Chlorobenzene
(d) Benzyl chloride
Answer: (d) Benzyl chloride
In simple words: Benzyl chloride undergoes substitution exclusively via the \( \text{S}_{\text{N}}1 \) route because it generates a highly stable carbocation that is protected by resonance.

Exam Tip: Carbocation stability is the key factor that determines whether a reaction prefers the \( \text{S}_{\text{N}}1 \) mechanism.

Question 13. Which alkyl halide from the given options will undergo \( \text{S}_{\text{N}}1 \) reaction faster?
(a) \( (\text{CH}_3)_3\text{C}-\text{Br} \)
(b) \( (\text{CH}_3)_2\text{CH}-\text{Br} \)
(c) \( \text{CH}_3-\text{CH}_2-\text{Br} \)
(d) \( (\text{CH}_3)_3\text{C}-\text{CH}_2-\text{Br} \)
Answer: (a) \( (\text{CH}_3)_3\text{C}-\text{Br} \)
In simple words: Tertiary butyl bromide forms a highly stable tertiary carbocation, allowing it to undergo \( \text{S}_{\text{N}}1 \) hydrolysis at the fastest speed.

Exam Tip: For \( \text{S}_{\text{N}}1 \) reactions, the reactivity order of alkyl halides is tertiary > secondary > primary.

Question 14. The order of reactivity of following alcohols with halogen acids is ...
(A) \( \text{CH}_3\text{CH}_2-\text{CH}_2-\text{OH} \)
(B) \( \text{CH}_3\text{CH}_2-\text{CH(OH)}-\text{CH}_3 \)
(C) \( (\text{CH}_3)_2\text{C(OH)}-\text{CH}_2\text{CH}_3 \)
(a) (A) > (B) > (C)
(b) (C) > (B) > (A)
(c) (B) > (A) > (C)
(d) (A) > (C) > (B)
Answer: (b) (C) > (B) > (A)
In simple words: Alcohols react with halogen acids through a carbocation intermediate. Since tertiary carbocations are the most stable, tertiary alcohols react the fastest.

Exam Tip: The reactivity of alcohols towards Lucas reagent or halogen acids follows the order of carbocation stability: 3° > 2° > 1°.

Question 15. Alkyl halides undergoing nucleophilic bimolecular substitution reaction involve
(a) retention of configuration
(b) formation of racemic mixture
(c) inversion of configuration
(d) formation of carbocation
Answer: (c) inversion of configuration
In simple words: In a nucleophilic bimolecular substitution (\( \text{S}_{\text{N}}2 \)) reaction, the nucleophile attacks from the opposite side of the leaving group, leading to Walden inversion.

Exam Tip: Always associate \( \text{S}_{\text{N}}2 \) reactions with backside attack and complete inversion of configuration.

Question 16. The table given below shows some of the features of \( \text{S}_{\text{N}}1 \) and \( \text{S}_{\text{N}}2 \) reaction mechanism.

Which of the rows shows an incorrect feature for at least one of the mechanisms?
(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)
Answer: (c) (iii)
In simple words: A good leaving group makes it easier for both \( \text{S}_{\text{N}}1 \) and \( \text{S}_{\text{N}}2 \) reactions to happen, so row (iii) contains an incorrect statement.
Exam Tip: Remember that a better leaving group always accelerates both substitution mechanisms because it departs more easily.

Question 17. Which of the following molecules has a chiral centre correctly labelled with an asterisk (*)?
(a) \( \text{CH}_3\text{C}^*\text{HBrCH}_3 \)
(b) \( \text{CH}_3\text{C}^*\text{HClCH}_2\text{Br} \)
(c) \( \text{HOCH}_2\text{C}^*\text{H(OH)CH}_2\text{OH} \)
(d) \( \text{CH}_3\text{C}^*\text{Br}_2\text{CH}_3 \)
Answer: (b) \( \text{CH}_3\text{C}^*\text{HClCH}_2\text{Br} \)
In simple words: For a carbon to be chiral, it must be bonded to four different groups. In option (b), the carbon with the asterisk is bonded to a hydrogen, a chlorine, a methyl group, and a \( -\text{CH}_2\text{Br} \) group.

Exam Tip: Check that all four attachments on the candidate carbon are completely distinct to confirm chirality.

Question 18. Which of the following compounds will undergo racemisation when solution of KOH hydrolyses?
(i) \( \text{C}_6\text{H}_4(\text{CH}_3)(\text{CH}_2\text{Cl}) \)
(ii) \( \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} \)
(iii) \( \text{CH}_3\text{CH(CH}_3)\text{CH}_2\text{Cl} \)
(iv) \( \text{H}-\text{C(Cl)(CH}_3)(\text{C}_2\text{H}_5) \)
Choose the correct option.
(a) (i) and (ii)
(b) (ii) and (iv)
(c) (iii) and (iv)
(d) Only (iv)
Answer: (d) Only (iv)
In simple words: Only chiral compounds that undergo nucleophilic substitution can produce a racemic mixture. Among the choices, only compound (iv) contains a chiral carbon atom.

Exam Tip: Racemisation occurs during \( \text{S}_{\text{N}}1 \) hydrolysis of optically active alkyl halides.

Question 19. Which of the following compounds is expected to be optically active?
(a) \( \text{CH}_3\text{CH}_2\text{CHBrCHO} \)
(b) \( \text{CH}_3\text{CH}_2\text{CBr}_2\text{CHO} \)
(c) \( \text{CH}_3\text{CH}_2\text{CH}_2\text{CHO} \)
(d) \( (\text{CH}_3)_2\text{CHCHO} \)
Answer: (a) \( \text{CH}_3\text{CH}_2\text{CHBrCHO} \)
In simple words: Option (a) has a chiral carbon atom bonded to four different groups (\( -\text{H} \), \( -\text{Br} \), \( -\text{CHO} \), and \( -\text{CH}_2\text{CH}_3 \)), which makes the molecule optically active.

Exam Tip: Look for a carbon with four unique groups to quickly identify optically active molecules.

Question 20. Identify A, B, C and D
\( \text{C}_2\text{H}_5\text{Cl} \) reactions:
- with \( \text{KCN} \rightarrow \text{D} \)
- with \( \text{AgCN} \rightarrow \text{C} \)
- with \( \text{Alc. KOH} \rightarrow \text{A} \)
- with \( \text{Aq. KOH} \rightarrow \text{B} \)
Options:
(a) A = C2H4, B = C2H5OH, C = C2H5NC, D = C2H5CN
(b) A = C2H5OH, B = C2H4, C = C2H5CN, D = C2H5NC
(c) A = C2H4, B = C2H5OH, C = C2H5CN, D = C2H5NC
(d) A = C2H5OH, B = C2H4, C = C2H5NC, D = C2H5CN
Answer: (a) A = C2H4, B = C2H5OH, C = C2H5NC, D = C2H5CN
In simple words: Alcoholic KOH causes elimination to give ethene, aqueous KOH replaces the chlorine with an alcohol group, KCN forms ethyl cyanide, and AgCN produces ethyl isocyanide.

Exam Tip: Remember that KCN is ionic and gives cyanides, while AgCN is covalent and gives isocyanides as the major product.

Question 21. Consider the following table.

'X', 'Y' and 'Z' in the above table are
(a) X = phenol, Y = Iodobenzene, Z = Silver fluoride
(b) X = Benzene, Y = Phenyl magnesium iodide, Z = Sodium fluoride
(c) X = Phenol, Y = Chlorobenzene, Z = Silver fluoride
(d) X = Benzene, Y = Iodobenzene, Z = Sodium fluoride
Answer: (a) X = phenol, Y = Iodobenzene, Z = Silver fluoride
In simple words: Phenol is produced via Dow's process, diphenyl is formed through Ullmann reaction using iodobenzene with copper, and fluoromethane is prepared using Swarts reaction with silver fluoride.
Exam Tip: Review named reactions like Dow's process, Ullmann reaction, and Swarts reaction, as they are frequently tested.

Question 22. Chlorination of toluene in presence of light and heat followed by treatment with aqueous NaOH gives
(a) benzoic acid
(b) 2,4-dihydroxytoluene
(c) p-cresol
(d) o-cresol
Answer: (a) benzoic acid
In simple words: Excess chlorine in the presence of sunlight converts the methyl group of toluene into a trichloromethyl group. Treating this intermediate with aqueous NaOH yields benzoic acid after the loss of water.

Exam Tip: Remember that side-chain chlorination of toluene with excess chlorine produces benzotrichloride, which hydrolyses to benzoic acid.

Question 25. Auto-oxidation of chloroform in air and light produces a poisonous gas known as
(a) phosphine
(b) mustard gas
(c) phosgene
(d) tear gas
Answer: (c) phosgene
In simple words: Chloroform slowly reacts with oxygen in the presence of light to form carbonyl chloride, which is commonly called phosgene, an extremely poisonous gas.

Exam Tip: To prevent this toxic oxidation, chloroform is stored in dark, closed bottles filled to the brim.

Question 26. Which of the following statement is incorrect for chloroform?
(a) Breathing about 900 ppm of chloroform can cause dizziness.
(b) Chloroform has no effect on central nervous system.
(c) People develop sores when skin is immersed in chloroform.
(d) Chloroform exposure causes damage to liver and kidney.
Answer: (b) Chloroform has no effect on central nervous system.
In simple words: Chloroform acts as an anesthetic and directly depresses the central nervous system, so saying it has no effect is completely incorrect.

Exam Tip: Chloroform was historically used as a general anesthetic but is now avoided due to its toxic effects on the liver and kidneys.

Assertion-Reason

Directions In the following questions, an Assertion (A) is followed by a corresponding. Reason (R).
Use the following keys to choose the appropriate answer,
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A),
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A),
(c) (A) is true, but (R) is false,
(d) (A) is false, but (R) is true.

Question 27. Assertion (A) Phosphorus chlorides (tri and penta) are preferred over thionyl chloride for the preparation of alkyl chlorides from alcohols.
Reason (R) Thionyl chloride give pure alkyl halides.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (d) (A) is false, but (R) is true.
In simple words: Thionyl chloride is actually preferred over phosphorus chlorides because its gaseous side products (\( \text{SO}_2 \) and \( \text{HCl} \)) escape easily, giving pure alkyl halides without extra purification steps.

Exam Tip: Thionyl chloride is the best reagent for converting alcohols to alkyl chlorides because the gaseous byproducts make isolation of the main product simple.

Question 28. Assertion (A) neo-pentyl chloride is formed when neo-pentyl alcohol reacts with HCl.
Reason (R) neo-pentyl alcohol is a primary alcohol.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (d) (A) is false, but (R) is true.
In simple words: Even though neo-pentyl alcohol is a primary alcohol, the carbocation formed undergoes a methyl shift rearrangement to produce 2-chloro-2-methylbutane instead of neo-pentyl chloride.

Exam Tip: Watch out for carbocation rearrangements when primary alcohols with adjacent branched carbons react with haloacids.

Question 29. Assertion (A) Aryl iodides can be prepared by reaction of arenes with iodine in the presence of an oxidising agent.
Reason (R) Oxidising agent oxidises \( \text{I}_2 \) into HI.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (c) (A) is true, but (R) is false.
In simple words: An oxidising agent is required to oxidise the byproduct \( \text{HI} \) back to \( \text{I}_2 \), which prevents the reversible reaction from moving backward. The oxidising agent does not oxidise \( \text{I}_2 \) to \( \text{HI} \).

Exam Tip: Strong oxidising agents like nitric acid or iodic acid are essential during iodination to destroy the reducing agent \( \text{HI} \) as soon as it forms.

Question 30. Assertion (A) The boiling points of alkyl halides decreases in the order RI > RBr > RCl > RF.
Reason (R) The boiling points of alkyl chlorides, bromides and iodides are considerably higher than that of the hydrocarbon of comparable molecular mass.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
In simple words: Both statements are correct, but the reason does not explain why iodides have higher boiling points than fluorides. That specific order is due to the increasing size and mass of the halogens, which strengthens van der Waals interactions.

Exam Tip: van der Waals forces increase with the polarizability and mass of the halogen atom, which directly dictates the boiling point trend.

Question 31. Assertion (A) Alkyl halides are insoluble in water.
Reason (R) Alkyl halides have halogen attached to \( \text{sp}^3 \)-hybridised carbon.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
In simple words: Both statements are correct, but the hybridisation of the carbon has nothing to do with water solubility. Alkyl halides are insoluble because they cannot form strong hydrogen bonds with water molecules.

Exam Tip: Water solubility requires the solute to break the original hydrogen-bonding network of water and build new hydrogen bonds, which alkyl halides cannot do.

Question 32. Assertion (A) KCN reacts with methyl chloride to give methyl isocyanide.
Reason (R) \( \text{CN}^- \) is an ambident nucleophile.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (d) (A) is false, but (R) is true.
In simple words: KCN is an ionic compound that releases cyanide ions, which attack through carbon to give methyl cyanide instead of methyl isocyanide.

Exam Tip: Cyanide is indeed an ambident nucleophile, but it prefers attacking through carbon because the resulting C-C bond is thermodynamically more stable than a C-N bond.

Question 33. Assertion (A) Nucleophilic substitution of iodoethane is easier than chloroethane.
Reason (R) Bond enthalpy of C-I bond is less than that of C-Cl bond.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
In simple words: The C-I bond is much longer and weaker than the C-Cl bond, meaning it breaks much more easily during a chemical reaction.

Exam Tip: The bond strength of carbon-halogen bonds decreases as the size of the halogen increases, making iodides the most reactive.

Question 34. Assertion (A) Hydrolysis of (-)-2-bromooctane proceeds with inversion of configuration.
Reason (R) This reaction proceeds through the formation of a carbocation.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (c) (A) is true, but (R) is false.
In simple words: The hydrolysis of secondary 2-bromooctane with strong base proceeds via the \( \text{S}_{\text{N}}2 \) pathway, which causes inversion. Since it is \( \text{S}_{\text{N}}2 \), no carbocation intermediate is formed.

Exam Tip: Watch out for mechanisms; inversion of configuration is the hallmark of the \( \text{S}_{\text{N}}2 \) process, which has a single-step transition state rather than a carbocation intermediate.

Question 35. Assertion (A) Inversion of configuration is observed in \( \text{S}_{\text{N}}2 \) reaction.
Reason (R) The reaction proceeds with the formation of carbocation.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (c) (A) is true, but (R) is false.
In simple words: Inversion occurs because the nucleophile attacks from the backside in a single step. No carbocation is generated during this process.

Exam Tip: Remember that carbocations are intermediates of \( \text{S}_{\text{N}}1 \) reactions, which generally lead to racemisation, not exclusive inversion.

Question 36. Assertion (A) Inversion of configuration is observed when 1-bromobutane is hydrolysed.
Reason (R) The reaction is \( \text{S}_{\text{N}}2 \) and proceeds with the formation of transition state.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
In simple words: 1-bromobutane is a primary alkyl halide and undergoes hydrolysis via the \( \text{S}_{\text{N}}2 \) pathway, which goes through a transition state and results in inversion of configuration.

Exam Tip: Primary alkyl halides consistently undergo nucleophilic substitution via the \( \text{S}_{\text{N}}2 \) mechanism due to minimal steric hindrance.

Question 37. Assertion (A) Nitration of chlorobenzene leads to the formation of m-nitrochlorobenzene.
Reason (R) \( -\text{NO}_2 \) group is a m-directing group.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (d) (A) is false, but (R) is true.
In simple words: Chlorine is an ortho/para director because of its resonance effect. Therefore, nitration of chlorobenzene yields ortho- and para-nitrochlorobenzene instead of the meta isomer.

Exam Tip: Halogens on a benzene ring are deactivating but ortho/para-directing due to the presence of lone pairs that can participate in resonance.

Question 38. Assertion (A) Presence of a nitro group at ortho or para position increases the reactivity of haloarenes towards nucleophilic substitution.
Reason (R) Nitro group, being an electron withdrawing group decreases the electron density over the benzene ring.
(a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
(c) (A) is true, but (R) is false.
(d) (A) is false, but (R) is true.
Answer: (a) Both (A) and (R) are true, and (R) is the correct explanation of (A).
In simple words: The strong electron-withdrawing nitro group pulls electron density away from the benzene ring, making it much easier for an incoming nucleophile to attack the ring.

Exam Tip: Electron-withdrawing groups make the benzene ring less electron-rich, thereby facilitating nucleophilic aromatic substitution reactions.

Case-Study

Question 39. Read the given passage and answer the questions that follow.
Observe the table in which physical data of halomethanes, polyhalogen derivatives and haloarenes is given. Answer the questions based on them.
Some physical data of halomethanes, polyhalogen derivatives and haloarenes.

(i) Why is melting point of p-dichlorobenzene higher than o-dichlorobenzene but boiling point of o-isomer is higher than p-isomer?
(ii) Why is dipole moment of \( \text{CH}_3\text{F} \) less than \( \text{CH}_3\text{Cl} \)? Or Why is \( \text{CH}_3\text{I} \) more reactive than \( \text{CH}_3\text{Cl} \)?
(iii) Why does \( \text{C}_6\text{H}_5\text{I} \) has higher dipole moment than \( \text{C}_6\text{H}_5\text{F} \)?
Answer:
(i) The para isomer is highly symmetrical, allowing it to pack more efficiently into the crystal lattice, which yields a higher melting point [39]. Conversely, the ortho isomer is more polar and has a higher dipole moment, resulting in stronger intermolecular attractions and a higher boiling point [39].
(ii) Dipole moment is calculated as the product of charge and bond distance [1, 39]. Even though fluorine is more electronegative than chlorine [1], the short length of the C-F bond makes the overall dipole moment of methyl fluoride slightly smaller than that of methyl chloride [1, 39].
Or: Methyl iodide is more reactive than methyl chloride because the C-I bond is much longer and has a lower bond dissociation energy than the C-Cl bond [39].
(iii) Dipole moment is determined by the equation \( \mu = e \times d \) [39]. Although fluorine is more electronegative, the carbon-iodine bond is significantly longer than the carbon-fluorine bond, which compensates for the charge difference and results in a larger overall dipole moment [39].
In simple words: Symmetrical shapes help molecules pack closely in solids, raising the melting point. Dipole moment depends on both electrical charge and the distance between atoms, so bond length is highly important.
Exam Tip: Always describe crystal lattice packing when explaining melting point differences among aromatic isomers, and address polarity when discussing boiling points.