101. Which statement is true for haloarenes in comparison with haloalkanes?
ⓐ. The carbon-halogen bond in haloarenes is less easily cleaved.
ⓑ. The carbon-halogen bond in haloarenes is always longer.
ⓒ. The carbon-halogen bond in haloarenes is purely ionic.
ⓓ. The carbon bonded to halogen in haloarenes is \( sp^3 \)-hybridised.
Correct Answer: The carbon-halogen bond in haloarenes is less easily cleaved.
Explanation: Haloarenes have halogen attached to an aromatic \( sp^2 \)-carbon, and resonance adds partial double-bond character to the \( C-X \) bond. This makes the bond shorter and stronger than the corresponding bond in haloalkanes. A stronger bond is less easily broken during reaction. That is why haloarenes are generally less reactive in bond-cleavage reactions.
102. The reduced tendency of chlorobenzene to undergo bond cleavage is not explained by:
ⓐ. partial double-bond character in the \( C-Cl \) bond
ⓑ. \( sp^2 \)-hybridisation of the ring carbon bonded to chlorine
ⓒ. shorter and stronger \( C-Cl \) bond
ⓓ. greater bond polarity alone
Correct Answer: greater bond polarity alone
Explanation: The major reasons for the lower reactivity of chlorobenzene are resonance, \( sp^2 \)-hybridisation, and the resulting shorter, stronger \( C-Cl \) bond. These features directly affect how easily the bond can break. Bond polarity by itself does not account for the unusual resistance of haloarenes. In fact, the bond remains difficult to cleave despite being polar.
103. Which comparison between chlorobenzene and benzyl chloride is correct?
ⓐ. chlorobenzene: \( sp^3 \); benzyl chloride: \( sp^3 \)
ⓑ. chlorobenzene: \( sp^2 \); benzyl chloride: \( sp^3 \)
ⓒ. chlorobenzene: \( sp^3 \); benzyl chloride: \( sp^2 \)
ⓓ. chlorobenzene: \( sp \); benzyl chloride: \( sp^2 \)
Correct Answer: chlorobenzene: \( sp^2 \); benzyl chloride: \( sp^3 \)
Explanation: In chlorobenzene, chlorine is attached directly to the aromatic ring carbon, which is \( sp^2 \)-hybridised. In benzyl chloride, chlorine is attached to the side-chain \( CH_2 \) carbon, which is \( sp^3 \)-hybridised. This difference strongly affects bond nature and reactivity. It is one of the clearest structural distinctions between haloarenes and benzylic halides.
104. Which factor directly contributes to the shorter \( C-X \) bond in haloarenes?
ⓐ. The halogen is attached to an \( sp^2 \)-hybridised carbon.
ⓑ. The aromatic ring destroys bond polarity.
ⓒ. Haloarenes always contain smaller halogen atoms.
ⓓ. The halogen atom loses all its lone pairs after bonding.
Correct Answer: The halogen is attached to an \( sp^2 \)-hybridised carbon.
Explanation: An \( sp^2 \)-hybridised carbon has greater \( s \)-character than an \( sp^3 \)-hybridised carbon. Greater \( s \)-character holds the bonding electrons closer to the nucleus, which helps shorten the bond. In haloarenes, this contributes to a shorter \( C-X \) bond than in haloalkanes. Resonance then further strengthens that bond.
105. Which statement best summarises the nature of the \( C-X \) bond in haloarenes?
ⓐ. It is a long, weak single bond that breaks more easily than in haloalkanes.
ⓑ. It is a nonpolar bond because the aromatic ring removes the halogen effect.
ⓒ. It is shorter and stronger because of \( sp^2 \) bonding and resonance.
ⓓ. It is identical in length and strength to the \( C-X \) bond in haloalkanes.
Correct Answer: It is shorter and stronger because of \( sp^2 \) bonding and resonance.
Explanation: In haloarenes, the halogen is bonded to an aromatic \( sp^2 \)-carbon, not to an alkyl \( sp^3 \)-carbon. The bond also gains partial double-bond character through resonance between the halogen lone pair and the ring. Together these effects make the bond shorter and stronger. This is why haloarenes show characteristic resistance toward reactions that require easy \( C-X \) bond cleavage.
106. In haloarenes, the halogen atom withdraws electron density from the benzene ring mainly through its:
ⓐ. \( +R \) effect
ⓑ. \( -I \) effect
ⓒ. hyperconjugative effect
ⓓ. \( +M \) effect
Correct Answer: \( -I \) effect
Explanation: Halogen atoms are more electronegative than carbon, so they pull electron density through the sigma bond framework. This electron-withdrawing effect is called the \( -I \) or negative inductive effect. Because of this, the aromatic ring becomes less reactive toward electrophiles than benzene itself. That is why haloarenes are generally deactivating in electrophilic substitution.
107. Which effect of halogen is responsible for increased electron density at the ortho and para positions of a haloarene?
ⓐ. \( -I \) effect
ⓑ. steric effect
ⓒ. bond fission effect
ⓓ. \( +R \) effect
Correct Answer: \( +R \) effect
Explanation: Halogen atoms possess lone pairs that can overlap with the \( \pi \)-system of the benzene ring. This delocalisation donates electron density into the ring by resonance, especially toward the ortho and para positions. That resonance donation is called the \( +R \) effect. It is this effect that makes halogens ortho- and para-directing despite being deactivating overall.
108. Why does the \( C-X \) bond in haloarenes show partial double-bond character?
ⓐ. The lone pair of halogen gets delocalised into the benzene ring.
ⓑ. The halogen atom forms two sigma bonds with carbon.
ⓒ. The aromatic ring becomes fully saturated after halogen substitution.
ⓓ. The halogen atom loses all nonbonding electrons after bonding.
Correct Answer: The lone pair of halogen gets delocalised into the benzene ring.
Explanation: In haloarenes, a lone pair on the halogen can participate in resonance with the aromatic ring. This delocalisation allows canonical forms in which the carbon-halogen bond is shown with some double-bond character. Because of that contribution, the bond is not a simple pure single bond. Its partial double-bond character helps explain its unusual strength and lower reactivity.
109. Haloarenes are deactivating in electrophilic substitution, but they still direct incoming electrophiles to the ortho and para positions. Which statement best explains this?
ⓐ. The \( -I \) effect activates the ortho and para positions more than the meta position.
ⓑ. Haloarenes form carbocations only at ortho and para positions.
ⓒ. \( -I \) deactivates the ring, while \( +R \) favours ortho and para attack.
ⓓ. Halogens donate electrons only through the sigma bond, not by resonance.
Correct Answer: \( -I \) deactivates the ring, while \( +R \) favours ortho and para attack.
Explanation: Two effects operate at the same time in haloarenes. The halogen withdraws electron density by its \( -I \) effect, so the ring becomes less reactive than benzene. At the same time, its lone pair can donate electron density by resonance, and that extra density appears mainly at the ortho and para positions. So the ring is deactivated overall, but orientation still favors ortho and para substitution.
110. In chlorobenzene, resonance causes increased electron density mainly at which ring positions?
ⓐ. only ortho
ⓑ. only para
ⓒ. meta and ortho
ⓓ. ortho and para
Correct Answer: ortho and para
Explanation: When the lone pair of chlorine participates in resonance with the benzene ring, the electron density is redistributed. The resonance forms place extra electron density particularly at the ortho and para positions. This is why electrophiles are directed mainly to those positions. The meta position does not receive the same resonance advantage.
111. Which statement about chlorobenzene is correct?
ⓐ. It is less reactive than benzene but mainly gives ortho and para products.
ⓑ. It is more reactive than benzene because chlorine donates electrons only by resonance.
ⓒ. It is meta-directing because chlorine withdraws electrons inductively.
ⓓ. It is equally reactive as benzene because the two effects of chlorine cancel exactly.
Correct Answer: It is less reactive than benzene but mainly gives ortho and para products.
Explanation: Chlorine withdraws electron density from the benzene ring through the \( -I \) effect, so chlorobenzene reacts more slowly than benzene in electrophilic substitution. However, chlorine also donates electron density by resonance from its lone pair. That resonance donation increases electron density at the ortho and para positions. So the ring is deactivated overall, yet the substitution orientation remains ortho/para.
112. Which one of the following is not a consequence of resonance between halogen lone pair and the aromatic ring?
ⓐ. partial double-bond character in the \( C-X \) bond
ⓑ. shorter and stronger \( C-X \) bond
ⓒ. easy \( C-X \) cleavage like an alkyl halide
ⓓ. electron donation toward ortho and para positions
Correct Answer: easy \( C-X \) cleavage like an alkyl halide
Explanation: Resonance in haloarenes strengthens the \( C-X \) bond rather than weakening it. The bond acquires partial double-bond character, becomes shorter and stronger, and is more difficult to break. The same resonance also helps increase electron density at ortho and para positions. So easy bond cleavage is not a consequence of this resonance effect.
113. Which substitution product is expected to predominate when chlorobenzene undergoes electrophilic substitution?
ⓐ. meta-substituted product mainly
ⓑ. only para-substituted product and no ortho product
ⓒ. mainly ortho- and para-substituted products
ⓓ. only meta- and para-substituted products
Correct Answer: mainly ortho- and para-substituted products
Explanation: The resonance effect of chlorine increases electron density at the ortho and para positions of the ring. As a result, those positions are more favorable for electrophilic attack than the meta position. Although the reaction is slower than in benzene because the ring is deactivated, the orientation still remains ortho/para. Hence ortho- and para-products are formed predominantly.
114. Which statement best describes the combined effect of halogen in a haloarene?
ⓐ. It shows only a \( -I \) effect and therefore must be meta-directing.
ⓑ. It shows only a \( +R \) effect and therefore strongly activates the ring.
ⓒ. It shows no electronic effect because the two effects cancel fully.
ⓓ. It is deactivating but ortho/para-directing.
Correct Answer: It is deactivating but ortho/para-directing.
Explanation: Halogens are unusual substituents on benzene because they exert two different electronic effects at once. Their \( -I \) effect withdraws electron density and reduces overall ring reactivity. Their \( +R \) effect donates electron density through resonance and favors attack at ortho and para positions. This combination explains the apparently contradictory behavior of haloarenes.
115. Which observation supports the view that halogen in a haloarene donates electron density by resonance?
ⓐ. The ring becomes completely saturated after substitution.
ⓑ. Electrophilic substitution occurs preferentially at the meta position.
ⓒ. The carbon-halogen bond behaves as a pure ionic bond.
ⓓ. Ortho and para positions become relatively electron-rich.
Correct Answer: Ortho and para positions become relatively electron-rich.
Explanation: Resonance donation from the halogen lone pair redistributes electron density in the aromatic ring. The resonance forms show that the ortho and para positions gain extra electron density compared with the meta position. That pattern helps explain the directing influence observed in electrophilic substitution. So the enhanced electron density at ortho and para positions is clear evidence of resonance donation.
116. Which haloalkane is expected to undergo nucleophilic substitution most readily when the alkyl group is the same in all cases?
ⓐ. Ethyl fluoride
ⓑ. Ethyl chloride
ⓒ. Ethyl bromide
ⓓ. Ethyl iodide
Correct Answer: Ethyl iodide
Explanation: When the alkyl group remains the same, the ease of nucleophilic substitution depends strongly on the carbon-halogen bond strength and leaving-group ability. The \( C-I \) bond is the weakest among the carbon-halogen bonds in alkyl halides. Iodide ion also leaves more easily than bromide, chloride, or fluoride. Therefore ethyl iodide is usually the most reactive in such substitutions.
117. Which order correctly represents the leaving-group ability of halide ions in nucleophilic substitution reactions?
ⓐ. \( I^- > Br^- > Cl^- > F^- \)
ⓑ. \( F^- > Cl^- > Br^- > I^- \)
ⓒ. \( Br^- > I^- > Cl^- > F^- \)
ⓓ. \( Cl^- > Br^- > I^- > F^- \)
Correct Answer: \( I^- > Br^- > Cl^- > F^- \)
Explanation: A better leaving group is one that can depart more easily and exist more stably after bond cleavage. Iodide ion is the best leaving group among the halides listed, while fluoride is the poorest. This trend matches the weakening of the \( C-X \) bond from \( C-F \) to \( C-I \). So the correct leaving-group order is \( I^- > Br^- > Cl^- > F^- \).
118. Which statement best explains why alkyl fluorides are generally less reactive than alkyl iodides in nucleophilic substitution?
ⓐ. Fluorine is less electronegative than iodine.
ⓑ. Alkyl fluorides contain an ionic \( C-F \) bond.
ⓒ. The \( C-F \) bond is strong and \( F^- \) leaves poorly.
ⓓ. Iodine forms a shorter bond with carbon than fluorine.
Correct Answer: The \( C-F \) bond is strong and \( F^- \) leaves poorly.
Explanation: Two related factors control this comparison. The \( C-F \) bond is very strong, so it is difficult to break during substitution. In addition, \( F^- \) is a poor leaving group compared with \( I^- \). Because alkyl iodides have a weaker bond and a better leaving group, they react more readily.
119. Which one of the following halide ions is the poorest leaving group?
ⓐ. \( I^- \)
ⓑ. \( F^- \)
ⓒ. \( Br^- \)
ⓓ. \( Cl^- \)
Correct Answer: \( F^- \)
Explanation: A poor leaving group resists departure from the carbon atom during substitution. Fluoride ion is the poorest leaving group among the common halides because the \( C-F \) bond is very strong and difficult to cleave. Iodide leaves most easily, followed by bromide and chloride. So \( F^- \) is the least effective leaving group in this series.
120. Which pair correctly matches bond-strength trend and leaving-group trend for alkyl halides?
ⓐ. Bond strength: \( C-I > C-Br > C-Cl > C-F \); Leaving group: \( F^- > Cl^- > Br^- > I^- \)
ⓑ. Bond strength: \( C-F > C-Cl > C-Br > C-I \); Leaving group: \( F^- > Cl^- > Br^- > I^- \)
ⓒ. Bond strength: \( C-F < C-Cl < C-Br Cl^- > Br^- > F^- \)
ⓓ. Bond strength: \( C-F > C-Cl > C-Br > C-I \); Leaving group: \( I^- > Br^- > Cl^- > F^- \)
Correct Answer: Bond strength: \( C-F > C-Cl > C-Br > C-I \); Leaving group: \( I^- > Br^- > Cl^- > F^- \)
Explanation: The carbon-halogen bond becomes weaker as the size of the halogen increases from fluorine to iodine. So the bond-strength order is \( C-F > C-Cl > C-Br > C-I \). At the same time, leaving-group ability improves in the opposite direction, giving \( I^- > Br^- > Cl^- > F^- \). Keeping these two trends separate helps avoid a very common confusion.
