401. The reason phenyl derivatives with electron-releasing groups often give ortho and para substitution is that these positions
ⓐ. are stabilised by electron donation
ⓑ. contain no carbon atoms
ⓒ. are outside the benzene ring
ⓓ. are converted into terminal alkynes
Correct Answer: are stabilised by electron donation
Explanation: Electron-releasing groups can donate electron density into the benzene ring. During electrophilic substitution, the sigma complex formed by ortho or para attack is often stabilised by this donation. This makes ortho and para pathways more favourable than the meta pathway for such substituents. The exact product ratio may also depend on steric effects, especially between ortho and para products. The directing effect is therefore electronic, with steric factors refining the final mixture.
402. Para substitution can be favoured over ortho substitution in some reactions of toluene because the para position has
ⓐ. no connection to the benzene ring
ⓑ. a terminal \( \mathrm{-C\equiv C-H} \) group
ⓒ. stronger ionic bonding than the ortho position
ⓓ. less crowding with the methyl group
Correct Answer: less crowding with the methyl group
Explanation: The methyl group in toluene directs incoming electrophiles mainly to ortho and para positions. Both positions are electronically favoured. However, the ortho positions are closer to the methyl group and may experience more steric crowding when the incoming group is bulky. The para position places the substituents farther apart on the ring. Product distribution therefore depends on both electronic directing effect and steric crowding.
403. A student says, “If a group is deactivating, it must always be meta-directing.” The best correction is that
ⓐ. all deactivating groups remove the benzene ring
ⓑ. halogens are deactivating but usually ortho-para directing
ⓒ. deactivating groups always convert benzene into alkanes
ⓓ. the word deactivating means no substitution can ever occur
Correct Answer: halogens are deactivating but usually ortho-para directing
Explanation: Many deactivating groups, such as \( \mathrm{-NO_2} \), are meta-directing. However, halogens are an important exception. They deactivate the ring by withdrawing electron density through inductive effect, but they direct incoming electrophiles mainly to ortho and para positions through resonance donation. Therefore, rate effect and orientation effect are related but not identical ideas. A substituent must be classified using both reactivity and directing behaviour.
404. The statement that correctly distinguishes activation and directing effect is
ⓐ. activation concerns molecular mass; directing effect concerns boiling point
ⓑ. activation concerns reaction rate; directing effect concerns substitution position
ⓒ. activation applies only to alkanes; directing effect applies only to alkynes
ⓓ. activation means addition; directing effect means combustion
Correct Answer: activation concerns reaction rate; directing effect concerns substitution position
Explanation: In substituted benzene chemistry, activation or deactivation refers to whether the ring reacts faster or slower than benzene toward electrophilic substitution. Directing effect refers to whether the incoming electrophile goes mainly to ortho, meta, or para positions. These ideas often correlate, but they are not exactly the same. Halogens show this clearly because they deactivate the ring but direct ortho-para. Separating rate effect from position effect helps explain many substituted-benzene product patterns.
405. A benzene ring has a \( \mathrm{-CH_3} \) group at position \(1\). During further nitration, the products formed mainly have \( \mathrm{-NO_2} \) at positions
ⓐ. \(2\) and \(4\)
ⓑ. \(3\) only
ⓒ. \(1\) and \(2\)
ⓓ. \(5\) only
Correct Answer: \(2\) and \(4\)
Explanation: A methyl group on benzene is an ortho-para directing group. If \( \mathrm{-CH_3} \) is at position \(1\), the ortho positions are \(2\) and \(6\), while the para position is \(4\). Positions \(2\) and \(6\) are equivalent in monosubstituted toluene, so they represent the same ortho product. The incoming \( \mathrm{-NO_2} \) group therefore appears mainly at ortho and para positions. The meta position \(3\) is not the favoured direction for a methyl-substituted ring.
406. A benzene ring already contains \( \mathrm{-NO_2} \) at position \(1\). If it undergoes chlorination under electrophilic substitution conditions, the major product is expected to have chlorine mainly at
ⓐ. position \(2\)
ⓑ. position \(4\)
ⓒ. position \(1\)
ⓓ. position \(3\)
Correct Answer: position \(3\)
Explanation: The nitro group is a strong electron-withdrawing group. It deactivates the benzene ring and directs incoming electrophiles mainly to the meta position. If \( \mathrm{-NO_2} \) is at position \(1\), the meta positions are \(3\) and \(5\), which are equivalent for monosubstitution. Chlorination therefore gives mainly the meta chloro product. Position \(1\) is already occupied, and ortho or para attack is less favoured because of the destabilising effect of \( \mathrm{-NO_2} \).
407. A substituted benzene has an electron-releasing group already attached. The most reasonable prediction for its reaction with an electrophile is that it will usually
ⓐ. slower than benzene and mainly meta-directing
ⓑ. faster than benzene and mainly ortho/para-directing
ⓒ. only addition across the ring
ⓓ. permanent loss of aromaticity after the first step
Correct Answer: faster than benzene and mainly ortho/para-directing
Explanation: Electron-releasing groups increase electron density in the benzene ring. A more electron-rich ring is generally more reactive toward electrophiles than benzene itself. Such groups commonly stabilise the sigma complex formed during ortho and para attack. Therefore, they usually activate the ring and direct substitution mainly to ortho and para positions. The final product regains aromaticity after loss of \( \mathrm{H^+} \), so the reaction remains substitution rather than permanent addition.
408. Use the arrangement described below for a disubstituted benzene prediction.
Case 1: A methyl group is already at position \(1\).
Case 2: A bulky electrophile attacks the ring.
Both ortho and para positions are electronically favoured. The para product may be favoured more strongly because
ⓐ. the para position gives less steric crowding with the methyl group
ⓑ. the para position is not part of the benzene ring
ⓒ. the methyl group becomes a meta director when the electrophile is bulky
ⓓ. the benzene ring changes into an alkyne before substitution
Correct Answer: the para position gives less steric crowding with the methyl group
Explanation: A methyl group is an ortho-para director, so both ortho and para attack are electronically favoured. However, the ortho positions lie adjacent to the methyl group. A bulky incoming group can experience more crowding at the ortho position. The para position places the two substituents farther apart on the ring. Product distribution can therefore reflect both electronic directing effects and steric crowding.
409. A table compares two substituted benzenes undergoing nitration.
| Starting compound | Group already present | Expected main orientation |
|---|
| P. Toluene | \( \mathrm{-CH_3} \) | ortho and para |
| Q. Nitrobenzene | \( \mathrm{-NO_2} \) | meta |
| R. Chlorobenzene | \( \mathrm{-Cl} \) | ortho and para, slower than benzene |
| S. Nitrobenzene | \( \mathrm{-NO_2} \) | strongly activated ortho and para |
The row that needs correction is
ⓐ. P
ⓑ. Q
ⓒ. R
ⓓ. S
Correct Answer: S
Explanation: Toluene gives mainly ortho and para products because \( \mathrm{-CH_3} \) is electron-releasing. Nitrobenzene gives mainly meta product because \( \mathrm{-NO_2} \) is strongly electron-withdrawing. Chlorobenzene is slower than benzene but still directs mainly to ortho and para positions because halogens show a special directing exception. Row S is incorrect because \( \mathrm{-NO_2} \) does not activate the ring and does not direct mainly ortho and para. The nitro group is a deactivating meta director.
410. Consider the following statements about halogens on benzene.
I. Halogens withdraw electron density by inductive effect.
II. Halogens can donate electron density by resonance to ortho and para positions.
III. Halogens are strongly activating meta directors.
ⓐ. I and III only
ⓑ. II and III only
ⓒ. I and II only
ⓓ. I, II, and III
Correct Answer: I and II only
Explanation: Halogens are electronegative and withdraw electron density from the ring by inductive effect. This makes halobenzenes less reactive than benzene toward electrophilic substitution. At the same time, halogens can donate electron density by resonance to ortho and para positions. This resonance donation explains their ortho-para directing behaviour. Statement III is incorrect because halogens are deactivating, not strongly activating, even though they are ortho-para directing.
411. Assertion: Chlorobenzene gives mainly ortho and para products in nitration, but reacts more slowly than benzene.
Reason: Chlorine withdraws electron density by inductive effect while donating by resonance to ortho and para positions.
ⓐ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Both Assertion and Reason are true, and Reason explains Assertion
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: Chlorobenzene is less reactive than benzene because chlorine withdraws electron density through its inductive effect. This deactivation lowers the rate of electrophilic substitution. However, chlorine can donate electron density by resonance to the ortho and para positions. That resonance effect makes ortho and para substitution more favourable than meta substitution. The Reason explains both parts of the Assertion: slower reaction and ortho-para orientation.
412. A reaction plan needs a major meta-substituted product from electrophilic substitution on an already substituted benzene ring. The most suitable directing group already present is
ⓐ. \( \mathrm{-CH_3} \)
ⓑ. \( \mathrm{-OH} \)
ⓒ. \( \mathrm{-NH_2} \)
ⓓ. \( \mathrm{-NO_2} \)
Correct Answer: \( \mathrm{-NO_2} \)
Explanation: The nitro group, \( \mathrm{-NO_2} \), is a strong electron-withdrawing group. It deactivates the benzene ring and directs incoming electrophiles mainly to the meta position. Groups such as \( \mathrm{-CH_3} \), \( \mathrm{-OH} \), and \( \mathrm{-NH_2} \) usually donate electron density and direct mainly to ortho and para positions. Therefore, choosing \( \mathrm{-NO_2} \) is consistent with a meta-oriented substitution plan. The existing substituent controls the orientation of the incoming electrophile.
413. Polynuclear aromatic hydrocarbons are best described as aromatic compounds that contain
ⓐ. fused or multiple aromatic ring systems
ⓑ. only one carbon atom and four hydrogen atoms
ⓒ. a single open-chain carbon-carbon triple bond only
ⓓ. sodium carboxylate groups as the main skeleton
Correct Answer: fused or multiple aromatic ring systems
Explanation: Polynuclear aromatic hydrocarbons contain two or more aromatic rings, often fused together. They are still hydrocarbons when they contain only carbon and hydrogen. Their extended aromatic systems make them different from simple benzene while keeping the aromatic-ring idea. These compounds are often discussed in connection with incomplete combustion and environmental chemistry. The term does not refer to methane, ordinary alkynes, or sodium salts.
414. Some polycyclic aromatic hydrocarbons are treated as health concerns because certain members are
ⓐ. harmless salts that dissolve completely in water
ⓑ. strong bases used to neutralise acids
ⓒ. carcinogenic or harmful after long exposure
ⓓ. ordinary saturated alkanes with no ring system
Correct Answer: carcinogenic or harmful after long exposure
Explanation: Some polycyclic aromatic hydrocarbons contain fused aromatic ring systems and can be harmful to health. Certain members are associated with carcinogenic risk, especially with repeated or long-term exposure. They may be produced during incomplete combustion of organic materials such as fuels, coal, tobacco, or biomass. This point is a health and environment note linked to hydrocarbon chemistry. The concern is not because they are water-soluble salts, but because of their persistent aromatic structures and biological effects.
415. A smoky flame from incomplete combustion of an aromatic-rich fuel may indicate formation of soot and complex aromatic products. This observation is most closely connected with
ⓐ. complete conversion of all carbon into \( \mathrm{CO_2} \)
ⓑ. high carbon content and incomplete oxidation
ⓒ. addition of \( \mathrm{H_2} \) across every benzene ring
ⓓ. reaction of sodium metal with dry ether
Correct Answer: high carbon content and incomplete oxidation
Explanation: Aromatic hydrocarbons generally have a high carbon percentage compared with many saturated hydrocarbons. When oxygen supply is limited, carbon may not be completely oxidised to \( \mathrm{CO_2} \). Soot, carbon particles, \( \mathrm{CO} \), and complex aromatic by-products may form under such conditions. A smoky flame is therefore linked with incomplete combustion. The observation is not a sign of hydrogenation or Wurtz coupling.
416. A table connects combustion conditions with possible products.
| Condition | Likely product pattern |
|---|
| P. Sufficient oxygen | mainly \( \mathrm{CO_2} \) and \( \mathrm{H_2O} \) |
| Q. Limited oxygen | \( \mathrm{CO} \), soot, or partially oxidised products may form |
| R. Aromatic-rich fuel with poor air supply | smokier flame is more likely |
| S. Sufficient oxygen | only metal acetylides form |
The row that needs correction is
ⓐ. P
ⓑ. S
ⓒ. Q
ⓓ. R
Correct Answer: S
Explanation: In sufficient oxygen, hydrocarbons burn completely to form \( \mathrm{CO_2} \) and \( \mathrm{H_2O} \), so row P is suitable. Limited oxygen can produce \( \mathrm{CO} \), soot, and other incomplete-combustion products, so row Q is suitable. Aromatic-rich fuels can burn with a smoky flame when air is insufficient, so row R is also suitable. Row S is incorrect because metal acetylides form from terminal alkynes with suitable metal-ion reagents, not from complete combustion. Combustion products depend mainly on oxygen supply.
417. A hydrocarbon sample decolourises bromine water and also decolourises cold dilute alkaline \( \mathrm{KMnO_4} \). The most likely conclusion is that the sample contains
ⓐ. only saturated \( \mathrm{C-C} \) single bonds
ⓑ. a sodium carboxylate group
ⓒ. carbon-carbon unsaturation
ⓓ. no carbon atoms
Correct Answer: carbon-carbon unsaturation
Explanation: Bromine water is decolourised when bromine adds across carbon-carbon multiple bonds under suitable conditions. Cold dilute alkaline \( \mathrm{KMnO_4} \) can also react with unsaturated bonds, such as \( \mathrm{C=C} \), during mild oxidation. If both tests are positive, the sample most likely contains carbon-carbon unsaturation. This observation may fit an alkene and can also be shown by alkynes under suitable conditions. The conclusion should be about unsaturation, not automatically about only one specific hydrocarbon family.
418. A terminal alkyne and an internal alkyne both decolourise bromine solution, but only the terminal alkyne gives a precipitate with ammoniacal silver nitrate. The difference is due to
ⓐ. the acidic hydrogen on terminal \(sp\)-carbon
ⓑ. the presence of oxygen in terminal alkynes
ⓒ. the absence of a triple bond in terminal alkynes
ⓓ. complete combustion of the internal alkyne
Correct Answer: the acidic hydrogen on terminal \(sp\)-carbon
Explanation: Both terminal and internal alkynes contain a carbon-carbon triple bond, so both can undergo addition reactions with bromine. The silver nitrate test, however, depends on the presence of an acidic terminal alkyne hydrogen. A terminal alkyne has the \( \mathrm{-C\equiv C-H} \) group, while an internal alkyne does not. The terminal hydrogen can be replaced by silver ion to form a silver acetylide precipitate. The two tests therefore detect different structural features.
419. A qualitative test record is given below.
Unknown \(X\) decolourises bromine water. Unknown \(X\) does not give a precipitate with ammoniacal \( \mathrm{AgNO_3} \). The molecular formula of \(X\) is \( \mathrm{C_4H_6} \), and it is known to be an alkyne.
The most suitable identity of \(X\) is
ⓐ. but-\(1\)-yne
ⓑ. butane
ⓒ. benzene
ⓓ. but-\(2\)-yne
Correct Answer: but-\(2\)-yne
Explanation: The formula \( \mathrm{C_4H_6} \) fits an open-chain monoalkyne with four carbon atoms. Bromine decolourisation supports the presence of unsaturation. The absence of a precipitate with ammoniacal \( \mathrm{AgNO_3} \) shows that the alkyne is not terminal. But-\(2\)-yne, \( \mathrm{CH_3C\equiv CCH_3} \), is an internal alkyne and lacks a terminal acidic hydrogen. But-\(1\)-yne would be expected to give the silver acetylide test.
420. Match the hydrocarbon observation with the most suitable inference.
| Observation | Inference |
|---|
| P. Bromine water decolourised | 1. Unsaturation may be present |
| Q. Cold alkaline \( \mathrm{KMnO_4} \) decolourised | 2. Mild oxidation of unsaturation may occur |
| R. Ammoniacal \( \mathrm{AgNO_3} \) gives precipitate | 3. Terminal alkyne group may be present |
ⓐ. P-2, Q-3, R-1
ⓑ. P-3, Q-1, R-2
ⓒ. P-1, Q-2, R-3
ⓓ. P-1, Q-3, R-2
Correct Answer: P-1, Q-2, R-3
Explanation: Bromine water decolourisation is commonly linked with addition to carbon-carbon multiple bonds. Cold alkaline \( \mathrm{KMnO_4} \) can react with unsaturation through mild oxidation and is used in Baeyer-type testing. Ammoniacal \( \mathrm{AgNO_3} \) gives a precipitate with terminal alkynes because of their acidic terminal hydrogen. The three observations do not all test the same feature. Interpreting them together gives a more reliable structural clue than using one observation alone.
