501. Which equation best represents bromination of aniline with bromine water?
ⓐ. \(C_6H_5NH_2 + 3Br_2 \rightarrow C_6H_2Br_3NH_2 + 3HBr\)
ⓑ. \(C_6H_5NH_2 + Br_2 \rightarrow C_6H_5Br + NH_3\)
ⓒ. \(C_6H_5NH_2 + 2Br_2 \rightarrow C_6H_3Br_2NH_2 + 2HBr\)
ⓓ. \(C_6H_5NH_2 + HBr \rightarrow C_6H_5Br + NH_4Br\)
Correct Answer: \(C_6H_5NH_2 + 3Br_2 \rightarrow C_6H_2Br_3NH_2 + 3HBr\)
Explanation: \(\textbf{Starting compound:}\)
Aniline is \(C_6H_5NH_2\).
\(\textbf{Reaction type:}\)
Bromination of the activated ring occurs by electrophilic substitution.
\(\textbf{Number of substitutions:}\)
The strongly activating \(-NH_2\) group causes substitution at two ortho positions and one para position.
\(\textbf{Hydrogen replacement:}\)
Three ring hydrogens are replaced by three bromine atoms.
\(\textbf{Final Answer:}\)
\[\boxed{C_6H_5NH_2 + 3Br_2 \rightarrow C_6H_2Br_3NH_2 + 3HBr}\]
502. Why does aniline give \(2,4,6\)-tribromoaniline with bromine water instead of mainly monobromoaniline?
ⓐ. Bromine water first converts aniline into benzene.
ⓑ. Strong \(-NH_2\) activation causes repeated substitution.
ⓒ. The ring in aniline is deactivated by the nitrogen lone pair.
ⓓ. The amino group directs bromine only to the meta position.
Correct Answer: Strong \(-NH_2\) activation causes repeated substitution.
Explanation: The \(-NH_2\) group donates electron density into the benzene ring. This makes the ring much more reactive than benzene itself. After one bromine atom enters the ring, the remaining activated positions can still react readily. Therefore, bromination does not stop at monobromination in aqueous bromine and gives \(2,4,6\)-tribromoaniline.
503. Which statement about bromination of aniline is accurate?
ⓐ. It requires anhydrous \(AlCl_3\) as an essential catalyst.
ⓑ. It gives only \(m\)-bromoaniline because \(-NH_2\) is meta-directing.
ⓒ. It occurs readily with bromine water.
ⓓ. It removes the amino group and forms bromobenzene directly.
Correct Answer: It occurs readily with bromine water.
Explanation: Aniline contains the strongly activating \(-NH_2\) group. The nitrogen lone pair increases electron density in the aromatic ring, especially at ortho and para positions. Because of this activation, bromination occurs readily with bromine water and does not require a Lewis acid catalyst. The reaction gives a highly substituted product rather than simple bromobenzene.
504. In the bromination of aniline with bromine water, which ring positions are substituted?
ⓐ. Only \(3\)-position
ⓑ. \(2\)- and \(6\)-positions only
ⓒ. \(3\)- and \(5\)-positions
ⓓ. \(2\)-, \(4\)-, and \(6\)-positions
Correct Answer: \(2\)-, \(4\)-, and \(6\)-positions
Explanation: \(\textbf{Directing group:}\)
The \(-NH_2\) group is ortho-para directing.
\(\textbf{Numbering:}\)
In aniline, the carbon attached to \(-NH_2\) is carbon \(1\).
\(\textbf{Activated positions:}\)
The ortho positions are \(2\) and \(6\), and the para position is \(4\).
\(\textbf{Product formed:}\)
Bromination with bromine water gives substitution at \(2\), \(4\), and \(6\).
\(\textbf{Final Answer:}\)
The substituted positions are \(2\), \(4\), and \(6\).
505. Which product is expected when aniline is treated with excess bromine water?
ⓐ. \(C_6H_2Br_3NH_2\)
ⓑ. \(C_6H_5Br\)
ⓒ. \(C_6H_5CH_2Br\)
ⓓ. \(C_6H_5N_2^+Br^-\)
Correct Answer: \(C_6H_2Br_3NH_2\)
Explanation: Excess bromine water reacts with aniline at multiple activated ring positions. The \(-NH_2\) group directs substitution to ortho and para positions. Three bromine atoms enter the ring at \(2\), \(4\), and \(6\). Therefore, the product formula is \(C_6H_2Br_3NH_2\).
506. Which comparison between benzene and aniline in bromination is correct?
ⓐ. Benzene reacts faster because it has an \(-NH_2\) group.
ⓑ. Aniline reacts readily.
ⓒ. Both react equally because substituents never affect benzene rings.
ⓓ. Aniline reacts only after conversion into nitrobenzene.
Correct Answer: Aniline reacts readily.
Explanation: Benzene is less reactive toward bromination and usually needs a catalyst such as \(FeBr_3\) for electrophilic bromination. Aniline is much more reactive because the \(-NH_2\) group donates electron density into the ring. This activation makes bromination possible with bromine water. The difference shows how strongly substituents can change aromatic reactivity.
507. Which statement best explains why direct bromination of aniline is difficult to stop at the monobromo stage?
ⓐ. The amino group is strongly activating and ortho-para directing.
ⓑ. The amino group is strongly deactivating and meta-directing.
ⓒ. The benzene ring is destroyed after the first substitution.
ⓓ. Bromine attacks only the nitrogen atom and not the ring.
Correct Answer: The amino group is strongly activating and ortho-para directing.
Explanation: The amino group donates electron density into the aromatic ring. This makes the ring highly reactive toward electrophilic bromination. The ortho and para positions are especially activated, so more than one substitution can occur under ordinary bromine-water conditions. Hence direct bromination of aniline commonly gives \(2,4,6\)-tribromoaniline instead of a monobromo product.
508. Which method helps obtain a controlled monobromo derivative from aniline?
ⓐ. Diazotise aniline and immediately reduce with \(H_3PO_2\).
ⓑ. Convert aniline to acetanilide, brominate, then hydrolyse.
ⓒ. Treat aniline with excess bromine water at room temperature.
ⓓ. Heat aniline with \(CHCl_3\) and alcoholic \(KOH\).
Correct Answer: Convert aniline to acetanilide, brominate, then hydrolyse.
Explanation: Direct bromination of aniline is too vigorous because \(-NH_2\) strongly activates the ring. Acetylation converts aniline into acetanilide, where the nitrogen lone pair is less available for ring activation. Bromination of acetanilide is more controlled and mainly gives the para bromo derivative. Hydrolysis then removes the acetyl group and regenerates the amino group.
509. Why does acetylation of aniline reduce excessive bromination?
ⓐ. Acetylation reduces \(-NH_2\) donation to the ring.
ⓑ. It converts the benzene ring into a saturated cyclohexane ring.
ⓒ. It removes nitrogen permanently from the molecule.
ⓓ. It changes the amino group into a strongly meta-directing nitro group.
Correct Answer: Acetylation reduces \(-NH_2\) donation to the ring.
Explanation: In aniline, the nitrogen lone pair strongly activates the aromatic ring. During acetylation, aniline becomes acetanilide, \(C_6H_5NHCOCH_3\). The nitrogen lone pair is partly involved with the carbonyl group, so its donation into the benzene ring is reduced. The ring remains ortho-para directing but is less excessively activated, making controlled substitution easier.
510. Which product is mainly formed when acetanilide is brominated under controlled conditions?
ⓐ. \(m\)-Bromoacetanilide
ⓑ. \(p\)-Bromoacetanilide
ⓒ. \(2,4,6\)-Tribromoaniline
ⓓ. Bromobenzene
Correct Answer: \(p\)-Bromoacetanilide
Explanation: Acetanilide contains the \(-NHCOCH_3\) group, which is still ortho-para directing but less activating than \(-NH_2\). Bromination is therefore more controlled than in aniline. The para product is usually favoured because the para position is less sterically crowded than the ortho positions. Hence \(p\)-bromoacetanilide is the major product.
511. Which sequence is suitable for preparing \(p\)-bromoaniline from aniline?
ⓐ. Acetylation, bromination, then hydrolysis
ⓑ. Bromination with excess bromine water only
ⓒ. Diazotisation, then treatment with \(CuBr/HBr\)
ⓓ. Carbylamine reaction, then acidification
Correct Answer: Acetylation, bromination, then hydrolysis
Explanation: \(\textbf{Protection step:}\)
Aniline is first acetylated to acetanilide to reduce excessive activation of the ring.
\(\textbf{Substitution step:}\)
Controlled bromination of acetanilide gives mainly \(p\)-bromoacetanilide.
\(\textbf{Deprotection step:}\)
Hydrolysis removes the acetyl group and restores \(-NH_2\).
\(\textbf{Final product:}\)
The final product is \(p\)-bromoaniline.
\(\textbf{Final Answer:}\)
The suitable sequence is acetylation, bromination, and hydrolysis.
512. Which statement correctly compares direct bromination of aniline and bromination after acetylation?
ⓐ. Both methods give only bromobenzene as the main product.
ⓑ. Direct bromination gives mainly \(p\)-bromoaniline, while acetylation gives no substitution.
ⓒ. Direct gives tribromoaniline; acetylation controls monobromination.
ⓓ. Direct bromination requires diazotisation, while acetylation gives only diazonium salts.
Correct Answer: Direct gives tribromoaniline; acetylation controls monobromination.
Explanation: The free \(-NH_2\) group in aniline activates the ring so strongly that bromine water gives multiple substitution. The product is \(2,4,6\)-tribromoaniline. Acetylation changes \(-NH_2\) into \(-NHCOCH_3\), reducing the activating effect. This allows bromination to be controlled, mainly giving a monobromo acetanilide that can be hydrolysed to bromoaniline.
513. Why is direct nitration of aniline difficult to control?
ⓐ. Protonation and oxidation occur in nitrating medium.
ⓑ. Aniline contains no aromatic ring for electrophilic substitution.
ⓒ. Aniline reacts only by replacing \(-NH_2\) with \(-NO_2\).
ⓓ. Aniline forms only benzene under nitrating conditions.
Correct Answer: Protonation and oxidation occur in nitrating medium.
Explanation: Direct nitration of aniline is complicated because the nitrating mixture is strongly acidic. The \(-NH_2\) group can be protonated to form the anilinium ion, \(C_6H_5NH_3^+\). This changes the directing and activating behaviour of the group. Aniline is also sensitive to oxidation under harsh acidic conditions, so direct nitration does not give a clean single product easily.
514. What is formed when aniline is protonated in strongly acidic medium?
ⓐ. \(C_6H_5N_2^+\)
ⓑ. \(C_6H_5NH_3^+\)
ⓒ. \(C_6H_5NO_2\)
ⓓ. \(C_6H_5CH_2NH_2\)
Correct Answer: \(C_6H_5NH_3^+\)
Explanation: Aniline has a lone pair on nitrogen, so it can accept a proton in acidic medium. Protonation changes \(C_6H_5NH_2\) into \(C_6H_5NH_3^+\), called the anilinium ion. Once protonated, the nitrogen lone pair is no longer available for strong resonance donation to the ring. This is why acidic conditions strongly affect electrophilic substitution reactions of aniline.
515. Which statement best describes the directing effect of the anilinium ion, \(C_6H_5NH_3^+\)?
ⓐ. It is strongly activating and ortho-para directing like \(-NH_2\).
ⓑ. It is deactivating and mainly meta directing.
ⓒ. It removes the benzene ring from the molecule.
ⓓ. It causes substitution only at the benzylic carbon.
Correct Answer: It is deactivating and mainly meta directing.
Explanation: In the anilinium ion, nitrogen carries a positive charge and has no free lone pair for donation into the ring. The group \(-NH_3^+\) withdraws electron density from the aromatic ring by its strong electron-withdrawing effect. This deactivates the ring toward electrophilic substitution. It also favours meta substitution compared with the free \(-NH_2\) group, which is ortho-para directing.
516. Direct nitration of aniline may give a mixture containing which types of nitroanilines?
ⓐ. Only \(m\)-nitroaniline
ⓑ. Only \(p\)-nitroaniline
ⓒ. \(o\)-, \(m\)-, and \(p\)-nitroanilines
ⓓ. Only trinitroaniline products
Correct Answer: \(o\)-, \(m\)-, and \(p\)-nitroanilines
Explanation: Free aniline has an \(-NH_2\) group, which directs electrophiles to ortho and para positions. In strongly acidic nitrating medium, some aniline is protonated to the anilinium ion, which is deactivating and meta directing. Because both forms can influence the reaction, a mixture of ortho, meta, and para nitroanilines may be obtained. This product mixture makes direct nitration less suitable when a single nitroaniline is desired.
517. Which sequence is suitable for preparing mainly \(p\)-nitroaniline from aniline?
ⓐ. Diazotisation, then treatment with \(CuCN/KCN\)
ⓑ. Acetylation, nitration, then hydrolysis
ⓒ. Carbylamine reaction, then acidification
ⓓ. Bromination with excess bromine water only
Correct Answer: Acetylation, nitration, then hydrolysis
Explanation: \(\textbf{Starting compound:}\)
Aniline is \(C_6H_5NH_2\).
\(\textbf{Protection step:}\)
Acetylation converts aniline into acetanilide, \(C_6H_5NHCOCH_3\).
\(\textbf{Nitration step:}\)
Acetanilide undergoes controlled nitration mainly at the para position.
\(\textbf{Deprotection step:}\)
Hydrolysis removes the acetyl group and regenerates \(-NH_2\).
\(\textbf{Final Answer:}\)
The sequence gives mainly \(p\)-nitroaniline.
518. Why is aniline converted into acetanilide before controlled nitration?
ⓐ. To remove the benzene ring before nitration
ⓑ. To protect and moderate \(-NH_2\) activation
ⓒ. To convert \(-NH_2\) into a permanently meta-directing group
ⓓ. To form benzenediazonium chloride directly
Correct Answer: To protect and moderate \(-NH_2\) activation
Explanation: The \(-NH_2\) group in aniline is strongly activating and can also be protonated in acidic nitrating medium. Acetylation converts it into \(-NHCOCH_3\), which is less activating because the nitrogen lone pair is partly involved with the carbonyl group. This makes electrophilic substitution more controlled. The acetyl group can later be removed by hydrolysis to regenerate the free amino group.
519. Which product is formed mainly when acetanilide is nitrated under controlled conditions?
ⓐ. \(m\)-Nitroacetanilide
ⓑ. \(p\)-Nitroacetanilide
ⓒ. \(2,4,6\)-Trinitroaniline
ⓓ. Nitrobenzene
Correct Answer: \(p\)-Nitroacetanilide
Explanation: Acetanilide contains the group \(-NHCOCH_3\), which is ortho-para directing. The para position is usually favoured over the ortho position because it is less sterically crowded. Controlled nitration therefore gives mainly \(p\)-nitroacetanilide. This product can be hydrolysed to obtain \(p\)-nitroaniline.
520. Which reaction converts \(p\)-nitroacetanilide into \(p\)-nitroaniline?
ⓐ. Hydrolysis
ⓑ. Diazotisation
ⓒ. Carbylamine reaction
ⓓ. Hoffmann degradation
Correct Answer: Hydrolysis
Explanation: \(p\)-Nitroacetanilide contains an acetyl-protected amino group. Hydrolysis breaks the amide bond and removes the acetyl group. The \(-NHCOCH_3\) group is converted back into \(-NH_2\). Since the nitro group remains on the ring, the product is \(p\)-nitroaniline.
