Q.Assertion: Phenols give o- and p-nitrophenol on nitration with dilute nitric acid.
Reason: -OH group present in phenol is o-, p-directing.

Showing the 12 most recent of 23 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.Chlorobenzene (A) on nitration gave B as major product. Reaction of B with NaOH at 443 K followed by acidification formed C. In another reaction of A with $CH_3COCl$ / anhy. $AlCl_3$ gave D (major product). What are C and D respectively? (A) 4-chlorophenol ($p\text{-}Cl$-$C_6H_4$-$OH$) ; 4-chloroacetophenone ($p\text{-}Cl$-$C_6H_4$-$COCH_3$) (B) 4-nitrophenol ($p\text{-}O_2N$-$C_6H_4$-$OH$) ; 4-chloroacetophenone ($p\text{-}Cl$-$C_6H_4$-$COCH_3$) (C) 4-chlorophenol ($p\text{-}Cl$-$C_6H_4$-$OH$) ; 3-chloroacetophenone ($m\text{-}Cl$-$C_6H_4$-$COCH_3$) (D) 4-nitrophenol ($p\text{-}O_2N$-$C_6H_4$-$OH$) ; 3-chloroacetophenone ($m\text{-}Cl$-$C_6H_4$-$COCH_3$)

   ›Reveal solutionSolution

   B (from nitration) = p-chloronitrobenzene; hydrolysing its (activated) C–Cl with hot NaOH then acidifying gives C = 4-nitrophenol; separately, Friedel–Crafts acylation of chlorobenzene gives the para product D = 4-chloroacetophenone.

   Concept and Intuition

   Two independent transformations of the same starting material, chlorobenzene: (i) nitration followed by an activated nucleophilic aromatic substitution of Cl by OH (only possible because the ring is now strongly deactivated/activated-for-SNAr by an ortho/para nitro group), and (ii) a direct Friedel–Crafts acylation, both controlled by the $o,p$-directing, weakly deactivating nature of Cl, with sterics favouring the para product in each case.

   Step-by-Step Solution

   1. Chlorobenzene nitration: Cl is $o,p$-directing (weak deactivator); despite modest steric differences, the major product is the para isomer, B = 1-chloro-4-nitrobenzene.
   2. Ordinary (unactivated) chlorobenzene needs extremely harsh conditions (Dow process: ~623 K, 300 atm) to convert Cl to OH. But once a strongly electron-withdrawing $NO_2$ group sits ortho/para to the Cl, it stabilises the Meisenheimer intermediate of nucleophilic aromatic substitution, so B reacts with NaOH at a much milder 443 K to displace Cl by OH, giving (after acidifying the phenoxide) C = 4-nitrophenol ($p\text{-}O_2N\text{-}C_6H_4\text{-}OH$). …

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2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.What are X and Y respectively in the following set of reactions? Anisole $\xrightarrow[CH_3CO_2H]{Br_2}$ X (major product) Toluene $\xrightarrow[(ii)\ OH^-]{(i)\ Br_2\ |\ UV\ Light}$ Y (major product) (A) $C_6H_5-O-CH_2CH_2Br$ ; $C_6H_5-CH(OH)-CH_3$ (B) 4-bromoanisole ($p\text{-}Br$-$C_6H_4$-$OCH_3$) ; benzyl alcohol ($C_6H_5-CH_2OH$) (C) $C_6H_5-O-CH(Br)-CH_3$ ; benzyl alcohol ($C_6H_5-CH_2OH$) (D) 2-bromoanisole ($o\text{-}Br$-$C_6H_4$-$OCH_3$) ; p-cresol ($p\text{-}HO$-$C_6H_4$-$CH_3$)

   ›Reveal solutionSolution

   Anisole undergoes ring (electrophilic) bromination to give mainly para-bromoanisole; toluene under UV light undergoes side-chain (radical) bromination at the benzylic carbon, and hydrolysis of that benzylic bromide gives benzyl alcohol.

   Concept and Intuition

   The conditions decide whether bromination happens on the ring or on the side chain: Br$_2$ in acetic acid (no light) with an activated ring (anisole) is classic electrophilic aromatic substitution; Br$_2$ with UV light on an alkyl-substituted arene (toluene) instead generates bromine radicals that abstract the weak benzylic C–H, giving side-chain substitution.

   Step-by-Step Solution

   1. Anisole: the $-OCH_3$ group is a powerful $o,p$-activator via resonance donation of a lone pair into the ring. With molecular $Br_2$ in acetic acid (ionic mechanism, electrophilic aromatic substitution), bromination occurs mainly at the position para to $-OCH_3$ (steric hindrance disfavours ortho): X = 4-bromoanisole ($p\text{-}Br\text{-}C_6H_4\text{-}OCH_3$).
   2. Toluene + $Br_2$/UV light: UV light homolyses $Br_2$ into radicals; the benzylic C–H bond of the methyl group is weak and gives a resonance-stabilised benzylic radical, so substitution happens on the side chain, not the ring, giving benzyl bromide ($C_6H_5CH_2Br$). …

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3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.What are X and Y respectively in the following reaction sequence? ($Me = -CH_3$) $Z + Y\ (Benzene\ derivative) \xleftarrow[\Delta]{HBr} Anisole \xrightarrow[anhy.\ AlCl_3]{CH_3Cl} X\ (major\ product)$ (A) 3-Methylanisole (–OMe and –Me in meta positions) ; Bromobenzene (B) 2-Methylanisole (–OMe and –Me in ortho positions) ; Phenol (C) 4-Methylanisole (–OMe and –Me in para positions) ; Bromobenzene (D) 4-Methylanisole (–OMe and –Me in para positions) ; Phenol

   ›Reveal solutionSolution

   Friedel–Crafts methylation of anisole gives predominantly the para product (X = 4-methylanisole); HBr cleaves anisole's ether linkage at the methyl carbon to give phenol (Y) and methyl bromide (Z).

   Concept and Intuition

   $-OCH_3$ on benzene is a strong activating, ortho/para-directing group. In Friedel–Crafts alkylation, both ortho and para products form, but the para isomer is favoured as the major product because of steric hindrance to ortho attack from the existing $-OCH_3$ group.

   Separately, aryl alkyl ethers like anisole are cleaved by strong acids such as HBr/HI on heating. Because the $C_{aryl}-O$ bond cannot easily undergo $S_N2$ (aryl carbons resist backside attack, and aryl cations are high energy), the nucleophile ($Br^-$) instead attacks the methyl carbon ($S_N2$ on $CH_3$), breaking the $O-CH_3$ bond. This releases the phenolic $-OH$ intact on the ring (giving phenol) and generates methyl bromide.

   Step-by-Step Solution

   1. Anisole + $CH_3Cl$, anhydrous $AlCl_3$ → Friedel–Crafts alkylation; $-OMe$ directs ortho/para; steric bulk favours para as the major product → X = 4-methylanisole (OMe, Me para).
   2. Anisole + HBr, $\Delta$ → ether cleavage; $Br^-$ attacks the less hindered methyl carbon ($S_N2$) → products are $CH_3Br$ (Z) and phenol (Y). …

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4. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.Which of the following does not involve in Friedel-Craft reaction? (A) $\mathrm{C_6H_5OCH_3}$ (anisole, drawn as a benzene ring with an $-OCH_3$ substituent) (B) $\mathrm{C_6H_5NH_2}$ (aniline, drawn as a benzene ring with an $-NH_2$ substituent) (C) $\mathrm{C_6H_5Cl}$ (chlorobenzene, drawn as a benzene ring with a $-Cl$ substituent) (D) $\mathrm{C_6H_6}$ (benzene, drawn as a plain benzene ring)

   ›Reveal solutionSolution

   Aniline is the classic exception that fails to undergo Friedel–Crafts reactions, because its $-\mathrm{NH_2}$ lone pair complexes with the $\mathrm{AlCl_3}$ catalyst, killing both the catalyst's activity and deactivating the ring.

   Concept and Intuition

   Friedel–Crafts alkylation/acylation needs a Lewis-acid catalyst (commonly anhydrous $\mathrm{AlCl_3}$) to generate the electrophile. This only works smoothly on arenes whose ring is not strongly deactivated and whose substituents don't sequester the catalyst. Aniline's amino group is a strong Lewis base — its nitrogen lone pair readily donates into $\mathrm{AlCl_3}$'s empty orbital, forming a stable acid–base adduct. This (a) uses up the catalyst so it can't generate the required electrophile, and (b) converts the ring substituent into an electron-withdrawing $-\mathrm{NH_2}\to\mathrm{AlCl_3}$ group (rather than the strongly activating free $-\mathrm{NH_2}$), strongly deactivating the ring toward electrophilic attack. As a result, aniline famously fails to give normal Friedel–Crafts products.

   Step-by-Step Solution

   1. Anisole ($\mathrm{C_6H_5OCH_3}$): the ether oxygen is a much weaker Lewis base than an amine nitrogen and still activates the ring strongly (o,p-director); anisole undergoes Friedel–Crafts reactions normally.
   2. Aniline ($\mathrm{C_6H_5NH_2}$): the amine nitrogen's lone pair is a strong Lewis base, complexing irreversibly with $\mathrm{AlCl_3}$, disabling the catalyst and deactivating the ring — Friedel–Crafts fails. …

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5. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.Consider the following Statement-I: In the nitration of aniline, more amount of m-nitroaniline is formed than expected. Statement-II: In the presence of a strongly acidic medium, aniline is protonated to form anilinium ion, which is meta directing. The correct answer is (A) Both statement-I and statement-II are correct (B) Both statement-I and statement-II are not correct (C) Statement-I is correct, but statement-II is not correct (D) Statement-I is not correct, but statement-II is correct

   ›Reveal solutionSolution

   Protonation of aniline to anilinium ion under strongly acidic nitration conditions explains the unexpectedly high m-nitroaniline yield; both statements are true. Answer: (A).

   Concept and Intuition

   Aniline's $\mathrm{-NH_2}$ group is a lone-pair donor and is normally a strong ortho/para director via resonance. However, nitration is carried out in strongly acidic media (conc. $\mathrm{HNO_3}$/conc. $\mathrm{H_2SO_4}$), and aniline is a reasonably basic amine — a substantial fraction of it is protonated under these conditions to the anilinium ion, $\mathrm{C_6H_5NH_3^+}$. The $\mathrm{-NH_3^+}$ group has no lone pair available to donate into the ring (it is used in the N–H bonds/protonation) and instead withdraws electron density inductively, behaving like other positively-charged substituents ($\mathrm{-NR_3^+}$) as a deactivating, meta-directing group. So nitration of aniline under these conditions is really a mixture of two competing pathways — direct nitration of the small residual free amine (ortho/para-directing) and nitration of the dominant anilinium ion (meta-directing) — and the meta contribution pushes the overall m-nitroaniline yield well above what pure amine-directed reasoning would predict.

   Step-by-Step Solution

   1. Aniline + conc. $\mathrm{HNO_3}$/conc. $\mathrm{H_2SO_4}$: strongly acidic medium protonates much of the aniline to anilinium ion.
   2. Anilinium ion ($\mathrm{-NH_3^+}$) is electron-withdrawing and meta-directing, unlike the free amine.
   3. Nitration therefore proceeds partly through the anilinium ion pathway, contributing meta product. …

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6. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.What are the major products X and Y respectively in the following set of reactions? (dark = dark) [FIGURE] (a benzene ring with COOH, reacting with $\mathrm{Br_2}$/Fe(dark) to give X) [FIGURE] (a benzene ring with $\mathrm{CH_3}$, reacting with $\mathrm{Br_2}$/Fe(dark) to give Y) (A) [FIGURE] (X = a benzene ring with COOH and Br meta to each other; Y = a benzene ring with $\mathrm{CH_3}$ and Br meta to each other) (B) [FIGURE] (X = a benzene ring with COOH and Br para to each other; Y = a benzene ring with $\mathrm{CH_3}$ and Br para to each other) (C) [FIGURE] (X = a benzene ring with COOH and Br para to each other; Y = a benzene ring with $\mathrm{CH_3}$ and Br meta to each other) (D) [FIGURE] (X = a benzene ring with COOH and Br meta to each other; Y = a benzene ring with $\mathrm{CH_3}$ and Br para to each other)

   ›Reveal solutionSolution

   COOH (meta director) sends Br to the meta position in benzoic acid; CH$_3$ (ortho/para director) sends Br predominantly to the para position in toluene — so X is meta-substituted and Y is para-substituted.

   Concept and Intuition

   Both reactions are electrophilic aromatic bromination catalysed by Fe (which generates the electrophile Br$^+$ from Br$_2$); "dark" conditions rule out the alternative photochemical free-radical side-chain bromination pathway, confirming these are ring substitutions. The regiochemistry is governed entirely by the directing effect of the substituent already on the ring:

   • $-$COOH is strongly electron-withdrawing (by resonance and induction), so it deactivates the ring but directs the incoming electrophile to the meta position (the position where the resulting arenium-ion intermediate avoids placing positive charge on the carbon bearing the electron-poor COOH group).
   • $-$CH$_3$ is electron-donating (hyperconjugation/+I), activating the ring and directing ortho/para; because Br is a moderately bulky electrophile, steric hindrance disfavours the ortho position relative to the CH$_3$ group, so the para isomer is the major product.

   Step-by-Step Solution

   1. Benzoic acid + Br$_2$/Fe: COOH is a meta director $\Rightarrow$ Br enters meta to COOH, giving X = 3-bromobenzoic acid (meta relationship between COOH and Br). …

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7. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.Which of the following will undergo methylation with $\mathrm{CH_3Cl}$ / anhy. $\mathrm{AlCl_3}$ ? a) Aniline b) Chlorobenzene c) Benzoic acid d) Anisole (A) a, d (B) b, c (C) b, d (D) a, c

   ›Reveal solutionSolution

   Friedel-Crafts alkylation with CH$_3$Cl/AlCl$_3$ works on chlorobenzene (weakly deactivated) and anisole (strongly activated) but fails on aniline (N lone pair complexes AlCl$_3$) and benzoic acid (too strongly deactivated).

   Concept and Intuition

   Friedel-Crafts reactions need the aromatic ring to be nucleophilic enough to attack the AlCl$_3$-generated electrophile ($^+$CH$_3$ here). Two situations cause the reaction to fail:

   1. Deactivation of the AlCl$_3$ catalyst itself: groups with a lone pair that is basic enough to coordinate directly with the Lewis acid — most notably $-$NH$_2$, $-$NHR, $-$NHCOR — tie up AlCl$_3$ as a salt/complex, leaving none to generate the electrophile. Aniline is the textbook example of this failure mode.
   2. Strong deactivation of the ring itself: strongly electron-withdrawing meta-directing groups (COOH, NO$_2$, CN, SO$_3$H, etc.) make the ring too electron-poor for the (only moderately electrophilic) Friedel-Crafts species to attack. Benzoic acid falls in this category.

   Halogens (as in chlorobenzene) are only weakly deactivating (net effect: deactivating overall by induction, but still ortho/para-directing by resonance donation of lone pairs) — the ring remains nucleophilic enough for alkylation to proceed. Ethers like anisole ($-$OCH$_3$) are strongly activating (electron-donating by resonance) and undergo Friedel-Crafts alkylation readily.

   Step-by-Step Solution

   1. Aniline (a): $-$NH$_2$ lone pair coordinates with AlCl$_3$, forming a complex that both consumes the catalyst and makes the ring effectively bear a positively-charged, strongly deactivating $-$NH$_2^+$AlCl$_3^-$-like group $\Rightarrow$ Friedel-Crafts fails. …

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8. AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
   Q.What are X and Y respectively in the following reactions? $\mathrm{C_6H_5CH_3} \xrightarrow{X} \mathrm{C_6H_5CHO} \xrightarrow{\text{Nitration}} Y$ (A) X = $\mathrm{KMnO_4}\,/\,\mathrm{OH^-}$ ; Y = 3-nitrobenzaldehyde ($m$-$\mathrm{O_2N}$-$\mathrm{C_6H_4}$-CHO) (B) X =

   (i) $\mathrm{CrO_2Cl_2}$,

   (ii) $\mathrm{H_3O^+}$ ; Y = 3-nitrobenzaldehyde ($m$-$\mathrm{O_2N}$-$\mathrm{C_6H_4}$-CHO) (C) X =

   (i) $\mathrm{CrO_2Cl_2}$,

   (ii) $\mathrm{H_3O^+}$ ; Y = 4-nitrobenzaldehyde ($p$-$\mathrm{O_2N}$-$\mathrm{C_6H_4}$-CHO) (D) X = $\mathrm{KMnO_4}\,/\,\mathrm{OH^-}$ ; Y = 4-nitrobenzaldehyde ($p$-$\mathrm{O_2N}$-$\mathrm{C_6H_4}$-CHO)

   ›Reveal solutionSolution

   This tests the Etard reaction (the selective route from toluene to benzaldehyde) and the meta-directing effect of the –CHO group during nitration.

   Concept and Intuition

   Strong oxidants like alkaline $\mathrm{KMnO_4}$ oxidise a toluene methyl group all the way to $-COOH$, overshooting the aldehyde stage. The Etard reaction ($\mathrm{CrO_2Cl_2}$ followed by aqueous hydrolysis) selectively stops the oxidation at the aldehyde. Once formed, the electron-withdrawing $-CHO$ group deactivates the ring and directs incoming electrophiles to the meta position.

   Step-by-Step Solution

   1. X: converting toluene directly to benzaldehyde (not benzoic acid) requires the Etard reaction — (i) treatment with $\mathrm{CrO_2Cl_2}$ to form a chromium complex, (ii) hydrolysis with $\mathrm{H_3O^+}$ to release $C_6H_5CHO$. $\mathrm{KMnO_4/OH^-}$ would instead push the oxidation all the way to $C_6H_5COOH$, so it is the wrong reagent here.
   2. Y: nitrating benzaldehyde. The $-CHO$ group is deactivating and meta-directing (like other carbonyl-bearing groups), so the nitro group is introduced predominantly at the meta position, giving 3-nitrobenzaldehyde ($m\text{-}O_2N\text{-}C_6H_4\text{-}CHO$). …

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9. AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
   Q.What is Y in the following reaction sequence? $\mathrm{C_6H_5OCH_3 \xrightarrow[anhy.\ AlCl_3]{CH_3Cl} X\ (Major) \xrightarrow[(ii)\ H_3O^+]{(i)\ KMnO_4/OH^-,\ \Delta} Y}$ (A) 3-methoxybenzoic acid (OCH3 meta to COOH on the benzene ring) (B) 4-methoxybenzoic acid (OCH3 para to COOH on the benzene ring) (C) 3-hydroxybenzoic acid (OH meta to COOH on the benzene ring) (D) 4-hydroxybenzoic acid (OH para to COOH on the benzene ring)

   ›Reveal solutionSolution

   Friedel-Crafts methylation of anisole gives the para isomer as the major product;

   oxidizing its methyl side chain to -COOH gives 4-methoxybenzoic acid, keeping the

   methoxy and carboxyl groups para to each other.

   Concept and Intuition

   The methoxy group (-OCH$_3$) is a powerful ortho, para-director (through resonance

   donation of a lone pair into the ring) in electrophilic aromatic substitution. In

   Friedel-Crafts alkylation, the bulky alkyl electrophile combined with steric hindrance

   near the -OCH$_3$ group means the para product dominates as the major isomer.

   Separately, hot alkaline $\mathrm{KMnO_4}$ is a powerful oxidant that converts any

   benzylic alkyl side chain (regardless of its length) all the way down to a -COOH group

   directly attached to the ring — it does not affect the ether linkage itself.

   Step-by-Step Solution

   1. $\mathrm{C_6H_5OCH_3}$ (anisole) + $\mathrm{CH_3Cl}$/anhydrous $\mathrm{AlCl_3}$ (Friedel-Crafts alkylation): methoxy directs the incoming methyl group ortho/para; the major product (less steric strain) is X = 1-methoxy-4-methylbenzene (p-cresol methyl ether), with $\mathrm{-OCH_3}$ and $\mathrm{-CH_3}$ para to each other.
   2. X + $\mathrm{KMnO_4/OH^-}$, $\Delta$, then $\mathrm{H_3O^+}$: vigorous oxidation …

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10. AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
    Q.The electrophile involved in sulphonation of benzene is (A) $\mathrm{S_2Cl_2}$ (B) $\mathrm{SO_3}$ (C) $\mathrm{SO_2}$ (D) $\mathrm{SCl_4}$

    ›Reveal solutionSolution

    Sulphonation of benzene is an electrophilic aromatic substitution in which the attacking electrophile is $\mathrm{SO_3}$ (or its protonated form), generated from fuming sulphuric acid.

    Concept and Intuition

    Electrophilic aromatic substitution needs a species that is electron-deficient enough to be attacked by the benzene ring's $\pi$-electron cloud. In sulphonation, benzene is treated with fuming sulphuric acid ($\mathrm{H_2SO_4}$ containing dissolved $\mathrm{SO_3}$, i.e. oleum). The sulphur atom in $\mathrm{SO_3}$ is highly electrophilic because it is bonded to three highly electronegative oxygens, leaving it electron-poor and eager to accept electron density from the aromatic ring.

    Step-by-Step Solution

    1. Fuming sulphuric acid supplies $\mathrm{SO_3}$ as the reactive species (sometimes considered as the protonated form $\mathrm{HSO_3^+}$ in strongly acidic medium).
    2. The ring's $\pi$-electrons attack the electrophilic sulphur of $\mathrm{SO_3}$, forming a sigma complex (arenium ion) with a $\mathrm{-SO_3^-}$ group attached to the ring carbon. …

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11. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
    Q.X and Y (major products) in the following reaction sequence are [FIGURE] (aniline - a benzene ring bearing an $\mathrm{NH_2}$ substituent - treated with acetic anhydride/pyridine to give X, then X treated with $\mathrm{Br_2}$/$\mathrm{CH_3COOH}$ to give Y) (A) [FIGURE] X: a benzene ring with $\mathrm{NH_2}$ and $\mathrm{COCH_3}$ substituents ortho to each other; [FIGURE] Y: the same ortho $\mathrm{NH_2}$/$\mathrm{COCH_3}$-substituted ring with an added Br substituent para to the $\mathrm{COCH_3}$ group (B) [FIGURE] X: a benzene ring with $\mathrm{NH_2}$ and $\mathrm{COCH_3}$ substituents in a meta relationship; [FIGURE] Y: the same meta-substituted ring but with the methyl of the acetyl group brominated, i.e. $\mathrm{COCH_2Br}$ instead of $\mathrm{COCH_3}$ (C) [FIGURE] X: acetanilide - a benzene ring bearing an $\mathrm{NHCOCH_3}$ substituent; [FIGURE] Y: N-bromo-N-phenylacetamide - the same structure but with Br on the nitrogen instead of H, i.e. $\mathrm{Br-N(COCH_3)}$-phenyl (D) [FIGURE] X: acetanilide - a benzene ring bearing an $\mathrm{NHCOCH_3}$ substituent; [FIGURE] Y: 4-bromoacetanilide - the same $\mathrm{NHCOCH_3}$-substituted ring with a Br substituent para to the $\mathrm{NHCOCH_3}$ group

    ›Reveal solutionSolution

    Tests the classic amine-protection strategy before electrophilic aromatic substitution. Answer: X = acetanilide, Y = 4-bromoacetanilide (option D).

    Concept and Intuition

    Free aniline is such a strong activator (via the lone pair on N) that direct bromination is hard to control — it gives 2,4,6-tribromoaniline even with dilute bromine water, because the ring is so highly activated. To get a clean mono-bromination product, the amine is first "protected" by acetylation (acetic anhydride/pyridine), converting -NH2 to the less-activating, sterically bulkier -NHCOCH3 group (acetanilide). This still directs ortho/para (via resonance donation from N), but the bulky acetamido group sterically disfavours the ortho position, so bromination occurs predominantly at the para position.

    Step-by-Step Solution

    1. Aniline + (CH₃CO)₂O / pyridine → acetanilide (X), $\mathrm{C_6H_5NHCOCH_3}$, via N-acetylation.
    2. Acetanilide + $\mathrm{Br_2}$/$\mathrm{CH_3COOH}$: the acetamido group is an ortho/para director but bulky, so bromination occurs mainly at the position para to it.
    3. Product Y = p-bromoacetanilide (4-bromoacetanilide), with Br para to the -NHCOCH3 group. …

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12. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
    Q.What are 'X' and 'Y' respectively in the following reactions? Anisole ($\mathrm{C_6H_5OCH_3}$) $\xrightarrow[\mathrm{CH_3COOH}]{\mathrm{Br_2}} X$ (major); Anisole $\xrightarrow[\Delta]{\mathrm{HI}} Y$ (A) X = a benzene ring bearing $\mathrm{OCH_2Br}$ (i.e. anisole's methyl hydrogen replaced by Br; ring itself unsubstituted); Y = phenol (benzene-OH) + $\mathrm{CH_3I}$ (B) X = 4-bromoanisole (structure: benzene ring with $\mathrm{OCH_3}$ and Br para to each other); Y = phenol (benzene-OH) + $\mathrm{CH_3I}$ (C) X = 2-bromoanisole (structure: benzene ring with $\mathrm{OCH_3}$ and Br ortho to each other); Y = iodobenzene (benzene-I) + $\mathrm{CH_3OH}$ (D) X = 4-bromoanisole (structure: benzene ring with $\mathrm{OCH_3}$ and Br para to each other); Y = iodobenzene (benzene-I) + $\mathrm{CH_3I}$

    ›Reveal solutionSolution

    Anisole undergoes para-selective electrophilic bromination (X = 4-bromoanisole) and, with HI, ether cleavage at the methyl carbon (not the aryl carbon), giving phenol + $\mathrm{CH_3I}$ (Y).

    Concept and Intuition

    • Bromination (X): The methoxy group is a powerful activating, ortho/para-directing substituent. Electrophilic aromatic substitution with $\mathrm{Br_2}$ in acetic acid proceeds readily; the para product dominates over ortho because of steric crowding near the bulky $\mathrm{OCH_3}$ group, giving 4-bromoanisole as the major product.
    • Ether cleavage (Y): Aryl alkyl ethers resist cleavage at the aryl–oxygen bond because that bond has partial double-bond character from resonance delocalisation of the oxygen lone pair into the ring — breaking it would require attacking an $sp^2$ aromatic carbon (very unfavourable for $S_N2$, and $S_N1$ would give a highly unstable aryl cation). Instead, the iodide ion (from HI) performs an $S_N2$ attack on the far less hindered, non-aromatic methyl carbon, displacing the phenoxide, which is protonated to phenol. This is the basis of the Zeisel method for estimating methoxy groups.

    Step-by-Step Solution

    1. Identify the directing effect of $\mathrm{-OCH_3}$: strongly activating, o/p-director.
    2. Steric bulk favours para substitution as the major product $\Rightarrow$ X = 4-bromoanisole. …

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