Q.(a) Write reasons for the following :

Showing the 12 most recent of 13 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.Identify the set with only meta directing groups (A) $-CH_3$ , $-C(=O)-R$ , $-NH-C(=O)-CH_3$ (B) $-CN$ , $-CO-R$ , $-COOH$ (C) $-OCH_3$ , $-C_2H_5$ , $-NH_2$ (D) $-NHR$ , $-CHO$ , $-NO_2$

   ›Reveal solutionSolution

   Meta directors are electron-withdrawing groups that destabilise ortho/para carbocation intermediates by resonance; $-CN$, $-CO-R$, and $-COOH$ are all meta directors, matching option (B) exactly.

   Concept and Intuition

   In electrophilic aromatic substitution, a substituent directs the incoming electrophile based on how it affects the stability of the arenium-ion intermediate at ortho/para vs meta positions. Groups with a lone pair or hyperconjugation that can donate electron density into the ring by resonance ($-NH_2$, $-NHR$, $-OCH_3$, $-CH_3$, $-C_2H_5$, $-NH-COCH_3$) are ortho/para directors (mostly activating, some like halogens deactivating but still o,p). Groups with a $\pi$-bond to a more electronegative atom directly on the ring ($-CN$, $-CHO$, $-COR$, $-COOH$, $-NO_2$, $-SO_3H$) withdraw electron density by resonance and destabilise the ortho/para arenium ions more than the meta one, making them meta directors (all deactivating).

   Step-by-Step Solution

   1. Option (A): $-CH_3$ (o,p, activating), $-C(=O)R$ (meta), $-NH-COCH_3$ (o,p, activating, via N lone pair) — mixed set, not all meta.
   2. Option (B): $-CN$ (meta), $-CO-R$ (meta), $-COOH$ (meta) — all three are meta directors. …

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2. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.Which of the following compounds is most reactive towards electrophilic substitution reactions? (A) Phenol, $\mathrm{C_6H_5OH}$ (benzene ring with an $\mathrm{OH}$ substituent) (B) Toluene, $\mathrm{C_6H_5CH_3}$ (benzene ring with a $\mathrm{CH_3}$ substituent) (C) Chlorobenzene, $\mathrm{C_6H_5Cl}$ (benzene ring with a $\mathrm{Cl}$ substituent) (D) Nitrobenzene, $\mathrm{C_6H_5NO_2}$ (benzene ring with a $\mathrm{NO_2}$ substituent)

   ›Reveal solutionSolution

   Among OH, CH₃, Cl, NO₂ substituents, -OH is the strongest ring-activator by resonance, making phenol the most reactive towards electrophilic substitution.

   Concept and Intuition

   The rate of electrophilic aromatic substitution depends on how much electron density a substituent pushes into the ring. Groups with a lone pair adjacent to the ring (like -OH, -NH₂) donate strongly by resonance and are powerful activators; alkyl groups (-CH₃) donate weakly by hyperconjugation/induction; halogens (-Cl) are deactivating overall (though o,p-directing) because their strong -I effect outweighs weak resonance donation; -NO₂ is strongly electron-withdrawing (both -I and -M) and strongly deactivating.

   Step-by-Step Solution

   1. Rank the activating strength: $-\mathrm{OH}$ (strong activator, resonance donor) $> -\mathrm{CH_3}$ (weak activator) $> -\mathrm{Cl}$ (weak deactivator, net) $> -\mathrm{NO_2}$ (strong deactivator).
   2. Phenol's OH group increases electron density at ortho/para positions the most, lowering the activation energy for electrophilic attack far more than a methyl group does. …

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3. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.Identify the amide which gives propan-1-amine by Hoffmann bromamide reaction. (A) $\mathrm{CH_3-CH_2-C(=O)-NH_2}$ (B) $\mathrm{CH_3-CH(CH_3)-C(=O)-NH_2}$ (C) $\mathrm{CH_3-CH_2-CH_2-C(=O)-NH_2}$ (D) $\mathrm{CH_3-CH_2-C(=O)-NH-CH_3}$

   ›Reveal solutionSolution

   Hoffmann degradation removes the carbonyl carbon from a primary amide, so to land on

   propan-1-amine (3 carbons) the amide must be the straight-chain 4-carbon amide,

   butanamide.

   Concept and Intuition

   In the Hoffmann bromamide degradation, a primary amide $R{-}CO{-}NH_2$ reacts with

   $Br_2/NaOH$ to give the amine $R{-}NH_2$ — the alkyl/aryl group $R$ migrates directly

   onto nitrogen while the carbonyl carbon is expelled as carbon dioxide (via an isocyanate

   intermediate). The crucial consequence: the product amine has one carbon less than the starting amide, and the carbon skeleton of $R$ is otherwise carried over unchanged

   (no rearrangement of $R$ itself).

   Step-by-Step Solution

   1. Target product: propan-1-amine, $CH_3CH_2CH_2{-}NH_2$ — a straight 3-carbon chain with the amine on the terminal carbon.
   2. Since Hoffmann degradation removes exactly the carbonyl carbon, the parent amide must have $R = CH_3CH_2CH_2{-}$ (propyl), i.e. the amide is $CH_3CH_2CH_2{-}C(=O){-}NH_2$ = butanamide (4 carbons total).
   3. Check each option:
      • (A) $CH_3CH_2C(=O)NH_2$ (propanamide, $R=$ ethyl) → gives ethanamine ($CH_3CH_2NH_2$), not propan-1-amine.
      • (B) $CH_3CH(CH_3)C(=O)NH_2$ (2-methylpropanamide, $R=$ isopropyl) → gives isopropylamine ($(CH_3)_2CHNH_2$), a branched 3-carbon amine, not propan-1-amine.
      • (C) $CH_3CH_2CH_2C(=O)NH_2$ (butanamide, $R=$ n-propyl) → gives …

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4. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.What are X and Y respectively in the following reactions? $Y \xleftarrow[\text{ii. }H_2O]{\text{i. }LiAlH_4} \mathrm{C_6H_5CONH_2} \xrightarrow[NaOH]{Br_2} X$ (A) aniline ($C_6H_5NH_2$) , aniline ($C_6H_5NH_2$) (B) aniline ($C_6H_5NH_2$) , benzylamine ($C_6H_5CH_2NH_2$) (C) benzylamine ($C_6H_5CH_2NH_2$) , benzylamine ($C_6H_5CH_2NH_2$) (D) benzylamine ($C_6H_5CH_2NH_2$) , aniline ($C_6H_5NH_2$)

   ›Reveal solutionSolution

   Br2/NaOH degrades benzamide to aniline (Hofmann degradation, loses a carbon); LiAlH4 simply reduces benzamide to benzylamine (keeps all carbons) — so X = aniline, Y = benzylamine.

   Concept and Intuition

   Benzamide, $C_6H_5CONH_2$, can be converted to an amine in two very different ways, and the key distinguishing feature is carbon count:

   • $Br_2/NaOH$ (Hofmann bromamide degradation) removes the carbonyl carbon entirely (as $CO_2$ after rearrangement), so the amine formed has one carbon fewer than the amide.
   • $LiAlH_4$ is a simple reducing agent: it reduces the $C=O$ of the amide down to a $CH_2$ group, keeping all the original carbons.

   Step-by-Step Solution

   1. Right-hand arrow: $C_6H_5CONH_2 \xrightarrow{Br_2/NaOH} X$. This is the classic Hofmann bromamide degradation: the amide is converted (via an isocyanate intermediate) into a primary amine with one carbon less than the starting amide — the phenyl group is retained but the carbonyl carbon is lost as carbonate/$CO_2$. So $X = C_6H_5NH_2$ (aniline). …

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5. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.Number of deactivating groups of the following is $\mathrm{-Cl, -SO_3H, -OH, -NHC_2H_5, -COOCH_3, -CH_3}$ (A) 4 (B) 3 (C) 2 (D) 1

   ›Reveal solutionSolution

   Of the six substituents, three are deactivating toward electrophilic aromatic substitution: $-Cl$, $-SO_3H$, and $-COOCH_3$.

   Concept and Intuition

   Groups that donate electron density into the benzene ring (by resonance or induction) activate it toward electrophilic substitution; groups that withdraw electron density deactivate it. Halogens are a special case: they withdraw inductively (deactivating overall) but still donate a lone pair by resonance (hence they remain o/p-directors despite being deactivating).

   Step-by-Step Solution

   1. $-Cl$: strong $-I$ effect dominates over weak $+M$ donation → net deactivating (o/p-director).
   2. $-SO_3H$: strongly electron-withdrawing (both $-I$ and $-M$) → deactivating (m-director).
   3. $-OH$: lone pair strongly donated into ring by resonance → activating.
   4. $-NHC_2H_5$: amine lone pair strongly donated by resonance → activating (even stronger than $-OH$). …

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6. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.In the reaction sequence Y is $\mathrm{CH_3CO_2H \xrightarrow[(2)\ \Delta]{(1)\ NH_3} P \xrightarrow{Br_2/NaOH} Y}$ (A) a primary amine with same number of carbons as in P (B) a primary amine with one carbon less than in P (C) a secondary amine with same number of carbons as in P (D) a secondary amine with one carbon less than in P

   ›Reveal solutionSolution

   Acetic acid → acetamide (P) → Hofmann bromamide degradation strips one carbon, giving methylamine as Y — a primary amine with one carbon fewer than P: option (B).

   Concept and Intuition

   The Hofmann bromamide degradation is a name reaction that converts an amide $\mathrm{R{-}CONH_2}$ directly into a primary amine $\mathrm{R{-}NH_2}$ using bromine and concentrated NaOH — crucially, the carbonyl carbon is lost in the process (extruded as carbonate/CO₂ via an isocyanate intermediate), so the product amine has exactly one carbon fewer than the parent amide, and it is always a primary amine regardless of the amide's structure.

   Step-by-Step Solution

   1. $\mathrm{CH_3COOH + NH_3} \to \mathrm{CH_3COONH_4}$ (ammonium acetate, an acid-base salt).
   2. On heating ($\Delta$), ammonium acetate loses water (dehydration) to give acetamide, $\mathrm{CH_3CONH_2}$ — this is P (2 carbons).
   3. P undergoes the Hofmann bromamide reaction with $\mathrm{Br_2/NaOH}$: mechanistically, this proceeds via an N-bromoamide → nitrene/isocyanate rearrangement (the alkyl/aryl group migrates from carbon to nitrogen) → hydrolysis of the isocyanate to a carbamic acid → spontaneous decarboxylation, releasing $\mathrm{CO_2}$ and leaving the amine.
   4. Net result: $\mathrm{CH_3CONH_2} \to \mathrm{CH_3NH_2}$ (methylamine) + $\mathrm{CO_2}$ + salts — Y is methylamine, a primary amine with one carbon less than P (P has 2 C, Y has 1 C). …

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7. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.Two statements are given below Statement I: Chlorobenzene on nitration gives 1-chloro-4-nitrobenzene as major product Statement II: Chlorobenzene undergoes nitration slowly than benzene Identify the correct answer (A) Statements I, II are correct (B) Statements I, II are incorrect (C) Statement I correct but statement II is incorrect (D) Statement II correct but statement I is incorrect

   ›Reveal solutionSolution

   Chlorobenzene is an ortho/para director (so Statement I is correct) but it deactivates the ring, making nitration slower than benzene (so Statement II is also correct). Thus both statements are true.

   Concept & Intuition: Ortho/Para Directing and Activation/Deactivation

   When a substituent is already on a benzene ring, it influences two things:

   1. Where the next substituent goes (orientation).
   2. How fast the reaction happens (reactivity).

   Chlorine is a fascinating case: it is ortho/para directing because it can donate electrons through resonance (lone pairs on chlorine can delocalize into the ring, stabilizing the intermediate carbocation at ortho/para positions). However, chlorine is also highly electronegative, so it withdraws electrons inductively (through sigma bonds), which deactivates the ring overall. The net effect: chlorobenzene reacts slower than benzene, but when it does react, the new group goes ortho or para.

   Now let’s check each statement.

   ---

   1. Statement I: Chlorobenzene on nitration gives 1-chloro-4-nitrobenzene as major product

      • Nitration is an electrophilic aromatic substitution. The electrophile is the nitronium ion ($\text{NO}_2^+$).
      • Chlorine’s resonance donation makes the ortho and para positions more electron-rich than the meta position.
      • The para product (1-chloro-4-nitrobenzene) is often the major one because ortho substitution can be slightly hindered by the chlorine atom’s size.
      • So Statement I is correct.

   2. Statement II: Chlorobenzene undergoes nitration slower than benzene

      • Benzene itself has no substituent; its reactivity is the baseline.
      • Chlorine withdraws electron density inductively (due to high electronegativity), making the ring less electron-rich overall. …

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8. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.[FIGURE] (a benzene ring bearing a $\mathrm{NO_2}$ substituent reacting with an electrophile $E^+$ to give a product benzene ring with $\mathrm{NO_2}$ and the electrophile E attached at the meta position relative to $\mathrm{NO_2}$) In the above reaction electrophile is substituted at meta position only, due to I. Electron density is more at ortho & para position II. Electron density is relatively less at ortho & para position III. Electron density is less at meta position IV. Electron density is relatively more at meta position correct answer is (A) I, III only (B) II, IV only (C) I only (D) III only

   ›Reveal solutionSolution

   $-NO_2$ withdraws electron density strongly from the ortho/para positions by resonance, so the meta position is comparatively electron-rich and that's where the electrophile attacks.

   Concept and Intuition

   In electrophilic aromatic substitution, the position attacked is the one with the most residual electron density (most nucleophilic carbon), not the one with the least. $-NO_2$ is meta-directing precisely because its strong $-M$ (resonance) and $-I$ (inductive) effects pull electron density away from the ring, concentrating the depletion at the ortho and para carbons (where resonance structures place formal positive charge on the ring carbon). This leaves the meta carbon relatively less deactivated — i.e., comparatively electron-richer — so the electrophile bonds there.

   Step-by-Step Solution

   1. Draw the resonance structures of nitrobenzene: positive charge on the ring appears at ortho and para carbons when $-NO_2$'s lone pair/π-system withdraws density.
   2. This means ortho/para carbons are the most electron-poor (statement I, claiming they have more electron density, is wrong; statement II, that they have relatively less, is right). …

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9. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.Identify the ortho and para directing groups towards aromatic electrophilic substitution reactions from the following list -OH (I) -CN (II) -CO2H (III) -OCH3 (IV) -NHCOCH3 (V) -CHO (VI) (A) I, IV, V (B) II, III, VI (C) I, II, IV (D) IV, V, VI

   ›Reveal solutionSolution

   Groups with a lone pair that can donate into the aromatic ring by resonance (–OH, –OCH$_3$, –NHCOCH$_3$) are ortho/para directors; groups with an electron-withdrawing $\pi$-system attached directly to the ring (–CN, –CO$_2$H, –CHO) are meta directors.

   Concept and Intuition

   In electrophilic aromatic substitution, a substituent already on the ring determines where the next electrophile attacks by how it distributes electron density around the ring through resonance. Groups bonded to the ring via an atom bearing a lone pair (O, N) can donate that lone pair into the ring's $\pi$ system, building up electron density specifically at the ortho and para positions, so they are ortho/para directors (and ring-activating). Groups bonded to the ring via a carbon that is itself part of an electron-withdrawing multiple bond (C≡N, C=O of an acid, C=O of an aldehyde) pull electron density away from the ring by resonance, leaving the meta position comparatively most electron-rich (least destabilized in the transition state), so they are meta directors (and ring-deactivating).

   Step-by-Step Solution

   1. –OH (I): oxygen lone pair donates into the ring — ortho/para director, activating.
   2. –CN (II): the C≡N group withdraws electron density — meta director, deactivating.
   3. –CO$_2$H (III): the carboxyl carbon is electron-poor (C=O), withdraws by resonance — meta director, deactivating.
   4. –OCH$_3$ (IV): like –OH, oxygen lone pair donates into ring — ortho/para director, activating. …

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10. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
    Q.Identify the major product of the following reaction $\mathrm{PhCONH_2}$ (benzamide) reacts with $\mathrm{Br_2 + NaOH}$ to give (A) 4-Br-$\mathrm{C_6H_4}$-$\mathrm{CH_2NH_2}$ (4-bromobenzylamine) (B) $\mathrm{PhCH_2NH_2}$ (benzylamine) (C) 4-Br-$\mathrm{C_6H_4}$-$\mathrm{CONH_2}$ (4-bromobenzamide) (D) $\mathrm{PhNH_2}$ (aniline)

    ›Reveal solutionSolution

    Br$_2$/NaOH on a primary amide is the classic Hofmann bromamide degradation, which shortens the chain by one carbon to give the amine. Answer: PhNH$_2$ (aniline).

    Concept and Intuition

    The Hofmann bromamide (Hofmann degradation) reaction converts a primary amide, $\mathrm{R-CONH_2}$, into a primary amine, $\mathrm{R-NH_2}$, with the loss of the carbonyl carbon as carbon dioxide/carbonate. Mechanistically: Br$_2$/NaOH first brominates the amide nitrogen (N-bromoamide), base-induced rearrangement then migrates the R group from carbon to nitrogen with simultaneous loss of bromide, forming an isocyanate (R$-$N=C=O), which is rapidly hydrolysed under the basic aqueous conditions to the amine plus carbonate. The key structural consequence is that the amine formed has one fewer carbon than the starting amide, and the R group (here, phenyl) ends up directly attached to nitrogen.

    Step-by-Step Solution

    1. Start: $\mathrm{PhCONH_2}$ (benzamide), where Ph is attached to the carbonyl carbon.
    2. Br$_2$/NaOH brominates nitrogen, then base-promoted rearrangement migrates the phenyl group from the carbonyl carbon directly to nitrogen (with loss of the carbonyl carbon as isocyanate then carbonate/CO$_2$).
    3. Net result: the phenyl group is now bonded straight to $-$NH$_2$, i.e. $\mathrm{Ph-NH_2}$ (aniline) — one carbon shorter than the starting amide. …

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11. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
    Q.Identify the major product of the following reaction: chlorobenzene + $Br_2$ $\xrightarrow{Anhyd.\ FeCl_3}$ ? (A) [FIGURE: 1-bromo-2-chlorobenzene — a benzene ring with Cl and Br substituents in the ortho (1,2) positions] (B) [FIGURE: 1-bromo-2-chlorobenzene — a benzene ring with Cl and Br substituents in the ortho (1,2) positions] (C) [FIGURE: 2,4,6-tribromochlorobenzene — a benzene ring with Cl at one carbon and Br at each of the other three alternating carbons] (D) [FIGURE: 1-bromo-4-chlorobenzene — a benzene ring with Br and Cl in the para (1,4) positions]

    ›Reveal solutionSolution

    Chlorobenzene undergoes electrophilic aromatic substitution with Br₂/FeCl₃. The chlorine atom is an ortho/para director, so the major product is the para isomer (1-bromo-4-chlorobenzene) due to steric hindrance at the ortho positions. The correct option is (D).

    ---

    Concept and Intuition

    This problem tests your understanding of directing effects in electrophilic aromatic substitution (EAS). Chlorine (Cl) on a benzene ring is a unique substituent: it is deactivating (due to its strong inductive electron withdrawal) but ortho/para directing (because its lone pairs can donate electron density via resonance to the ortho and para positions). When we add a second substituent (here, Br), the incoming electrophile (Br⁺, generated by Br₂/FeCl₃) will preferentially attack the positions that are most electron-rich — the ortho and para positions relative to the Cl.

    However, the ortho positions are sterically crowded (adjacent to the bulky Cl atom), so the para product is usually the major one. The reaction is a classic example of electrophilic bromination of a deactivated aromatic ring, catalyzed by FeCl₃ (a Lewis acid that polarizes Br₂).

    ---

    Step-by-Step Reasoning

    1. Identify the catalyst’s role Anhydrous FeCl₃ acts as a Lewis acid. It coordinates with Br₂, polarizing the Br–Br bond and generating a stronger electrophile:

    $\text{Br}_2 + \text{FeCl}_3 \rightarrow \text{Br}^\delta+ \cdots \text{FeCl}_3\text{Br}^\delta-$

    This effectively creates a Br⁺ species that can attack the benzene ring.

    2. Analyze the directing effect of Cl

       Chlorine has two opposing effects:

       • Inductive withdrawal (due to high electronegativity) makes the ring less reactive overall (deactivating).
       • Resonance donation (lone pairs on Cl can delocalize into the ring) increases electron density at the ortho and para positions. The resonance structures show that the ortho and para carbons carry partial negative charge, making them the only sites for electrophilic attack.

    3. Consider steric hindrance

       The ortho positions are adjacent to the Cl atom. The Cl atom is relatively large, so an incoming Br atom at the ortho position would experience steric repulsion. The para position is farther away and much less hindered. Therefore, the para product is kinetically favored (major product).

    4. Evaluate the options

       • (A) and (B) both show ortho products (1-bromo-2-chlorobenzene). They are essentially identical drawings; these are minor products. …

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12. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
    Q.The reagent P used for the reaction is Cyclopentanecarboxamide (a cyclopentane ring with a $C(=O)NH_2$ substituent) $\xrightarrow{P}$ cyclopentylamine (a cyclopentane ring with an $NH_2$ substituent) (A) $Zn - Hg/HCl$ (B) $HCl / SnCl_2$ (C) $Br_2 / NaOH$ (D) $ZnCl / HCl$

    ›Reveal solutionSolution

    Cyclopentanecarboxamide → cyclopentylamine is a one-carbon degradation, achieved by the Hofmann bromamide reaction using $Br_2$/NaOH.

    Concept and Intuition

    The transformation given — an amide, $R-CONH_2$, converting into an amine with one less carbon, $R-NH_2$ — is the signature outcome of the Hofmann bromamide degradation reaction:

    $R-CONH_2 \xrightarrow{Br_2, NaOH} R-NH_2 + Na_2CO_3 + 2NaBr + 2H_2O$

    Mechanistically, $Br_2$/NaOH first brominates the amide nitrogen to give an N-bromoamide; base-mediated rearrangement (loss of bromide, migration of the R group from carbon to nitrogen) generates an isocyanate ($R-N=C=O$), which is then hydrolyzed under the basic aqueous conditions to give the primary amine $R-NH_2$ directly (with the original carbonyl carbon lost as carbonate).

    The other reagents don't fit this transformation:

    • $Zn-Hg/HCl$ (Clemmensen reduction) reduces a ketone/aldehyde carbonyl to $-CH_2-$, not applicable to an amide going to an amine with carbon loss.
    • $HCl/SnCl_2$ (Stephen reduction) reduces nitriles to aldehydes.
    • $ZnCl_2/HCl$ (Lucas reagent) is a test/reaction for alcohols, not related here.

    Step-by-Step Solution …

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