Q.Draw the structures of major monohalo products in each of the following reactions:

Showing the 12 most recent of 51 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Observe the following I and II reactions Rate determining step in the reactions I, II respectively is (A) Cleavage of C–Cl bond in both I, II (B) Cleavage of C–Cl bond in I, attack of OH$^-$ in II (C) Attack of OH$^-$ in I, C–Cl bond cleavage in II (D) Attack of OH$^-$ in both I and II

   ›Reveal solutionSolution

   The key is identifying the rate-determining step (RDS) from the reaction mechanism. In I (SN1), the RDS is the cleavage of the C–Cl bond; in II (SN2), the RDS is the attack of OH⁻. The correct option is (B).

   The question asks you to compare the rate-determining steps in two reactions, labelled I and II. Without seeing the exact structures, the pattern is unmistakable: Reaction I follows an SN1 mechanism (tertiary alkyl halide, polar protic solvent, etc.), while Reaction II follows an SN2 mechanism (primary alkyl halide, strong nucleophile, etc.). The rate-determining step in each case is the slowest step — the one that determines the overall rate law.

   1. Reaction I — SN1 mechanism

      In an SN1 reaction, the first step is the ionization of the alkyl halide: the C–Cl bond breaks heterolytically to form a carbocation and a chloride ion. This step is slow because it involves bond breaking and charge separation. The subsequent attack of OH⁻ on the carbocation is fast.

      Hence, the rate-determining step is the cleavage of the C–Cl bond.

   2. Reaction II — SN2 mechanism

      In an SN2 reaction, the nucleophile (OH⁻) attacks the carbon centre from the back side simultaneously as the leaving group (Cl⁻) departs. This is a single, concerted step. The rate depends on the concentrations of both the alkyl halide and OH⁻. …

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2. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Observe the statements given about $\mathrm{C_6H_5N_2BF_4}$ (X) I. Reaction of X with NaF gave $\mathrm{C_6H_5F}$ II. Heating X with $\mathrm{NaNO_2}$ | Cu yielded $\mathrm{C_6H_5NO_2}$ III. Heating X gave $\mathrm{C_6H_5F}$ IV. Reaction of X with $\mathrm{HNO_3}$ gave $\mathrm{C_6H_5NO_2}$ Correct statements are (only) (A) I, II only (B) II, III only (C) I, IV only (D) III, IV only

   ›Reveal solutionSolution

   $\mathrm{C_6H_5N_2BF_4}$ gives fluorobenzene on simple heating (Balz–Schiemann, statement III) and nitrobenzene with $\mathrm{NaNO_2}/\mathrm{Cu}$ (statement II). Statements I and IV are wrong — option (B).

   The concept first

   A diazonium group, $-\overset{+}{\mathrm{N}}\!\equiv\!\mathrm{N}$, is the finest leaving group in aromatic chemistry: it departs as nitrogen gas, which is thermodynamically irresistible. That is why the diazonium salt is the master intermediate for putting almost any group onto a benzene ring — you first make $\mathrm{Ar\!-\!N_2^+}$, then let a nucleophile trap the aryl cation/radical.

   But fluorine is the awkward case. Ordinary Sandmeyer chemistry ($\mathrm{CuCl}$, $\mathrm{CuBr}$, $\mathrm{KI}$) fails for $\mathrm{F^-}$, because fluoride is a poor nucleophile in these conditions. The trick is to build the fluorine into the counter-ion: precipitate the diazonium salt as its tetrafluoroborate, then heat the dry solid. Fluoride is delivered intramolecularly from $\mathrm{BF_4^-}$ — this is the Balz–Schiemann reaction. So the whole point of using $\mathrm{BF_4^-}$ is that no external fluoride source is needed.

   Step-by-step

   1. Statement III — heating X gives $\mathrm{C_6H_5F}$.

   $\mathrm{C_6H_5\overset{+}{N_2}BF_4^- \xrightarrow{\ \Delta\ } C_6H_5F + N_2\uparrow + BF_3}$

   This is precisely the Balz–Schiemann reaction. TRUE.

   2. Statement I — X + NaF gives $\mathrm{C_6H_5F}$. Adding $\mathrm{NaF}$ is neither necessary nor effective; free fluoride does not substitute the diazonium group (that failure is the very reason the fluoroborate route exists). FALSE. …

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3. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.$\ce{C6H5N2Cl} \xrightarrow{\text{Cu|HCl}} X \xrightarrow{\text{(i) NaOH}} Y$ $673\,\text{K},\,300\,\text{atm}$ $\text{(ii) } \ce{H3O^+}$ The incorrect statement about X and Y is (A) X undergoes Fittig reaction (B) Y gives o-hydroxybenzaldehyde with CHCl$_3$ and NaOH (C) Y forms salt with NaHCO$_3$ solution (D) X is chemically inert at room temperature

   ›Reveal solutionSolution

   The reaction sequence is the Sandmeyer reaction (conversion of diazonium salt to chlorobenzene, X) followed by high‑pressure/high‑temperature hydrolysis to phenol (Y). The incorrect statement is (C) because phenol does not react with NaHCO₃.

   Concept & Intuition

   This problem tests your knowledge of two classic organic transformations:

   1. Sandmeyer reaction – replacing a diazonium group with chlorine using CuCl/HCl.
   2. High‑temperature, high‑pressure hydrolysis – converting chlorobenzene to phenol under drastic conditions (NaOH, 673 K, 300 atm).

   Once you identify X and Y, you evaluate each statement about their chemical behavior. The trick is remembering that phenol is a weaker acid than carbonic acid, so it does not liberate CO₂ from NaHCO₃ – a common pitfall.

   ---

   Step‑by‑step reasoning

   1. Identify X
      • Starting material: benzenediazonium chloride ($\ce{C6H5N2Cl}$).
      • Reagent: CuCl/HCl (Sandmeyer conditions).
      • The diazonium group is replaced by chlorine:

   $\ce{C6H5N2Cl ->[CuCl/HCl] C6H5Cl + N2}$

    So **X = chlorobenzene**.
   

   2. Identify Y

   • Chlorobenzene is treated with NaOH at 673 K and 300 atm (the Dow process).
   • The chlorine is displaced by hydroxide, forming sodium phenoxide.
   • Acidification with $\ce{H3O+}$ gives phenol:

   $\ce{C6H5Cl ->[NaOH, 673 K, 300 atm] C6H5ONa ->[H3O+] C6H5OH}$

    So **Y = phenol**.
   

   3. Evaluate statement (A): “X undergoes Fittig reaction”

   • The Fittig reaction is a coupling of two aryl halides with sodium metal to give a biaryl.
   • Chlorobenzene (X) does undergo this reaction (e.g., with Na in dry ether to give biphenyl).
   • True.

   4. Evaluate statement (B): “Y gives o‑hydroxybenzaldehyde with CHCl₃ and NaOH”

      • This is the Reimer–Tiemann reaction of phenol with chloroform in base.
      • The major product is indeed salicylaldehyde (o‑hydroxybenzaldehyde).
      • True.

   5. Evaluate statement (C): “Y forms salt with NaHCO₃ solution”

      • Phenol is acidic (pKa ≈ 10) but weaker than carbonic acid (pKa₁ ≈ 6.4).
      • NaHCO₃ is a weaker base; it can only deprotonate acids stronger than H₂CO₃. …

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4. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.What is the end product P in the given sequence of reactions? (A) p-Hydroxybenzaldehyde (B) m-Hydroxybenzaldehyde (C) o-Hydroxybenzaldehyde (D) o-Hydroxybenzoic acid

   ›Reveal solutionSolution

   The reaction sequence is a classic Reimer–Tiemann reaction on phenol followed by a Cannizzaro-type workup; the final product is o-hydroxybenzaldehyde, so the correct option is (C).

   The key concept here is the Reimer–Tiemann reaction, which introduces a formyl group (–CHO) ortho to the –OH group on phenol. The reaction uses chloroform (CHCl₃) in the presence of a strong base (like NaOH) to generate a dichlorocarbene (:CCl₂) intermediate. This carbene attacks the electron-rich ortho position of the phenoxide ion, leading to an aldehyde after hydrolysis. The para product is also possible but is usually minor due to steric hindrance; the ortho product is the major one.

   Now, let’s walk through the sequence step by step:

   1. Step 1: Phenol + NaOH

      Phenol (C₆H₅OH) reacts with NaOH to form the phenoxide ion (C₆H₅O⁻). This ion is much more nucleophilic than phenol itself, especially at the ortho and para positions, because the negative charge on oxygen donates electron density into the ring.

   2. Step 2: Reimer–Tiemann reaction (CHCl₃ + NaOH, heat)

      Chloroform (CHCl₃) in the presence of strong base generates dichlorocarbene (:CCl₂). This highly reactive electrophile attacks the ortho position of the phenoxide ion (the para position is also attacked but less favored). The intermediate then undergoes hydrolysis to give an aldehyde group at the ortho position. The product after this step is salicylaldehyde (o-hydroxybenzaldehyde).

   3. Step 3: Dilute H₂SO₄ workup …

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5. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.An alkene, X ($\mathrm{C_5H_{10}}$) exhibits cis/trans isomerism. Bromination of 'X' followed by reaction with reagent/s (Y) gave the product 'Z'. What are 'Y' and 'Z'? (A) $\mathrm{NaNH_2}$ ; $(\mathrm{CH_3})_2\mathrm{CHC} \equiv \mathrm{CH}$ (B) $\mathrm{KOH}$ (aq) ; $\mathrm{CH_3CH_2CH_2C} \equiv \mathrm{CH}$ (C) alc.$\mathrm{KOH, NaNH_2}$ ; $\mathrm{CH_3CH_2C} \equiv \mathrm{CCH_3}$ (D) alc.$\mathrm{KOH, NaNH_2}$ ; $\mathrm{CH_3CH_2CH_2C} \equiv \mathrm{CH}$

   ›Reveal solutionSolution

   The alkene X must be pent-2-ene (the only C₅H₁₀ alkene with cis/trans isomerism). Bromination gives a vicinal dibromide; double dehydrohalogenation with alc. KOH then NaNH₂ yields pent-2-yne. The correct option is (C).

   Concept & Intuition

   The key is to identify the alkene first. For an alkene C₅H₁₀ to show cis/trans isomerism, each doubly bonded carbon must have two different substituents. The only straight-chain pentene satisfying this is pent-2-ene (CH₃–CH=CH–CH₂–CH₃). (Pent-1-ene has a terminal =CH₂ group, so no cis/trans; branched isomers like 2-methylbut-2-ene have a carbon with two identical methyl groups, also no cis/trans.)

   Bromination adds Br₂ across the double bond, giving a vicinal dibromide. To get an alkyne, we need two successive eliminations of HBr. The first elimination (with a strong base like alc. KOH) gives a bromoalkene; the second elimination (with a stronger base like NaNH₂) gives the alkyne. The product is pent-2-yne, not pent-1-yne, because the triple bond forms where the original double bond was.

   Step-by-step reasoning

   1. Identify alkene X

      • Formula C₅H₁₀, must be an alkene (one degree of unsaturation).
      • For cis/trans isomerism, each sp² carbon must have two different groups.
      • Pent-1-ene (CH₂=CH–CH₂–CH₂–CH₃): one sp² carbon has two H’s → no cis/trans.
      • 2-Methylbut-1-ene (CH₂=C(CH₃)–CH₂–CH₃): terminal =CH₂ → no cis/trans.
      • 2-Methylbut-2-ene ((CH₃)₂C=CH–CH₃): one sp² carbon has two methyls → no cis/trans.
      • Pent-2-ene (CH₃–CH=CH–CH₂–CH₃): each sp² carbon has H and an alkyl group → cis/trans possible.
      • Hence X = pent-2-ene.

   2. Bromination of X

      • Br₂ adds across the double bond:

   $\mathrm{CH_3CH=CHCH_2CH_3 + Br_2 \rightarrow CH_3CHBrCHBrCH_2CH_3}$

    This is a vicinal dibromide (2,3-dibromopentane).
   

   3. First elimination (reagent Y part 1: alc. KOH)

   • Alcoholic KOH is a strong base that promotes dehydrohalogenation.
   • The vicinal dibromide undergoes elimination of one HBr to give a bromoalkene. The more substituted alkene (Saytzeff rule) is favoured:

   $\mathrm{CH_3CHBrCHBrCH_2CH_3 \xrightarrow{alc.\ KOH} CH_3CBr=CHCH_2CH_3}$

    (The double bond forms between C2 and C3, with Br on the more substituted carbon.)
   

   4. Second elimination (reagent Y part 2: NaNH₂)

   • NaNH₂ is a very strong base (amide ion) needed to remove the second HBr from a bromoalkene.
   • This gives an alkyne: $$$$

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6. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.What are X and Y respectively in the following reaction sequence? Y $\xleftarrow[\text{(ii) KMnO}_4\,|\,\mathrm{OH^-},\ \mathrm{H_3O^+}]{\text{(i) Br}_2\,|\,\mathrm{Fe}}$ Ethylbenzene ($\mathrm{C_6H_5-C_2H_5}$) $\xrightarrow[\text{(ii) Mg | dry ether ;

   (iii) CO}_2,\ \mathrm{H_3O^+}]{\text{(i) Br}_2,|,h\nu}$X (A)$\mathrm{C_6H_5-CH_2CH_2-COOH}$(3-phenylpropanoic acid) ; 4-bromophenylacetic acid ($\mathrm{Br-C_6H_4-CH_2-COOH}$) (B)$\mathrm{C_6H_5-CH(CH_3)-COOH}$(2-phenylpropanoic acid) ;$p$-bromobenzoic acid ($\mathrm{Br-C_6H_4-COOH}$) (C)$p$-Ethylbenzoic acid ($\mathrm{C_2H_5-C_6H_4-COOH}$) ;$\mathrm{C_6H_5-CH(Br)-COOH}$(2-bromo-2-phenylacetic acid) (D)$m$-Ethylbenzoic acid ($\mathrm{C_2H_5-C_6H_4-COOH}$, meta) ;$p$-bromobenzoic acid ($\mathrm{Br-C_6H_4-COOH}$)

   ›Reveal solutionSolution

   Light-induced bromination hits the benzylic carbon (giving, after Grignard carboxylation, 2-phenylpropanoic acid = $X$), while $\mathrm{Br_2/Fe}$ hits the ring para and $\mathrm{KMnO_4}$ then burns the ethyl group down to $-\mathrm{COOH}$ (giving $p$-bromobenzoic acid = $Y$). Option (B).

   The concept first

   Three separate rules are being combined.

   1. $\mathrm{Br_2}/h\nu$ versus $\mathrm{Br_2}/\mathrm{Fe}$ — the single most useful contrast in aromatic chemistry.

   • $h\nu$ (light/heat, no catalyst) $\Rightarrow$ a free-radical chain reaction that attacks the side chain, specifically the benzylic C–H, because the benzyl radical is resonance-stabilised by the ring.
   • $\mathrm{Fe}$ / $\mathrm{FeBr_3}$ (a Lewis acid) $\Rightarrow$ an electrophilic substitution on the ring.

   2. Side-chain oxidation by $\mathrm{KMnO_4}$. Alkaline $\mathrm{KMnO_4}$ (then acid work-up) oxidises any alkyl side chain that possesses a benzylic hydrogen right back to a single $-\mathrm{COOH}$ group, however long the chain:

   $\mathrm{C_6H_5\!-\!CH_2CH_3}\ \xrightarrow[\mathrm{H_3O^{+}}]{\mathrm{KMnO_4/OH^{-}}}\ \mathrm{C_6H_5\!-\!COOH}$

   3. Grignard carboxylation adds exactly one carbon.

   $\mathrm{R\!-\!Br}\ \xrightarrow{\mathrm{Mg,\ dry\ ether}}\ \mathrm{R\!-\!MgBr}\ \xrightarrow{\mathrm{CO_2}}\ \mathrm{R\!-\!COOMgBr}\ \xrightarrow{\mathrm{H_3O^{+}}}\ \mathrm{R\!-\!COOH}$

   Step-by-step — the right-hand branch ($X$)

   Step 1 — benzylic bromination. Ethylbenzene is $\mathrm{C_6H_5\!-\!CH_2\!-\!CH_3}$. Under $\mathrm{Br_2}/h\nu$, a bromine radical abstracts the benzylic hydrogen (the one on the $\mathrm{CH_2}$ next to the ring), because the radical formed there is delocalised into the ring:

   $\mathrm{C_6H_5CH_2CH_3}\ \xrightarrow{\mathrm{Br_2},\,h\nu}\ \mathrm{C_6H_5\!-\!CH(Br)\!-\!CH_3}\quad(\text{1-bromo-1-phenylethane})$

   Step 2 — make the Grignard.

   $\mathrm{C_6H_5CH(Br)CH_3}\ \xrightarrow{\mathrm{Mg,\ dry\ ether}}\ \mathrm{C_6H_5CH(MgBr)CH_3}$

   Step 3 — carboxylate it. The carbanion-like carbon attacks $\mathrm{CO_2}$; acid work-up liberates the acid:

   $\mathrm{C_6H_5CH(MgBr)CH_3}\ \xrightarrow[\mathrm{H_3O^{+}}]{\mathrm{CO_2}}\ \mathrm{C_6H_5\!-\!CH(CH_3)\!-\!COOH}$

   $X=\textbf{2-phenylpropanoic acid}$

   The carboxyl group lands exactly where the bromine was — on the benzylic carbon — so the branch point is retained.

   Step-by-step — the left-hand branch ($Y$) …

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7. TG EAPCET 2026Set ap-2026-05-04-FN1 markMCQ
   Q.Identify the reactions, in which products are correctly given I. $\mathrm{I Br} \xrightarrow{\mathrm{H}_2\mathrm{O}} \mathrm{HI} + \mathrm{HOBr}$ II. $\mathrm{ClF}_3 \xrightarrow{\mathrm{H}_2\mathrm{O}} \mathrm{HF} + \mathrm{HOClO}$ III. $\mathrm{IF}_5 \xrightarrow{\mathrm{H}_2\mathrm{O}} \mathrm{HF} + \mathrm{HIO}_3$ The correct answer is (A) I, II only (B) II, III only (C) I, III only (D) I, II, III

   ›Reveal solutionSolution

   Hydrolysis of interhalogen compounds involves the more electronegative halogen forming a hydrohalic acid and the less electronegative halogen forming an oxyacid, retaining its oxidation state. Based on this, reactions II and III are correctly given. The correct option is (B).

   The hydrolysis of interhalogen compounds is a reaction with water where the compound breaks down. The key principle governing these reactions is the difference in electronegativity between the two halogens. The more electronegative halogen will form a hydrohalic acid (HX), while the less electronegative halogen will form an oxyacid (HXO$_n$). The oxidation state of the less electronegative halogen in the interhalogen compound is typically preserved in the resulting oxyacid.

   Let's analyze each reaction:

   1. Reaction I: $\mathrm{IBr} \xrightarrow{\mathrm{H}_2\mathrm{O}} \mathrm{HI} + \mathrm{HOBr}$

      • Concept: In the interhalogen compound IBr, bromine (Br) is more electronegative than iodine (I). Therefore, upon hydrolysis, bromine should form hydrobromic acid ($\mathrm{HBr}$), and iodine should form an oxyacid.
      • Oxidation States: In IBr, iodine has an oxidation state of $+1$, and bromine has an oxidation state of $-1$.
      • Expected Products: The more electronegative bromine forms $\mathrm{HBr}$. The less electronegative iodine, retaining its $+1$ oxidation state, forms hypoiodous acid ($\mathrm{HOI}$).
      • Comparison: The given products are $\mathrm{HI}$ and $\mathrm{HOBr}$. In $\mathrm{HI}$, iodine is in the $-1$ oxidation state, and in $\mathrm{HOBr}$, bromine is in the $+1$ oxidation state. This implies that iodine is more electronegative than bromine, which is incorrect. The roles of the halogens are swapped in the given products.
      • Conclusion: Reaction I is incorrect. The correct products would be $\mathrm{HBr} + \mathrm{HOI}$.

   2. Reaction II: $\mathrm{ClF}_3 \xrightarrow{\mathrm{H}_2\mathrm{O}} \mathrm{HF} + \mathrm{HOClO}$

      • Concept: In $\mathrm{ClF}_3$, fluorine (F) is more electronegative than chlorine (Cl). Thus, fluorine should form hydrofluoric acid ($\mathrm{HF}$), and chlorine should form an oxyacid.
      • Oxidation States: In $\mathrm{ClF}_3$, chlorine has an oxidation state of $+3$, and fluorine has an oxidation state of $-1$.
      • Expected Products: The more electronegative fluorine forms $\mathrm{HF}$. The less electronegative chlorine, retaining its $+3$ oxidation state, forms chlorous acid ($\mathrm{HClO}_2$). …

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8. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.Consider the following reaction sequence. (A) and (C) are \ce{CH3CHO} \xrightarrow{\text{(i) CH3MgBr}} \xrightarrow{\text{(ii) H2O / H^+}} (\text{A}) \xrightarrow{\text{H2SO4, } \Delta} (\text{B}) \xrightarrow{\text{(i) B2H6}} \xrightarrow{\text{(ii) H2O, H2O2 \mid OH^-}} (\text{C}) (A) Functional isomers (B) Metamers (C) Optical isomers (D) Position isomers

   ›Reveal solutionSolution

   The reaction sequence converts acetaldehyde to 2-butanol via a Grignard addition and then hydroboration-oxidation; the starting material and final product are structural isomers that differ in the position of the hydroxyl group, making them position isomers.

   The key here is to identify the structures of (A), (B), and (C) step by step, then compare the starting material (acetaldehyde) with the final product (C) to determine the type of isomerism.

   1. First step: Grignard reaction on acetaldehyde Acetaldehyde ($\ce{CH3CHO}$) reacts with $\ce{CH3MgBr}$ (methylmagnesium bromide). The Grignard reagent adds to the carbonyl carbon, forming an alkoxide after the nucleophilic attack. Upon workup with $\ce{H2O/H+}$, the alkoxide is protonated to give a secondary alcohol.
      • The reaction:

   $\ce{CH3CHO + CH3MgBr -> CH3CH(OMgBr)CH3 ->[H2O/H+] CH3CH(OH)CH3}$

   • So (A) is 2-propanol (isopropyl alcohol), $\ce{CH3CH(OH)CH3}$.

   2. Second step: Dehydration of (A) to form (B) Heating 2-propanol with concentrated $\ce{H2SO4}$ causes elimination of water (dehydration), yielding an alkene. Since it’s a secondary alcohol, the major product follows Zaitsev’s rule: the more substituted alkene is favored.
      • Dehydration:

   $\ce{CH3CH(OH)CH3 ->[H2SO4, \Delta] CH3CH=CH2 + H2O}$

   • So (B) is propene, $\ce{CH3CH=CH2}$.

   3. Third step: Hydroboration-oxidation of (B) Propene undergoes hydroboration with $\ce{B2H6}$ (diborane), followed by oxidation with $\ce{H2O2/OH-}$. This reaction adds water across the double bond in an anti-Markovnikov fashion — the hydroxyl group ends up on the less substituted carbon.
      • For propene: $\ce{CH3CH=CH2}$ → the boron adds to the terminal carbon (less hindered), so after oxidation, the OH is on the terminal carbon.
      • Product:

   $\ce{CH3CH=CH2 ->[1. B2H6][2. H2O2/OH-] CH3CH2CH2OH}$

   • So (C) is 1-propanol (n-propyl alcohol), $\ce{CH3CH2CH2OH}$.

   4. Comparing starting material and final product The question asks about the relationship between (A) and (C). Wait — careful: The problem states “(A) and (C) are” followed by the isomer types. So we compare (A) = 2-propanol and (C) = 1-propanol.
      • Both have the molecular formula $\ce{C3H8O}$ (same molecular weight, same atoms).
      • They differ in the position of the hydroxyl group: on carbon 2 vs. carbon 1. …

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9. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.Which of the following is an example of electrophilic substitution reaction? (A) $\mathrm{CH_3CHO + HCN \rightarrow CH_3CH(OH)CN}$ (B) $(\mathrm{CH_3})_3\mathrm{CX} + \mathrm{H_2O} \rightarrow (\mathrm{CH_3})_3\mathrm{C-OH} + \mathrm{HX}$ (C) $\mathrm{C_6H_6 + CH_3COCl} \xrightarrow{\mathrm{AlCl_3}} \mathrm{C_6H_5(COCH_3)} + \mathrm{HCl}$ (D) $\mathrm{BrCH_2CH_2Br + Zn} \xrightarrow[\text{alcohol}]{\Delta} \mathrm{CH_2=CH_2} + \mathrm{ZnBr_2}$

   ›Reveal solutionSolution

   The key idea is that electrophilic substitution involves an electrophile replacing a hydrogen on an aromatic ring; only reaction (C) fits this pattern, making it the correct answer.

   Concept and Intuition

   Electrophilic substitution reactions are characteristic of aromatic compounds like benzene. In these reactions, an electron‑deficient species (an electrophile) attacks the electron‑rich aromatic ring, replacing one of the hydrogen atoms. The other options represent different reaction types: nucleophilic addition (A), nucleophilic substitution (B), and elimination (D). Recognizing the hallmark of aromatic substitution—a benzene ring reacting with an electrophile in the presence of a Lewis acid catalyst—immediately points to option (C).

   Step‑by‑Step Reasoning

   1. Identify the reaction type in (A)

      $\mathrm{CH_3CHO + HCN \rightarrow CH_3CH(OH)CN}$

      Here, the carbonyl carbon of acetaldehyde is attacked by the cyanide ion (a nucleophile). This is a nucleophilic addition reaction, not electrophilic substitution.

   2. Identify the reaction type in (B)

      $(\mathrm{CH_3})_3\mathrm{CX} + \mathrm{H_2O} \rightarrow (\mathrm{CH_3})_3\mathrm{C-OH} + \mathrm{HX}$

      A tertiary alkyl halide reacts with water to form an alcohol. The leaving group (X⁻) departs, and water acts as a nucleophile. This is an S_N1 nucleophilic substitution (or, less commonly, E1 if elimination occurs, but here the product is an alcohol). No aromatic ring is involved.

   3. Identify the reaction type in (C)

      $\mathrm{C_6H_6 + CH_3COCl} \xrightarrow{\mathrm{AlCl_3}} \mathrm{C_6H_5(COCH_3)} + \mathrm{HCl}$

      Benzene reacts with acetyl chloride in the presence of AlCl₃. The Lewis acid AlCl₃ generates the acylium ion ($\mathrm{CH_3CO^+}$), a strong electrophile. This electrophile attacks the benzene ring, replacing a hydrogen atom to form acetophenone. This is the classic Friedel–Crafts acylation, a prime example of electrophilic aromatic substitution.

   4. Identify the reaction type in (D) …

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10. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.Identify the reaction related to Deacon’s process (A) $2\mathrm{H}_2\mathrm{O} + 2\mathrm{Cl}_2 \xrightarrow{\text{sunlight}} 4\mathrm{HCl} + \mathrm{O}_2$ (B) $4\mathrm{HCl} + \mathrm{O}_2 \xrightarrow[\text{723 K}]{\text{CuCl}_2} 2\mathrm{Cl}_2 + 2\mathrm{H}_2\mathrm{O}$ (C) $2\mathrm{NaCl} + \mathrm{H}_2\mathrm{SO}_4 \xrightarrow{\text{823 K}} \mathrm{Na}_2\mathrm{SO}_4 + 2\mathrm{HCl}$ (D) $\mathrm{Na}_2\mathrm{S}_2\mathrm{O}_3 + \mathrm{Cl}_2 + \mathrm{H}_2\mathrm{O} \rightarrow \mathrm{Na}_2\mathrm{SO}_4 + 2\mathrm{HCl} + \mathrm{S}$

    ›Reveal solutionSolution

    Deacon’s process is an industrial method to produce chlorine by oxidizing hydrogen chloride with oxygen over a copper(II) chloride catalyst at about 723 K. The correct reaction is option (B).

    Concept & Intuition

    Deacon’s process was developed in the 19th century to recover chlorine from byproduct HCl, which was abundant in the Leblanc soda process. Instead of wasting HCl, chemists found that passing it with air over a hot catalyst (CuCl₂) converts it back to chlorine gas. The key insight: this is a catalytic oxidation of HCl, not a photochemical or simple displacement reaction. The catalyst lowers the temperature needed, making the process economically viable.

    Step-by-step reasoning

    1. Identify the defining features of Deacon’s process

       • Reactants: hydrogen chloride (HCl) and oxygen (O₂).
       • Products: chlorine (Cl₂) and water (H₂O).
       • Catalyst: copper(II) chloride (CuCl₂).
       • Temperature: around 723 K (450 °C).
       • Overall reaction: $4\mathrm{HCl} + \mathrm{O}_2 \rightarrow 2\mathrm{Cl}_2 + 2\mathrm{H}_2\mathrm{O}$.

    2. Examine each option against these criteria

       • (A) $2\mathrm{H}_2\mathrm{O} + 2\mathrm{Cl}_2 \xrightarrow{\text{sunlight}} 4\mathrm{HCl} + \mathrm{O}_2$

         This is the reverse of Deacon’s process — it uses sunlight to decompose chlorine into HCl. Not correct.

       • (B) $4\mathrm{HCl} + \mathrm{O}_2 \xrightarrow[\text{723 K}]{\text{CuCl}_2} 2\mathrm{Cl}_2 + 2\mathrm{H}_2\mathrm{O}$

         Matches exactly: reactants, products, catalyst (CuCl₂), and temperature (723 K). This is the textbook Deacon process.

       • (C) $2\mathrm{NaCl} + \mathrm{H}_2\mathrm{SO}_4 \xrightarrow{\text{823 K}} \mathrm{Na}_2\mathrm{SO}_4 + 2\mathrm{HCl}$

         This is the salt-cake or Mannheim process for making HCl, not chlorine. No oxygen or catalyst involved. …

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11. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.Identify the product 'P' in the given reaction sequence $\ce{(CH3)2C=C(CH3)2 ->[(1) O3][(2) Zn/H2O] A ->[(1) Ba(OH)2][(2) \Delta] P}$ (A) $\ce{(CH3)2C(OH)-CH2-CO-CH3}$ (4-hydroxy-4-methylpentan-2-one) (B) $\ce{(CH3)2CH-CH(OH)-CO-CH3}$ (3-hydroxy-4-methylpentan-2-one) (C) $\ce{(CH3)2C=CH-CO-CH3}$ (4-methylpent-3-en-2-one) (D) $\ce{(CH3)2C=C(OH)-CHO}$

    ›Reveal solutionSolution

    Ozonolysis gives acetone; $\ce{Ba(OH)2}$ then aldol-condenses it to diacetone alcohol, which on heating loses water to give mesityl oxide — option (C).

    The concept first

    Reductive ozonolysis ($\ce{O3}$, then $\ce{Zn}/\ce{H2O}$) cuts a $\ce{C=C}$ and caps each carbon with $\ce{=O}$, keeping every substituent. The zinc is there to destroy the $\ce{H2O2}$ that would otherwise oxidise the products further.

    Aldol condensation requires an $\alpha$-hydrogen. A base removes it to make a carbanion/enolate, which adds to the carbonyl carbon of a second molecule (aldol addition, giving a $\beta$-hydroxy carbonyl). Warming then eliminates water to form an $\alpha,\beta$-unsaturated carbonyl — the product is stabilised by conjugation, and that stabilisation is the driving force for the dehydration.

    Step 1 — Find A

    $\ce{(CH3)2C=C(CH3)2 ->[(1) O3][(2) Zn/H2O] 2\ CH3-CO-CH3}$

    A tetrasubstituted alkene cleaves symmetrically, so A = propanone (acetone).

    Step 2 — Aldol addition with $\ce{Ba(OH)2}$

    Acetone has six $\alpha$-hydrogens. The base makes the enolate, which attacks the carbonyl of another acetone:

    $\ce{CH3COCH3 + CH3COCH3 ->[Ba(OH)2] (CH3)2C(OH)-CH2-CO-CH3}$ …

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12. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.The correct statements about the products B and C in the given reactions are (Anhy = anhydrous, ethanolic) $\text{CH}_3\text{CH}_2\text{OH} \xrightarrow[\text{Anhy ZnCl}_2]{\text{HCl}} \text{A} \xrightarrow{\text{ethanolic AgCN}} \text{B (Minor)} + \text{C (Major)}$ I. B and C are functional isomers II. With $H_2$ | Catalyst B gives 1° amine and C gives 2° amine III. B on acid hydrolysis gives formic acid and C gives $C_3H_6O_2$ IV. C forms isocyanate with HgO (A) I & III (B) II & III (C) I, II & IV (D) II, III & IV

    ›Reveal solutionSolution

    $\mathrm{AgCN}$ is covalent, so its nitrogen attacks: the major product C is ethyl isocyanide and the minor B is the nitrile. Checking the four statements, I, II and IV are true and III is reversed — option (C).

    The concept first

    The cyanide ion is ambident: it can bond through carbon (giving a nitrile, $\mathrm{R\!-\!C\!\equiv\!N}$) or through nitrogen (giving an isocyanide, $\mathrm{R\!-\!N\!\equiv\!C}$). Which end wins depends on the metal salt:

    • KCN is ionic. The free $\mathrm{CN^-}$ attacks through carbon (the better nucleophilic centre) → nitrile is major.
    • AgCN is largely covalent. The carbon is tied up with silver, so only the nitrogen lone pair is available → isocyanide is major.

    That single fact drives the whole question.

    Step-by-step

    1. Make A. $\mathrm{CH_3CH_2OH + HCl \xrightarrow{anhy.\ ZnCl_2} CH_3CH_2Cl}$ (Groves' process). A = ethyl chloride.
    2. React with ethanolic AgCN.

    $\mathrm{C_2H_5Cl \xrightarrow{AgCN} \underbrace{C_2H_5NC}_{C\ (major)} + \underbrace{C_2H_5CN}_{B\ (minor)}}$

    3. Statement I — functional isomers? Both are $\mathrm{C_3H_5N}$. Same molecular formula, different functional group (cyanide vs isocyanide). TRUE.
    4. Statement II — reduction products?
       • $\mathrm{C_2H_5C\!\equiv\!N \xrightarrow{H_2/cat} C_2H_5CH_2NH_2}$ (propan-1-amine): the nitrogen carries two H's → primary amine ✓ (B gives $1^\circ$).
       • $\mathrm{C_2H_5\!-\!N\!\equiv\!C \xrightarrow{H_2/cat} C_2H_5\!-\!NH\!-\!CH_3}$ ($N$-methylethanamine): the nitrogen already carries the ethyl group, and the reduced carbon becomes a methyl → secondary amine ✓ (C gives $2^\circ$). TRUE. …

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