Q.A compound Z with molecular formula $C_3H_9N$ reacts with $C_6H_5SO_2Cl$ to give a solid, insoluble in alkali. Identify Z.

Showing the 12 most recent of 15 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Benzene gets converted to ‘X’ in a reaction (A) and to ‘Y’ in another reaction (B). X gets oxidized by ammoniacal silver nitrate solution but not Y. Reactions A and B respectively are (A) Stephen; Fittig (B) Fittig; Stephen (C) Gatterman-Koch; Friedel-Crafts (D) Friedel-Crafts; Gatterman-Koch

   ›Reveal solutionSolution

   The key is that X must be an aldehyde (oxidizable by Tollens’ reagent) and Y must be a ketone (not oxidizable). Benzene to aldehyde is Gatterman–Koch; benzene to ketone is Friedel–Crafts acylation. So reaction A = Gatterman–Koch, reaction B = Friedel–Crafts. The correct option is (C).

   The problem tests your knowledge of two classic benzene reactions and the chemical distinction between aldehydes and ketones. The clue is that X reacts with ammoniacal silver nitrate (Tollens’ reagent) — that means X is an aldehyde (or another easily oxidized group like a formyl group). Y does not react, so Y is a ketone. Now we just need to match which reaction on benzene gives an aldehyde and which gives a ketone.

   1. Identify the functional group of X and Y

      • Tollens’ reagent oxidizes aldehydes to carboxylic acids, producing a silver mirror. Ketones are not oxidized under these mild conditions.
      • Therefore, X must be an aldehyde (e.g., benzaldehyde) and Y must be a ketone (e.g., acetophenone).

   2. Recall the reactions that introduce a –CHO group onto benzene

      • Gatterman–Koch reaction: Benzene + CO + HCl in the presence of AlCl₃ and CuCl gives benzaldehyde.

   $\text{C}_6\text{H}_6 + \text{CO} + \text{HCl} \xrightarrow{\text{AlCl}_3/\text{CuCl}} \text{C}_6\text{H}_5\text{CHO}$

   • This is a direct formylation, producing an aldehyde.

   3. Recall the reaction that introduces a –COR group onto benzene
      • Friedel–Crafts acylation: Benzene + acyl chloride (or anhydride) in the presence of AlCl₃ gives a ketone.

   $\text{C}_6\text{H}_6 + \text{RCOCl} \xrightarrow{\text{AlCl}_3} \text{C}_6\text{H}_5\text{COR} + \text{HCl}$

   • The product is a ketone, which does not react with Tollens’ reagent.

   4. Match the reactions to A and B
      • Reaction A produces X (aldehyde) → must be Gatterman–Koch. …

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2. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.Identify the functional group Y in the end product of the reaction sequence? Heptane $\xrightarrow{\text{Mo}_2\text{O}_3 \\ 773\,\text{K} \\ 10-20\,\text{atm}}$ A $\xrightarrow{\text{(i) (CH}_3\text{CO)}_2\text{O}+\text{CrO}_3 \\ 273-283\,\text{K}}$ $\xrightarrow{\text{(ii) H}_3\text{O}^+,\,\Delta}$ $\longrightarrow$ \chemfig{*6(-=-=-=)}-Y (A) $-\mathrm{OH}$ (B) $\mathrm{O} \atop{\|} -\mathrm{C}-\mathrm{CH}_3$ (C) $\mathrm{O} \atop{\|} -\mathrm{C}-\mathrm{OH}$ (D) $\mathrm{O} \atop{\|} -\mathrm{C}-\mathrm{H}$

   ›Reveal solutionSolution

   The reaction sequence involves the aromatization of heptane to toluene, followed by the Etard reaction which oxidizes the methyl group of toluene to an aldehyde group. The functional group Y is therefore an aldehyde. The final answer is (D).

   The problem asks us to identify the functional group Y in the final product of a two-step reaction sequence starting from heptane. This requires understanding two key organic reactions: aromatization of alkanes and the Etard reaction.

   The first step converts a straight-chain alkane into an aromatic compound. This process, known as aromatization or catalytic reforming, involves cyclization, dehydrogenation, and isomerization. For a 7-carbon alkane like heptane, it typically forms toluene (methylbenzene).

   The second step involves the selective oxidation of the methyl group attached to the aromatic ring. The reagents $\text{(CH}_3\text{CO)}_2\text{O} + \text{CrO}_3$ at low temperature are characteristic of the Etard reaction, which specifically converts a benzylic methyl group into an aldehyde group.

   Let's break down the reaction sequence step-by-step.

   1. First Reaction: Aromatization of Heptane
      • Starting Material: Heptane ($\text{CH}_3(\text{CH}_2)_5\text{CH}_3$), a straight-chain alkane with 7 carbon atoms.
      • Reagents/Conditions: $\text{Mo}_2\text{O}_3$ (molybdenum oxide catalyst), $773\,\text{K}$ (high temperature), $10-20\,\text{atm}$ (high pressure).
      • Concept: These conditions are characteristic of aromatization (catalytic reforming). Alkanes with six or more carbon atoms, when heated to high temperatures and pressures in the presence of catalysts like $\text{Cr}_2\text{O}_3$, $\text{V}_2\text{O}_5$, or $\text{Mo}_2\text{O}_3$ supported on alumina, undergo cyclization and dehydrogenation to form aromatic compounds.
      • Product A: For heptane (7 carbons), the product is toluene (methylbenzene).

   $\text{CH}_3(\text{CH}_2)_5\text{CH}_3 \xrightarrow{\text{Mo}_2\text{O}_3 \\ 773\,\text{K} \\ 10-20\,\text{atm}} \chemfig{*6(-=-=-=)\text{-CH}_3} + 4\text{H}_2$

       So, **A is Toluene**.
   

   2. Second Reaction: Etard Reaction of Toluene

   * Starting Material: Toluene (A).

   * Reagents/Conditions:

   * (i) $(\text{CH}_3\text{CO})_2\text{O} + \text{CrO}_3$ (acetic anhydride and chromium trioxide) at $273-283\,\text{K}$ (low temperature).

   * (ii) $\text{H}_3\text{O}^+,\,\Delta$ (acidic hydrolysis with heating).

   * Concept: This is the Etard reaction. It is a selective oxidation reaction that converts a methyl group directly attached to an aromatic ring (a benzylic methyl group) into an aldehyde group. The reaction proceeds via the formation of an intermediate chromium complex (Etard complex), which is then hydrolyzed under acidic conditions to yield the aldehyde.

   * Product: The methyl group of toluene is oxidized to an aldehyde group.

   $\chemfig{*6(-=-=-=)\text{-CH}_3} \xrightarrow{\text{(i) (CH}_3\text{CO)}_2\text{O}+\text{CrO}_3 \\ 273-283\,\text{K}} \text{Etard Complex} \xrightarrow{\text{(ii) H}_3\text{O}^+,\,\Delta} \chemfig{*6(-=-=-=)\text{-CHO}}$

       The final product is **Benzaldehyde**. …
   

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3. TG EAPCET 2024Set ap-2024-05-08-FN1 markMCQ
   Q.Which of the following sequence of reagents convert benzoic acid to benzaldehyde? (A) $\mathrm{C_2H_5OH, H^+;

   (i) DIBAL-H\ (ii)\ H_2O}$(B)$\mathrm{SOCl_2; H_2|Ni}$(C)$\mathrm{C_2H_5OH, H^+; LiAlH_4, H_2O}$(D)$\mathrm{LiAlH_4, H_2O; KMnO_4|H^+}$

   ›Reveal solutionSolution

   Benzoic acid is first converted to an ester, ethyl benzoate, which is then selectively reduced to benzaldehyde using DIBAL-H at low temperatures. The correct sequence of reagents is (A).

   The challenge in converting a carboxylic acid to an aldehyde lies in the fact that aldehydes are generally more reactive towards reduction than carboxylic acids themselves. If you try to reduce a carboxylic acid directly, it's very difficult to stop the reaction at the aldehyde stage; the aldehyde will almost always be further reduced to a primary alcohol.

   To overcome this, a common strategy is to first convert the carboxylic acid into a less reactive derivative, such as an ester or an acid chloride. This derivative can then be reduced using a mild, selective reducing agent that stops precisely at the aldehyde stage.

   Let's evaluate each option based on this principle:

   1. Analyze Option (A): $\mathrm{C_2H_5OH, H^+; (i) DIBAL-H\ (ii)\ H_2O}$
      • Step 1: $\mathrm{C_2H_5OH, H^+}$ Benzoic acid undergoes Fischer esterification with ethanol in the presence of an acid catalyst ($\mathrm{H^+}$) to form ethyl benzoate.

   $\mathrm{C_6H_5COOH + C_2H_5OH \xrightarrow{H^+} C_6H_5COOC_2H_5 + H_2O}$

       This converts the carboxylic acid into an ester, which is a suitable derivative for controlled reduction.
   *   **Step 2: $\mathrm{(i) DIBAL-H\ (ii)\ H_2O}$**
       Diisobutylaluminium hydride (DIBAL-H) is a bulky, mild reducing agent. When used in stoichiometric amounts at low temperatures (typically $-78^\circ\mathrm{C}$), it selectively reduces esters to aldehydes. The reaction proceeds via an intermediate hemiacetal, which is stable at low temperatures and then hydrolyzes to the aldehyde upon aqueous workup.
   

   $\mathrm{C_6H_5COOC_2H_5 \xrightarrow{(i) DIBAL-H, -78^\circ C (ii) H_2O} C_6H_5CHO + C_2H_5OH}$

       This sequence successfully converts benzoic acid to benzaldehyde.
   

   2. Analyze Option (B): $\mathrm{SOCl_2; H_2|Ni}$

   * Step 1: $\mathrm{SOCl_2}$

   Benzoic acid reacts with thionyl chloride ($\mathrm{SOCl_2}$) to form benzoyl chloride. This is a standard method to convert carboxylic acids to acid chlorides.

   $\mathrm{C_6H_5COOH + SOCl_2 \rightarrow C_6H_5COCl + SO_2 + HCl}$

   *   **Step 2: $\mathrm{H_2|Ni}$**
       Hydrogenation with a nickel catalyst ($\mathrm{H_2|Ni}$) is a strong reducing condition. While acid chlorides can be reduced to aldehydes using hydrogen with a poisoned palladium catalyst (Rosenmund reduction, e.g., $\mathrm{Pd/BaSO_4}$), $\mathrm{H_2|Ni}$ is not selective for stopping at the aldehyde stage. It would likely reduce the acid chloride further, potentially to a primary alcohol ($\mathrm{C_6H_5CH_2OH}$), or it might not be the appropriate catalyst for this specific transformation. Therefore, this option is unlikely to yield benzaldehyde selectively.
   

   3. Analyze Option (C): $\mathrm{C_2H_5OH, H^+; LiAlH_4, H_2O}$

   * Step 1: $\mathrm{C_2H_5OH, H^+}$

   As in option (A), this step forms ethyl benzoate.

   $\mathrm{C_6H_5COOH + C_2H_5OH \xrightarrow{H^+} C_6H_5COOC_2H_5 + H_2O}$

   *   **Step 2: $\mathrm{LiAlH_4, H_2O}$** …
   

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4. TG EAPCET 2024Set ap-2024-05-07-AN1 markMCQ
   Q.A carbonyl compound $X(\mathrm{C}_3\mathrm{H}_6\mathrm{O})$ on oxidation gave carboxylic acid $Y(\mathrm{C}_3\mathrm{H}_6\mathrm{O}_2)$. Oxime of X is (A) $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH} = \mathrm{NNH}_2$ (B) $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH} = \mathrm{NOH}$ (C) $(\mathrm{CH}_3)_2\mathrm{C} = \mathrm{N} - \mathrm{NH}_2$ (D) $(\mathrm{CH}_3)_2\mathrm{C} = \mathrm{N} - \mathrm{OH}$

   ›Reveal solutionSolution

   The carbonyl compound X (C₃H₆O) is propanal, which forms an oxime with hydroxylamine; the oxime is CH₃CH₂CH=NOH, corresponding to option (B).

   Concept & Intuition

   We are told that a carbonyl compound X with formula C₃H₆O (an aldehyde or ketone) is oxidized to a carboxylic acid Y with formula C₃H₆O₂. The key clue: oxidation of an aldehyde gives a carboxylic acid with the same number of carbon atoms, while oxidation of a ketone breaks the carbon chain (giving smaller acids). Since Y has exactly 3 carbons, X must be an aldehyde — specifically propanal (CH₃CH₂CHO). The oxime of an aldehyde or ketone is formed by reaction with hydroxylamine (NH₂OH), giving a C=NOH group. So the oxime of propanal is CH₃CH₂CH=NOH.

   Step-by-step reasoning

   1. Identify the functional group of X

      The molecular formula C₃H₆O corresponds to either an aldehyde (propanal) or a ketone (acetone). Both are carbonyl compounds.

   2. Use the oxidation result

      Oxidation of an aldehyde yields a carboxylic acid with the same number of carbons. Oxidation of a ketone cleaves the carbon chain, typically giving a mixture of smaller acids. Here Y is C₃H₆O₂, a three-carbon carboxylic acid (propanoic acid). This is only possible if X is an aldehyde — propanal (CH₃CH₂CHO).

   3. Confirm the identity of X

      Propanal: CH₃CH₂CHO. Oxidation gives propanoic acid: CH₃CH₂COOH (C₃H₆O₂). Perfect match.

   4. Form the oxime of X

      An oxime is formed when a carbonyl compound reacts with hydroxylamine (NH₂OH), replacing the C=O with C=NOH.

      For propanal:

      $$\text{CH}_3\text{CH}_2\text{CHO} + \text{NH}_2\text{OH} \rightarrow \text{CH}_3\text{CH}_2\text{CH}=\text{NOH} + \text{H}_2\text{O} …$$

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5. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.Which of the following compound has no reaction with sodium metal? (A) Phenol (B) Ethanol (C) Benzoic acid (D) Anisole

   ›Reveal solutionSolution

   The key idea is that sodium metal reacts with compounds containing a labile (acidic) hydrogen atom, typically from –OH or –COOH groups. Anisole (methoxybenzene) has no such hydrogen, so it does not react. The correct option is (D).

   Concept and Intuition

   Sodium metal is a strong reducing agent, but its classic reaction with organic compounds is a single displacement where it replaces a hydrogen atom that is bonded to a highly electronegative atom (like oxygen). This happens when the hydrogen is acidic enough to be displaced as H⁺, forming sodium alkoxides, phenoxides, or carboxylates, plus hydrogen gas.

   • Phenol has an –OH group attached to an aromatic ring; the hydrogen is acidic (pKa ≈ 10).
   • Ethanol has an –OH group; the hydrogen is weakly acidic (pKa ≈ 16) but still reacts with sodium.
   • Benzoic acid has a –COOH group; the hydrogen is strongly acidic (pKa ≈ 4.2).
   • Anisole (C₆H₅–O–CH₃) is an ether; it has no O–H bond — only C–O–C. No labile hydrogen exists, so sodium cannot displace anything.

   Step-by-step reasoning

   1. Identify the reactive functional group Sodium metal reacts with compounds that have a hydrogen atom attached to oxygen (or sometimes nitrogen or sulfur) that can be removed as H⁺. The general reaction is:

   $2\text{R–OH} + 2\text{Na} \rightarrow 2\text{R–ONa} + \text{H}_2\uparrow$

   This requires an O–H bond.

   2. Examine each option
      • (A) Phenol: Structure is C₆H₅–OH. Contains an O–H bond. Reacts:

   $2\text{C}_6\text{H}_5\text{OH} + 2\text{Na} \rightarrow 2\text{C}_6\text{H}_5\text{ONa} + \text{H}_2$

   • (B) Ethanol: Structure is CH₃CH₂–OH. Contains an O–H bond. Reacts:

   $2\text{CH}_3\text{CH}_2\text{OH} + 2\text{Na} \rightarrow 2\text{CH}_3\text{CH}_2\text{ONa} + \text{H}_2$

   • (C) Benzoic acid: Structure is C₆H₅–COOH. Contains an O–H bond in the carboxyl group. Reacts: $$$$

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6. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.Which of the following represents Gatterman-Koch reaction? (A) [FIGURE] Benzoyl chloride ($\mathrm{C_6H_5COCl}$) $\xrightarrow[\mathrm{Pd\text{-}BaSO_4}]{\mathrm{H_2}}$ benzaldehyde ($\mathrm{C_6H_5CHO}$) (B) [FIGURE] Toluene $\xrightarrow[\text{(ii) } \mathrm{H_3O^+}]{\text{(i) } \mathrm{CrO_2Cl_2}/\mathrm{CS_2}}$ benzaldehyde ($\mathrm{C_6H_5CHO}$) (C) [FIGURE] Benzene $\xrightarrow[\text{Anh. } \mathrm{AlCl_3}/\mathrm{CuCl}]{\mathrm{CO},\ \mathrm{HCl}}$ benzaldehyde ($\mathrm{C_6H_5CHO}$) (D) [FIGURE] Benzene $\xrightarrow[\text{Anh. } \mathrm{AlCl_3}]{\mathrm{C_2H_5COCl}}$ propiophenone ($\mathrm{C_6H_5CO-C_2H_5}$)

   ›Reveal solutionSolution

   Gatterman–Koch formylates benzene with $\mathrm{CO}$ and $\mathrm{HCl}$ over anhydrous $\mathrm{AlCl_3}/\mathrm{CuCl}$ to give benzaldehyde. That is option (C); the other three are Rosenmund, Étard and Friedel–Crafts acylation. Option (C).

   The concept: why we need a special reaction at all

   To attach a $-\mathrm{CHO}$ group to benzene by a Friedel–Crafts acylation you would need formyl chloride, $\mathrm{HCOCl}$ — but that compound is unstable and decomposes instantly to $\mathrm{CO} + \mathrm{HCl}$.

   Gatterman–Koch simply runs that decomposition backwards in situ: pass $\mathrm{CO}$ and $\mathrm{HCl}$ gases into benzene in the presence of anhydrous $\mathrm{AlCl_3}$ with $\mathrm{CuCl}$ as a co-catalyst; the electrophilic formylating species is generated on the spot:

   $\mathrm{C_6H_6} + \mathrm{CO} + \mathrm{HCl} \xrightarrow[\mathrm{CuCl}]{\text{anhyd. } \mathrm{AlCl_3}} \mathrm{C_6H_5CHO}$

   The $\mathrm{CuCl}$ helps absorb the $\mathrm{CO}$ and keep the reactive species available.

   Step 1 — Identify each option

   • (A) $\mathrm{C_6H_5COCl} \xrightarrow{\mathrm{H_2},\ \mathrm{Pd\text{-}BaSO_4}} \mathrm{C_6H_5CHO}$ — this is the Rosenmund reduction: a poisoned palladium catalyst reduces an acyl chloride only as far as the aldehyde. Not our reaction. ✗ …

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7. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.The major products of Reimer-Tiemann reaction and Kolbe reaction are respectively (A) [FIGURE] Phenol ($\mathrm{C_6H_5OH}$), and benzaldehyde ($\mathrm{C_6H_5CHO}$) (B) [FIGURE] Salicylaldehyde (benzene ring with $-\mathrm{OH}$ and, ortho to it, $-\mathrm{CHO}$), and salicylic acid (benzene ring with $-\mathrm{OH}$ and, ortho to it, $-\mathrm{CO_2H}$) (C) [FIGURE] o-Cresol (benzene ring with $-\mathrm{OH}$ and, ortho to it, $-\mathrm{CH_3}$), and salicylic acid (benzene ring with $-\mathrm{OH}$ and, ortho to it, $-\mathrm{CO_2H}$) (D) [FIGURE] 4-Nitrophenol (benzene ring with $-\mathrm{OH}$ and, para to it, $-\mathrm{NO_2}$), and 1,4-benzoquinone

   ›Reveal solutionSolution

   Reimer–Tiemann $\Rightarrow$ salicylaldehyde; Kolbe $\Rightarrow$ salicylic acid. Both are ortho products from the phenoxide ion. Option (B).

   The concept: the phenoxide ion is the real nucleophile

   Phenol on its own is a modest nucleophile. But with $\mathrm{NaOH}$ it becomes the phenoxide ion, whose negative charge is delocalised onto the ortho and para ring carbons. That extra electron density lets the ring attack even weak electrophiles. The two reactions differ only in which weak electrophile is offered.

   Reaction 1 — Reimer–Tiemann (the electrophile is dichlorocarbene)

   $\mathrm{CHCl_3} + \mathrm{OH^{-}} \longrightarrow :\!\mathrm{CCl_2} + \mathrm{Cl^{-}} + \mathrm{H_2O}$

   The phenoxide attacks $:\!\mathrm{CCl_2}$ at the ortho carbon; rearomatisation gives an $\mathrm{ArCHCl_2}$ group, which the alkaline medium hydrolyses:

   $\mathrm{Ar\!-\!CHCl_2} \xrightarrow{\mathrm{OH^{-}},\ \text{then } \mathrm{H_3O^{+}}} \mathrm{Ar\!-\!CHO}$

   $\textbf{Product: salicylaldehyde (2-hydroxybenzaldehyde)}$

   Reaction 2 — Kolbe (the electrophile is $\mathrm{CO_2}$)

   $\mathrm{C_6H_5OH} \xrightarrow{\mathrm{NaOH}} \mathrm{C_6H_5O^{-}Na^{+}} \xrightarrow[\text{pressure}]{\mathrm{CO_2}} \text{sodium salicylate} \xrightarrow{\mathrm{H_3O^{+}}} \textbf{salicylic acid}$

   The phenoxide attacks the electrophilic carbon of $\mathrm{CO_2}$, again at the ortho position (the sodium ion chelates the phenoxide oxygen and the incoming $\mathrm{CO_2}$, holding it next door). …

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8. TG EAPCET 2023Set ap-2023-05-11-FN1 markMCQ
   Q.phenol $\xrightarrow{\ce{CH_3CO,H}}$ X $\xrightarrow{\ce{AlCl_3}}$ Y (Major) The incorrect statement about Y is (A) It undergoes reduction with $\ce{H_2N-NH_2}$ / KOH, glycol, $\Delta$ (B) It gives iodoform test (C) It liberates hydrogen with Na metal (D) Conversion of X to Y is Friedel-Crafts reaction

   ›Reveal solutionSolution

   Phenol reacts with acetic anhydride to form phenyl acetate (X), which undergoes Fries rearrangement with AlCl₃ to give ortho- and para-hydroxyacetophenone (Y, major = para). The incorrect statement is (D) — the conversion is a Fries rearrangement, not a Friedel-Crafts reaction.

   The reaction sequence begins with phenol and acetic anhydride. Acetic anhydride acetylates the phenolic –OH group to form an ester, phenyl acetate (X). When this ester is treated with a Lewis acid like AlCl₃, the acetyl group migrates from the oxygen to the aromatic ring in what is known as the Fries rearrangement. The major product (Y) is para-hydroxyacetophenone, though some ortho-isomer also forms.

   The structure of Y is:

   $\ce{HO-C6H4-CO-CH3}$

   (the –OH and –COCH₃ groups are para to each other).

   Now let's examine each statement about Y:

   1. Statement (A): Reduction with hydrazine/KOH (Wolff-Kishner) or glycol/Δ (Clemmensen-like conditions) Y contains a ketone group (–CO–CH₃). The Wolff-Kishner reduction converts a carbonyl to a methylene group:

   $\ce{-CO-CH3 -> -CH2-CH3}$

   This is a standard reaction for ketones, so Y will indeed undergo this reduction. Statement (A) is correct.

   2. Statement (B): Iodoform test

      The iodoform test is positive for compounds containing the structural unit $\ce{CH3-CO-}$ (a methyl ketone) or $\ce{CH3-CH(OH)-}$ (secondary alcohol with a methyl group). Y is para-hydroxyacetophenone, which has the $\ce{-CO-CH3}$ group directly attached to the benzene ring. When treated with iodine and base, the methyl ketone oxidizes and cleaves to give iodoform (CHI₃, a yellow precipitate). Statement (B) is correct.

   3. Statement (C): Liberation of hydrogen with Na metal

      Y has a phenolic –OH group. Phenols are weakly acidic and react with sodium metal to liberate hydrogen gas:

   $\ce{Ar-OH + Na -> Ar-O^- Na^+ + 1/2 H2 ^}$

   Statement (C) is correct.

   4. Statement (D): Conversion of X to Y is a Friedel-Crafts reaction …

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9. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.I. $(\mathrm{CH}_3)_2\mathrm{C} = \mathrm{CH}_2 \xrightarrow{\mathrm{KMnO}_4/\mathrm{H}^+} X + \mathrm{CO}_2 + \mathrm{H}_2\mathrm{O}$ II. $\mathrm{CH}_3 - \mathrm{CH} = \mathrm{CH} - \mathrm{CH}_3 \xrightarrow{\mathrm{KMnO}_4/\mathrm{H}^+} Y$ The functional groups in X and Y are respectively Options : (A) $\begin{matrix} \mathrm{O} \\ \vert \\ \mathrm{C} - \mathrm{CH}_3 \end{matrix} , \quad \begin{matrix} \mathrm{O} \\ \vert \\ \mathrm{C} - \mathrm{H} \end{matrix}$ (B) $\begin{matrix} \mathrm{O} \\ \vert \\ \mathrm{C} - \mathrm{CH}_3 \end{matrix} , \quad \begin{matrix} \mathrm{O} \\ \vert \\ \mathrm{C} - \mathrm{OH} \end{matrix}$ (C) $\begin{matrix} \mathrm{O} \\ \vert \\ \mathrm{C} - \mathrm{H} \end{matrix} , \quad \begin{matrix} \mathrm{O} \\ \vert \\ \mathrm{C} - \mathrm{O} \end{matrix}$ (D) $\begin{matrix} \mathrm{O} \\ \vert \\ \mathrm{C} - \mathrm{O} \end{matrix} , \quad \begin{matrix} \mathrm{O} \\ \vert \\ \mathrm{C} - \mathrm{H} \end{matrix}$

   ›Reveal solutionSolution

   The key idea is that hot acidic KMnO₄ cleaves alkenes at the double bond, oxidising each vinylic carbon to a carbonyl group (ketone, aldehyde, or CO₂ depending on substitution). For (CH₃)₂C=CH₂, the more substituted side gives acetone (a ketone) and the terminal =CH₂ gives CO₂; for CH₃–CH=CH–CH₃, both sides give acetic acid (a carboxylic acid). So X has a ketone group and Y has a carboxylic acid group, which matches option (B).

   ---

   Concept & Intuition

   Hot acidic KMnO₄ is a strong oxidising agent that breaks carbon‑carbon double bonds completely. Each carbon that was part of the double bond ends up as part of a carbonyl group (C=O). The exact product depends on how many hydrogens that carbon originally had:

   • A carbon with two alkyl groups (disubstituted) becomes a ketone.
   • A carbon with one alkyl group and one hydrogen (monosubstituted) becomes a carboxylic acid.
   • A carbon with two hydrogens (terminal =CH₂) becomes CO₂ (plus water).

   This is a classic “oxidative cleavage” reaction — think of it as cutting the alkene in half and oxidising the cut ends.

   ---

   Step‑by‑step reasoning

   1. Identify the alkene in reaction I

      The alkene is (CH₃)₂C=CH₂.

      • Left side of the double bond: the carbon is attached to two methyl groups (and the other carbon) → it is a disubstituted vinylic carbon.
      • Right side: the carbon is attached to two hydrogens (and the other carbon) → it is a terminal (=CH₂) carbon.

   2. Predict the products for reaction I

      • The disubstituted carbon (with two alkyl groups) oxidises to a ketone: acetone, (CH₃)₂C=O.
      • The terminal carbon (with two H’s) oxidises all the way to CO₂ (and H₂O). So X is acetone, which contains the ketone functional group (–C(=O)–).

   3. Identify the alkene in reaction II

      The alkene is CH₃–CH=CH–CH₃ (but‑2‑ene).

      • Both vinylic carbons are identical: each is attached to one methyl group and one hydrogen → they are monosubstituted (one alkyl, one H).

   4. Predict the products for reaction II

      Each monosubstituted carbon oxidises to a carboxylic acid.

      • CH₃–CH= becomes CH₃–COOH (acetic acid). …

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10. TG EAPCET 2022Set ap-2022-07-30-FN1 markMCQ
    Q.The major product of the following reaction sequence is [FIGURE: 1-[2-(hydroxymethyl)phenyl]propan-1-one — a benzene ring carrying a $-C(=O)CH_2CH_3$ group and, at the ortho position, a $-CH_2OH$ group] $\xrightarrow{\text{i) PCC; ii) NaOH, }\Delta}$ (A) 2-methylindane-1,3-dione — bicyclic indane with two C=O groups and a $CH_3$ on the carbon between them (B) 2-methyl-1H-inden-1-one — indenone with the $CH_3$ on the carbon adjacent to the C=O (C) 3-methyl-1H-inden-1-one — indenone with the $CH_3$ on the carbon away from the C=O (position 3) (D) Isobenzofuran-1(3H)-one (phthalide) — bicyclic five-membered lactone with a ring oxygen and one C=O

    ›Reveal solutionSolution

    PCC oxidises the $-\mathrm{CH_2OH}$ to $-\mathrm{CHO}$; NaOH/$\Delta$ then drives an intramolecular aldol condensation between the ketone's $\alpha$-carbon and that aldehyde, closing a five-membered ring. The product is 2-methyl-1H-inden-1-one — option (B).

    The concept first

    Two classical reactions, run back to back.

    PCC (pyridinium chlorochromate) is the mild chromium oxidant. In an anhydrous medium ($\mathrm{CH_2Cl_2}$) it converts:

    $\mathrm{1^\circ\ alcohol} \longrightarrow \textbf{aldehyde} \quad(\text{it stops here — no carboxylic acid})$

    $\mathrm{2^\circ\ alcohol} \longrightarrow \text{ketone}$

    Aldol condensation requires a carbonyl with an $\alpha$-hydrogen (to make the nucleophilic enolate) and a carbonyl that is electrophilic (to be attacked). Warm base then dehydrates the $\beta$-hydroxy carbonyl to a conjugated enone. When both carbonyls sit in the same molecule and geometry permits, the aldol becomes intramolecular and forges a ring — and rings of size five or six form fastest.

    Step-by-step

    1. Read the substrate. A benzene ring carries:
       • a propanoyl group, $-\mathrm{C(=O)CH_2CH_3}$, and
       • ortho to it, a primary alcohol, $-\mathrm{CH_2OH}$.
    2. Step (i) — PCC. The primary alcohol is oxidised, and PCC stops cleanly at the aldehyde:

    $\mathrm{Ar}\!-\!\mathrm{CH_2OH} \xrightarrow{\ \mathrm{PCC}\ } \mathrm{Ar}\!-\!\mathrm{CHO}$

    We now have a 1,2-disubstituted benzene bearing an aldehyde and an ethyl ketone side by side.

    3. Step (ii) — identify the enolate. Which carbonyl has an $\alpha$-hydrogen?

    • The aryl aldehyde ($\mathrm{ArCHO}$) has no $\alpha$-H — the carbon next to it is an aromatic carbon. It can only be the electrophile.
    • The ketone has an $\alpha$-$\mathrm{CH_2}$ (from the ethyl group). Hydroxide removes one of those protons to give the enolate. It is the nucleophile. This asymmetry is what makes the reaction clean — no self-condensation confusion.

    4. Close the ring. The enolate carbon attacks the aldehyde carbon. Trace the atoms of the new ring: $\underbrace{\mathrm{C}_{ar}}_{1} - \underbrace{\mathrm{C}{=}\mathrm{O}}_{2} - \underbrace{\alpha\mathrm{C(CH_3)}}_{3} - \underbrace{\mathrm{CH(OH)}}_{4} - \underbrace{\mathrm{C}_{ar}}_{5}$ …

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11. TG EAPCET 2022Set ap-2022-07-31-AN1 markMCQ
    Q.The total number of aromatic (benzenoid compounds) isomers for the molecular formula of $\mathrm{C_7H_8O}$ is (A) 2 (B) 4 (C) 5 (D) 6

    ›Reveal solutionSolution

    We systematically place a $\ce{-CH3}$ and $\ce{-OH}$ group on a benzene ring, then consider all possible positions. The total number of benzenoid aromatic isomers is 5.

    The molecular formula $\ce{C7H8O}$ with a benzene ring ($\ce{C6H4}$) leaves us with $\ce{CH4O}$ to distribute as substituents. For aromatic compounds, we need to attach groups that maintain the benzene core while accounting for all atoms.

    The degree of unsaturation is $\frac{2(7) + 2 - 8}{2} = 4$, which matches a benzene ring (4 degrees). So we're looking at a disubstituted benzene with one carbon and one oxygen in the substituents.

    The possible substitution patterns are:

    • A methyl group ($\ce{-CH3}$) and a hydroxyl group ($\ce{-OH}$): methylphenols (cresols)
    • A methoxy group ($\ce{-OCH3}$): anisole

    Let me work through each case.

    Methylphenols (Cresols): $\ce{-CH3}$ and $\ce{-OH}$ on benzene

    When two different groups occupy a benzene ring, we have three positional isomers:

    1. ortho (1,2-positions): The $\ce{-OH}$ and $\ce{-CH3}$ are adjacent → o-cresol

    2. meta (1,3-positions): The groups are separated by one carbon → m-cresol

    3. para (1,4-positions): The groups are opposite each other → p-cresol

    All three cresols are aromatic and satisfy $\ce{C7H8O}$.

    Methoxybenzene (Anisole): $\ce{-OCH3}$ on benzene …

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12. TG EAPCET 2021Set ap-2021-08-10-AN1 markMCQ
    Q.The molecule that undergo self oxidation and reduction (disproportionation) reaction intramolecularly upon heating with concentrated alkali is (A) 3-Bromobenzaldehyde (B) 4-Methoxybenzaldehyde (C) Phthalic acid (D) Phthalaldehyde

    ›Reveal solutionSolution

    The key is that only an aldehyde lacking an alpha‑hydrogen can undergo the Cannizzaro reaction (intramolecular disproportionation) in concentrated alkali. Phthalaldehyde, with two aldehyde groups on the same benzene ring, can undergo an intramolecular Cannizzaro reaction upon heating with concentrated alkali. The correct option is (D).

    The concept here is the Cannizzaro reaction — a disproportionation of an aldehyde without an α‑hydrogen, where one molecule is oxidized to a carboxylic acid and another is reduced to an alcohol, under concentrated alkali. When two aldehyde groups are present on the same molecule (like in phthalaldehyde), the reaction can occur intramolecularly: one –CHO gets oxidized, the other gets reduced, forming a hydroxy acid. This is a classic test for aromatic aldehydes without α‑hydrogens.

    Let’s examine each option:

    1. 3‑Bromobenzaldehyde – This is a simple aromatic aldehyde with no α‑hydrogen. It can undergo the Cannizzaro reaction, but only intermolecularly (between two separate molecules). The question specifies “intramolecularly” — so this does not fit.

    2. 4‑Methoxybenzaldehyde – Same as above: it has no α‑hydrogen, so it undergoes intermolecular Cannizzaro, not intramolecular. Not the answer.

    3. Phthalic acid – This is a dicarboxylic acid, not an aldehyde. It cannot undergo a Cannizzaro reaction at all (no aldehyde group). Incorrect. …

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