Q.The reagent which does not react with both acetone and benzaldehyde is __________.

Showing the 12 most recent of 20 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.Consider the following reaction Cumene $\xrightarrow[(ii)\ H_3O^+]{(i)\ O_2}$ A + B A is acidic in nature and forms salt with aq. NaOH. A and B are converted to corresponding hydrocarbons with reagents X and Y respectively. What are X and Y? (A) $X = Sn, \Delta$ ; $Y = NaBH_4$ (B) $X = NaBH_4$ ; $Y = Sn, \Delta$ (C) $X = Zn, \Delta$ ; $Y = Zn|Hg, HCl$ (D) $X = Zn|Hg, HCl$ ; $Y = Zn, \Delta$

   ›Reveal solutionSolution

   The cumene process gives phenol (A) and acetone (B). Phenol → benzene needs zinc-dust distillation; acetone → propane needs Clemmensen reduction (Zn-Hg/HCl). So X = Zn,Δ and Y = Zn|Hg,HCl.

   Concept and Intuition

   The cumene process is the industrial route to phenol: cumene (isopropylbenzene) is air-oxidised to cumene hydroperoxide, which is then cleaved by dilute acid ($\mathrm{H_3O^+}$) via the Hock rearrangement to give phenol (A) and acetone (B) simultaneously. Since A (phenol) is acidic (reacts with NaOH to form sodium phenoxide, confirming it's phenol) — this fixes A = phenol, B = acetone.

   To convert each oxygen-containing product back to its parent hydrocarbon:

   • Phenol → benzene: distillation with zinc dust removes the $-\mathrm{OH}$ group (reduction/dehydroxylation), giving benzene.
   • Acetone (a ketone) → propane: the carbonyl group of a ketone is fully reduced to $-\mathrm{CH_2}-$ by the Clemmensen reduction, using zinc amalgam ($\mathrm{Zn(Hg)}$) and concentrated HCl.

   Step-by-Step Solution

   1. Cumene $\xrightarrow{(i)\ O_2}$ cumene hydroperoxide $\xrightarrow{(ii)\ H_3O^+}$ phenol (A) + acetone (B) (Hock cleavage).
   2. A is stated to be acidic and to form a salt with aqueous NaOH — this confirms A = phenol (forms sodium phenoxide).
   3. By elimination, B = acetone.
   4. Converting A (phenol) to its hydrocarbon (benzene): reagent X = zinc dust distillation, $\mathrm{Zn}, \Delta$.
   5. Converting B (acetone) to its hydrocarbon (propane): reagent Y = Clemmensen reduction, $\mathrm{Zn(Hg)}, \mathrm{HCl}$. …

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2. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.Identify C in the given reaction sequence $C_6H_5CHO \xrightarrow{Conc.\ KOH} A + B \xrightarrow[\Delta]{H^+} C$ (A) $H-CO_2C_6H_5$ (B) $C_6H_5CO_2CH_3$ (C) $C_6H_5CO_2CH_2C_6H_5$ (D) $C_6H_5CO_2C_6H_5$

   ›Reveal solutionSolution

   Benzaldehyde + conc. KOH is the Cannizzaro reaction (no $\alpha$-H), giving benzoate + benzyl alcohol; acid-catalysed heating of that pair is Fischer esterification, giving benzyl benzoate.

   Concept and Intuition

   Cannizzaro's reaction is the disproportionation seen only in aldehydes lacking an $\alpha$-hydrogen (so they cannot enolize/undergo aldol): under concentrated base, one molecule's carbonyl carbon is attacked by hydroxide, and a hydride is transferred to a second molecule of aldehyde. This simultaneously oxidizes one molecule to the carboxylate (A) and reduces the other to the alcohol (B). Once you have both a carboxylic acid (after acidifying the salt) and an alcohol, heating them with an acid catalyst is the standard Fischer esterification, forming an ester and water.

   Step-by-Step Solution

   1. $C_6H_5CHO$ has no $\alpha$-H (the carbon next to $CHO$ is aromatic), so conc. KOH causes Cannizzaro disproportionation: one $C_6H_5CHO$ is oxidized to $C_6H_5COO^-K^+$ (A, potassium benzoate) and another is reduced to $C_6H_5CH_2OH$ (B, benzyl alcohol).
   2. $A + B \xrightarrow[\Delta]{H^+}$: acid protonates the benzoate to benzoic acid $C_6H_5COOH$, and under acid catalysis with heat, benzoic acid and benzyl alcohol undergo Fischer esterification, losing water.
   3. Product C = $C_6H_5CO_2CH_2C_6H_5$, benzyl benzoate.

   Common Mistakes …

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3. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.Reaction of benzonitrile (A) with the reagent X gave B. In another reaction of A with the reagent Y gave C. Reaction of B and C in the presence of dil. NaOH at 293 K gave an $\alpha, \beta$-unsaturated carbonyl compound. What are X and Y respectively? (A) DIBAL-H, $H_2O$; $(CH_3)_2Cd$ (B) $SnCl_2 + HCl, H_3O^+$; $CH_3MgBr, H_2O$ (C) DIBAL-H, $H_2O$; $H_2/Ni$ (D) $SnCl_2 + HCl, H_2O$; $(CH_3)_2Cd$

   ›Reveal solutionSolution

   This tests two named nitrile transformations — Stephen reduction (nitrile → aldehyde) and Grignard addition to a nitrile (nitrile → ketone) — that are then combined in a Claisen–Schmidt condensation to give an α,β-unsaturated carbonyl compound.

   Concept and Intuition

   A nitrile carbon is electrophilic like a carbonyl carbon. Two classic ways to convert it into a carbonyl compound:

   • Stephen reduction: $RCN \xrightarrow{SnCl_2/HCl} RCH=NH\cdot HCl \xrightarrow{H_3O^+} RCHO$ — stops cleanly at the aldehyde.
   • Grignard addition: $RCN + R'MgX \rightarrow RC(=NMgX)R' \xrightarrow{H_2O} RCOR'$ — gives a ketone after hydrolysis of the intermediate imine (metalated imine → ketone on aqueous work-up).

   An aldehyde with no α-hydrogens (like benzaldehyde) cannot self-condense, but it readily undergoes a crossed aldol (Claisen–Schmidt) condensation with a ketone that does have α-hydrogens, under dilute base at low/room temperature, forming an α,β-unsaturated ketone (chalcone-type product).

   Step-by-Step Solution

   1. A = benzonitrile, $C_6H_5CN$.
   2. Reagent X = $SnCl_2 + HCl$ then $H_3O^+$: Stephen reduction converts A to the imine hydrochloride, then hydrolysis gives B = benzaldehyde, $C_6H_5CHO$.
   3. Reagent Y = $CH_3MgBr$ then $H_2O$: the methyl Grignard adds to the nitrile carbon, and aqueous work-up hydrolyses the resulting ketimine to give C = acetophenone, $C_6H_5COCH_3$. …

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4. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.In the given reaction sequence, conversion of X to Y is an example of benzene $\xrightarrow[\text{Anhy. } AlCl_3]{CH_3COCl} X \xrightarrow[\Delta]{(1)\ N_2H_4\quad (2)\ KOH/glycol} Y$ (A) Clemmensen reduction (B) Stephen reduction (C) Wolff-Kishner reduction (D) Rosenmund reduction

   ›Reveal solutionSolution

   $X$ (acetophenone, from Friedel-Crafts acylation of benzene) is converted to $Y$ (ethylbenzene) by removing the carbonyl oxygen entirely and replacing $C=O$ with $CH_2$ — the classic Wolff-Kishner reduction, identified by its exact reagents: hydrazine then KOH/glycol with heat.

   Concept and Intuition

   Reducing a ketone's carbonyl carbon all the way down to a $CH_2$ group (full deoxygenation) can be done by a few named methods, each with a signature reagent set: Clemmensen (Zn-Hg/HCl, acidic), Wolff-Kishner (hydrazine then strong base/heat, basic), Stephen reduction (a nitrile to aldehyde, unrelated here), and Rosenmund reduction (an acid chloride to aldehyde using $H_2$/Pd-BaSO$_4$, also unrelated). The reagents given — $N_2H_4$ followed by KOH in ethylene glycol under heat — are the textbook signature of Wolff-Kishner.

   Step-by-Step Solution

   1. Benzene undergoes Friedel-Crafts acylation with acetyl chloride ($CH_3COCl$) and anhydrous $AlCl_3$ (Lewis acid catalyst), installing an acetyl group on the ring: $X = C_6H_5\text{-CO-CH}_3$ (acetophenone).
   2. Step (1): $X$ is treated with hydrazine ($N_2H_4$), which condenses with the ketone carbonyl to form a hydrazone, $C_6H_5\text{-C(=N-NH}_2\text{)-CH}_3$.
   3. Step (2): heating the hydrazone with KOH in a high-boiling solvent (ethylene glycol) decomposes it, releasing $N_2$ gas and reducing the carbon to a $CH_2$ group: $Y = C_6H_5\text{-CH}_2\text{-CH}_3$ (ethylbenzene). …

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5. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.Which of the following compounds give salt of an acid and alcohol on heating with concenrated KOH ? (A) Formaldehyde (B) Acetaldehyde (C) Acetone (D) Acetophenone

   ›Reveal solutionSolution

   The Cannizzaro reaction (base-induced disproportionation into an alcohol + a

   carboxylate salt) requires an aldehyde with no $\alpha$-hydrogen — formaldehyde is the

   only such compound among the options.

   Concept and Intuition

   The Cannizzaro reaction is a base-mediated redox disproportionation of an aldehyde

   lacking $\alpha$-hydrogens: hydroxide attacks the carbonyl carbon, and a hydride is

   transferred from this tetrahedral intermediate to a second aldehyde molecule. One

   molecule ends up oxidised to a carboxylate (salt of the acid), the other reduced to the

   alcohol. This pathway is only available when there is no $\alpha$-H, because if

   $\alpha$-H is present, the base instead deprotonates it to form an enolate, which favours

   aldol condensation (a completely different, much faster pathway) over Cannizzaro.

   Step-by-Step Solution

   1. Check each option for $\alpha$-hydrogens:
      • Formaldehyde ($HCHO$): the carbonyl carbon has no attached carbon at all, so there is no $\alpha$-carbon and hence no $\alpha$-H. Cannizzaro is the only option.
      • Acetaldehyde ($CH_3CHO$): has $\alpha$-H on the $CH_3$ group — undergoes aldol condensation with conc. base, not Cannizzaro.
      • Acetone ($CH_3COCH_3$): has $\alpha$-H on both methyl groups — undergoes aldol condensation.
      • Acetophenone ($C_6H_5COCH_3$): has $\alpha$-H on its methyl group — undergoes …

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6. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
   Q.What is the end product 'R' in the reaction sequence ? Phenol (benzene ring with $\mathrm{OH}$) $\xrightarrow{Zn/\Delta}$ P $\xrightarrow[\text{Anhy. }AlCl_3]{CO,\ HCl}$ Q $\xrightarrow[\text{conc.}HCl]{Zn-Hg}$ R (A) $\mathrm{C_6H_5Cl}$ (benzene ring with a single $\mathrm{Cl}$ substituent, chlorobenzene) (B) benzene ring with $\mathrm{Cl}$ and $\mathrm{CH_3}$ substituents at para positions ($\mathrm{4\text{-}Cl\text{-}C_6H_4\text{-}CH_3}$) (C) $\mathrm{C_6H_5CH_3}$ (benzene ring with a single $\mathrm{CH_3}$ substituent, toluene) (D) $\mathrm{C_6H_5CH_2Cl}$ (benzene ring with a single $\mathrm{CH_2Cl}$ substituent, benzyl chloride)

   ›Reveal solutionSolution

   Phenol → benzene (Zn-dust distillation) → benzaldehyde (Gattermann–Koch formylation) → toluene (Clemmensen reduction) = R.

   Concept and Intuition

   This chains three named reactions that a student should recognise by their exact reagents: (i) Zn/heat distillation of phenol reductively removes the phenolic $-\mathrm{OH}$, giving the parent arene;

   (ii) $\mathrm{CO + HCl}$ with anhydrous $\mathrm{AlCl_3}$ (or $\mathrm{CuCl}$) on benzene is the Gattermann–Koch reaction, a formylation that installs a $-\mathrm{CHO}$ group directly onto the ring;

   (iii) Zn-amalgam with concentrated HCl (Clemmensen reduction) reduces a carbonyl group all the way to a methylene/methyl group, converting an aldehyde into the corresponding alkylbenzene.

   Step-by-Step Solution

   1. Phenol $\xrightarrow{Zn/\Delta}$ P: Zn dust distillation removes the $-\mathrm{OH}$ (as $\mathrm{ZnO}$), giving benzene. P = $\mathrm{C_6H_6}$.
   2. P $\xrightarrow[\text{anhy. } AlCl_3]{CO,\ HCl}$ Q: this is the Gattermann–Koch reaction — formylation of benzene to benzaldehyde. Q = $\mathrm{C_6H_5CHO}$. …

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7. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.Consider the following sequence of reactions. (anhy = anhydrous; conc. = concentrated) $\mathrm{C_6H_5COONa} \xrightarrow[\Delta]{\mathrm{NaOH/CaO}} X \xrightarrow[\mathrm{Anhy.\ AlCl_3}]{\mathrm{CO + HCl}} Y \xrightarrow[\mathrm{NaOH}]{\text{conc.}} A + B$ If A is the reduction product of Y, what is B? (A) Sodium formate (B) Sodium phenoxide (C) Sodium salt of benzoic acid (D) Sodium salt of salicylic acid

   ›Reveal solutionSolution

   This traces sodium benzoate through decarboxylation, Gattermann–Koch formylation, and a Cannizzaro reaction; $B$ is sodium benzoate. Answer: (C).

   Concept and Intuition

   Three classical named reactions are stacked here. (1) Sodium salts of carboxylic acids, heated with soda-lime, lose $\mathrm{CO_2}$ (decarboxylation) to give the corresponding hydrocarbon — sodium benzoate gives benzene. (2) The Gattermann–Koch reaction formylates an aromatic ring using $\mathrm{CO}$ and $\mathrm{HCl}$ with anhydrous $\mathrm{AlCl_3}$ (and often $\mathrm{CuCl}$) as catalyst, generating an aldehyde directly on the ring — benzene gives benzaldehyde. (3) An aldehyde with no $\alpha$-hydrogen (like benzaldehyde, whose only adjacent carbon is the aromatic ring) cannot undergo aldol-type self-condensation; instead, treatment with concentrated base disproportionates it via the Cannizzaro reaction: one molecule is reduced (hydride transfer) to the primary alcohol, the other is oxidised to the carboxylate salt.

   Step-by-Step Solution

   1. $\mathrm{C_6H_5COONa} \xrightarrow[\Delta]{\mathrm{NaOH/CaO}}$ decarboxylation $\to$ $X = \mathrm{C_6H_6}$ (benzene).
   2. $\mathrm{C_6H_6} \xrightarrow[\mathrm{Anhy.\,AlCl_3}]{\mathrm{CO+HCl}}$ Gattermann–Koch $\to$ $Y = \mathrm{C_6H_5CHO}$ (benzaldehyde).
   3. $\mathrm{C_6H_5CHO}$ has no $\alpha$-H, so conc. NaOH drives a Cannizzaro reaction: $2\,\mathrm{C_6H_5CHO} + \mathrm{NaOH} \to \mathrm{C_6H_5CH_2OH} + \mathrm{C_6H_5COONa}$. …

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8. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.The final product (C) in the given reaction sequence is (anhy = anhydrous) $\mathrm{C_6H_5COOH} \xrightarrow{\mathrm{SOCl_2}} (A) \xrightarrow[\text{anhy. }AlCl_3]{\mathrm{C_6H_6}} (B) \xrightarrow[\Delta]{\text{(i) }\mathrm{NH_2-NH_2}\ \text{(ii) KOH/}(\mathrm{CH_2OH})_2} (C)$ (A) Benzophenone (B) Diphenyl methane (C) Diphenylmethanol (D) Benzoic acid

   ›Reveal solutionSolution

   $\mathrm{SOCl_2}$ makes the acid chloride, Friedel–Crafts gives benzophenone, and Wolff–Kishner reduces its carbonyl completely to $\mathrm{CH_2}$, giving diphenylmethane.

   Concept and Intuition

   $\mathrm{SOCl_2}$ is the standard reagent to convert a carboxylic acid to its acid chloride (releasing $\mathrm{SO_2}$ and $\mathrm{HCl}$, both leaving as gases — a very clean conversion). An acid chloride is an excellent acylating agent for Friedel–Crafts acylation of an aromatic ring. Wolff–Kishner reduction (hydrazine, then base and heat) removes a carbonyl oxygen entirely, converting a ketone (or aldehyde) directly to a $\mathrm{CH_2}$ group — the "deoxygenating" counterpart to Clemmensen reduction, useful when acid-sensitive groups are present.

   Step-by-Step Solution

   1. $\mathrm{C_6H_5COOH} \xrightarrow{\mathrm{SOCl_2}}$ A $= \mathrm{C_6H_5COCl}$ (benzoyl chloride).
   2. A + $\mathrm{C_6H_6}$, anhy. $\mathrm{AlCl_3}$ → Friedel–Crafts acylation: B $= \mathrm{C_6H_5COC_6H_5}$ (benzophenone). …

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9. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.A carbonyl compound X ($C_8H_8O$) undergoes disproportionation with conc. KOH on heating. Product of X with Zn-Hg/HCl is Y and product of X with $\mathrm{NaBH_4}$ is Z. What are Y and Z respectively ? (A) $\mathrm{C_6H_5CH_2CH_3}$ (ethylbenzene) , $\mathrm{C_6H_5CH_2CH_2OH}$ (2-phenylethanol) (B) 1,4-dimethylbenzene (ring with $CH_3$ and $H_3C$ para to each other) , 4-methylbenzyl alcohol (ring with $H_3C$ and $CH_2OH$ para to each other) (C) 4-methylbenzyl alcohol (ring with $H_3C$ and $CH_2OH$ para to each other) , 1,4-dimethylbenzene (ring with $H_3C$ and $CH_3$ para to each other) (D) $\mathrm{C_6H_5CH_2CH_2OH}$ (2-phenylethanol) , $\mathrm{C_6H_5CH_2CH_3}$ (ethylbenzene)

   ›Reveal solutionSolution

   X must be an aldehyde with no $\alpha$-H (since it disproportionates with KOH — Cannizzaro), i.e. p-methylbenzaldehyde; Clemmensen (Zn-Hg/HCl) fully reduces $-CHO\to -CH_3$ giving p-xylene (Y), while $NaBH_4$ only reduces to $-CH_2OH$ giving 4-methylbenzyl alcohol (Z). Answer: (B).

   Concept and Intuition

   The Cannizzaro reaction is a base-mediated disproportionation (self oxidation-reduction) unique to aldehydes lacking an $\alpha$-hydrogen (so they cannot instead undergo aldol condensation). Aromatic aldehydes such as benzaldehyde and its ring-substituted derivatives are the classic examples — the carbon bearing $-CHO$ is attached directly to the ring, so there is no $\alpha$-carbon with abstractable hydrogens in the usual aldol sense. With formula $C_8H_8O$, a ring-methyl-substituted benzaldehyde ($CH_3-C_6H_4-CHO$, a tolualdehyde) fits perfectly (7 ring+methyl carbons + 1 carbonyl carbon = 8 C; H: 4 ring H + 3 methyl H + 1 CHO H = 8 H; one O) and undergoes Cannizzaro on heating with conc. KOH.

   Two separate, independent reductions are then applied to the same starting aldehyde X (not to Cannizzaro products):

   • Clemmensen reduction (Zn-Hg amalgam / conc. HCl) is a strong reduction that converts a carbonyl all the way to a methylene/methyl group: $-CHO \to -CH_3$. Applied to p-tolualdehyde, this gives 1,4-dimethylbenzene (p-xylene) = Y.
   • $NaBH_4$ is a mild hydride reducing agent that reduces aldehydes only to the corresponding primary alcohol, stopping at $-CH_2OH$ (it does not touch the aromatic ring or over-reduce). Applied to p-tolualdehyde, this gives 4-methylbenzyl alcohol = Z.

   Step-by-Step Solution

   1. Deduce X: $C_8H_8O$ that undergoes Cannizzaro (disproportionation) ⇒ must have no $\alpha$-H ⇒ an aromatic (ring) aldehyde: 4-methylbenzaldehyde, $CH_3C_6H_4CHO$. …

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10. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
    Q.Which of the following sequence of reagents convert 3-hexene to propane? (A) $\mathrm{KMnO_4} \mid \mathrm{H^+}$; NaOH, CaO (B)

    (i) $\mathrm{O_3}$

    (ii) Zn, $\mathrm{H_2O}$; $\mathrm{NaBH_4}$ (C)

    (i) $\mathrm{O_3}$,

    (ii) Zn, $\mathrm{H_2O}$; Zn-Hg, HCl (D) $\mathrm{KMnO_4} \mid \mathrm{H^+}$; $\mathrm{LiAlH_4}$, $\mathrm{H_2O}$

    ›Reveal solutionSolution

    Ozonolysis of symmetric 3-hexene gives propanal, and Clemmensen reduction (Zn-Hg/HCl) then converts the aldehyde carbonyl fully to $\mathrm{CH_2}$, yielding propane.

    Concept and Intuition

    3-Hexene is symmetric about its central C=C double bond, so oxidative/reductive cleavage there gives two identical 3-carbon fragments. To go all the way to an alkane (propane) from the resulting carbonyl compound (propanal), the carbonyl oxygen must be completely removed — this is exactly what Clemmensen reduction does (converts $\mathrm{C{=}O}$ to $\mathrm{CH_2}$ under acidic conditions using zinc amalgam and HCl), unlike NaBH$_4$ or LiAlH$_4$ which only reduce the carbonyl to an alcohol.

    Step-by-Step Solution

    1. 3-Hexene: $\mathrm{CH_3CH_2{-}CH{=}CH{-}CH_2CH_3}$ (double bond between C3 and C4, symmetric).
    2. Step (i) O$_3$ then (ii) Zn, H$_2$O: reductive ozonolysis cleaves the C=C bond, giving two molecules of propanal, $\mathrm{CH_3CH_2CHO}$ (3 carbons each).
    3. Step Zn-Hg, HCl (Clemmensen reduction): reduces the aldehyde carbonyl directly to a $\mathrm{CH_2}$ group, converting $\mathrm{CH_3CH_2CHO}$ to $\mathrm{CH_3CH_2CH_3}$ — propane. …

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11. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
    Q.What are X, Y, Z in the following reaction sequence respectively? Benzene $\xrightarrow{X}$ [benzene ring with a -CHO group] $\xrightarrow[\Delta]{\text{Conc. NaOH}}$ Y + Z (A) CO, HCl, $\mathrm{AlCl_3}$; Y = benzene ring with $\mathrm{CH_2ONa}$, Z = benzene ring with COOH (B) CO, HCl, CuCl; Y = benzene ring with $\mathrm{CH_2OH}$, Z = benzene ring with COONa (C) HCHO, HCl, anhy. $\mathrm{AlCl_3}$; Y = benzene ring with $\mathrm{CH_2OH}$, Z = benzene ring with COOH (D) CO, HCl; Y = benzene ring with $\mathrm{CH_2COOH}$, Z = benzene ring with OH

    ›Reveal solutionSolution

    Gattermann-Koch formylation (CO/HCl/AlCl3 or CuCl) gives benzaldehyde, which then undergoes Cannizzaro disproportionation in hot conc. NaOH into benzyl alcohol and sodium benzoate.

    Concept and Intuition

    Benzene has no easy route to an aldehyde by simple oxidation, so formylation is done via the Gattermann-Koch reaction: CO and HCl, activated by anhydrous $\mathrm{AlCl_3}$ (Lewis acid) with $\mathrm{CuCl}$ (cuprous chloride) as catalyst, generate an electrophilic formyl cation equivalent that substitutes onto the ring, giving benzaldehyde directly. Benzaldehyde, having no alpha-hydrogen, cannot undergo aldol-type self-condensation; instead, concentrated base makes it undergo Cannizzaro reaction — intermolecular disproportionation where one molecule is reduced (hydride transfer) and the other oxidised.

    Step-by-Step Solution

    1. $\mathrm{C_6H_6 \xrightarrow[\text{AlCl}_3/\text{CuCl}]{\text{CO, HCl}} C_6H_5CHO}$ (benzaldehyde) — this is X.
    2. Hot conc. NaOH on benzaldehyde: two molecules disproportionate — $2\,\mathrm{C_6H_5CHO} + \mathrm{NaOH} \rightarrow \mathrm{C_6H_5CH_2OH} + \mathrm{C_6H_5COONa}$. …

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12. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
    Q.The number of products obtained in the following reaction is [FIGURE] (acetaldehyde, drawn as $\mathrm{CH_3CHO}$) + [FIGURE] (acetophenone, drawn as a benzene ring bearing a $\mathrm{-COCH_3}$ substituent) $\xrightarrow[\Delta]{NaOH}$ ? (A) 1 (B) 3 (C) 2 (D) 4

    ›Reveal solutionSolution

    In a crossed-aldol reaction between acetaldehyde and acetophenone under basic, heated conditions, both self-aldol and crossed-aldol products form, but only the conjugated enones survive after dehydration; the total number of distinct products is 3.

    The key idea here is that both aldehydes and ketones can act as enolate donors in the presence of a strong base like NaOH. Acetaldehyde (CH₃CHO) has α-hydrogens and can form an enolate; acetophenone (C₆H₅COCH₃) also has α-hydrogens on its methyl group. Under heat (Δ), the initial aldol addition products dehydrate to give α,β-unsaturated carbonyl compounds. We must count all distinct, stable products.

    ---

    1. Identify possible enolate donors and acceptors

       • Acetaldehyde: can be both donor (via its enolate) and acceptor (electrophilic carbonyl).
       • Acetophenone: can also be both donor (enolate from the methyl group) and acceptor (carbonyl). This gives four possible combinations for the aldol addition step.

    2. List all crossed and self-aldol combinations

       • Self-aldol of acetaldehyde: Two molecules of CH₃CHO give 3-hydroxybutanal, which dehydrates to crotonaldehyde (CH₃CH=CHCHO).
       • Self-aldol of acetophenone: Two molecules of C₆H₅COCH₃ give a β-hydroxyketone, which dehydrates to 1,3-diphenyl-2-buten-1-one (C₆H₅COCH=C(CH₃)C₆H₅).
       • Crossed-aldol (acetaldehyde as donor, acetophenone as acceptor): Enolate from acetaldehyde attacks the carbonyl of acetophenone → 4-hydroxy-4-phenyl-2-butanone, which dehydrates to 4-phenyl-3-buten-2-one (C₆H₅CH=CHCOCH₃).
       • Crossed-aldol (acetophenone as donor, acetaldehyde as acceptor): Enolate from acetophenone attacks acetaldehyde → 3-hydroxy-1-phenyl-1-butanone, which dehydrates to 1-phenyl-2-buten-1-one (C₆H₅COCH=CHCH₃).

    3. Check for duplicates

       The two crossed products are different:

       • 4-phenyl-3-buten-2-one has the double bond between the phenyl and the carbonyl (conjugation with the ring).
       • 1-phenyl-2-buten-1-one has the double bond between the carbonyl and the methyl group. They are not the same compound. So we have:
       • Product A: crotonaldehyde (from acetaldehyde self-aldol)
       • Product B: 1,3-diphenyl-2-buten-1-one (from acetophenone self-aldol)
       • Product C: 4-phenyl-3-buten-2-one (crossed, acetaldehyde donor)
       • Product D: 1-phenyl-2-buten-1-one (crossed, acetophenone donor) That’s 4 possible products.

    4. Consider the reaction conditions: NaOH and heat …

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