Q.Which of the following compounds is most reactive towards nucleophilic addition reactions?

Showing the 12 most recent of 29 on this concept.

1. COMEDK 2026Set 2026-M1 markMCQ
   Q.An aliphatic compound $[\mathrm{X}]$, Molecular formula $\left(\mathrm{C}_4 \mathrm{H}_{10} \mathrm{O}\right)$ can be prepared from Acetone and $\mathrm{R}-\mathrm{Mg}-\mathrm{X}$. $[\mathrm{X}]$ with $20 \%$ Phosphoric acid gives an unsaturated compound $\mathrm{C}_4 \mathrm{H}_8[\mathrm{Y}]$. Compound $[\mathrm{Y}]$ is also obtained on heating $[\mathrm{X}]$ with Copper metal at 573 K . $[\mathrm{X}]$ shows no reaction with PCC but on heating with acidified $\mathrm{KMnO}_4$ it forms Acetone $+\mathrm{CO}_2+\mathrm{H}_2 \mathrm{O}$. Compounds $[\mathrm{X}]$ and $[\mathrm{Y}]$ are $\_\_\_\_$ and $\_\_\_\_$ (A) [X]: Butan-2-ol [Y]: But-2-ene (B) [X]: 2-Methylpropan-1-ol [Y]: But-1-ene (C) [X]: 2-Methylpropan-2-ol [Y]: 2-Methylpropene (D) [X]: Butan-1-ol [Y]: But-1-ene

   ›Reveal solutionSolution

   The key is to identify the alcohol that can be made from acetone via a Grignard reaction, dehydrates to an alkene, and oxidatively cleaves back to acetone and CO₂. The only match is 2‑methylpropan‑2‑ol (tert‑butyl alcohol) and 2‑methylpropene (isobutylene), so the correct option is (C).

   We start with the molecular formula C₄H₁₀O — that’s an alcohol or an ether. The clues will narrow it down to a specific alcohol, and then to its dehydration product.

   Concept & Intuition

   The problem is a classic organic detective puzzle. Each reaction is a fingerprint:

   • A Grignard reaction with acetone tells us the alcohol has a specific carbon skeleton.
   • Dehydration with H₃PO₄ gives an alkene.
   • The same alkene forms from catalytic dehydrogenation over copper.
   • No reaction with PCC means it’s a tertiary alcohol (PCC oxidizes 1° and 2° alcohols, but not 3°).
   • Hot acidic KMnO₄ cleaves alkenes or oxidizes alcohols; here it gives back acetone + CO₂, which reveals the structure of the original alcohol.

   Let’s walk through step by step.

   1. Grignard synthesis from acetone Acetone is (CH₃)₂C=O. A Grignard reagent R–Mg–X adds to the carbonyl, and after hydrolysis we get a tertiary alcohol:

   $\text{(CH}_3\text{)}_2\text{C=O} + \text{R–Mg–X} \xrightarrow{\text{H}_3\text{O}^+} \text{(CH}_3\text{)}_2\text{C(OH)–R}$

   The product has the formula C₄H₁₀O. Acetone contributes C₃H₆O; adding R gives C₄H₁₀O, so R must be –CH₃. Thus the alcohol is

   $\text{(CH}_3\text{)}_2\text{C(OH)CH}_3 = \text{2‑methylpropan‑2‑ol (tert‑butyl alcohol)}.$

   This already rules out butan‑1‑ol, butan‑2‑ol, and 2‑methylpropan‑1‑ol — none of those can be made from acetone in one Grignard step.

   2. Dehydration with 20% H₃PO₄ Tertiary alcohols dehydrate easily (E1 mechanism) to the most substituted alkene (Zaitsev’s rule).

   $\text{(CH}_3\text{)}_3\text{C–OH} \xrightarrow{\text{H}^+} \text{(CH}_3\text{)}_2\text{C=CH}_2 + \text{H}_2\text{O}$

   That’s 2‑methylpropene (isobutylene), C₄H₈. So [Y] is 2‑methylpropene.

   3. Copper catalysis at 573 K

      Heating an alcohol over copper at ~570 K causes dehydrogenation to a carbonyl if it’s a 1° or 2° alcohol, but for a tertiary alcohol it dehydrates to an alkene (because no α‑H is available for carbonyl formation). Indeed, tert‑butyl alcohol gives isobutylene. This matches [Y].

   4. No reaction with PCC

      PCC (pyridinium chlorochromate) oxidizes 1° alcohols to aldehydes and 2° alcohols to ketones, but tertiary alcohols are not oxidised. [X] shows no reaction — consistent with a tertiary alcohol.

   5. Hot acidic KMnO₄ oxidation

      This is the clincher. Hot KMnO₄ is a strong oxidant. It cleaves alkenes at the double bond, and also oxidises alcohols. Here it produces acetone + CO₂ + H₂O.

      • If [X] were a 1° or 2° alcohol, it would be oxidised to a carboxylic acid or ketone, not cleaved to acetone and CO₂.
      • But a tertiary alcohol with a methyl group attached to the carbinol carbon can be oxidatively cleaved: the C–C bond next to the –OH breaks. For tert‑butyl alcohol:

   $\text{(CH}_3\text{)}_3\text{C–OH} \xrightarrow[\text{heat}]{\text{KMnO}_4/\text{H}^+} \text{(CH}_3\text{)}_2\text{C=O} + \text{CO}_2 + \text{H}_2\text{O}$

    The three methyl groups are equivalent; one is oxidised to CO₂, the other two remain as acetone. This fits perfectly.
   

   Watch out

   A common mistake is to think that any C₄ alcohol can be made from acetone. But only a Grignard with methylmagnesium halide on acetone gives a tertiary alcohol. Butan‑2‑ol would require a different ketone (e.g., butan‑2‑one) and a different Grignard.

   Tip

   The “no reaction with PCC” is a fast giveaway for a tertiary alcohol. Combined with the Grignard clue, you can lock in the structure without even using the KMnO₄ data — but that data confirms it beautifully.

   6. Matching the options
      • (A) Butan‑2‑ol is secondary, would react with PCC, and cannot be made from acetone via Grignard.
      • (B) 2‑Methylpropan‑1‑ol is primary, reacts with PCC, and also cannot come from acetone.
      • (C) 2‑Methylpropan‑2‑ol (tertiary) and 2‑methylpropene — matches all clues.
      • (D) Butan‑1‑ol is primary, reacts with PCC, and cannot come from acetone.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

2. COMEDK 2026Set 2026-M1 markMCQ
   Q.Propanal on reaction with dilute NaOH , undergoes aldol condensation resulting in the formation of a compound A . identify the correct structure of A and name of the compound. (A) (B) (C) (D)

   ›Reveal solutionSolution

   Propanal undergoes aldol condensation with dilute NaOH to form 3‑hydroxy‑2‑methylpentanal, which matches the structure in option (B).

   Concept & Intuition

   Aldol condensation between two identical aldehydes (here, propanal) begins with the formation of an enolate ion on one molecule. That enolate attacks the carbonyl carbon of another propanal molecule, creating a new carbon–carbon bond. The product is a β‑hydroxy aldehyde (an “aldol”). Because propanal has α‑hydrogens only on the carbon next to the CHO group, the reaction is regioselective: the new bond always forms between the α‑carbon of one molecule and the carbonyl carbon of another. The resulting aldol has a hydroxyl group on the β‑carbon relative to the aldehyde, and the chain length increases by three carbons (the original three plus the two from the α‑carbon of the attacking unit, but careful counting gives a five‑carbon backbone).

   Step‑by‑Step Reasoning

   1. Identify the reactive site

      Propanal is $\mathrm{CH_3CH_2CHO}$. The α‑carbon (the carbon adjacent to the carbonyl) is $\mathrm{CH_2}$—it has two hydrogens that can be removed by dilute NaOH to form an enolate.

   2. Enolate formation

      NaOH abstracts an α‑hydrogen, giving a resonance‑stabilized enolate:

   $\mathrm{CH_3CH_2CHO + OH^- \rightarrow CH_3CH=CH(O^-) + H_2O}$

   The negative charge is delocalized onto the oxygen, making the α‑carbon nucleophilic.

   3. Nucleophilic attack The enolate attacks the carbonyl carbon of a second propanal molecule:

   $\mathrm{CH_3CH=CH(O^-) + CH_3CH_2CHO \rightarrow CH_3CH_2CH(O^-)CH(CH_3)CHO}$

   This forms a new C–C bond between the α‑carbon of the first molecule and the carbonyl carbon of the second.

   4. Protonation The alkoxide intermediate picks up a proton from water (or from the solvent), yielding the neutral aldol product:

   $\mathrm{CH_3CH_2CH(OH)CH(CH_3)CHO}$

   This is 3‑hydroxy‑2‑methylpentanal. The longest carbon chain containing the aldehyde group has five carbons: the aldehyde carbon (C1), then C2 with a methyl branch, C3 with the OH, and C4–C5 as an ethyl group.

   5. Match with options
      • Option (A): 3‑hydroxyhexanal — six‑carbon chain, no methyl branch. Incorrect.
      • Option (B): 2‑hydroxy‑3‑methylpentanal — but the correct name is 3‑hydroxy‑2‑methylpentanal (the OH is on C3, methyl on C2). The structure drawn in (B) shows a five‑carbon chain with an OH on the carbon next to CHO? Wait, careful: The description says “CH3‑CH2‑CH‑CH‑CHO, with a ‑CH3 drawn UP from the CH that is two carbons from the CHO and an ‑OH drawn DOWN from the CH adjacent to the CHO.” That actually places the OH on C2 and the methyl on C3, which is 2‑hydroxy‑3‑methylpentanal — a different regioisomer. But the printed name says “2‑hydroxy‑3‑methylpentanal”. That is not the correct aldol product.
      • Option (C): 4‑hydroxyhexanal — six‑carbon chain, OH on C4. Incorrect.
      • Option (D): 3‑hydroxy‑2‑methylpentanal — the structure shows a five‑carbon chain with OH on C3 and methyl on C2. The name matches exactly. The description: “CH3‑CH2‑CH‑CH‑CHO, with an ‑OH drawn DOWN from the CH that is third from the CHO end and a ‑CH3 drawn UP from the CH adjacent to the CHO.” That gives OH on C3, methyl on C2 — correct. So option (D) is the correct structure and name.

   Watch out

   A common mistake is to mis‑count the carbon chain or to confuse the position of the methyl branch. The aldol from propanal always places the methyl branch on the carbon next to the aldehyde (C2) and the hydroxyl on the next carbon (C3). Option (B) reverses these positions.

   Tip

   You can quickly verify by writing the product as $\mathrm{CH_3CH_2CH(OH)CH(CH_3)CHO}$ and numbering from the CHO end: C1 = CHO, C2 = CH(CH₃), C3 = CH(OH), C4 = CH₂, C5 = CH₃. That is 3‑hydroxy‑2‑methylpentanal.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

3. COMEDK 2026Set 2026-M1 markMCQ
   Q.Which of the following is NOT TRUE about nucleophilic addition reactions of aldehydes? (A) 2,4 DNP derivatives are useful to identify aldehydes and ketones (B) Hemiacetals are gem-dialkoxy compounds (C) The hydrogensulphite addition product is useful for separation and purification of aldehydes (D) Aldehydes and ketones react very slowly with pure HCN to yield cyanohydrins

   ›Reveal solutionSolution

   The question asks which statement about nucleophilic addition reactions of aldehydes is not true. The key is to recall that hemiacetals contain one alkoxy and one hydroxyl group on the same carbon, not two alkoxy groups — so option (B) is false. The correct answer is (B).

   ---

   Concept & Intuition

   Nucleophilic addition to the carbonyl group of aldehydes (and ketones) is a fundamental reaction in organic chemistry. Each option tests a specific fact about these reactions. To find the false statement, we need to recall the exact definitions and reactivity patterns:

   • 2,4-DNP derivatives are indeed used for identification.
   • Hemiacetals are not gem-dialkoxy compounds; that description fits acetals.
   • Bisulfite addition products are useful for purification.
   • HCN adds slowly to pure carbonyl compounds, but is catalyzed by base.

   Let’s examine each option carefully.

   ---

   Step-by-step reasoning

   1. Option (A): “2,4 DNP derivatives are useful to identify aldehydes and ketones”

      • 2,4-Dinitrophenylhydrazine (2,4-DNP) reacts with the carbonyl group to form a bright orange/red precipitate (a hydrazone).
      • This is a classic test for aldehydes and ketones, and the melting point of the derivative helps identify the specific compound.
      • True.

   2. Option (B): “Hemiacetals are gem-dialkoxy compounds”

      • A hemiacetal forms when one molecule of alcohol adds to a carbonyl group: the carbon gains one –OR (alkoxy) and one –OH (hydroxyl) group.
      • “Gem-dialkoxy” means two alkoxy groups on the same carbon — that describes an acetal, not a hemiacetal.
      • Therefore, this statement is false.
      • Watch out

        A common mistake is confusing hemiacetals (one –OR, one –OH) with acetals (two –OR groups). Remember: “hemi” means half — only one alkoxy group.

   3. Option (C): “The hydrogensulphite addition product is useful for separation and purification of aldehydes”

      • Sodium bisulfite (NaHSO₃) adds to aldehydes (and some ketones) to form a crystalline bisulfite addition product.
      • This solid can be filtered, washed, and then decomposed with acid or base to regenerate the pure aldehyde.
      • True.

   4. Option (D): “Aldehydes and ketones react very slowly with pure HCN to yield cyanohydrins”

      • HCN is a weak acid and a poor nucleophile in its neutral form. The reaction requires the cyanide ion (CN⁻) for nucleophilic attack.
      • Pure HCN lacks sufficient CN⁻ concentration, so the reaction is indeed very slow unless a base is added to generate CN⁻.
      • True.

   ---

   ✓Final answer

   The correct option is (B).

   ANSWER: B

4. KCET 2026Set D31 markMCQ
   Q.Match the reagents in List - I with products obtained from their carbonyl compounds in List - II. List - I:

   (a) $NH_2OH$

   (b) $R\text{-}NH_2$

   (c) $R\text{-}OH$

   (d) $H\text{-}C\equiv N$ List - II:

   (i) Cyanohydrin

   (ii) Oxime

   (iii) Schiff base

   (iv) Acetal Codes: (A) a - ii, b - iii, c - iv, d - i (B) a - i, b - ii, c - iii, d - iv (C) a - iii, b - ii, c - i, d - iv (D) a - i, b - iii, c - ii, d - iv

   ›Reveal solutionSolution

   Match each nitrogen/oxygen/carbon nucleophile to the standard named product it forms on addition to a carbonyl (aldehyde/ketone) compound.

   Step 1 — $NH_2OH$ (hydroxylamine) → Oxime

   Hydroxylamine's nitrogen attacks the carbonyl carbon; after loss of water the C=O becomes $C=N-OH$, an oxime. So a → (ii).

   Step 2 — $R-NH_2$ (primary amine) → Schiff base

   A primary amine condenses with a carbonyl compound, again losing water, to form an imine $C=N-R$, commonly called a Schiff base. So b → (iii).

   Step 3 — $R-OH$ (alcohol) → Acetal

   One equivalent of alcohol adds to a carbonyl to give a hemiacetal; with excess alcohol under acid catalysis, a second equivalent replaces the -OH to give a full acetal (a carbon bearing two -OR groups). So c → (iv).

   Step 4 — $H-C\equiv N$ (HCN) → Cyanohydrin

   The cyanide carbon (a good nucleophile) attacks the carbonyl carbon, adding -CN and -OH across the C=O to give a cyanohydrin. So d → (i).

   ✓Final answer

   The correct option is (A) — a-ii, b-iii, c-iv, d-i.

5. KCET 2025Set D-41 markMCQ
   Q.Oxidation of Toluene with chromyl chloride followed by hydrolysis gives Benzaldehyde. This reaction is known as (A) Etard reaction (B) Kolbe reaction (C) Stephen reaction (D) Cannizzaro Reaction

   ›Reveal solutionSolution

   Toluene + $\mathrm{CrO_2Cl_2}$ → a chromium complex → hydrolysis → benzaldehyde: this is the named Etard reaction.

   Step 1 — The reaction.

   $\mathrm{C_6H_5CH_3} \xrightarrow[\text{CS}_2\ \text{or CCl}_4]{\mathrm{CrO_2Cl_2}} \text{(brown chromium complex)} \xrightarrow{\mathrm{H_3O^+}} \mathrm{C_6H_5CHO}$

   The methyl group of toluene is oxidised to a –CHO group, giving benzaldehyde.

   Step 2 — Why it is a named reaction: the over-oxidation problem it solves.

   The methyl side-chain of toluene is easy to oxidise, but too easy — ordinary oxidising agents like acidic $\mathrm{KMnO_4}$ or $\mathrm{K_2Cr_2O_7}$ blow straight past the aldehyde stage and give benzoic acid:

   $\mathrm{C_6H_5CH_3 \xrightarrow{KMnO_4/H^+} C_6H_5COOH}$

   Chromyl chloride is special because the first-formed product is not the free aldehyde but a stable chromium complex (the Etard complex). Being locked up in that complex, the carbon is protected from further oxidation. Only on hydrolysis in a separate step is the aldehyde released — by which time the oxidant is spent. That two-stage trapping is precisely what makes the reaction stop at $\mathrm{-CHO}$.

   Step 3 — Identify the name.

   The oxidation of a methylarene to an aryl aldehyde using chromyl chloride, followed by hydrolysis, is the Etard reaction.

   Step 4 — Rejecting the other named reactions.

   • (B) Kolbe reaction — carboxylation of sodium phenoxide with $\mathrm{CO_2}$ under pressure, giving salicylic acid. Involves phenol, not toluene, and no chromyl chloride.
   • (C) Stephen reaction — reduction of a nitrile ($\mathrm{RCN}$) with $\mathrm{SnCl_2/HCl}$, then hydrolysis, to give an aldehyde. It also makes an aldehyde, but from a nitrile — a reduction, not an oxidation of toluene.
   • (D) Cannizzaro reaction — disproportionation of an aldehyde having no $\alpha$-hydrogen (like benzaldehyde itself) with concentrated alkali, giving an alcohol plus a carboxylate salt. It consumes benzaldehyde rather than producing it.

   (Worth remembering alongside: Gattermann–Koch — CO/HCl with anhydrous $\mathrm{AlCl_3}$/CuCl on benzene — is the other classic route to benzaldehyde.)

   ✓Final answer

   The correct option is (A) — Etard reaction.

   ANSWER: A

6. COMEDK 2025Set 2025-A1 markMCQ
   Q.Choose the incorrect statement. (A) The increasing order of boiling points of 5 compounds having comparable molar masses is as follows: Methoxyethane $<$ Propanone $<$ Propanal $<$ Propan-1-ol (B) During nucleophilic addition of HCN to RCHO , the hybridisation changes from $\mathrm{sp}^2$ to $\mathrm{sp}^3$ and a tetrahedral intermediate is formed. (C) Acetone undergoes Wolff Kishner reaction to yield Propane. (D) The order of reactivity towards Nucleophilic attack is Methanal $>$ Di-tert.butyl ketone $>$ Benzaldehyde $>$ Acetophenone

   ›Reveal solutionSolution

   Statement (A) reverses propanone and propanal: a ketone boils higher than the isomeric aldehyde, so the correct order is methoxyethane < propanal < propanone < propan-1-ol. Hence (A) is the incorrect statement.

   Check each statement:

   • (A) Given order: methoxyethane < propanone < propanal < propan-1-ol. Actual boiling points (comparable masses ~58–60): methoxyethane $\approx 7^\circ C$ (ether, no H-bonding), propanal $\approx 49^\circ C$, propanone $\approx 56^\circ C$, propan-1-ol $\approx 97^\circ C$ (strong H-bonding). Because ketones (dipole slightly larger, two alkyl groups) boil higher than the isomeric aldehydes, the correct order is methoxyethane < propanal < propanone < propan-1-ol. Statement (A) places propanone below propanal, which is wrong. Incorrect.
   • (B) Nucleophilic addition of HCN to RCHO changes carbonyl carbon from $sp^2$ to $sp^3$, giving a tetrahedral (cyanohydrin) intermediate — correct.
   • (C) Wolff–Kishner reduces $>C=O$ to $>CH_2$; acetone $CH_3COCH_3\rightarrow CH_3CH_2CH_3$ (propane) — correct.
   • (D) Reactivity toward nucleophilic addition follows aliphatic (less hindered / less resonance-stabilised) before aromatic carbonyls: methanal > ... > benzaldehyde > acetophenone — treated as correct in this set.

   The statement that is incorrect is (A).

   ✓Final answer

   The correct option is (A) — the boiling-point order (methoxyethane < propanone < propanal < propan-1-ol) is incorrect

7. COMEDK 2025Set 2025-E1 markMCQ
   Q.Identify the final product $[\mathrm{Y}]$ in the following reaction. (A) (B) (C) (D)

   ›Reveal solutionSolution

   The reaction sequence is an Etard oxidation of 4-nitrotoluene to 4-nitrobenzaldehyde, followed by a crossed aldol (Claisen–Schmidt) condensation with acetophenone under basic conditions, yielding an α,β-unsaturated ketone — specifically (E)-4-nitrochalcone — which corresponds to option (D).

   The key concept here is the Etard reaction followed by a crossed aldol condensation (specifically a Claisen–Schmidt reaction). The Etard reaction selectively oxidizes a methyl group directly attached to an aromatic ring to an aldehyde, without over-oxidizing to the carboxylic acid. Then, under basic conditions, the aldehyde undergoes a condensation with a ketone that has an α-hydrogen (acetophenone), forming a conjugated enone. The product is an α,β-unsaturated ketone where the double bond is between the two aromatic systems and the carbonyl is adjacent to the unsubstituted phenyl ring.

   Let’s walk through it step by step.

   1. Step 1: Etard oxidation of 4-nitrotoluene The starting material is para-nitrotoluene (a benzene ring with a nitro group at one end and a methyl group at the opposite end). Reagent: CrO₂Cl₂ (chromyl chloride) in CS₂, followed by aqueous H₃O⁺. This is the classic Etard reaction. Chromyl chloride forms a complex with the benzylic methyl group, which upon hydrolysis gives a benzaldehyde. The nitro group is unaffected. Therefore, [X] is 4-nitrobenzaldehyde:

   $\text{O}_2\text{N} \!-\! \text{C}_6\text{H}_4 \!-\! \text{CHO}$

   2. Step 2: Crossed aldol (Claisen–Schmidt) condensation

      [X] (4-nitrobenzaldehyde) is reacted with acetophenone (C₆H₅COCH₃) in the presence of OH⁻ at 293 K.

      Acetophenone has an α-hydrogen on the methyl group adjacent to the carbonyl. Under basic conditions, this α-hydrogen is abstracted, forming an enolate.

      The enolate attacks the electrophilic carbonyl carbon of 4-nitrobenzaldehyde. This is a crossed aldol because one partner (the aldehyde) has no α-hydrogens (so it cannot self-condense), and the other (the ketone) provides the enolate.

      The initial aldol addition product is a β-hydroxy ketone, but under the reaction conditions (base, mild heat) it immediately undergoes dehydration to form a conjugated enone.

      The dehydration occurs because the resulting α,β-unsaturated system is highly stabilized by conjugation with both the nitro-substituted ring and the carbonyl group.

   3. Structure of the final product [Y]

      The dehydration removes a water molecule from the β-hydroxy ketone, creating a carbon-carbon double bond between the α and β carbons relative to the carbonyl.

      The double bond is formed between the carbon that was the aldehyde carbon (now β to the carbonyl) and the α-carbon of the original acetophenone.

      Thus, the product is:

   $\text{O}_2\text{N} \!-\! \text{C}_6\text{H}_4 \!-\! \text{CH}\!=\!\text{CH}\!-\!\text{CO}\!-\!\text{C}_6\text{H}_5$

   This is 4-nitrochalcone. The nitro group is on one benzene ring, the carbonyl is on the other, and the two are linked by a trans (usually) carbon-carbon double bond.

   Looking at the options:

   • (A) has a saturated CH₂–CH₂ bridge — no double bond, so incorrect.
   • (B) has a double bond but no carbonyl, and an extra methyl on the second ring — incorrect.
   • (C) has an ester linkage (CO–O) — not formed here.
   • (D) shows exactly the structure: nitro-phenyl–CH=CH–CO–phenyl, with the carbonyl on the plain-phenyl side. This matches.

   Watch out

   A common mistake is to think the Etard reaction gives a carboxylic acid or that the aldol condensation stops at the β-hydroxy stage. Remember: Etard stops at aldehyde, and under basic conditions the aldol product dehydrates readily because of conjugation.

   Tip

   The Claisen–Schmidt condensation is a specific type of crossed aldol where one reactant is an aromatic aldehyde (no α-H) and the other is a ketone. The product is always an α,β-unsaturated ketone, often called a chalcone when both rings are aromatic.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

8. COMEDK 2025Set 2025-M1 markMCQ
   Q.Which one of the following reactions does not give the correct combination of reactants and the major products formed in the reaction? (A) $\mathrm{CH}_3-\mathrm{CO}-\mathrm{CH}_3+\mathrm{NH}_2-\mathrm{NH}-\mathrm{CO}-\mathrm{NH}_2($ weakly basic $) \rightarrow\left(\mathrm{CH}_3\right)_2 \mathrm{C}=\mathrm{N}-\mathrm{NH}-\mathrm{CO}-\mathrm{NH}_2$ (B) $\mathrm{CH}_3 \mathrm{CHO}+2 \mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}($ dry HCl gas $) \leftrightharpoons \mathrm{CH}_3-\mathrm{CH}-\left(\mathrm{OC}_2 \mathrm{H}_5\right)_2$ (C) $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CH}-\mathrm{CH}_2-\mathrm{CH}_2-\mathrm{CN}$ $\xrightarrow[\text { (ii). } \mathrm{H}_2 \mathrm{O}]{\text { (i). DIBAL-H }} \mathrm{CH}_3-\left(\mathrm{CH}_2\right)_4-\mathrm{CHO}$ (D) $\mathrm{C}_6 \mathrm{H}_5-\mathrm{CO}-\mathrm{CH}_3+\mathrm{I}_2+\mathrm{Na}_2 \mathrm{CO}_3($ Heat $) \rightarrow \mathrm{CHI}_3+\mathrm{C}_6 \mathrm{H}_5 \mathrm{COO}^{--}$

   ›Reveal solutionSolution

   The key is to check each reaction for the correct major product under the given conditions. Only option (C) fails because DIBAL-H reduces a nitrile to an aldehyde but stops at the aldehyde stage only when the reaction is quenched under controlled conditions; here the product shown is a saturated aldehyde, but the starting material has a double bond that would also be reduced, so the given product is not the major one.

   Concept & Intuition

   Each reaction tests a specific named reaction or functional-group transformation. To spot the incorrect one, you must recall the exact conditions and the regioselectivity/chemoselectivity of the reagents. The pitfall in (C) is that DIBAL-H (diisobutylaluminium hydride) reduces a nitrile to an aldehyde, but it also reduces carbon‑carbon double bonds if they are present — and here the starting material contains an alkene. The product shown ignores that reduction, so it is not the major product.

   Step‑by‑Step Reasoning

   1. Option (A): Acetone reacts with semicarbazide (NH₂–NH–CO–NH₂) under weakly basic conditions to form a semicarbazone. The carbonyl oxygen is replaced by =N–NH–CO–NH₂. This is the classic semicarbazone formation; the product shown is correct. ✓

   2. Option (B): Acetaldehyde reacts with two equivalents of ethanol in the presence of dry HCl gas to form an acetal. The mechanism: protonation of the carbonyl, nucleophilic attack by ethanol, loss of water, then a second ethanol attack gives the geminal diether. The product CH₃–CH(OC₂H₅)₂ is the correct acetal. ✓

   3. Option (C): The starting material is an unsaturated nitrile: CH₃–CH=CH–CH₂–CH₂–CN. DIBAL‑H at low temperature reduces the nitrile to an imine, which upon aqueous work‑up gives an aldehyde. However, DIBAL‑H also reduces isolated alkenes (it is a strong hydride donor). The double bond would be hydrogenated under these conditions. Therefore the major product after work‑up would be the saturated aldehyde CH₃–(CH₂)₄–CHO, but the reaction as written shows the same saturated aldehyde — wait, that is exactly what is shown! Let’s re‑examine: the product given is CH₃–(CH₂)₄–CHO, which is indeed the fully saturated aldehyde. So why is this wrong? Because the starting material has a double bond between C3 and C4 (numbering from the methyl end). After reduction, the double bond is gone, so the product should be CH₃–(CH₂)₄–CHO. That matches. But the problem says “does not give the correct combination”. The catch: DIBAL‑H is typically used at –78 °C to selectively reduce nitriles to aldehydes without touching other reducible groups. At that low temperature, the alkene might survive. However, the reaction conditions here are not specified with a low temperature; they simply say “(i) DIBAL‑H,

      (ii) H₂O”. In standard practice, DIBAL‑H reduction of a nitrile to an aldehyde is done at low temperature to avoid over‑reduction. But even at low temperature, an isolated alkene is not reduced by DIBAL‑H — DIBAL‑H does not reduce simple alkenes; it reduces only conjugated alkenes or certain activated double bonds. So the alkene should remain intact. Then the product should be an unsaturated aldehyde: CH₃–CH=CH–CH₂–CH₂–CHO. The product shown is the saturated aldehyde, which is incorrect. Therefore (C) is the wrong combination. ✗

   4. Option (D): Acetophenone (C₆H₅–CO–CH₃) undergoes the iodoform reaction with I₂ and Na₂CO₃ upon heating. The methyl ketone is cleaved to give iodoform (CHI₃) and benzoate ion (C₆H₅COO⁻). This is correct. ✓

   Watch out

   A common mistake is to think DIBAL‑H reduces all alkenes. In fact, DIBAL‑H is selective for nitriles, esters, and some carbonyls; isolated alkenes are not reduced under the usual conditions. The error here is that the product shown has lost the double bond, which would not happen.

   Tip

   For DIBAL‑H reductions of nitriles, always check whether the molecule contains other reducible groups. If the alkene is isolated, it survives; if conjugated to the nitrile, it may be reduced.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

9. COMEDK 2025Set 2025-M1 markMCQ
   Q.Identify $[\mathrm{X}]$, the final product formed when 2 moles of Ethanal undergoes the following series of reactions with reagents [(i) to (iv)] (A) But-3-enoic acid (B) But-2-enoic acid (C) Propanoic acid (D) Ethanoic acid

   ›Reveal solutionSolution

   The reaction sequence is an aldol condensation of acetaldehyde followed by oxidation of the resulting α,β-unsaturated aldehyde to the corresponding carboxylic acid. The final product is but-2-enoic acid (crotonic acid), so the correct option is (B).

   ---

   Concept & Intuition

   This problem tests your understanding of two classic carbonyl reactions in sequence:

   1. Aldol condensation – When two molecules of acetaldehyde are treated with dilute NaOH and then heated, they undergo a base-catalyzed aldol addition followed by dehydration. The result is an α,β-unsaturated aldehyde (crotonaldehyde).
   2. Oxidation of an aldehyde to a carboxylic acid – The aldehyde group in crotonaldehyde is then oxidized by a chromium-based reagent (like chromic acid, H₂CrO₄, or K₂Cr₂O₇/H⁺) to give the corresponding carboxylic acid. The carbon–carbon double bond remains intact under these mild oxidative conditions.

   The key insight: the double bond formed in the aldol step is conjugated with the carbonyl, and it survives the oxidation step. So the final product is an unsaturated carboxylic acid with four carbons.

   ---

   Step-by-step reasoning

   1. Aldol addition Two molecules of acetaldehyde (CH₃CHO) react in the presence of dilute NaOH. One molecule acts as the enolate (nucleophile) and attacks the carbonyl carbon of the other. This gives 3-hydroxybutanal (aldol):

   $\text{CH}_3\text{CHO} + \text{CH}_3\text{CHO} \xrightarrow{\text{dil. NaOH}} \text{CH}_3\text{CH(OH)CH}_2\text{CHO}$

   2. Dehydration (heat) On heating, the β-hydroxy aldehyde undergoes elimination of water to form a conjugated α,β-unsaturated aldehyde. The more stable trans isomer is the major product:

   $\text{CH}_3\text{CH(OH)CH}_2\text{CHO} \xrightarrow{\Delta} \text{CH}_3\text{CH=CHCHO} + \text{H}_2\text{O}$

   This compound is crotonaldehyde (but-2-enal).

   3. Oxidation with Cr(VI) reagent The aldehyde group is oxidized to a carboxylic acid by a chromium-based oxidant (e.g., H₂CrO₄ or K₂Cr₂O₇ in acidic medium). The double bond is not affected under these conditions:

   $\text{CH}_3\text{CH=CHCHO} \xrightarrow{\text{Cr(VI), H}^+} \text{CH}_3\text{CH=CHCOOH}$

   The product is but-2-enoic acid (also called crotonic acid).

   4. Check the options
      • (A) But-3-enoic acid – double bond at the wrong position (between C3 and C4).
      • (B) But-2-enoic acid – correct.
      • (C) Propanoic acid – only three carbons, not possible from two acetaldehydes.
      • (D) Ethanoic acid – only two carbons, would require no carbon–carbon bond formation.

   Watch out

   A common mistake is to think that the double bond gets oxidized or that the product is a saturated acid. Remember: Cr(VI) oxidizes aldehydes to acids but does not attack isolated or conjugated C=C bonds under typical conditions.

   Tip

   The aldol condensation of acetaldehyde always gives a four-carbon chain. Counting carbons is a quick sanity check: two C₂ units → C₄ product. That immediately eliminates options (C) and (D).

   ---

   ✓Final answer

   The correct option is (B).

   ANSWER: B

10. COMEDK 2024Set 2024-A1 markMCQ
    Q.Which of the following compounds undergo Aldol condensation followed by dehydration to give 3-phenylprop-2-en-1-al? (A) Benzaldehyde and Acetophenone (B) Two molecules of Benzaldehyde (C) Benzaldehyde and Acetone (D) Benzaldehyde and Ethanal

    ›Reveal solutionSolution

    The target, 3-phenylprop-2-en-1-al (cinnamaldehyde), is the crossed-aldol/dehydration product of benzaldehyde and ethanal: benzaldehyde lacks α-H, so the enolate of ethanal (CH$_3$CHO) attacks it, and loss of water gives $\mathrm{C_6H_5CH=CH-CHO}$.

    The product 3-phenylprop-2-en-1-al is cinnamaldehyde, $\mathrm{C_6H_5-CH=CH-CHO}$ (3 carbons in the propenal chain + phenyl).

    A crossed aldol needs one carbonyl without α-hydrogen (benzaldehyde) and one with α-hydrogen (ethanal):

    $\mathrm{C_6H_5CHO + CH_3CHO \xrightarrow{OH^-} C_6H_5-CH(OH)-CH_2-CHO \xrightarrow{-H_2O} C_6H_5-CH=CH-CHO}$

    The enolate of ethanal adds to benzaldehyde; dehydration of the aldol gives the conjugated $\alpha,\beta$-unsaturated aldehyde, cinnamaldehyde. (Acetone or acetophenone would give a longer/ketonic product, and two benzaldehydes cannot self-condense — no α-H.)

    ✓Final answer

    The correct option is (D) — Benzaldehyde and Ethanal

11. COMEDK 2024Set 2024-E1 markMCQ
    Q.What is/are the product/s formed when Benzaldehyde and Ethanal react in presence of dil. $\mathrm{NaOH}$ followed by heating the intermediate product formed? (A) 2-Methylpent-2-enal & But-2-enal (B) 3-Phenylprop-2-enal & But-2-enal (C) Only product is But-2-enal (D) Only product is 2-Phenylprop-2-enal

    ›Reveal solutionSolution

    The reaction is a crossed aldol condensation between benzaldehyde and ethanal in dilute NaOH, followed by dehydration. The major products are 3-phenylprop-2-enal (cinnamaldehyde) from the crossed reaction and but-2-enal from self-condensation of ethanal. The correct option is (B).

    ---

    The key concept here is the crossed (mixed) aldol condensation between an aromatic aldehyde (benzaldehyde) and an aliphatic aldehyde (ethanal) under basic conditions. Benzaldehyde has no alpha-hydrogens, so it cannot form an enolate; it can only act as the electrophile. Ethanal has alpha-hydrogens and can form an enolate, acting as the nucleophile. In dilute NaOH, both self-condensation of ethanal and crossed condensation with benzaldehyde occur. Upon heating, the intermediate β-hydroxy aldehydes dehydrate to form α,β-unsaturated aldehydes.

    1. Identify the reactants and their roles

       • Benzaldehyde (C₆H₅CHO): no α-hydrogens → cannot form enolate; acts only as the electrophile (carbonyl carbon).
       • Ethanal (CH₃CHO): has α-hydrogens → forms enolate in base; acts as the nucleophile.

    2. Self-condensation of ethanal

       Two molecules of ethanal undergo aldol addition:

    $\text{CH}_3\text{CHO} + \text{CH}_3\text{CHO} \xrightarrow{\text{dil. NaOH}} \text{CH}_3\text{CH(OH)CH}_2\text{CHO}$

    This is 3-hydroxybutanal. On heating, it dehydrates:

    $\text{CH}_3\text{CH(OH)CH}_2\text{CHO} \xrightarrow{\Delta} \text{CH}_3\text{CH=CHCHO} + \text{H}_2\text{O}$

    The product is but-2-enal (crotonaldehyde).

    3. Crossed condensation (benzaldehyde + ethanal) The enolate of ethanal attacks the carbonyl carbon of benzaldehyde:

    $\text{C}_6\text{H}_5\text{CHO} + \text{CH}_3\text{CHO} \xrightarrow{\text{dil. NaOH}} \text{C}_6\text{H}_5\text{CH(OH)CH}_2\text{CHO}$

    This is 3-hydroxy-3-phenylpropanal. On heating, it dehydrates:

    $\text{C}_6\text{H}_5\text{CH(OH)CH}_2\text{CHO} \xrightarrow{\Delta} \text{C}_6\text{H}_5\text{CH=CHCHO} + \text{H}_2\text{O}$

    The product is 3-phenylprop-2-enal (cinnamaldehyde).

    4. Why no other crossed product?

       Benzaldehyde cannot form an enolate, so it cannot attack another aldehyde. Thus, the only possible crossed product is the one where ethanal enolate attacks benzaldehyde.

    5. Heating step

       The intermediate β-hydroxy aldehydes are not stable under heat; they undergo E1cb elimination to give conjugated enals. Both products are formed simultaneously because both self- and crossed-condensation occur in the same pot.

    Watch out

    A common mistake is to think that only the crossed product forms. In reality, ethanal self-condenses readily, so both products appear unless one reactant is in large excess. Here, dilute NaOH and equimolar or near-equimolar amounts yield a mixture.

    Tip

    Benzaldehyde is often used in crossed aldol reactions precisely because it cannot self-condense, simplifying the product mixture. But the other aldehyde (here ethanal) always self-condenses to some extent.

    6. Match with options
       • Option (A): 2-Methylpent-2-enal (not formed here; that would require propanal or longer chain) & But-2-enal → incorrect.
       • Option (B): 3-Phenylprop-2-enal (cinnamaldehyde) & But-2-enal → correct.
       • Option (C): Only but-2-enal → ignores crossed product.
       • Option (D): Only 2-phenylprop-2-enal (wrong name; that would be a different structure) → incorrect.

    ✓Final answer

    The correct option is (B).

    ANSWER: B

12. COMEDK 2024Set 2024-E1 markMCQ
    Q.Compounds $\mathrm{A}$ and $\mathrm{B}$, having the same molecular formula $(\mathrm{C}_4 \mathrm{H}_8 \mathrm{O})$, react separately with $\mathrm{CH}_3 \mathrm{MgBr}$, followed by reaction with dil. $\mathrm{HCl}$ to form compounds $\mathrm{X}$ and $\mathrm{Y}$ respectively. Compound $\mathrm{Y}$ undergoes acidic dehydration in presence of Conc. $\mathrm{H}_2 \mathrm{SO}_4$ much more readily than $\mathrm{X}$. Compound $\mathrm{Y}$ also reacts with Lucas reagent, much more readily than $\mathrm{X}$, with appearance of turbidity. Identify $\mathrm{X}$ and $\mathrm{Y}$. (A) $\mathrm{X}=\text { Pentan }-1-\mathrm{ol} \quad \mathrm{Y}=\text { Pentan }-2-\mathrm{ol}$ (B) $\mathrm{X}=\text { Butan-2-ol } \quad \mathrm{Y}=\text { 2-Methylpropan-2-ol }$ (C) $\mathrm{X}=\text { Pentan-1-ol } \quad \mathrm{Y}=2-\text { Methylpentan-2-ol }$ (D) $\mathrm{X}=\text { Pentan-2-ol } \quad \mathrm{Y}=2 \text {-Methylbutan-2-ol }$

    ›Reveal solutionSolution

    The key is that A and B are C₄H₈O carbonyl compounds (likely an aldehyde and a ketone) that react with CH₃MgBr to give alcohols X and Y. Y dehydrates and reacts with Lucas reagent much faster, so Y must be a tertiary alcohol. The only option where Y is tertiary and X is primary/secondary from C₄H₈O precursors is (B).

    Concept & Intuition

    The molecular formula C₄H₈O suggests either an aldehyde or a ketone (or possibly an enol, but here they react with Grignard reagent). When a carbonyl compound reacts with CH₃MgBr, the Grignard adds a methyl group, forming an alcohol after acidic workup. The product’s structure depends on whether the starting carbonyl was an aldehyde (giving a secondary alcohol) or a ketone (giving a tertiary alcohol).

    The key difference: tertiary alcohols dehydrate easily under acidic conditions (via a stable carbocation) and react instantly with Lucas reagent (ZnCl₂/HCl) to form a cloudy alkyl chloride. Primary alcohols do neither readily; secondary alcohols do so slowly. So Y must be tertiary, X must be primary or secondary.

    We also know A and B have the same formula C₄H₈O — so they are isomers. After adding CH₃MgBr, the alcohols X and Y will have one more carbon (C₅H₁₂O). Let’s check each option.

    Step-by-step reasoning

    1. Identify the starting compounds A and B

       Each is C₄H₈O. After reaction with CH₃MgBr (adds CH₃) and then dilute HCl, we get an alcohol with formula C₅H₁₂O. So X and Y are pentanol isomers (or methyl-substituted butanols). The options list various C₅ alcohols.

    2. Analyze the reactivity clues

       • Y undergoes acidic dehydration much more readily than X.
       • Y reacts with Lucas reagent much more readily than X (turbidity appears quickly). Both point to Y being a tertiary alcohol (forms a stable tertiary carbocation). X is either primary or secondary.

    3. Check each option

       • (A) X = pentan-1-ol (primary), Y = pentan-2-ol (secondary). Secondary alcohol dehydrates and reacts with Lucas faster than primary, but not “much more readily” — both are slow compared to tertiary. Also, pentan-2-ol is not tertiary. So this doesn’t fit the “much more readily” clue.
       • (B) X = butan-2-ol (secondary, C₄), Y = 2-methylpropan-2-ol (tertiary, C₄). Wait — these are C₄ alcohols, but we need C₅ alcohols after Grignard? Let’s check: Starting from C₄H₈O, adding CH₃ gives C₅H₁₂O. But butan-2-ol is C₄H₁₀O — that would mean the starting compound was C₃H₆O? No, that’s inconsistent. Actually, option (B) says X = butan-2-ol (C₄) and Y = 2-methylpropan-2-ol (C₄). That would mean the starting A and B were C₃H₆O? But the problem says A and B are C₄H₈O. So (B) is impossible — the alcohols must have 5 carbons. So (B) is out.
       • (C) X = pentan-1-ol (primary, C₅), Y = 2-methylpentan-2-ol (tertiary, C₆?). 2-Methylpentan-2-ol has 6 carbons — that would require starting from a C₅ carbonyl, not C₄. So (C) is wrong.
       • (D) X = pentan-2-ol (secondary, C₅), Y = 2-methylbutan-2-ol (tertiary, C₅). This fits: both are C₅ alcohols. Starting from C₄H₈O:
         • To get pentan-2-ol (secondary), the starting carbonyl must be butan-2-one (a ketone) — but that would give a tertiary alcohol after adding CH₃MgBr? Wait: If A is butan-2-one (CH₃COCH₂CH₃), adding CH₃MgBr gives 2-methylbutan-2-ol (tertiary), not pentan-2-ol. So pentan-2-ol would come from butanal (aldehyde) — but butanal is C₄H₈O, and adding CH₃MgBr gives pentan-2-ol (secondary). That works.
         • To get 2-methylbutan-2-ol (tertiary), the starting carbonyl must be butan-2-one (ketone) — adding CH₃MgBr gives exactly that. So A = butanal, B = butan-2-one. Then X = pentan-2-ol (secondary), Y = 2-methylbutan-2-ol (tertiary). This matches: tertiary alcohol dehydrates and reacts with Lucas much faster than secondary. So (D) is correct.

    4. Confirm the answer

       Option (D) is the only one where Y is tertiary, X is secondary, both have 5 carbons, and the starting compounds are C₄H₈O isomers.

    Watch out

    A common mistake is to forget that the Grignard adds a carbon — so the product alcohols have one more carbon than the starting carbonyl. Options (B) and (C) give alcohols with wrong carbon counts.

    Tip

    Lucas test and dehydration ease are both governed by carbocation stability: tertiary > secondary > primary. So the “much more readily” clue instantly tells you Y is tertiary.

    ✓Final answer

    The correct option is (D).

    ANSWER: D