Q.How will you convert ethanoic acid into benzene?

Showing the 12 most recent of 34 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.Isobutyl alcohol can be prepared from which of the following reactions? I. $\mathrm{(CH_3)_2C=CH_2}$ (isobutylene) $\xrightarrow{\text{(i) } HBr/(C_6H_5CO)_2O_2 \text{

   (ii) } OH^-}$ II. $\mathrm{(CH_3)_2C=O}$ (acetone) $\xrightarrow{\text{(i) } MeMgBr \text{

   (ii) } H_2O}$ III. $\mathrm{(CH_3)_2CH-MgBr}$ (isopropylmagnesium bromide) $\xrightarrow{\text{(i) } HCHO \text{

   (ii) } H_2O}$ Correct answer is (A) I, II only (B) II, III only (C) I, III only (D) I, II, III

   ›Reveal solutionSolution

   This tests which classic routes actually deliver isobutyl alcohol, $\mathrm{(CH_3)_2CHCH_2OH}$: the peroxide-effect hydration of isobutylene and the Grignard-formaldehyde route both work; the acetone + methyl Grignard route gives a tertiary alcohol instead.

   Concept and Intuition

   Isobutyl alcohol is a primary alcohol: $\mathrm{(CH_3)_2CH\text{-}CH_2\text{-}OH}$. To build it you need to end up with a $\mathrm{-CH_2OH}$ group attached to an isopropyl carbon. Grignard reagents reacting with formaldehyde ($\mathrm{HCHO}$) always give a primary alcohol (one carbon longer than the Grignard's carbon skeleton, with a $\mathrm{CH_2OH}$ end), which is exactly what's needed here. But a Grignard reacting with a ketone like acetone gives a tertiary alcohol, not primary — so route II can never give isobutyl alcohol regardless of which Grignard is used, because the ketone carbon becomes the fully-substituted alcohol carbon.

   Step-by-Step Solution

   1. Route I: Isobutylene $\mathrm{(CH_3)_2C=CH_2}$ + $\mathrm{HBr}$ with peroxide $\mathrm{(C_6H_5CO)_2O_2}$ → anti-Markovnikov addition (radical mechanism) puts Br on the less substituted (terminal, $\mathrm{CH_2}$) carbon: $\mathrm{(CH_3)_2CH\text{-}CH_2Br}$. Then $\mathrm{OH^-}$ (SN2) replaces Br with OH: $\mathrm{(CH_3)_2CH\text{-}CH_2OH}$ = isobutyl alcohol. Works.
   2. Route II: Acetone $\mathrm{(CH_3)_2C=O}$ + $\mathrm{CH_3MgBr}$ → addition across C=O gives $\mathrm{(CH_3)_2C(CH_3)\text{-}OMgBr}$, and after $\mathrm{H_2O}$ workup: $\mathrm{(CH_3)_3C\text{-}OH}$, tert-butyl alcohol — a tertiary alcohol, not isobutyl alcohol. Fails. …

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2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.What are A and B respectively in the following reaction sequence? $C_2H_4 \xrightarrow[A]{273\ K} HO-CH_2CH_2-OH \xrightarrow[(ii)\ B]{(i)\ CH_3COCH_3}$ [cyclic ketal product: 2,2-dimethyl-1,3-dioxolane] (A) $KMnO_4/H^+$ ; dil HCl (B) $KMnO_4/H^+$ ; dry HCl (C) dil $KMnO_4$ ; dil HCl (D) dil $KMnO_4$ ; dry HCl

   ›Reveal solutionSolution

   Cold dilute KMnO$_4$ converts ethylene to ethylene glycol by syn-dihydroxylation (A); reacting the glycol with acetone under anhydrous (dry HCl) acid catalysis then forms the cyclic ketal (B).

   Concept and Intuition

   This sequence tests two classic named transformations: (i) cold, dilute, alkaline permanganate as a mild oxidant that adds two -OH groups across a C=C double bond without further oxidative cleavage (unlike hot/concentrated/acidic permanganate, which cleaves the alkene), and (ii) protection of a ketone as a cyclic ketal using a diol and an anhydrous acid catalyst — anhydrous conditions are essential because ketal formation is an equilibrium that water reverses.

   Step-by-Step Solution

   1. A: $C_2H_4 \xrightarrow{A,\ 273K} HOCH_2CH_2OH$. This is the standard test/preparation: cold, dilute (Baeyer's) $KMnO_4$ syn-adds two hydroxyls across the double bond, giving ethylene glycol without breaking the C–C bond. So A = dilute $KMnO_4$ (not the hot/acidic $KMnO_4/H^+$ form, which would cleave the alkene to smaller acids).
   2. Ethylene glycol then reacts with acetone, $CH_3COCH_3$. …

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3. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.The product Z of the given reaction sequence is $C_2H_4 \xrightarrow[(ii)\ H^+/KMnO_4]{(i)\ H^+/H_2O} X \xrightarrow{SOCl_2} Y \xrightarrow{(C_2H_5)_2Cd} Z$ (A) An acid chloride (B) A ketone (C) An aldehyde (D) An ester

   ›Reveal solutionSolution

   Diorganocadmium reagents are the classic way to stop an acid-chloride reaction cleanly

   at the ketone stage, so Z here is a ketone. Answer: (B).

   Concept and Intuition

   This sequence walks through a standard functional-group conversion chain:

   1. $H^+/H_2O$ hydrates ethylene (Markovnikov addition of water) to ethanol.
   2. $H^+/KMnO_4$ (acidic potassium permanganate) is a strong oxidiser; a primary alcohol like ethanol is oxidised all the way through the aldehyde to the carboxylic acid (acetic acid), so $X = CH_3COOH$.
   3. $SOCl_2$ is the standard reagent to convert a carboxylic acid cleanly into the acid chloride (with only gaseous byproducts $SO_2$ and $HCl$), so $Y = CH_3COCl$ (acetyl chloride).
   4. Dialkylcadmium reagents, $R_2Cd$, are a classically taught way to convert an acid chloride selectively into a ketone: they are much less reactive than Grignard or organolithium reagents, so they react once with the acid chloride and then stop — they do not attack the resulting ketone further (whereas a Grignard reagent would add twice, over-reacting to give a tertiary alcohol). So $CH_3COCl + (C_2H_5)_2Cd \rightarrow CH_3COC_2H_5$ (butan-2-one), a ketone.

   Step-by-Step Solution

   1. $C_2H_4 \xrightarrow{H^+/H_2O} CH_3CH_2OH$ (ethanol).
   2. $CH_3CH_2OH \xrightarrow{H^+/KMnO_4} CH_3COOH = X$ (acetic acid; full oxidation of a 1° alcohol under acidic $KMnO_4$). …

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4. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.What are X and Y respectively in the following set of reactions? $Y \xleftarrow[(ii)\ (CH_3CO)_2O]{(i)\ NH_2OH} C_6H_5CHO \xrightarrow[(iii)\ KCN]{(i)\ LiAlH_4,\ H_2O\ (ii)\ PBr_3} X\ (major\ product)$ (A) $C_6H_5CN$ ; $C_6H_5NC$ (B) $C_6H_5CN$ ; $C_6H_5CN$ (C) $C_6H_5CH_2NC$ ; $C_6H_5NC$ (D) $C_6H_5CH_2CN$ ; $C_6H_5CN$

   ›Reveal solutionSolution

   One arm reduces the aldehyde, brominates, then displaces with cyanide to give benzyl

   cyanide ($X$); the other arm forms the oxime and dehydrates it with acetic anhydride

   to give benzonitrile ($Y$). Answer: (D).

   Concept and Intuition

   Benzaldehyde is the common starting point for two different named transformations here:

   Path to X (reduce → halogenate → substitute):

   • $LiAlH_4$ reduces the aldehyde to a primary alcohol: $C_6H_5CHO \to C_6H_5CH_2OH$ (benzyl alcohol).
   • $PBr_3$ swaps the $-OH$ for $-Br$: $C_6H_5CH_2OH \to C_6H_5CH_2Br$ (benzyl bromide), a reactive, unhindered primary/benzylic halide.
   • $KCN$ performs a clean $S_N2$ displacement of bromide by cyanide (benzylic + primary = ideal $S_N2$ substrate): $C_6H_5CH_2Br \to C_6H_5CH_2CN$.
   • So $X = C_6H_5CH_2CN$ (phenylacetonitrile / benzyl cyanide) — note the extra $CH_2$ compared to a direct nitrile on the ring carbon.

   Path to Y (oxime formation, then dehydration to nitrile):

   • $NH_2OH$ converts the aldehyde directly to its oxime: $C_6H_5CHO \to C_6H_5CH=N-OH$ (benzaldoxime).
   • Acetic anhydride, $(CH_3CO)_2O$, is the classic reagent to dehydrate an aldoxime to a nitrile (it acetylates the oxime $-OH$, and elimination of acetic acid then forms the $C\equiv N$ triple bond): $C_6H_5CH=NOH \to C_6H_5C\equiv N$.
   • So $Y = C_6H_5CN$ (benzonitrile) — directly on the ring carbon, no extra $CH_2$.

   Step-by-Step Solution

   1. Reduce: $C_6H_5CHO \xrightarrow{LiAlH_4,H_2O} C_6H_5CH_2OH$.
   2. Brominate: $C_6H_5CH_2OH \xrightarrow{PBr_3} C_6H_5CH_2Br$.
   3. Substitute: $C_6H_5CH_2Br \xrightarrow{KCN} C_6H_5CH_2CN = X$. …

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5. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.What are X and Y in the following reaction sequence? Ethanal $\xleftarrow[\text{(ii) } Zn + H_2O]{\text{(i) } O_3}$ $C_4H_8$ $\xrightarrow[\text{(ii) X}]{\text{(i) } Br_2/CCl_4}$ Y (A) X = alcoholic KOH, $NaNH_2$; Y = $CH_3CH_2C\equiv CH$ (B) X = alcoholic KOH, $NaNH_2$; Y = $CH_3C\equiv CCH_3$ (C) X = alcoholic KOH; Y = $CH_3CH_2C\equiv CH$ (D) X = aq KOH; Y = $CH_3C\equiv CCH_3$

   ›Reveal solutionSolution

   Ozonolysis pins $C_4H_8$ as the symmetric but-2-ene; double dehydrohalogenation of its Br2-addition product (alcoholic KOH then NaNH2) gives the symmetric alkyne but-2-yne. Answer: X = alc. KOH, NaNH2; Y = $CH_3C\equiv CCH_3$.

   Concept and Intuition

   Reductive ozonolysis ($O_3$ then $Zn/H_2O$) cleaves a C=C bond into two carbonyl fragments. Getting only ethanal ($CH_3CHO$) as product means the alkene must be symmetric with a $CH_3CH=CHCH_3$ skeleton (but-2-ene) — each half of the cleaved double bond is $CH_3CH=$, giving $CH_3CHO$ on both sides. Separately, this same alkene undergoes anti addition of $Br_2$ across the double bond to give a vicinal dihalide (2,3-dibromobutane). Converting a vicinal dihalide fully to an alkyne requires two eliminations: alcoholic KOH removes the first HBr (giving a bromoalkene), and because the remaining vinylic C–H is much less acidic/harder to eliminate with a mild base, a much stronger base like sodamide ($NaNH_2$) is needed to complete the second elimination to the alkyne.

   Step-by-Step Solution

   1. Ozonolysis of $C_4H_8$ gives only ethanal $\Rightarrow C_4H_8 = CH_3-CH=CH-CH_3$ (but-2-ene, symmetric alkene).
   2. But-2-ene + $Br_2/CCl_4$ (anti addition) $\rightarrow$ 2,3-dibromobutane, $CH_3CHBr-CHBrCH_3$.
   3. Treat with X = alcoholic KOH: first dehydrohalogenation removes one HBr, forming 2-bromobut-2-ene (a vinylic bromide).
   4. Treat further with $NaNH_2$ (strong base): removes the second (vinylic) H–C–Br unit, completing elimination to the alkyne. …

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6. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.Identify the major product 'P' in the following sequence of reactions (Conc. = concentrated, major = major, dil = dilute) $CH_3COCH_2CH_3 \xrightarrow[\text{(ii) Conc. } H_2SO_4]{\text{(i) } NaBH_4}$ X (major) $\xrightarrow[\text{(ii) } Zn/H_2O]{\text{(i) } O_3}$ Y $\xrightarrow[C_6H_5CHO, \Delta]{\text{dil NaOH}}$ P (A) $\mathrm{C_6H_5-CH=CH-CHO}$ (cinnamaldehyde) (B) $\mathrm{C_6H_5-CH(OH)-CH_2-CHO}$ (C) $\mathrm{CH_3-CH=CH-C_6H_5}$ (D) [FIGURE] (a benzene ring bearing a $-CHO$ group and, at the meta position, a $-CH=CH_2$ vinyl group)

   ›Reveal solutionSolution

   Butan-2-one → (NaBH₄) butan-2-ol → (H₂SO₄, Zaitsev) but-2-ene → (ozonolysis) two acetaldehyde molecules → (crossed Claisen–Schmidt aldol with benzaldehyde, heat) cinnamaldehyde.

   Concept and Intuition

   This is a classic multi-step synthesis chain that combines four separate named ideas: chemoselective ketone reduction (NaBH₄ leaves other functionality alone), acid-catalysed E1 dehydration favouring the Zaitsev (more substituted, more stable) alkene, reductive ozonolysis (which cleaves a C=C bond into two carbonyl fragments without further oxidising them to acids), and finally a crossed aldol condensation: when only one of the two carbonyl partners has α-hydrogens (acetaldehyde does; benzaldehyde does not), the enolate of the α-H-bearing partner attacks the other carbonyl, and base + heat drives the reaction through dehydration to the thermodynamically stable, conjugated α,β-unsaturated carbonyl product.

   Step-by-Step Solution

   1. $CH_3COCH_2CH_3$ (butan-2-one) $\xrightarrow{NaBH_4}$ selective reduction of the ketone (NaBH₄ doesn't touch other groups) gives butan-2-ol, $CH_3CH(OH)CH_2CH_3$.
   2. Butan-2-ol $\xrightarrow{\text{conc. } H_2SO_4}$ X: acid-catalysed E1 dehydration. Two alkenes are possible (but-1-ene, terminal, and but-2-ene, internal); by Zaitsev's rule the more substituted, more stable alkene is the major product: but-2-ene, $CH_3CH=CHCH_3$.
   3. But-2-ene $\xrightarrow{(i) O_3\ (ii) Zn/H_2O}$ Y: reductive ozonolysis cleaves the C=C bond. Since but-2-ene is symmetric (methyl on each end of the double bond), cleavage gives two identical molecules of acetaldehyde, $CH_3CHO$ — so Y = acetaldehyde. …

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7. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.Consider the following reaction sequence Vinylbenzene $\xrightarrow[\Delta]{\mathrm{KMnO_4+KOH}} X \xrightarrow[\Delta]{\mathrm{NaOH+CaO}} Y$ 'Y' can also be formed from (A) Polymerisation of ethylene (B) Polymerisation of propyne (C) Aromatisation of n-hexane (D) Aromatisation of n-heptane

   ›Reveal solutionSolution

   This tests recognizing the oxidative-cleavage + decarboxylation sequence from styrene to benzene, and knowing that catalytic aromatisation of n-hexane (a 6-carbon chain) also produces benzene.

   Concept and Intuition

   Vinylbenzene (styrene, $C_6H_5-CH=CH_2$) has a reactive alkene side chain. Hot alkaline $KMnO_4$ vigorously oxidizes alkenes attached to an aromatic ring, cleaving the double bond: the carbon directly attached to the benzene ring is oxidized up to a carboxylic acid (giving benzoic acid), while the terminal $=CH_2$ carbon is fully oxidized away (as $CO_2$/formic acid, lost from the mixture). So $X$ = benzoic acid, $C_6H_5COOH$. Heating a sodium carboxylate salt with soda lime ($NaOH+CaO$) causes decarboxylation, replacing $-COONa$ with $-H$: benzoate → benzene. So $Y$ = benzene. The question then asks for another synthetic route to benzene: catalytic reforming (aromatisation/dehydrocyclization) of n-hexane, a straight six-carbon alkane, cyclizes and dehydrogenates over a catalyst (e.g. Pt/Cr₂O₃) to directly give benzene, releasing hydrogen gas — this exactly matches Y.

   Step-by-Step Solution

   1. Identify X: styrene + hot alkaline $KMnO_4$ → oxidative cleavage of the $-CH=CH_2$ group → benzoic acid ($C_6H_5COOH$) after acidification.
   2. Identify Y: benzoic acid (as its sodium salt) + soda lime, heat → decarboxylation → benzene ($C_6H_6$).
   3. Evaluate the options for alternate synthesis of benzene (Y):
      • (A) Polymerisation of ethylene gives polyethylene, a saturated polymer — not benzene. …

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8. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.What are X, Y, Z in the following reaction sequence? But-2-ene $\xrightarrow{X}$ Ethanoic acid $\xrightarrow{Y}$ Ethanoyl chloride $\xrightarrow[\text{Anhy. } AlCl_3]{\text{Benzene}}$ Z (A) $\mathrm{KMnO_4/H^+}$; $\mathrm{SOCl_2}$; Acetophenone (B) $\mathrm{KMnO_4/H^+}$; $\mathrm{Cl_2}$; Propiophenone (C) Cold $\mathrm{KMnO_4}$; $\mathrm{SOCl_2}$; Propiophenone (D) Cold $\mathrm{KMnO_4}$; $\mathrm{Cl_2}$; Acetophenone

   ›Reveal solutionSolution

   Hot acidic $\mathrm{KMnO_4}$ cleaves but-2-ene's double bond to two moles of ethanoic acid; $\mathrm{SOCl_2}$ converts the acid to ethanoyl chloride; Friedel–Crafts acylation on benzene then gives acetophenone. The answer is (A).

   Concept and Intuition

   Each step here tests a classic named transformation:

   • Oxidative cleavage of alkenes: hot, acidified $\mathrm{KMnO_4}$ is a strong oxidant that completely cleaves a C=C bond. A =CH–R carbon (one H on the double-bond carbon) is oxidised all the way to –COOH (R–COOH), while a =CR₂ carbon (no H) is oxidised only to a ketone. Cold, dilute $\mathrm{KMnO_4}$ is much milder — it only syn-dihydroxylates the double bond to a vicinal diol, it does not cleave it.
   • Acid → acid chloride: $\mathrm{SOCl_2}$ (thionyl chloride) is the standard, clean reagent for converting a carboxylic acid to its acyl chloride, releasing only gases ($\mathrm{SO_2}$ and HCl) that leave the product pure. $\mathrm{Cl_2}$ is not a reagent for this conversion.
   • Friedel–Crafts acylation: an acyl chloride + arene, catalysed by anhydrous $\mathrm{AlCl_3}$, installs an acyl (–CO–R) group onto the ring, giving an aryl ketone.

   Step-by-Step Solution

   1. But-2-ene: $\mathrm{CH_3-CH=CH-CH_3}$ — a symmetric internal alkene; each alkene carbon carries exactly one H (a mono-substituted vinylic carbon).
   2. Treat with hot, acidified $\mathrm{KMnO_4}$ (X): the double bond is oxidatively cleaved, and since each carbon bears one H, each half becomes a –COOH group: $\mathrm{CH_3-CH=CH-CH_3 \xrightarrow{KMnO_4/H^+} 2\ CH_3COOH}$ (ethanoic acid). This rules out the "Cold $\mathrm{KMnO_4}$" options (C, D), since cold $\mathrm{KMnO_4}$ would only give butane-2,3-diol, not cleave the double bond.
   3. Convert ethanoic acid to ethanoyl chloride using Y = $\mathrm{SOCl_2}$: $\mathrm{CH_3COOH + SOCl_2 \rightarrow CH_3COCl + SO_2\uparrow + HCl\uparrow}$. This rules out option B, which proposes $\mathrm{Cl_2}$ (not a reagent for this conversion). …

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9. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.The IUPAC name of the product Z in the reaction sequence is $\mathrm{C_3H_6} \xrightarrow{\mathrm{H_2O/H^+}} X \xrightarrow{\mathrm{CrO_3}} Y\ (\text{major}) \underset{(2)\ \Delta}{\overset{(1)\ \mathrm{Ba(OH)_2}}{\rightleftharpoons}} Z$ (A) But-2-enal (B) 4-Methylpent-3-enal (C) 4-Methylpent-3-en-2-one (D) 2-Methylpent-2-enal

   ›Reveal solutionSolution

   This chains Markovnikov hydration, alcohol oxidation, and base-catalyzed acetone self-aldol condensation to arrive at mesityl oxide — answer (C).

   Concept and Intuition

   Propene's acid-catalyzed hydration follows Markovnikov's rule (H adds to the carbon with more H's already), giving the secondary alcohol isopropanol, not the primary alcohol. Oxidizing a secondary alcohol with $\mathrm{CrO_3}$ stops cleanly at the ketone (no further oxidation possible, since there's no H left on that carbon after forming the C=O). Finally, acetone — having α-hydrogens — undergoes a classic base-catalyzed aldol condensation with itself: two acetone molecules combine to a β-hydroxy ketone (diacetone alcohol), and heating drives dehydration to the conjugated enone, mesityl oxide.

   Step-by-Step Solution

   1. $\mathrm{CH_2=CHCH_3} \xrightarrow{\mathrm{H_2O/H^+}}$ Markovnikov addition places OH on the more substituted carbon: $X = (\mathrm{CH_3})_2\mathrm{CHOH}$ (isopropanol).
   2. $X \xrightarrow{\mathrm{CrO_3}}$ oxidation of a 2° alcohol gives a ketone: $Y = (\mathrm{CH_3})_2\mathrm{CO}$ (acetone), the major product.
   3. Acetone + Ba(OH)₂ (base catalyst): self-aldol addition — the enolate of one acetone molecule attacks the carbonyl carbon of another, forming diacetone alcohol, $(\mathrm{CH_3})_2\mathrm{C(OH)CH_2COCH_3}$. …

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10. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.An alcohol X($\mathrm{C_4H_{10}O}$) on dehydration gave alkene ($\mathrm{C_4H_8}$) as major product, which on bromination followed by treatment with Y gave alkyne $\mathrm{C_4H_6}$. Alkyne $\mathrm{C_4H_6}$ does not react with sodium metal. What are X and Y? (alc. = alcoholic, aq. = aqueous) (A) $\mathrm{CH_3-CH(OH)-CH_2-CH_3}$ ; aq. KOH (B) $\mathrm{CH_3-CH(OH)-CH_2-CH_3}$ ;

    (i) alc.KOH

    (ii) $\mathrm{NaNH_2}$ (C) $\mathrm{CH_3-CH_2-CH_2-CH_2-OH}$ ; alc. KOH (D) $\mathrm{CH_3-CH_2-CH_2-CH_2-OH}$ ;

    (i) alc.KOH

    (ii) $\mathrm{NaNH_2}$

    ›Reveal solutionSolution

    The clue "alkyne $\mathrm{C_4H_6}$ does not react with sodium" identifies it as internal 2-butyne (no acidic terminal H); tracing that back through dehydration → bromination → double dehydrohalogenation identifies X as 2-butanol and Y as alc. KOH then $\mathrm{NaNH_2}$.

    Concept and Intuition

    Terminal alkynes ($\mathrm{R-C\equiv C-H}$) have an acidic $\equiv\mathrm{C-H}$ (sp carbon, high s-character) and react with active metals like sodium, releasing $\mathrm{H_2}$ and forming an acetylide. Internal alkynes ($\mathrm{R-C\equiv C-R'}$) have no such H and are inert to Na. This distinction is the key to identifying which $\mathrm{C_4H_6}$ isomer is meant, and that in turn tells us which alcohol/alkene/dihalide the synthesis must have passed through.

    Step-by-Step Solution

    1. $\mathrm{C_4H_6}$ alkynes: but-1-yne ($\mathrm{CH\equiv C-CH_2-CH_3}$, terminal) and but-2-yne ($\mathrm{CH_3-C\equiv C-CH_3}$, internal). Since the product does not react with Na, it must be the internal isomer, but-2-yne (2-butyne).
    2. To make an internal alkyne symmetrically by double dehydrohalogenation of a vicinal dihalide, the dihalide must be 2,3-dibromobutane, $\mathrm{CH_3-CHBr-CHBr-CH_3}$.
    3. That dibromide comes from bromination of the alkene 2-butene, $\mathrm{CH_3-CH=CH-CH_3}$ (adding $\mathrm{Br_2}$ across the double bond).
    4. 2-Butene, in turn, is the Zaitsev (major, more-substituted) product of acid-catalysed dehydration of 2-butanol, $\mathrm{CH_3-CH(OH)-CH_2-CH_3}$ — matching alcohol X = $\mathrm{C_4H_{10}O}$. …

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11. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
    Q.What are B and C respectively in the following set of reactions ? $C \xleftarrow[\Delta]{Zn} 1,2\text{-dibromopropane} \xrightarrow[\text{(ii) } NaNH_2]{\text{(i) alc.KOH}} A \xrightarrow[\text{Catalyst}]{\text{Lindlar}} B$ (A) [FIGURE] (two three-carbon skeletal alkene structures drawn side by side: the first has its double bond on the left segment of the zigzag; the second has its double bond on the right segment of the zigzag) (B) [FIGURE] (two three-carbon skeletal chains drawn side by side with no double bond shown on either) (C) [FIGURE] (a three-carbon skeletal structure with a double bond on the left segment, next to a three-carbon skeletal chain with no double bond) (D) [FIGURE] (a three-carbon skeletal chain with no double bond, next to a three-carbon skeletal structure with a double bond on the right segment)

    ›Reveal solutionSolution

    Both routes from 1,2-dibromopropane — direct Zn debromination (giving C) and alc. KOH/NaNH$_2$ then Lindlar hydrogenation (giving B) — converge on the same terminal alkene, propene; so B and C are both alkene structures, just drawn with the double bond at opposite ends of the chain.

    Concept and Intuition

    Zinc dust with heat is the classic reagent for debrominating a vicinal dihalide: it removes the two halogens from adjacent carbons and reforms the C=C bond exactly where they were, regenerating the parent alkene (this is the reverse of the bromine-addition test for unsaturation). Separately, alcoholic KOH under heat performs double dehydrohalogenation on a vicinal dihalide, eliminating two equivalents of HX to build a triple bond — giving an alkyne. Since 1,2-dibromopropane's two bromines are on C1–C2, this produces propyne, which is already a terminal alkyne; treating a terminal alkyne with the very strong base NaNH$_2$ merely removes its acidic $\equiv C\text{–}H$ proton (forming the sodium acetylide), it does not change the carbon skeleton. Finally, Lindlar's poisoned catalyst partially hydrogenates a triple bond to a cis-alkene — but because this alkyne is terminal, the product carbon bears two H atoms and there is no cis/trans distinction: hydrogenation simply gives propene again.

    Step-by-Step Solution

    1. 1,2-Dibromopropane: $CH_2Br\text{-}CHBr\text{-}CH_3$.
    2. Zn/$\Delta$ (debromination): removes both Br atoms, forms the double bond between C1–C2 → $C = CH_2{=}CH{-}CH_3$ (propene).
    3. (i) alc. KOH, heat: eliminates 2 HBr in succession to form the triple bond at the same position → propyne, $CH_3{-}C{\equiv}CH$ (already a terminal alkyne).
    4. (ii) NaNH$_2$: deprotonates the terminal $\equiv C\text{-}H$ (forms sodium propynide); the carbon skeleton and unsaturation position are unchanged, so this intermediate is effectively still propyne (call it A) once reprotonated. …

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12. AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
    Q.What are X and Y in the following reactions? benzyl alcohol ($\mathrm{C_6H_5CH_2OH}$) $\xleftarrow{X}$ benzoic acid ($\mathrm{C_6H_5COOH}$) $\xrightarrow[\text{(ii) Y}]{\text{(i) } SOCl_2}$ benzaldehyde ($\mathrm{C_6H_5CHO}$) (A) X = $\mathrm{B_2H_6}$; Y = $\mathrm{H_2/Pd}$ (B) X = $\mathrm{B_2H_6}$; Y = $\mathrm{H_2,\ Pd\text{-}BaSO_4}$ (C) X =

    (i) $\mathrm{NaBH_4}$

    (ii) $\mathrm{H_2O}$; Y = $\mathrm{H_2,\ Pd\text{-}BaSO_4}$ (D) X =

    (i) $\mathrm{NaBH_4}$

    (ii) $\mathrm{H_2O}$; Y = $\mathrm{H_2/Pd}$

    ›Reveal solutionSolution

    Benzoic acid → benzyl alcohol needs a reagent strong enough to reduce $-\mathrm{COOH}$ ($\mathrm{B_2H_6}$, not $\mathrm{NaBH_4}$); benzoic acid → benzaldehyde via the acid chloride needs Rosenmund's poisoned-catalyst hydrogenation ($\mathrm{H_2, Pd\text{-}BaSO_4}$) to avoid over-reduction to the alcohol.

    Concept and Intuition

    Reducing agents differ sharply in strength: $\mathrm{NaBH_4}$ is a mild reducing agent that reduces aldehydes/ketones but is generally too weak to touch carboxylic acids or esters; $\mathrm{LiAlH_4}$ and $\mathrm{B_2H_6}$ (diborane) are strong enough to reduce $-\mathrm{COOH}$ all the way to $-\mathrm{CH_2OH}$. Separately, converting an acid to an aldehyde (rather than over-reducing to the alcohol) requires stopping hydrogenation exactly at the aldehyde stage — achieved industrially/academically via Rosenmund's reduction: the acid chloride is hydrogenated over Pd poisoned with $\mathrm{BaSO_4}$ (or quinoline-S), which prevents further reduction to the alcohol.

    Step-by-Step Solution

    1. Benzoic acid → benzyl alcohol: this is a reduction of $-\mathrm{COOH}$ to $-\mathrm{CH_2OH}$. $\mathrm{NaBH_4}$ cannot reduce carboxylic acids, ruling out options (C) and (D). $\mathrm{B_2H_6}$ (diborane) selectively reduces $-\mathrm{COOH}$ to $-\mathrm{CH_2OH}$ — so X = $\mathrm{B_2H_6}$.
    2. Benzoic acid + $\mathrm{SOCl_2}$ → benzoyl chloride ($\mathrm{C_6H_5COCl}$), releasing $\mathrm{SO_2}$ and HCl — a clean, standard way to make an acid chloride. …

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