Q.How will you bring about the following conversions?

Showing the 12 most recent of 17 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The IUPAC name of the end product Z in the given reaction sequence is $\mathrm{CH_2=CH-CH_3}$ (propene) $\xrightarrow{\text{(i) } Br_2/CCl_4 \text{

   (ii) alc. KOH}, \Delta} X \xrightarrow{NaNH_2, \Delta} Y \xrightarrow{\text{Excess HBr}} Z$ (A) 1, 1 – Dibromopropane (B) 1, 2 – Dibromopropane (C) 1, 3 – Dibromopropane (D) 2, 2 – Dibromopropane

   ›Reveal solutionSolution

   The sequence builds propyne from propene, converts it to its sodium acetylide, and then adds excess HBr twice with Markovnikov regiochemistry onto the same carbon, giving Z = 2,2-dibromopropane.

   Concept and Intuition

   Alkyne chemistry mirrors alkene chemistry but doubled: an unsymmetrical alkyne undergoing addition of 2 equivalents of $\mathrm{HX}$ follows Markovnikov's rule at each step, and because the intermediate vinyl halide already has a halogen on the more-substituted carbon (stabilising the next carbocation-like transition state through induction/hyperconjugation from the existing substituents), the second $\mathrm{HX}$ addition lands on the same carbon as the first — giving a geminal dihalide, not a vicinal one. Terminal alkynes are also weakly acidic (the sp-hybridised C–H is more acidic than sp²/sp³ C–H) and are deprotonated cleanly by strong bases like $\mathrm{NaNH_2}$.

   Step-by-Step Solution

   1. Propene $\to$ 1,2-dibromopropane: $\mathrm{Br_2/CCl_4}$ adds across the propene double bond: $\mathrm{CH_2=CH-CH_3} \to \mathrm{CH_2Br-CHBr-CH_3}$.
   2. $\to X$ (alc. KOH, $\Delta$): Excess alcoholic KOH with heat causes double dehydrohalogenation (elimination of 2 HBr) from the vicinal dibromide, forming the triple bond: $X = \mathrm{CH_3-C\equiv CH}$ (propyne).
   3. $X \to Y$ ($\mathrm{NaNH_2}, \Delta$): Sodamide is a very strong, non-nucleophilic-toward-carbon base; it deprotonates the acidic terminal alkynyl hydrogen of propyne: $Y = \mathrm{CH_3-C\equiv C^-\,Na^+}$ (sodium propynylide/prop-1-ynide).
   4. $Y \to Z$ (excess HBr): The acetylide $Y$ is first protonated by $\mathrm{HBr}$ back to propyne (simple acid-base neutralisation, since $Y$ is a very strong base). With HBr in excess, the resulting propyne then undergoes two sequential Markovnikov additions of $\mathrm{H-Br}$ across the triple bond:
      • First addition: H goes to the terminal CH, Br goes to the internal carbon (more substituted, more stable intermediate) $\Rightarrow \mathrm{CH_3-CBr=CH_2}$ (2-bromopropene). …

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2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.Consider the given sequence of reactions $CH_3-C\equiv CH \xrightarrow{HBr} X \xrightarrow{HBr} Y$ Total number of isomers possible for the product $Y$ is (A) 1 (B) 2 (C) 4 (D) 3

   ›Reveal solutionSolution

   Two successive HBr additions to propyne can give $2,2$-, $1,1$- or $1,2$-dibromopropane, and the vicinal isomer is chiral, so $Y$ has 4 possible isomers.

   Concept and Intuition

   Adding HBr to a terminal alkyne is an electrophilic addition. The first HBr converts the triple bond to a double bond (a bromopropene); the second HBr saturates that double bond to a dibromopropane. Because each addition can place bromine on either of the two triple-bond carbons ($C_1$ or $C_2$), several constitutional dibromides are reachable, and any carbon left with four different groups makes the product optically active.

   Step-by-Step Solution

   1. In propyne $CH_3-C\equiv CH$, only the two triple-bond carbons ($C_1$ terminal, $C_2$ internal) can receive the incoming H and Br; the methyl carbon is untouched, so a $1,3$-dibromide is impossible.
   2. Both bromines on $C_2$ (Markovnikov twice): $CH_3-CBr_2-CH_3$, i.e. $2,2$-dibromopropane.
   3. Both bromines on $C_1$: $CH_3-CH_2-CHBr_2$, i.e. $1,1$-dibromopropane.
   4. One bromine on each of $C_1$ and $C_2$: $CH_3-CHBr-CH_2Br$, i.e. $1,2$-dibromopropane. Here $C_2$ carries four different groups ($CH_3$, $H$, $Br$, $CH_2Br$), so it is a chiral centre giving two enantiomers ($R$ and $S$). …

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3. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.What are X and Y in the following set of reactions respectively? I. 2-Methylpropene $\xrightarrow[H^+]{H_2O}$ X II. 2-Methylpropane $\xrightarrow{KMnO_4}$ Y (A) $(CH_3)_2CHCH_2OH$ ; $(CH_3)_2CHCOOH$ (B) $(CH_3)_2CHCH_2OH$ ; $(CH_3)_3COH$ (C) $(CH_3)_3COH$ ; $(CH_3)_3COH$ (D) $(CH_3)_3COH$ ; $(CH_3)_2CHCOOH$

   ›Reveal solutionSolution

   Both reactions converge on the same product, tert-butanol: acid-catalysed Markovnikov hydration of 2-methylpropene, and KMnO₄ oxidation of the weak tertiary C–H bond in 2-methylpropane.

   Concept and Intuition

   Reaction I is acid-catalysed hydration of an alkene, which follows Markovnikov's rule: the proton $H^+$ adds first to the alkene carbon that will generate the more stable carbocation. For 2-methylpropene, $(CH_3)_2C=CH_2$, protonating the terminal $=CH_2$ carbon produces a tertiary carbocation $(CH_3)_3C^+$, which is far more stable than the alternative primary carbocation. Water then attacks this tertiary carbocation, and after loss of a proton the product is the tertiary alcohol $(CH_3)_3COH$ (tert-butyl alcohol).

   Reaction II is oxidation of an alkane by hot $KMnO_4$. Alkanes are normally resistant to $KMnO_4$, but a tertiary C–H bond (as found in 2-methylpropane / isobutane, $(CH_3)_3CH$) is comparatively weak and more easily abstracted, so vigorous oxidation selectively converts that one C–H bond into a C–OH bond, giving the tertiary alcohol $(CH_3)_3COH$. Crucially, a tertiary alcohol has no more hydrogen on that carbon, so it cannot be oxidised further to a ketone or acid — the oxidation naturally stops at the alcohol stage.

   Both pathways funnel to the same tertiary alcohol.

   Step-by-Step Solution

   1. Identify 2-methylpropene's structure: $(CH_3)_2C=CH_2$.
   2. Apply Markovnikov addition of $H_2O/H^+$: $H^+$ to $=CH_2$ (forms 3° carbocation) → $OH$ attaches to the central carbon → $X = (CH_3)_3COH$. …

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4. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.An alkyne X ($C_4H_6$) does not form sodium alkynide. Reaction of X with $HBr$ gave Y. Another reaction of X with $Na/liq.\ NH_3$ gave Z. Identify Y and Z (A) Y = geminal dibromide ; Z = non-polar compound (B) Y = geminal dibromide ; Z = polar compound (C) Y = vicinal dibromide ; Z = polar compound (D) Y = vicinal dibromide ; Z = non-polar compound

   ›Reveal solutionSolution

   X is the symmetrical internal alkyne but-2-yne; $HBr$ addition gives a geminal dibromide (Y) via two Markovnikov additions on the same carbon, while $Na/liq.\ NH_3$ reduction gives the non-polar trans-alkene (Z).

   Concept and Intuition

   An alkyne that fails to form a sodium alkynide with $NaNH_2$/Na has no acidic terminal $\equiv C-H$ — it must be an internal alkyne. The only $C_4H_6$ internal alkyne is but-2-yne, $CH_3-C\equiv C-CH_3$, which is also symmetric.

   When $HX$ adds twice to a symmetrical internal alkyne, the first addition gives a vinylic halide $CH_3-CBr=CH-CH_3$. On the second addition, Markovnikov's rule directs the new proton/halide so that the incoming carbocation forms on the carbon that already carries the halogen — a halogen atom can stabilise an adjacent positive charge through lone-pair donation (like a bridged halonium ion), even though inductively it is electron-withdrawing. The net outcome is that both halogens end up on the same carbon, giving a geminal dihalide rather than a vicinal one.

   Separately, dissolving-metal reduction of an internal alkyne with $Na$ in liquid $NH_3$ proceeds by a free-radical/anion mechanism that adds hydrogens from opposite faces (anti addition), which for an internal alkyne always delivers the thermodynamically favoured trans-alkene. For but-2-yne this gives trans-2-butene, which — being symmetric about its centre — has its C–CH$_3$ bond dipoles pointing in opposite directions and largely cancelling, making it essentially non-polar (unlike the cis isomer, whose dipoles do not cancel).

   Step-by-Step Solution

   1. X = but-2-yne, $CH_3-C\equiv C-CH_3$ (internal, symmetric, no acidic H).
   2. $X + HBr$ (excess/2 equiv, Markovnikov both times) → both Br end up on the same carbon → Y = 2,2-dibromobutane, a geminal dibromide.
   3. $X + Na/liq.\ NH_3$ → anti addition of two H atoms across the triple bond → Z = trans-2-butene. …

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5. AP EAPCET 2026Set ap-2026-05-20-FN1 markMCQ
   Q.A and B are two position isomers of an alkene $C_5H_{10}$. Both A and B do not exhibit cis-trans isomerism. Addition of HBr with A forms X (major product) and B adds HBr to form Y (major product). What are X and Y respectively? (A) [FIGURE] (two skeletal bromoalkane structures separated by a semicolon: the first is a 4-carbon zigzag chain ending in a carbon bearing two methyl branches plus a Br substituent, i.e. a tertiary bromide $CH_3CH_2C(CH_3)_2Br$; the second is a shorter zigzag chain with one methyl branch near one end and a Br substituent near the other end) (B) [FIGURE] (two skeletal bromoalkane structures separated by a semicolon: the first is a 5-carbon zigzag chain with a Br substituent hanging from a middle carbon; the second is a 5-carbon zigzag chain with a Br substituent, labelled above the chain, on a carbon near one end) (C) [FIGURE] (two skeletal bromoalkane structures separated by a semicolon: the first is the same 5-carbon chain with a mid-chain Br as in option B; the second is the same tertiary-bromide structure as the first structure in option A, $CH_3CH_2C(CH_3)_2Br$, drawn in a mirrored orientation) (D) [FIGURE] (two skeletal bromoalkane structures separated by a semicolon, the same two structures as option A but drawn in mirrored/reversed left-right orientation)

   ›Reveal solutionSolution

   Markovnikov addition of HBr places bromine on the more substituted (more stable carbocation) carbon. A and B are the two position-isomeric pentenes on the 2-methylbutane skeleton that lack cis-trans isomerism — 2-methyl-2-butene and 3-methyl-1-butene — and they give a tertiary and a secondary bromide respectively, matching option (A).

   Concept and Intuition

   For an alkene to lack cis-trans (geometric) isomerism, at least one carbon of the C=C must carry two identical substituents (e.g., two H's, or a terminal =CH₂). Addition of HBr follows Markovnikov's rule: H⁺ adds to the carbon that already has more hydrogens, generating the more stable (more substituted) carbocation, to which Br⁻ then adds.

   Step-by-step solution

   1. Identify A and B. Both are C₅H₁₀ alkenes without cis-trans isomerism. The pair that fits the pictured products is 2-methyl-2-butene, (CH₃)₂C=CHCH₃ (no cis-trans: one alkene carbon bears two identical methyl groups), and 3-methyl-1-butene, (CH₃)₂CH-CH=CH₂ (no cis-trans: the terminal =CH₂ carbon has two identical H's).

   2. HBr addition to 2-methyl-2-butene. The double bond is between a carbon bearing two methyls and a carbon bearing one methyl and one H. Markovnikov addition puts H⁺ on the CH (giving H two neighbours already) and generates the tertiary carbocation at the other carbon; Br⁻ attacks there. Product: 2-bromo-2-methylbutane, CH₃CH₂C(CH₃)₂Br — a tertiary bromide.

   3. HBr addition to 3-methyl-1-butene. The double bond is terminal: (CH₃)₂CH-CH=CH₂. Markovnikov addition puts H⁺ on the terminal CH₂ and generates a secondary carbocation at the adjacent CH (flanked by an isopropyl group and the new CH₃); Br⁻ attacks there. Product: 2-bromo-3-methylbutane, (CH₃)₂CH-CHBr-CH₃ — a secondary bromide. …

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6. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.What are X and Z in the following reaction sequence ? (X forms sodium alkynide.) $X\,(\mathrm{C_4H_6}) \xrightarrow{\text{partial reduction}} Y \xrightarrow{H_2O/H^+} Z\,(\text{major})$ (A) $\mathrm{CH_3C\equiv CCH_3}$ , $\mathrm{CH_3CH_2CH(OH)CH_3}$ (B) $\mathrm{CH_3C\equiv CCH_3}$ , $\mathrm{CH_3CH_2CH_2CH_2OH}$ (C) $\mathrm{CH_3CH_2C\equiv CH}$ , $\mathrm{CH_3CH_2CH(OH)CH_3}$ (D) $\mathrm{CH_3CH_2C\equiv CH}$ , $\mathrm{CH_3CH_2CH_2CH_2OH}$

   ›Reveal solutionSolution

   The alkynide clue fixes X as the terminal alkyne 1-butyne; partial (Lindlar-type)

   reduction gives 1-butene, and Markovnikov hydration of that alkene gives butan-2-ol as

   the major product Z.

   Concept and Intuition

   Only a terminal alkyne ($-C{\equiv}CH$) has an acidic proton on the sp carbon (pKa

   ~25), acidic enough to be removed by a strong base like sodium amide to form a sodium

   alkynide (acetylide) salt. Of the two possible $C_4H_6$ alkyne isomers (1-butyne,

   terminal, and 2-butyne, internal), only 1-butyne can do this — 2-butyne has no terminal

   $\equiv C{-}H$.

   "Partial reduction" of an alkyne (using $H_2$ with Lindlar's poisoned catalyst, or

   Na/liquid $NH_3$) stops at the alkene stage rather than going all the way to the alkane.

   Acid-catalysed hydration of an alkene proceeds through a carbocation intermediate and

   therefore follows Markovnikov's rule: the proton adds to the carbon that already has

   more hydrogens, and $-OH$ ends up on the carbon that can form the more stable

   (more substituted) carbocation.

   Step-by-Step Solution

   1. $X$ ($C_4H_6$) forms a sodium alkynide ⟹ $X$ must have a terminal $\equiv C{-}H$ ⟹ $X = CH_3CH_2C{\equiv}CH$ (1-butyne).
   2. Partial reduction of $X$ (e.g. Lindlar $H_2$/Pd-BaSO$_4$/quinoline) reduces the triple bond to a double bond only: $Y = CH_3CH_2CH{=}CH_2$ (1-butene).
   3. $Y \xrightarrow{H_2O/H^+}$: the alkene is protonated at the terminal ($CH_2$) carbon (giving the more stable secondary carbocation on C2), then water attacks that secondary carbon. …

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7. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.What is 'Z' in the following reaction sequence? (alcohol = ఆల్కహాల్; colourless = రంగులేనిది) $\mathrm{C_3H_6 \xrightarrow[CCl_4]{Br_2} X\ (colourless) \xrightarrow[(ii)\ NaNH_2,\ \Delta]{(i)\ KOH/alcohol} Y \xrightarrow[Hg^{2+},\ H^{+}]{H_2O,\ 333\ K} Z}$ (A) Acetone (B) Propanal (C) Propanol-2 (D) Methoxy ethane

   ›Reveal solutionSolution

   A three-step conversion from propene to a vicinal dibromide, then to a terminal alkyne (propyne) via double dehydrohalogenation with sodalime's 'zipper' effect, then Markovnikov hydration of the alkyne to a methyl ketone. Z = acetone. Answer: (A).

   Concept and Intuition

   This sequence tests three classic named transformations in one chain:

   1. Anti-addition of $\mathrm{Br_2}$ to an alkene gives a colourless vicinal dihalide (the decolourisation of bromine water/Br2-CCl4 is the standard test for unsaturation).
   2. Double dehydrohalogenation with $\mathrm{NaNH_2}$ — sodamide is a very strong, non-nucleophilic base. When a vicinal dihalide is treated first with alcoholic KOH (one elimination to a bromoalkene) and then with excess $\mathrm{NaNH_2}$/heat, a second elimination occurs to give an alkyne, and $\mathrm{NaNH_2}$'s strongly basic conditions also isomerise any internal alkyne toward the thermodynamically favoured terminal (acetylenic) position — the 'alkyne zipper' reaction — because the terminal alkynyl proton is acidic enough to be removed, locking the triple bond there as its conjugate-base acetylide until work-up.
   3. Markovnikov hydration of a terminal alkyne ($\mathrm{H_2O}$/$\mathrm{Hg^{2+}}$, $\mathrm{H_2SO_4}$) proceeds via Markovnikov addition of $\mathrm{-OH}$ to the more substituted alkyne carbon, giving an unstable enol that tautomerises to a methyl ketone.

   Step-by-Step Solution

   1. $\mathrm{C_3H_6}=\mathrm{CH_3-CH=CH_2}$ (propene).
   2. $\mathrm{Br_2}/\mathrm{CCl_4}$ adds across the double bond: $\mathrm{CH_3-CHBr-CH_2Br}$ = X (1,2-dibromopropane), colourless (the orange bromine colour is consumed) — matches the clue 'colourless'.
   3. (i) Alcoholic KOH removes one HBr (E2) from X, giving a bromopropene (e.g. $\mathrm{CH_2=CBr-CH_3}$ or $\mathrm{CH_3-CH=CHBr}$).
   4. (ii) $\mathrm{NaNH_2}$, heat, removes the remaining HBr to form the carbon–carbon triple bond, and drives the triple bond to the terminal position (zipper effect): Y = propyne, $\mathrm{CH_3-C\equiv CH}$. …

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8. AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
   Q.Observe the following set of reactions $\mathrm{C_3H_4 \xrightarrow[\Delta]{Hg^{2+}/H^+} X \xrightarrow[(ii)\ H_2O]{(i)\ MeMgBr} Y}$ $\mathrm{C_3H_6 \xrightarrow{(BH_3)_2} A \xrightarrow{H_2O_2/OH^-} B}$ Correct statement regarding Y and B is (A) Both Y and B are dehydrated with Conc. $\mathrm{H_2SO_4}$/443 K (B) Both Y and B are dehydrated with 20% $\mathrm{H_3PO_4}$/358 K (C) Y is dehydrated with Conc. $\mathrm{H_2SO_4}$/443 K and B with 20% $\mathrm{H_3PO_4}$/358 K (D) B is dehydrated with Conc. $\mathrm{H_2SO_4}$/443 K and Y with 20% $\mathrm{H_3PO_4}$/358 K

   ›Reveal solutionSolution

   Y turns out to be tert-butanol (3° alcohol, easy to dehydrate — mild conditions) and B

   turns out to be n-propanol (1° alcohol, needs harsh conditions) — matching option (D).

   Concept and Intuition

   This problem chains two independent synthetic routes to two different alcohols, then

   tests the well-known rule that ease of acid-catalysed dehydration follows 3° > 2° > 1° (more substituted carbocation intermediates form more easily). Because

   1° alcohols form the least stable carbocation, they need the harshest conditions

   (concentrated acid, high temperature) to dehydrate, while 3° alcohols dehydrate readily

   even with a dilute acid at a lower temperature.

   Step-by-Step Solution

   1. Route 1: $\mathrm{C_3H_4}$ (propyne) + $\mathrm{Hg^{2+}/H^+}$ (acid-catalysed, Markovnikov hydration of an alkyne) → the enol tautomerizes to the ketone, giving X = acetone ($\mathrm{CH_3COCH_3}$).
   2. X + (i) $\mathrm{MeMgBr}$ (ii) $\mathrm{H_2O}$: Grignard addition to the ketone carbonyl, followed by aqueous workup, gives a tertiary alcohol: Y = $(\mathrm{CH_3})_3\mathrm{C-OH}$ (tert-butanol).
   3. Route 2: $\mathrm{C_3H_6}$ (propene) + $(\mathrm{BH_3})_2$ (hydroboration, boron adds to the less hindered/terminal carbon, anti-Markovnikov) → A = a trialkylborane with boron on C1.
   4. A + $\mathrm{H_2O_2/OH^-}$ (oxidation, retention of configuration, anti-Markovnikov overall): gives a primary alcohol, B = $\mathrm{CH_3CH_2CH_2OH}$ (n-propanol).
   5. Since Y is a tertiary alcohol, it dehydrates under mild conditions: 20% $\mathrm{H_3PO_4}$ at 358 K. …

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9. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.An alkene X ($\mathrm{C_4H_8}$) on reaction with HBr gave Y ($\mathrm{C_4H_9Br}$). Reaction of Y with benzene in the presence of anhydrous $\mathrm{AlCl_3}$ gave Z which is resistant to oxidation with $\mathrm{KMnO_4-KOH}$. What are X, Y, Z respectively? (A) X = $\mathrm{(CH_3)_2C=CH_2}$ (isobutylene); Y = $\mathrm{(CH_3)_3CBr}$ (tert-butyl bromide); Z = $\mathrm{C_6H_5C(CH_3)_3}$ (tert-butylbenzene) (B) X = $\mathrm{(CH_3)_2C=CH_2}$ (isobutylene); Y = $\mathrm{(CH_3)_2CHCH_2Br}$ (isobutyl bromide); Z = $\mathrm{C_6H_5CH_2CH(CH_3)_2}$ (isobutylbenzene) (C) X = $\mathrm{CH_3CH_2CH=CH_2}$ (1-butene); Y = $\mathrm{CH_3CH_2CH(Br)CH_3}$ (sec-butyl bromide); Z = $\mathrm{C_6H_5CH(CH_3)CH_2CH_3}$ (sec-butylbenzene) (D) X = $\mathrm{CH_3CH_2CH=CH_2}$ (1-butene); Y = $\mathrm{CH_3CH_2CH_2CH_2Br}$ (n-butyl bromide); Z = $\mathrm{C_6H_5CH_2CH_2CH_2CH_3}$ (n-butylbenzene)

   ›Reveal solutionSolution

   The clue "resistant to $\mathrm{KMnO_4}$-KOH oxidation" identifies Z as tert-butylbenzene (no benzylic hydrogen), which traces back to isobutylene (X) and tert-butyl bromide (Y) — answer (A).

   Concept and Intuition

   $\mathrm{KMnO_4}$ (with KOH, i.e. under alkaline oxidative conditions) oxidizes an alkyl side chain on benzene only if it has at least one benzylic hydrogen (a C–H bond directly on the carbon attached to the ring); the oxidation proceeds via that benzylic C–H, ultimately converting the entire side chain to $-\mathrm{COOH}$. If the ring-attached carbon has no hydrogen (i.e. is fully substituted/quaternary, as in a tert-butyl group), there is no benzylic C–H to abstract, and the side chain survives oxidation intact.

   Step-by-Step Solution

   1. Z resists $\mathrm{KMnO_4}$-KOH oxidation $\Rightarrow$ Z's ring-attached carbon has no benzylic hydrogen $\Rightarrow$ Z must be tert-butylbenzene, $\mathrm{C_6H_5C(CH_3)_3}$ (the carbon attached to the ring bears three methyls and no H).
   2. Z is made by Friedel–Crafts alkylation of benzene with Y in presence of anhydrous $\mathrm{AlCl_3}$; for the tert-butyl group to end up on the ring, Y must be tert-butyl bromide, $(\mathrm{CH_3})_3\mathrm{CBr}$ (its ionization gives the stable 3° carbocation that alkylates benzene). …

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10. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
    Q.Identify X and Y in the following reaction $\mathrm{CH_2=CH_2} \xrightarrow[\text{(ii) Y}]{\text{(i) X}} \mathrm{CH_3CH_2I}$ (A) $\mathrm{HBr,\ NaI/dry\ CH_3COCH_3}$ (B) $\mathrm{HBr,\ I_2/dry\ CH_3COCH_3}$ (C) $\mathrm{Br_2,\ NaI/dry\ CH_3COCH_3}$ (D) $\mathrm{Br_2,\ I_2/dry\ CH_3COCH_3}$

    ›Reveal solutionSolution

    Ethylene is first hydrohalogenated with HBr to bromoethane, then converted to iodoethane via the Finkelstein reaction with NaI in dry acetone.

    Concept and Intuition

    Alkyl iodides are usually not made by direct addition of I2/HI to alkenes; the standard route is: add HBr to the alkene to get the bromide, then perform halide exchange (Finkelstein) using NaI in dry acetone. Acetone dissolves NaI but not NaBr, pulling the reaction forward.

    Step-by-Step Solution

    1. $\mathrm{CH_2=CH_2} + \mathrm{HBr} \to \mathrm{CH_3CH_2Br}$ (addition; unambiguous since ethylene is symmetric).
    2. $\mathrm{CH_3CH_2Br} + \mathrm{NaI} \xrightarrow{\text{dry acetone}} \mathrm{CH_3CH_2I} + \mathrm{NaBr}\downarrow$ - an $S_N2$ halide exchange. …

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11. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
    Q.Which of the following reactions is not correct? (A) $\mathrm{CH_2Br-CH_2Br \xrightarrow{Zn} CH_2=CH_2}$ (B) $\mathrm{CH_3-C\equiv C-CH_3 \xrightarrow[Na]{Liq.\ NH_3} trans\text{-}CH_3CH=CHCH_3}$ (drawn as $\mathrm{H_3C}$ and $\mathrm{H}$ on the left carbon, $\mathrm{H}$ and $\mathrm{CH_3}$ on the right carbon of the double bond) (C) $\mathrm{CH_3-C\equiv CH \xrightarrow[Hg^{+2},\ H^+]{H_2O} CH_3-CH_2-CHO}$ (D) $\mathrm{Ph-CH_2-Br \xrightarrow[Dry\ ether]{Na} Ph-(CH_2)_2-Ph}$

    ›Reveal solutionSolution

    Mercuric-ion-catalyzed hydration of a terminal alkyne follows Markovnikov's rule and gives a methyl ketone, not the aldehyde shown in (C); so (C) is the wrong reaction.

    Concept and Intuition

    Alkyne hydration under $Hg^{2+}/H^+$ catalysis proceeds via Markovnikov addition of water: the OH adds to the more substituted carbon of the triple bond, producing an enol that tautomerizes to the more stable ketone (for terminal alkynes, this is always a methyl ketone, never an aldehyde, because an aldehyde would require anti-Markovnikov addition).

    Step-by-Step Solution

    1. Check (A): 1,2-dibromoethane with Zn dust undergoes debromination (removal of vicinal halogens) to give ethene — this is a standard, correct reaction.
    2. Check (B): internal alkynes with Na/liquid $NH_3$ undergo dissolving-metal reduction which stereospecifically gives the trans-alkene — this is correctly represented.
    3. Check (C): propyne + $H_2O$ with $Hg^{2+}/H^+$ catalyst should add water to the more substituted (internal, $C2$) carbon by Markovnikov's rule, giving the enol $CH_3-C(OH)=CH_2$, which tautomerizes to acetone, $CH_3-CO-CH_3$ — NOT propanal ($CH_3-CH_2-CHO$), which would require anti-Markovnikov addition (only seen with hydroboration-oxidation, not this acid-catalyzed method). So (C) as written is incorrect. …

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12. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
    Q.Among the following the appropriate reactants for the preparation of 1-ethyl cyclohexanol are (A) Ethylidenecyclohexane (structure: a cyclohexane ring with an exocyclic $=\mathrm{CHCH_3}$ double bond) + $\mathrm{H_3O^+}$ (B) Vinylcyclohexane (structure: a cyclohexane ring bearing a $-\mathrm{CH=CH_2}$ substituent) + $\mathrm{BH_3}$ (C) Cyclohexylmagnesium bromide (structure: a cyclohexane ring bearing a $-\mathrm{MgBr}$ substituent) + $\mathrm{CH_3COCH_3}$ (D) Cyclohexyl methyl ketone (structure: a cyclohexane ring bonded to $-\mathrm{C(=O)CH_3}$) + $\mathrm{NaBH_4}$

    ›Reveal solutionSolution

    1-Ethylcyclohexanol needs OH and ethyl on the same ring carbon; Markovnikov (acid-catalysed) hydration of ethylidenecyclohexane delivers exactly that product via the more stable tertiary carbocation.

    Concept and Intuition

    The target, 1-ethylcyclohexan-1-ol, is a tertiary alcohol: the ring carbon (C-1) carries both –OH and an ethyl group, plus its two ring bonds.

    • Option (A): Ethylidenecyclohexane has an exocyclic double bond, ring-C$=$CH–CH$_3$. Protonating this alkene under Markovnikov control places $\mathrm{H^+}$ on the less substituted alkene carbon (the exocyclic $=$CHCH$_3$ carbon), completing it into a full $-CH_2CH_3$ (ethyl) group, and generating the carbocation on the more substituted ring carbon — a tertiary carbocation (bonded to two ring carbons + the new ethyl carbon). Water then attacks this cation, installing –OH exactly where the ethyl group also sits: 1-ethylcyclohexanol. This matches perfectly.
    • Option (B): Anti-Markovnikov hydroboration–oxidation of vinylcyclohexane puts OH on the terminal carbon, giving 2-cyclohexylethanol — wrong connectivity.
    • Option (C): Grignard addition of cyclohexylMgBr to acetone gives ring–C(CH$_3$)$_2$OH (two methyls on the carbinol carbon, not one ethyl on the ring) — a different, isomeric tertiary alcohol.
    • Option (D): NaBH$_4$ reduction of cyclohexyl methyl ketone gives ring–CH(OH)–CH$_3$, a secondary alcohol with the OH on an exocyclic carbon, not 1-ethylcyclohexanol.

    Step-by-Step Solution

    1. Draw the target: ring carbon bearing both OH and $\mathrm{-CH_2CH_3}$. …

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