Q.The following reaction is classified as: CH3CH2I + KOH(aq) → CH3CH2OH + KI.

Showing the 12 most recent of 29 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.Consider the following reactions in which 'P' is the major product $\mathrm{CH_3-CH=CH_2} \xrightarrow[\text{Benzoyl peroxide}]{HBr} P \xrightarrow{\text{Finkelstein reaction}} Q$ $P \xrightarrow{\text{Swarts reaction}} R$ The correct order of reactivity of $P, Q$ and $R$ towards $S_N2$ reaction is (A) $P > Q > R$ (B) $R > Q > P$ (C) $Q > P > R$ (D) $Q > R > P$

   ›Reveal solutionSolution

   This tests the peroxide (anti-Markovnikov) effect, halogen-exchange reactions, and leaving-group trends in $S_N2$; the answer is $Q>P>R$.

   Concept and Intuition

   The peroxide effect reverses the regiochemistry of HBr addition to an alkene via a radical mechanism, placing Br on the LESS substituted carbon. Once we have the alkyl bromide P, the Finkelstein reaction (NaI in dry acetone, driven by NaBr's insolubility in acetone) swaps Br for I, while the Swarts reaction (using a metal fluoride like $AgF$/$Hg_2F_2$/$SbF_3$) swaps a halogen for F. $S_N2$ reaction rates on the same carbon skeleton are governed almost entirely by how good a leaving group the halide is — better leaving groups (weaker, more polarisable C–X bonds) react faster.

   Step-by-Step Solution

   1. $CH_3-CH=CH_2 + HBr \xrightarrow{\text{peroxide}}$: anti-Markovnikov addition via radical mechanism gives $CH_3CH_2CH_2Br$ (1-bromopropane) = P.
   2. $P \xrightarrow{\text{Finkelstein (NaI/acetone)}} CH_3CH_2CH_2I$ (1-iodopropane) = Q — Br is replaced by I.
   3. $P \xrightarrow{\text{Swarts}} CH_3CH_2CH_2F$ (1-fluoropropane) = R — Br is replaced by F. …

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2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.Observe the following reactions I. $CH_3CH_2CH_2CH_2OH \xrightarrow[95\%\ H_3PO_4]{KI}$ II. $CH_3CH_2CH_2CH_2OH \xrightarrow[273\ K]{HCl}$ III. $CH_3CH_2CH_2CH_2Cl \xrightarrow[dry\ CH_3COCH_3]{NaI}$ Feasible reactions are (dry = dry; only = only) (A) I, II only (B) I only (C) I, III only (D) I, II, III

   ›Reveal solutionSolution

   (I) KI/95% H$_3$PO$_4$ converts a 1° alcohol to the iodide — feasible; (II) plain HCl at 273 K on a 1° alcohol without ZnCl$_2$ is not; (III) NaI in dry acetone (Finkelstein) converts the chloride to the iodide — feasible.

   Concept and Intuition

   Alcohol → alkyl halide conversions and halide-exchange reactions each need the right combination of reagent, catalyst and conditions; simply naming an acid/halide isn't enough — the specific catalyst (or its absence) decides whether the reaction actually goes.

   Step-by-Step Solution

   1. I: $ROH + KI + 95\%\,H_3PO_4 \rightarrow RI$. Phosphoric acid generates HI in situ without oxidising it to $I_2$ (unlike conc. $H_2SO_4$, which does oxidise HI) — this is the standard, reliable lab preparation of alkyl iodides from alcohols. Feasible.
   2. II: Conversion of an alcohol to the chloride by plain HCl requires anhydrous $ZnCl_2$ (the Lucas reagent) as a Lewis-acid catalyst, and even then primary alcohols react extremely slowly, needing heat, not a cold 273 K bath. With no catalyst mentioned and a low temperature specified, this reaction is not expected to proceed. Not feasible. …

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3. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.Consider the two chlorides given below (I) (R)-2-Chloropentane (II) (R)-3-Chloroheptane Chloride (I) undergoes nucleophilic substitution reaction (X) in single step Chloride (II) undergoes nucleophilic substitution reaction (Y) in two steps Stereochemistry involved in reactions X and Y respectively is (A) Retention ; racemisation (B) Inversion ; retention (C) Inversion ; racemisation (D) Racemisation ; inversion

   ›Reveal solutionSolution

   Single-step substitution = $S_N2$ = inversion; two-step substitution = $S_N1$ = racemisation via a planar carbocation.

   Concept and Intuition

   The number of "steps" in a nucleophilic substitution directly tells you the mechanism and hence the stereochemical outcome. $S_N2$ is a single concerted step: the nucleophile attacks from the side opposite the leaving group (backside attack), so the configuration at that carbon inverts (like an umbrella flipping inside out) — this is the Walden inversion. $S_N1$ proceeds in two steps: first the leaving group departs to form a planar, $sp^2$-hybridised carbocation, then the nucleophile attacks; because the carbocation is planar, the nucleophile can attack from either face with roughly equal probability, so the product is (partially or fully) racemic.

   Step-by-Step Solution

   1. (R)-2-Chloropentane undergoes reaction X in a single step $\Rightarrow$ this is $S_N2$ $\Rightarrow$ backside attack $\Rightarrow$ configuration at that carbon inverts. …

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4. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.Consider the following organic halides (I) $\mathrm{CH_3-CH_2-CH_2-Br}$ (II) $\mathrm{CH_2=CH-CH_2-Br}$ (III) $\mathrm{CH_3-CH=CH-Br}$ The correct order of reactivity towards $S_N2$ reaction is (A) I > II > III (B) III > II > I (C) II > I > III (D) III > I > II

   ›Reveal solutionSolution

   Allylic halides are the fastest $S_N2$ substrates (resonance-stabilised TS), plain

   primary halides are intermediate, and vinylic halides essentially do not undergo

   $S_N2$ at all. Answer: (C).

   Concept and Intuition

   $S_N2$ reactivity is governed by (a) how open backside attack is and (b) how stable the

   transition state (partial carbanion-like character at the reacting carbon) is.

   • Allylic systems ($CH_2=CH-CH_2-Br$): the transition state's developing negative character can delocalise into the adjacent $\pi$ bond (resonance stabilisation), which lowers the activation energy — allylic halides are more reactive toward $S_N2$ than simple unactivated primary halides.
   • Simple primary alkyl halides ($CH_3CH_2CH_2Br$): normal, unassisted $S_N2$ rate — no rate-enhancing or -retarding electronic effect, but no steric hindrance either.
   • Vinylic halides ($CH_3-CH=CH-Br$): the C–Br bond here is an $sp^2$–halogen bond, shorter and stronger than an $sp^3$ C–X bond, and the halogen lone pairs / $\pi$ system make backside approach and $sp^3$-like transition-state geometry essentially impossible. Vinylic (and aryl) halides are effectively inert to $S_N2$.

   Step-by-Step Solution

   1. Rank by transition-state stabilisation: allylic (resonance-assisted) > plain primary …

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5. AP EAPCET 2026Set ap-2026-05-19-AN1 markMCQ
   Q.Consider the following reaction and statements given about X 1-Ethylcyclohexene + HBr $\rightarrow$ X (major product) I. It undergoes nucleophilic substitution generally in polar protic solvents II. C-Br bond is cleaved in rate determining step in its nucleophilic substitution reaction III. The rate of its nucleophilic substitution depends on concentration of nucleophile Correct statements are (only) (A) I, III only (B) I, II only (C) II only (D) II, III only

   ›Reveal solutionSolution

   The major product of HBr addition to 1-ethylcyclohexene is a tertiary bromide, which reacts by $S_N1$; only statements I and II (protic-solvent preference, and C–Br cleavage in the RDS) are true of $S_N1$ — nucleophile concentration does not affect its rate.

   Concept and Intuition

   Markovnikov's rule: in HX addition to an unsymmetrical alkene, H adds to the carbon that already has more hydrogens, and X adds to the more substituted carbon (via the more stable carbocation intermediate). A tertiary halide formed this way reacts via $S_N1$: a two-step mechanism where the rate-limiting step is unimolecular ionisation of the substrate to form a carbocation, so the rate law is $Rate = k[RX]$ only — the nucleophile attacks in a fast, non-rate-determining second step.

   Step-by-Step Solution

   1. 1-Ethylcyclohexene has its ring double bond between C1 (bearing the ethyl substituent) and C2. Protonation (Markovnikov) puts H on C2 (more H's already), generating the more stable 3° carbocation at C1 (stabilised by the ring + ethyl group).
   2. $Br^-$ then attacks this cation, giving 1-bromo-1-ethylcyclohexane, X — a tertiary alkyl halide.
   3. Tertiary halides react via $S_N1$ (steric hindrance disfavours back-side $S_N2$ attack, and the cation is stable enough to form).
   4. Statement I: $S_N1$ needs a solvent that can solvate/stabilise both the developing carbocation and the leaving group — polar protic solvents do this well (H-bonding to $Br^-$, dipole stabilisation of the cation). TRUE. …

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6. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.In which of the following, the given four compounds (I to IV) are correctly arranged in the decreasing order of reactivity towards $S_N2$ reaction? 1-Bromobutane (I) 1-Bromo-2,2-dimethylpropane (II) 1-Bromo-2-methylbutane (III) 1-Bromo-3-methylbutane (IV) (A) $I > IV > III > II$ (B) $I > II > III > IV$ (C) $I > III > IV > II$ (D) $IV > III > II > I$

   ›Reveal solutionSolution

   This tests how β- vs γ-branching affects SN2 rate through steric hindrance to backside attack. All four are primary bromides, so the ranking comes purely from how much alkyl bulk sits near the reacting carbon.

   Concept and Intuition

   An SN2 reaction requires the nucleophile to attack the carbon bearing the leaving group from the back side, opposite the C–Br bond. Anything that crowds that back side — especially bulky groups on the β-carbon (the carbon next to the one bearing Br) — dramatically slows the reaction, because the incoming nucleophile has to squeeze past those groups. Branching further away, on the γ-carbon, still slows things down but far less, since it's one more bond removed from the transition state.

   The four bromides, drawn out:

   • I: 1-Bromobutane — $CH_3CH_2CH_2CH_2Br$ — no branching anywhere near C1.
   • II: 1-Bromo-2,2-dimethylpropane (neopentyl bromide) — $BrCH_2C(CH_3)_3$ — the β-carbon (C2) carries three methyl groups, making it maximally crowded right next to the reaction site.
   • III: 1-Bromo-2-methylbutane — $BrCH_2CH(CH_3)CH_2CH_3$ — one methyl branch on the β-carbon (C2).
   • IV: 1-Bromo-3-methylbutane (isoamyl bromide) — $BrCH_2CH_2CH(CH_3)CH_3$ — the branch is on the γ-carbon (C3), not the β-carbon.

   Step-by-Step Solution

   1. Rank by hindrance at the β-carbon first, since that's what dominates the SN2 transition state.
   2. I has zero β-branching → most reactive. …

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7. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.What are X and Y in the following set of reactions? (dry acetone) $\mathrm{CH_3CH_2CH_2F} \xleftarrow{Y} \mathrm{CH_3CH_2CH_2Br} \xrightarrow{X} \mathrm{CH_3CH_2CH_2I}$ (A) NaI / dry acetone; $\mathrm{SbF_3}$ (B) NaI / $\mathrm{H_2O}$; NaF (C) NaI / $\mathrm{H_2O}$; $\mathrm{SbF_3}$ (D) NaI / dry acetone; NaF

   ›Reveal solutionSolution

   This tests the two classic reagent-specific halogen-exchange reactions on alkyl halides — Finkelstein (→ iodide) and Swarts (→ fluoride); answer is (A).

   Concept and Intuition

   Converting one alkyl halide into another by simple halide exchange relies on choosing a reagent/solvent combination that drives the equilibrium in the desired direction, since the incoming and outgoing halide ions could in principle both act as nucleophiles.

   • Finkelstein reaction: NaI dissolved in dry acetone reacts with an alkyl chloride/bromide. NaI is soluble in acetone, but the NaBr (or NaCl) byproduct is NOT — it precipitates out of solution. Removing a product drives the $S_N2$ equilibrium continually toward the iodide, even though I⁻ is also a good leaving group.
   • Swarts reaction: direct nucleophilic displacement of Cl/Br by F⁻ is inefficient because F⁻ is a poor nucleophile and C–F bond formation needs a different activation route. Instead, treating the alkyl halide with an inorganic metal fluoride (AgF, Hg₂F₂, CoF₂, or SbF₃) selectively installs fluorine.

   Step-by-Step Solution

   1. Identify the two transformations shown: $\mathrm{CH_3CH_2CH_2Br} \to \mathrm{CH_3CH_2CH_2I}$ (via X) and $\mathrm{CH_3CH_2CH_2Br} \to \mathrm{CH_3CH_2CH_2F}$ (via Y).
   2. Br → I is the textbook Finkelstein reaction: reagent $\mathrm{NaI}$ in dry acetone. So $X = \text{NaI/dry acetone}$. …

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8. AP EAPCET 2025Set ap-2025-05-19-AN1 markMCQ
   Q.The $S_N2$ reactivity of the following compounds will be in the order I) $\mathrm{C_6H_5CH(CH_3)Br}$ II) $(\mathrm{C_6H_5})_2\mathrm{CHBr}$ III) $(\mathrm{C_6H_5})_2\mathrm{C(CH_3)Br}$ IV) $\mathrm{C_6H_5CH_2Br}$ (A) III < II < I < IV (B) III < I < II < IV (C) II < III < IV < I (D) II < IV < I < III

   ›Reveal solutionSolution

   SN2 reactivity falls as steric bulk around the reacting carbon rises; ranking these four benzylic halides by hindrance gives III < II < I < IV — answer (A).

   Concept and Intuition

   $S_N2$ is a single concerted step where the nucleophile attacks from the side opposite the leaving group. Anything that crowds that backside approach — extra alkyl/aryl groups on the reacting carbon — slows the reaction sharply, regardless of any electronic (resonance) stabilization that might matter for $S_N1$. So for a purely $S_N2$ ranking, count and identify substituents on the carbon bearing Br.

   Step-by-Step Solution

   1. IV ($\mathrm{C_6H_5CH_2Br}$): primary carbon, only one phenyl + two H's — least hindered, reacts fastest by $S_N2$.
   2. I ($\mathrm{C_6H_5CH(CH_3)Br}$): secondary carbon, one phenyl + one methyl + one H.
   3. II ($(\mathrm{C_6H_5})_2\mathrm{CHBr}$): secondary carbon, but now TWO phenyl groups + one H — bulkier than I, since two aromatic rings crowd the transition state more than one phenyl + one small methyl. …

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9. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.What are X and Y respectively in the following set of reactions ? (red = red) $\mathrm{CH_3CH_2CH_2OH} \xrightarrow[\text{red P}]{\mathrm{Br_2}} X$ $\mathrm{CH_3CH_2COOH} \xrightarrow[\text{(ii) } H_2O]{\text{(i) } Br_2/\text{red P}} Y$ (A) $\mathrm{CH_3CH_2CH_2Br}$ , $\mathrm{CH_3CH_2COBr}$ (B) $\mathrm{CH_3CH_2CH_2Br}$ , $\mathrm{CH_3CHBrCOOH}$ (C) $\mathrm{CH_3CHBrCH_2OH}$ , $\mathrm{CH_3CH_2COBr}$ (D) $(\mathrm{CH_3})_2\mathrm{CHBr}$ , $\mathrm{CH_3CHBrCOOH}$

   ›Reveal solutionSolution

   $Br_2$/red P converts an alcohol to the corresponding alkyl bromide by simple

   substitution, but converts a carboxylic acid (with an $\alpha$-H) into an

   $\alpha$-bromo acid via the Hell–Volhard–Zelinsky reaction.

   Concept and Intuition

   Red phosphorus with $Br_2$ generates phosphorus tribromide ($PBr_3$) in situ. $PBr_3$

   reacts with an alcohol's $-OH$ group by simple nucleophilic substitution, replacing

   $-OH$ with $-Br$ — no $\alpha$-carbon chemistry is involved here because the substrate

   is an alcohol, not a carbonyl compound.

   With a carboxylic acid instead, $Br_2$/red P behaves completely differently: it is the

   Hell–Volhard–Zelinsky (HVZ) reaction. The acid first forms a small amount of acid

   bromide (via $PBr_3$), which enolises far more readily than the acid itself; $Br_2$ then

   adds at the $\alpha$-carbon of this enol. Aqueous work-up hydrolyses the acid bromide

   back to the free acid, now carrying a bromine on the $\alpha$-carbon.

   Step-by-Step Solution

   1. Top reaction: $CH_3CH_2CH_2OH$ (1-propanol) + $Br_2$/red P → the red P + $Br_2$ combination is simply a way of forming $PBr_3$, which substitutes the $-OH$ for $-Br$. Product $X = CH_3CH_2CH_2Br$ (1-bromopropane), with no change to the carbon skeleton.
   2. Bottom reaction: $CH_3CH_2COOH$ (propanoic acid) + (i) $Br_2$/red P — this is the HVZ reaction. The $\alpha$-carbon (the $CH_2$ next to $-COOH$) is brominated via the acid bromide/enol intermediate, giving the acid bromide $CH_3CHBrCOBr$. …

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10. AP EAPCET 2025Set ap-2025-05-20-AN1 markMCQ
    Q.Observe the following reactions I) $\mathrm{CH_3CH_2C(CH_3)_2Cl}$ (2-chloro-2-methylbutane) $\xrightarrow{OH^-}$ $\mathrm{CH_3CH_2C(CH_3)_2OH}$ (2-methylbutan-2-ol) II) 1-chloro-4-nitrobenzene (benzene ring with $\mathrm{Cl}$ at C1 and $\mathrm{NO_2}$ at C4) $\xrightarrow{(i)\ OH^-,\ 443K\ \ (ii)\ H^+}$ 4-nitrophenol (benzene ring with $\mathrm{OH}$ at C1 and $\mathrm{NO_2}$ at C4) The correct statement regarding the mechanism involved in the above reactions is (A) In both I, II C – Cl bond is cleaved in slow step of the reaction (B) In both I, II C – Cl bond is cleaved in fast step of the reaction (C) In I C – Cl bond is cleaved in slow step and in II fast step of the reaction (D) In I C – Cl bond is cleaved in fast step and in II slow step of the reaction

    ›Reveal solutionSolution

    Reaction I is $\mathrm{S_N1}$ (C–Cl cleaves in the rate-determining, slow, ionisation step); Reaction II is nucleophilic aromatic substitution (addition–elimination), where C–Cl cleaves only after the slow nucleophilic-addition step, i.e., in the fast step.

    Concept and Intuition

    Mechanism dictates when the leaving group departs relative to the rate-determining step. A tertiary substrate reacting with a hydroxide nucleophile goes by $\mathrm{S_N1}$: the substrate first ionises to a stabilised tertiary carbocation (this ionisation, involving C–Cl bond cleavage, IS the slow, rate-determining step), and the nucleophile then attacks rapidly. An activated aryl halide (electron-withdrawing $-\mathrm{NO_2}$ para to the leaving group) instead reacts by nucleophilic aromatic substitution: the nucleophile first adds to the ring (slow step, forming a resonance-stabilised Meisenheimer/anionic intermediate) and only afterwards does the halide leave (fast step) to restore aromaticity.

    Step-by-Step Solution

    1. Reaction I: $\mathrm{CH_3CH_2C(CH_3)_2Cl} \xrightarrow{OH^-} \mathrm{CH_3CH_2C(CH_3)_2OH}$. The substrate is a 3° alkyl chloride, and 3° substrates with a weak-to-moderate nucleophile like $\mathrm{OH^-}$ typically proceed via $\mathrm{S_N1}$: rate-determining ionisation to the 3° carbocation (C–Cl bond breaks HERE, in the slow step), followed by fast capture of $\mathrm{OH^-}$. …

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11. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
    Q.An alcohol $X(\mathrm{C_5H_{12}O})$ on dehydration gives Y (major product). Reaction of Y with HBr gave $Z(\mathrm{C_5H_{11}Br}$, major product). Z undergoes nucleophilic substitution in two steps. What are X and Y? (A) [FIGURE] (skeletal structure of a straight-chain 5-carbon alcohol with OH on the middle (3rd) carbon, i.e. pentan-3-ol) ; [FIGURE] (skeletal structure of a straight-chain 5-carbon alkene with the C=C double bond between the 2nd and 3rd carbons, i.e. pent-2-ene) (B) [FIGURE] (skeletal structure of a branched 5-carbon alcohol: an ethyl chain attached to a carbon bearing a methyl branch and the OH group, a tertiary alcohol) ; [FIGURE] (skeletal structure of a branched 5-carbon alkene with the more-substituted internal C=C double bond, drawn with a methyl branch at the alkene carbon) (C) [FIGURE] (skeletal structure of a branched 5-carbon alcohol with the OH on a carbon flanked by two methyl-bearing branch points) ; [FIGURE] (skeletal structure of a branched 5-carbon alkene with a terminal $=CH_2$ and a methyl branch near it) (D) [FIGURE] (skeletal structure of a branched 5-carbon alcohol: an ethyl chain attached to a carbon bearing a methyl branch and the OH group) ; [FIGURE] (skeletal structure of a branched 5-carbon alkene with a terminal $=CH_2$ and a methyl branch, an ethyl group on the other side)

    ›Reveal solutionSolution

    This tests tracing a dehydration/HBr-addition sequence and recognising that 'substitution in two steps' signals an SN1 mechanism, which requires Z to be a tertiary halide. The answer is (B): X = 2-methylbutan-2-ol, Y = 2-methyl-2-butene.

    Concept and Intuition

    Nucleophilic substitution can proceed by two mechanisms: SN2 (one concerted step — backside attack and leaving-group departure happen simultaneously) or SN1 (two steps — first the leaving group departs to form a carbocation, then the nucleophile attacks it). SN1 is favoured when the resulting carbocation is highly stabilized, i.e., for tertiary (and benzylic/allylic) halides. So 'Z undergoes nucleophilic substitution in two steps' is the key clue telling us Z must be a stable, typically tertiary, alkyl halide.

    Step-by-Step Solution

    1. Work backward from the clue: Z (C$_5$H$_{11}$Br) reacts via a two-step (SN1) mechanism, so Z should be a tertiary bromide, since tertiary carbocations are the most stabilized and SN1 is strongly favoured there.
    2. Test option (B): X = 2-methylbutan-2-ol, $\mathrm{CH_3CH_2C(CH_3)(OH)CH_3}$ — a tertiary alcohol (C5H12O ✓).
    3. Dehydrate X (E1, acid-catalyzed, Zaitsev's rule: more substituted alkene is major): removing water and an H from the adjacent CH$_2$ gives the more substituted (trisubstituted) alkene, Y = 2-methyl-2-butene, $\mathrm{(CH_3)_2C=CHCH_3}$. Since X is already tertiary at the OH carbon, no carbocation rearrangement is needed — this is a clean, direct major product.
    4. Add HBr to Y (Markovnikov addition): the proton adds to the carbon that gives the more stable carbocation. Protonating the =CH– carbon (attached to CH$_3$ and H) leaves the positive charge on the (CH$_3$)$_2$C= carbon, forming a stabilized tertiary carbocation. Br$^-$ then attacks this carbon, giving Z = 2-bromo-2-methylbutane, $\mathrm{(CH_3)_2C(Br)CH_2CH_3}$ (C5H11Br ✓, a tertiary bromide).
    5. Z being tertiary reacts via SN1 (2 steps: ionize to the tertiary carbocation, then nucleophile attacks) — exactly matching the clue. …

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12. AP EAPCET 2024Set ap-2024-05-17-FN1 markMCQ
    Q.What is X in the following reaction? $\mathrm{(CH_3)_3C-Br + OH^- \rightarrow X}$ (A) Butan-2-ol (B) 2-methyl propan-2-ol (C) 2-Methyl propan-1-ol (D) Butan-1-ol

    ›Reveal solutionSolution

    Substitution of Br by OH$^-$ on the tertiary carbon of tert-butyl bromide gives

    tert-butyl alcohol, i.e. 2-methylpropan-2-ol, with the same carbon skeleton.

    Concept and Intuition

    $(\mathrm{CH_3})_3\mathrm{C-Br}$ is a tertiary alkyl halide (the carbon bearing Br

    is attached to three methyl groups). Tertiary halides react with nucleophiles/hydroxide

    predominantly via the $\mathrm{S_N1}$ mechanism: the halide leaves to form a stable

    tertiary carbocation, which is then captured by $\mathrm{OH^-}$ (or water) to directly

    give the alcohol with the same carbon skeleton — no rearrangement is needed since

    the carbocation is already the most stable (tertiary) form.

    Step-by-Step Solution

    1. Starting material: $(\mathrm{CH_3})_3\mathrm{C-Br}$, i.e. 2-bromo-2-methylpropane (tert-butyl bromide).
    2. $\mathrm{OH^-}$ substitutes for $\mathrm{Br^-}$ at the same (tertiary) carbon: $(\mathrm{CH_3})_3\mathrm{C-Br} + \mathrm{OH^-} \rightarrow (\mathrm{CH_3})_3\mathrm{C-OH} + \mathrm{Br^-}$.
    3. $(\mathrm{CH_3})_3\mathrm{C-OH}$ is named 2-methylpropan-2-ol (tert-butanol) — the OH is on the same central carbon that held Br, still bonded to three methyl groups.
    4. This rules out Butan-1-ol/Butan-2-ol (those imply a straight 4-carbon chain, but the …

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