Q.The process of converting alkyl halides into alcohols involves ____________.

Showing the 12 most recent of 13 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.With reference to the two statements Assertion and Reason, choose the correct option. Assertion: The order of reactivity towards $\mathrm{S}_{\mathrm{N}} 1$ reaction is: $\mathrm{C}_6 \mathrm{H}_5-\mathrm{CH}_2 \mathrm{Br}>\left(\mathrm{CH}_3\right)_3-\mathrm{C}-\mathrm{Br}>\left(\mathrm{CH}_3\right)_2-\mathrm{CH}-\mathrm{Br}$. Reason: Among the given 3 compounds, the Benzyl carbocation formed is the most stable while Isopropyl carbocation is the least stable one. (A) Reason is correct but Assertion is wrong (B) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion (C) Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion (D) Assertion is correct but Reason is wrong

   ›Reveal solutionSolution

   The assertion correctly orders SN1 reactivity by carbocation stability (benzyl > tert-butyl > isopropyl), and the reason correctly identifies that stability order, so both are true and the reason explains the assertion. The correct option is (B).

   Concept & Intuition

   SN1 reactions proceed via a carbocation intermediate. The rate of an SN1 reaction depends on how easily the leaving group departs and, crucially, on the stability of the carbocation formed. More stable carbocations form faster and lead to higher reactivity. Here, we compare three carbocations: benzyl (from $\ce{C6H5CH2Br}$), tert-butyl (from $\ce{(CH3)3CBr}$), and isopropyl (from $\ce{(CH3)2CHBr}$). Benzyl carbocation is exceptionally stable due to resonance delocalization of the positive charge into the aromatic ring. Tert-butyl carbocation is stabilized by hyperconjugation from nine C–H bonds. Isopropyl carbocation has only six hyperconjugative C–H bonds, making it the least stable of the three. Thus the reactivity order given in the assertion matches the stability order given in the reason.

   Step-by-step reasoning

   1. Identify the reaction type

      The question concerns $\mathrm{S_N1}$ reactivity. In an $\mathrm{S_N1}$ reaction, the rate-determining step is the formation of a carbocation after the leaving group departs. Therefore, the relative rates depend directly on the relative stabilities of the carbocation intermediates.

   2. Analyze the three carbocations

      • Benzyl carbocation ($\ce{C6H5CH2+}$): The positive charge on the $\ce{CH2}$ group is delocalized into the benzene ring via resonance. This gives it exceptional stability — comparable to or even greater than a tertiary carbocation.
      • tert-Butyl carbocation ($\ce{(CH3)3C+}$): A tertiary carbocation stabilized by hyperconjugation from nine $\alpha$ C–H bonds. It is very stable, but not as stable as a resonance-stabilized benzyl carbocation.
      • Isopropyl carbocation ($\ce{(CH3)2CH+}$): A secondary carbocation with only six hyperconjugative C–H bonds. It is less stable than a tertiary carbocation.

   3. Order the carbocation stabilities

      Benzyl > tert-butyl > isopropyl. This is a well-established order in organic chemistry.

   4. Relate stability to SN1 reactivity …

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2. COMEDK 2026Set 2026-M1 markMCQ
   Q.Identify the wrong statement from the following (A) $\mathrm{S}_{\mathrm{N}} 1$ reaction of Alkyl halides is favoured by the presence of bulky substituents, protic solvents and the reaction products are a racemic mixture (B) Vinyl halides do not undergo nucleophilic substitution reaction easily c. (C) The product formed when (-)-2-Methylbutan-1-OI is heated with Conc. HCl is (+)-1-Chloro-2-methylbutane with Retention of configuration (D) $\mathrm{S}_{\mathrm{N}} 2$ reaction of optically active halides are accompanied by Retention of configuration

   ›Reveal solutionSolution

   The question asks to identify the wrong statement among four about organic reaction mechanisms. The key is that SN2 reactions invert configuration, not retain it, so statement (D) is false. The correct answer is (D).

   Let’s examine each statement carefully, using the principles of nucleophilic substitution and stereochemistry.

   1. Statement (A): “SN1 reaction of Alkyl halides is favoured by the presence of bulky substituents, protic solvents and the reaction products are a racemic mixture.”

      • Reasoning: SN1 reactions proceed via a carbocation intermediate. Bulky substituents stabilize the carbocation (by hyperconjugation and inductive effects) and also hinder the backside attack needed for SN2, favoring SN1. Protic solvents stabilize the carbocation and the leaving group through solvation. The planar carbocation allows attack from either side, giving a racemic mixture (though sometimes with slight inversion due to ion-pairing). This statement is correct.

   2. Statement (B): “Vinyl halides do not undergo nucleophilic substitution reaction easily.”

      • Reasoning: In vinyl halides (e.g., CH₂=CH–Cl), the carbon–halogen bond has partial double-bond character due to resonance, making it stronger. Also, the SN2 transition state would require a planar backside attack that is geometrically unfavorable, and SN1 would form an unstable vinyl carbocation. Thus, they are indeed resistant to nucleophilic substitution. This statement is correct.

   3. Statement (C): “The product formed when (-)-2-Methylbutan-1-Ol is heated with Conc. HCl is (+)-1-Chloro-2-methylbutane with Retention of configuration.” …

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3. COMEDK 2026Set 2026-M1 markMCQ
   Q.Two statements, Assertion and Reason are given. With reference to them, choose the correct option. Assertion: Hydrolysis of 2-Bromo-2-methylpropane follows $S_N 1$ mechanism which involves 2 steps and the first step is: Reason: The energy required to cleave $\mathrm{C}-\mathrm{Br}$ is taken from the energy released due to Inductive effect and hyper-conjugative effect. (A) Assertion is correct but Reason is wrong (B) Reason is correct but Assertion is wrong (C) Both Assertion and Reason are correct but Reason is not the correct explanation of Assertion (D) Both Assertion and Reason are correct and Reason is the correct explanation of Assertion

   ›Reveal solutionSolution

   The Assertion correctly describes the two-step $S_N1$ hydrolysis of 2‑bromo‑2‑methylpropane, but the Reason misidentifies the source of activation energy — it is not supplied by inductive or hyperconjugative effects, which only stabilize the carbocation after it forms. Hence the Assertion is correct, the Reason is wrong.

   Concept and intuition

   The $S_N1$ mechanism proceeds via a rate‑determining ionization of the C–Br bond to form a carbocation. The energy needed to break this bond comes from thermal collisions (heat) in the solvent, not from internal electronic effects. Inductive and hyperconjugative effects stabilize the resulting carbocation, lowering the activation energy for its formation, but they do not release energy that directly cleaves the bond. The Reason confuses stabilization of a product with the source of activation energy.

   Step‑by‑step reasoning

   1. Identify the mechanism

      The substrate is 2‑bromo‑2‑methylpropane, a tertiary alkyl halide. Tertiary halides favor $S_N1$ over $S_N2$ because of steric hindrance and the stability of the tertiary carbocation. The Assertion correctly states that hydrolysis follows $S_N1$, which occurs in two steps:

      • Step 1 (slow, rate‑determining): ionization to a tert‑butyl carbocation and bromide ion.
      • Step 2 (fast): nucleophilic attack by water on the carbocation, followed by deprotonation.

   2. Examine the first step shown in the figure

      The figure shows the equilibrium:

   $(\text{CH}_3)_3\text{C–Br} \rightleftharpoons (\text{CH}_3)_3\text{C}^+ + \text{Br}^-$

   labelled “Slow” and “Reversible”. This is exactly the correct depiction of the rate‑determining ionization. So the Assertion is factually correct.

   3. Analyze the Reason

      The Reason claims: “The energy required to cleave C–Br is taken from the energy released due to Inductive effect and hyper‑conjugative effect.”

      • Inductive and hyperconjugative effects are stabilizing interactions that occur after the carbocation forms. They lower the potential energy of the carbocation relative to the transition state, thereby reducing the activation energy ($E_a$).
      • However, the actual energy to break the C–Br bond comes from the kinetic energy of molecular collisions (thermal energy from the surroundings). No “energy release” from these electronic effects directly supplies the bond‑breaking energy; rather, they make the transition state easier to reach by stabilizing the developing positive charge.

   4. Distinguish between stabilization and energy source …

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4. COMEDK 2025Set 2025-A1 markMCQ
   Q.Identify the incorrect statement. (A) When (一)-2-bromooctane is reacted with aqueous KOH , the substitution follows $\mathrm{S}_{\mathrm{N}} 2$ mechanism and product formed is (一) -Octan-2-ol. (B) Benzylic halides show higher reactivity towards $S_N 1$ reaction. (C) The reaction of $\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2-\mathrm{CHCl}-\mathrm{C}_6 \mathrm{H}_5$ with alc. KOH on heating yields two products. (D) Optically active (-)-2-Methylbutan-1-ol on reaction with HCl yields a product $(+)$-1-Chloro-2-methylbutane.

   ›Reveal solutionSolution

   The key is to analyze each statement for stereochemical and mechanistic consistency. Only statement (A) is incorrect because the $\mathrm{S}_{\mathrm{N}}2$ reaction of (−)-2-bromooctane with aqueous KOH inverts configuration, giving (+)-octan-2-ol, not (−)-octan-2-ol.

   Concept and Intuition

   This question tests your understanding of substitution mechanisms ($\mathrm{S}_{\mathrm{N}}1$ vs $\mathrm{S}_{\mathrm{N}}2$), stereochemistry (inversion vs racemization), and the special reactivity of benzylic and allylic halides. Each statement must be checked against known principles:

   • $\mathrm{S}_{\mathrm{N}}2$ proceeds with inversion of configuration.
   • $\mathrm{S}_{\mathrm{N}}1$ proceeds with racemization (or partial retention if the carbocation is asymmetric).
   • Benzylic halides stabilize carbocations, favoring $\mathrm{S}_{\mathrm{N}}1$.
   • Alcoholic KOH promotes elimination (E2), not substitution.
   • The reaction of an optically active alcohol with HCl involves an $\mathrm{S}_{\mathrm{N}}1$ mechanism via a carbocation, leading to racemization or inversion depending on the structure.

   Let’s examine each statement step by step.

   ---

   1. Statement (A):

      (−)-2-Bromooctane is a secondary alkyl halide. Aqueous KOH is a strong nucleophile and a weak base, favoring $\mathrm{S}_{\mathrm{N}}2$ over E2 for secondary halides. In $\mathrm{S}_{\mathrm{N}}2$, the nucleophile attacks from the back, inverting the stereochemistry.

      • Starting material: (−)-2-bromooctane (specific rotation negative).
      • Product: octan-2-ol. If the reaction is pure $\mathrm{S}_{\mathrm{N}}2$, the product should have the opposite configuration, i.e., (+)-octan-2-ol.
      • The statement claims the product is (−)-octan-2-ol, which would require retention — impossible for $\mathrm{S}_{\mathrm{N}}2$. Conclusion: Statement (A) is incorrect.

   2. Statement (B):

      Benzylic halides (e.g., $\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2\mathrm{Cl}$) form a resonance-stabilized benzylic carbocation in $\mathrm{S}_{\mathrm{N}}1$ reactions. This stabilization lowers the activation energy, making them much more reactive than primary or secondary alkyl halides in $\mathrm{S}_{\mathrm{N}}1$.

      Conclusion: Statement (B) is correct.

   3. Statement (C):

      The compound $\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2-\mathrm{CHCl}-\mathrm{C}_6\mathrm{H}_5$ is 1,2-diphenyl-1-chloroethane. Alcoholic KOH is a strong base in a protic solvent, favoring E2 elimination. The β-hydrogens are on both sides of the chlorine, leading to two possible alkenes:

      • Elimination toward the $\mathrm{CH}_2\mathrm{C}_6\mathrm{H}_5$ side gives $\mathrm{C}_6\mathrm{H}_5\mathrm{CH}=\mathrm{CH}-\mathrm{C}_6\mathrm{H}_5$ (stilbene, cis/trans possible).
      • Elimination toward the $\mathrm{C}_6\mathrm{H}_5$ side gives $\mathrm{C}_6\mathrm{H}_5\mathrm{CH}_2-\mathrm{C}(\mathrm{C}_6\mathrm{H}_5)=\mathrm{CH}_2$ (1,1-diphenylethene). So indeed two products form. Conclusion: Statement (C) is correct.

   4. Statement (D):

      (−)-2-Methylbutan-1-ol is a primary alcohol. Reaction with HCl proceeds via protonation of the OH, loss of water to form a primary carbocation, which then rearranges (via a 1,2-hydride shift) to a more stable secondary carbocation. The chloride ion attacks this secondary carbocation, giving 1-chloro-2-methylbutane. …

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5. COMEDK 2025Set 2025-M1 markMCQ
   Q.Which one of the following is a true statement with reference to reaction between 2- Bromo-2-methylpropane and aqueous KOH ? (A) The rate of reaction depends on the concentration of the haloalkane and the nucleophile $\mathrm{OH}^{-}$ (B) The reaction occurs at a fast rate since the substrate is a tertiary alkyl halide and follows $\mathrm{S}_{\mathrm{N}} 1$ mechanism. (C) The reaction is not favoured by the presence of polar protic solvent. (D) The reaction occurs at a slow rate since it follows $\mathrm{S}_{\mathrm{N}} 2$ mechanism.

   ›Reveal solutionSolution

   The reaction of 2‑bromo‑2‑methylpropane with aqueous KOH follows an S_N1 mechanism because it is a tertiary alkyl halide; the rate depends only on the haloalkane concentration, not on the nucleophile, and the polar protic solvent actually favours the reaction. The true statement is (B).

   The key here is to recognise the substrate: 2‑bromo‑2‑methylpropane is a tertiary alkyl halide. Tertiary halides are too sterically hindered for a back‑side attack (S_N2), so they react via an S_N1 mechanism. In S_N1, the rate‑determining step is the loss of the leaving group to form a carbocation — this step involves only the haloalkane, not the nucleophile. Therefore the rate is independent of the nucleophile concentration. Aqueous KOH provides both a polar protic solvent (water) and a strong nucleophile (OH⁻), but the nucleophile only attacks the carbocation in a fast second step.

   Let’s examine each option carefully.

   1. Option (A): “The rate depends on the concentration of the haloalkane and the nucleophile OH⁻.”

      This would be true for an S_N2 reaction, where both the alkyl halide and the nucleophile appear in the rate law. But for a tertiary halide, the mechanism is S_N1, and the rate law is first order in the haloalkane only. So (A) is false.

   2. Option (B): “The reaction occurs at a fast rate since the substrate is a tertiary alkyl halide and follows S_N1 mechanism.”

      Tertiary carbocations are relatively stable (due to hyperconjugation and inductive effects from three alkyl groups), so the S_N1 reaction proceeds at a moderate to fast rate under typical conditions. The statement correctly identifies both the mechanism and the reason for the rate. This looks correct.

   3. Option (C): “The reaction is not favoured by the presence of polar protic solvent.” …

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6. KCET 2024Set B-21 markMCQ
   Q.In the following scheme of reaction, $\begin{array}{l} \text{C}_2\text{H}_5\text{Cl} \xrightarrow{X} \text{C}_2\text{H}_5\text{F} \\ \text{C}_2\text{H}_5\text{Cl} \xrightarrow{Y} \text{CH}_2=\text{CH}_2 \\ \text{C}_2\text{H}_5\text{Cl} \xrightarrow{Z} \text{C}_4\text{H}_{10} \end{array}$ X, Y and Z respectively are : (A) AgF, alcoholic KOH and benzene (B) HF, aqueous KOH and Na in dry ether (C) Hg$_2$F$_2$, alcoholic KOH and Na in dry ether (D) CoF$_2$, aqueous KOH and benzene

   ›Reveal solutionSolution

   Identify the three named reactions — Swarts (metal fluoride), dehydrohalogenation (alcoholic KOH) and Wurtz (Na/dry ether) — and only one option gets all three right.

   1. Reaction X: $\mathrm{C_2H_5Cl \longrightarrow C_2H_5F}$ (halogen exchange)

   This is the Swarts reaction: an alkyl chloride/bromide is heated with a metallic fluoride, and the halogen is swapped:

   $\mathrm{C_2H_5Cl} + \mathrm{Hg_2F_2} \longrightarrow \mathrm{C_2H_5F} + \mathrm{Hg_2Cl_2}$

   The reagents that work are $\mathrm{AgF},\ \mathrm{Hg_2F_2},\ \mathrm{CoF_2},\ \mathrm{SbF_3}$. Note that HF (option B) does not do this — a hydrogen halide cannot displace a halide from an alkyl halide, so (B) is out on step X alone.

   2. Reaction Y: $\mathrm{C_2H_5Cl \longrightarrow CH_2{=}CH_2}$ (elimination)

   Making an alkene from an alkyl halide means removing HCl — β-elimination (E2), i.e. dehydrohalogenation. The decisive rule:

   • Alcoholic KOH ⇒ the base is the ethoxide ion $\mathrm{C_2H_5O^-}$, a strong, bulky base ⇒ it abstracts the β-hydrogen ⇒ elimination ⇒ alkene.
   • Aqueous KOH ⇒ $\mathrm{OH^-}$ is well-solvated and acts as a nucleophile ⇒ substitution ⇒ ethanol, not ethene.

   $\mathrm{CH_3CH_2Cl} \xrightarrow{\text{alc. KOH},\ \Delta} \mathrm{CH_2{=}CH_2} + \mathrm{KCl} + \mathrm{H_2O}$

   So (B) and (D), which both say aqueous KOH, are wrong here.

   3. Reaction Z: $\mathrm{C_2H_5Cl \longrightarrow C_4H_{10}}$ (chain doubling)

   The product $\mathrm{C_4H_{10}}$ (n-butane) has twice the carbon count of the $\mathrm{C_2}$ starting material — two alkyl fragments have been coupled. That is the Wurtz reaction: an alkyl halide with sodium metal in dry ether:

   $2\,\mathrm{C_2H_5Cl} + 2\,\mathrm{Na} \xrightarrow{\text{dry ether}} \mathrm{CH_3CH_2{-}CH_2CH_3} + 2\,\mathrm{NaCl}$ …

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7. COMEDK 2024Set 2024-E1 markMCQ
   Q.Arrange the compounds $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$ in the increasing order of their reactivity towards $\mathrm{S_N}1$ reaction. $\begin{aligned} & \mathrm{A}=\mathrm{C}_6 \mathrm{H}_5 \mathrm{CH}_2 \mathrm{Cl} \\ & \mathrm{B}=\mathrm{C}_6 \mathrm{H}_5 \mathrm{Cl} \\ & \mathrm{C}=\mathrm{CH}_2=\mathrm{CH}-\mathrm{Cl} \\ & \mathrm{D}=\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{Cl} \end{aligned}$ (A) $\mathrm{D}<\mathrm{B}<\mathrm{A}<\mathrm{C}$ (B) $\mathrm{D}<\mathrm{C}<\mathrm{B}<\mathrm{A}$ (C) $\mathrm{B}<\mathrm{D}<\mathrm{C}<\mathrm{A}$ (D) $\mathrm{C}<\mathrm{A}<\mathrm{D}<\mathrm{B}$

   ›Reveal solutionSolution

   $S_N1$ rate tracks carbocation stability: chlorobenzene gives the least stable cation and benzyl the most, giving the increasing order $B<D<C<A$ - option (C).

   In an $S_N1$ reaction the slow step is loss of $\text{Cl}^-$ to form a carbocation, so reactivity follows carbocation stability.

   • A, $\text{C}_6\text{H}_5\text{CH}_2\text{Cl}$ $\to$ a benzyl cation, strongly resonance-stabilised by the ring $\Rightarrow$ most reactive.
   • B, $\text{C}_6\text{H}_5\text{Cl}$ $\to$ the C-Cl bond has partial double-bond (resonance) character and a phenyl cation is extremely unstable $\Rightarrow$ least reactive.
   • D, $\text{CH}_3\text{CH}_2\text{Cl}$ $\to$ an ordinary primary $sp^3$ C-Cl bond. …

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8. COMEDK 2024Set 2024-M1 markMCQ
   Q.Arrange the following compounds in the decreasing order of reactivity towards $\mathrm{S}_{\mathrm{N}} 1$ reaction. (A) A > D > B > C (B) D > B > A > C (C) C > A > B > D (D) B > A > C > D

   ›Reveal solutionSolution

   S_N1 reactivity follows carbocation stability: tertiary > secondary > primary. Compound [C] (tertiary) is most reactive, [A] (secondary) next, and the two primary halides [B] and [D] are least reactive, with the less hindered straight-chain [B] slightly ahead of the branched [D]. The order is C > A > B > D, which is option (C).

   The key concept is that S_N1 reactions proceed through a carbocation intermediate. The rate-determining step is the departure of the leaving group (Cl⁻) to form that carbocation, so anything that stabilizes it — more alkyl groups supplying electron density through hyperconjugation and induction — speeds up the reaction. The more substituted the carbon bearing chlorine, the more stable the carbocation and the faster the S_N1 reaction.

   Let's analyze each compound:

   1. Identify the substitution level of the carbon attached to Cl:

      • [C]: chlorine on a carbon bonded to three other carbons → tertiary (3°), giving a highly stabilized tertiary carbocation.
      • [A]: chlorine on a carbon bonded to two other carbons → secondary (2°), less stable than tertiary but more stable than primary.
      • [B]: chlorine on a terminal carbon of a straight chain → primary (1°), an unstable primary carbocation.
      • [D]: chlorine on a terminal carbon of a branched chain → also primary (1°), an unstable primary carbocation.

   2. Compare the two primary compounds [B] and [D]: …

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9. COMEDK 2024Set 2024-M1 markMCQ
   Q.Match the compounds given in column I with the corresponding most stable Carbocations formed by each, as given in Column II, when they undergo acidic dehydration in presence of Conc. $\mathrm{H}_2 \mathrm{SO}_4$. .tg {border-collapse:collapse;border-spacing:0;} .tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal;} .tg th{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal;} .tg .tg-c3ow{border-color:inherit;text-align:center;vertical-align:top} .tg .tg-7btt{border-color:inherit;font-weight:bold;text-align:center;vertical-align:top} .tg .tg-0pky{border-color:inherit;text-align:left;vertical-align:top} No. Column I No. Column II A 2-Methylbutan-1-ol P ${(C{H_3})_2}\mathop C\limits^ + - CH{(C{H_3})_2}$ B Butan-2-ol Q $C{H_3} - C{H_2} - C{H_2} - \mathop C\limits^ + {H_2}$ C 3, 3-Dimethylbutan-2-ol R ${(C{H_3})_2}\mathop C\limits^ + - C{H_2} - C{H_3}$ D Butan-1-ol S $C{H_3} - \mathop C\limits^ + H - C{H_2} - C{H_3}$ (A) $\mathrm{A}=\mathrm{S} \quad \mathrm{B}=\mathrm{R} \quad \mathrm{C}=\mathrm{Q} \quad \mathrm{D}=\mathrm{P}$ (B) $\mathrm{A}=\mathrm{Q} \quad \mathrm{B}=\mathrm{R} \quad \mathrm{C}=\mathrm{S} \quad \mathrm{D}=\mathrm{P}$ (C) $\mathrm{A}=\mathrm{R} \quad \mathrm{B}=\mathrm{S} \quad \mathrm{C}=\mathrm{P} \quad \mathrm{D}=\mathrm{Q}$ (D) $\mathrm{A}=\mathrm{Q} \quad \mathrm{B}=\mathrm{S} \quad \mathrm{C}=\mathrm{P} \quad \mathrm{D}=\mathrm{R}$

   ›Reveal solutionSolution

   The key is to identify the most stable carbocation formed during acidic dehydration of each alcohol, considering carbocation rearrangements (hydride/methyl shifts) that lead to the most substituted, most stable intermediate. The correct matches are A→R, B→S, C→P, D→Q, which corresponds to option (C).

   Concept & Intuition

   When an alcohol is heated with concentrated $\mathrm{H}_2\mathrm{SO}_4$, it undergoes dehydration via an E1 mechanism: protonation of the –OH group, loss of water to form a carbocation, then loss of a proton to give an alkene. The key intermediate is the carbocation. However, carbocations can rearrange via 1,2-hydride or 1,2-alkyl shifts to form a more stable (more substituted) carbocation. The question asks for the most stable carbocation formed, not necessarily the first one. So we must trace possible rearrangements for each alcohol.

   Step-by-step reasoning

   1. Compound A: 2-Methylbutan-1-ol

      • Structure: $\mathrm{CH_3CH_2CH(CH_3)CH_2OH}$ (primary alcohol).
      • Protonation and loss of water gives a primary carbocation: $\mathrm{CH_3CH_2CH(CH_3)CH_2^+}$ (1°).
      • A 1,2-hydride shift from the adjacent carbon (the one bearing the methyl group) yields a more stable secondary carbocation: $\mathrm{CH_3CH_2C^+(CH_3)CH_3}$.
      • This is exactly $(CH_3)_2C^+CH_2CH_3$, which is R.
      • No further shift gives a tertiary carbocation here because the adjacent carbons are either primary or secondary; a methyl shift would not improve stability. So the most stable is R.

   2. Compound B: Butan-2-ol

      • Structure: $\mathrm{CH_3CH_2CH(OH)CH_3}$ (secondary alcohol).
      • Direct loss of water gives a secondary carbocation: $\mathrm{CH_3CH_2C^+HCH_3}$ (2°).
      • This is already quite stable. Could it rearrange? A 1,2-hydride shift from the terminal methyl group would give a primary carbocation (less stable), so no. A methyl shift from the ethyl group would also give a primary carbocation. So no rearrangement improves stability.
      • The carbocation is $\mathrm{CH_3C^+HCH_2CH_3}$, which is S.

   3. Compound C: 3,3-Dimethylbutan-2-ol

      • Structure: $(\mathrm{CH_3})_3\mathrm{CCH(OH)CH_3}$ (secondary alcohol).
      • Loss of water gives a secondary carbocation: $(\mathrm{CH_3})_3\mathrm{CCH^+CH_3}$ (2°).
      • Notice the adjacent carbon is a quaternary carbon (tert-butyl group). A 1,2-methyl shift from that tert-butyl group can occur: the methyl group moves to the carbocation center, and the positive charge moves to the former quaternary carbon.
      • Result: $(\mathrm{CH_3})_2\mathrm{C^+CH(CH_3)_2}$ — a tertiary carbocation (3°).
      • This is exactly P: $(CH_3)_2C^+CH(CH_3)_2$.
      • This is the most stable possible.

   4. Compound D: Butan-1-ol

      • Structure: $\mathrm{CH_3CH_2CH_2CH_2OH}$ (primary alcohol).
      • Loss of water gives a primary carbocation: $\mathrm{CH_3CH_2CH_2CH_2^+}$ (1°).
      • A 1,2-hydride shift from the adjacent carbon gives a secondary carbocation: $\mathrm{CH_3CH_2C^+HCH_3}$ (2°).
      • This is S again? Wait, but S is already used for B. Let's check: S is $\mathrm{CH_3C^+HCH_2CH_3}$, which is exactly the same as the secondary carbocation from butan-1-ol. However, we must see if a further shift is possible. A 1,2-hydride shift from the other side would give a primary carbocation again, so no. A methyl shift would give a primary carbocation. So the most stable is the secondary carbocation.
      • But note: The secondary carbocation from butan-1-ol is identical to that from butan-2-ol? Actually, careful: Butan-1-ol gives $\mathrm{CH_3CH_2C^+HCH_3}$ (the positive charge is on C2). Butan-2-ol gives $\mathrm{CH_3CH_2C^+HCH_3}$ as well (same structure). So both B and D would give the same carbocation? That can't be, because the options list different matches. Let's re-examine:
        • Butan-2-ol: direct loss gives $\mathrm{CH_3CH_2C^+HCH_3}$ (2°).
        • Butan-1-ol: after one hydride shift gives $\mathrm{CH_3CH_2C^+HCH_3}$ (2°).
        • So indeed both give the same secondary carbocation. But the question expects different matches. This suggests that for butan-1-ol, a further rearrangement might occur? Let's check: The secondary carbocation from butan-1-ol has a methyl group on the adjacent carbon (C1) and an ethyl group on the other side (C3). A 1,2-methyl shift from the ethyl group would give a tertiary carbocation? No, moving a methyl from the ethyl group would give $\mathrm{CH_3C^+(CH_3)CH_2CH_3}$? Actually, if we take $\mathrm{CH_3CH_2C^+HCH_3}$ and shift a methyl from the ethyl group (C3) to C2, we get $\mathrm{CH_3C^+(CH_3)CH_2CH_3}$ which is the same as R? Wait, R is $(CH_3)_2C^+CH_2CH_3$. That is exactly $\mathrm{CH_3C^+(CH_3)CH_2CH_3}$. So a methyl shift from the ethyl group gives a tertiary carbocation!
        • Is that possible? Yes, because the ethyl group has a methyl that can undergo a 1,2-shift. So the secondary carbocation can rearrange to a tertiary carbocation via a methyl shift.
        • Therefore, the most stable carbocation from butan-1-ol is actually the tertiary one: R.
      • But then B and D would both give R? That can't be. Let's check butan-2-ol again: The secondary carbocation from butan-2-ol is $\mathrm{CH_3CH_2C^+HCH_3}$. Can it undergo a methyl shift? The adjacent carbons are C1 (methyl) and C3 (methylene of ethyl). A methyl shift from C3 would give the same tertiary carbocation R. So butan-2-ol can also rearrange to R.
      • This is a problem: both B and D could give R. However, the question asks for the most stable carbocation formed. For butan-2-ol, the secondary carbocation is already quite stable, and a methyl shift would require breaking a C-C bond, which has a higher activation energy than a hydride shift. Typically, in acidic dehydration, secondary carbocations do not always rearrange to tertiary if the shift is a methyl shift (slower). But it is possible.
      • Let's look at the given options: In option (C), B is matched with S (secondary) and D with Q (primary). That suggests that for butan-1-ol, the most stable carbocation is actually primary? That seems wrong.
      • Wait, Q is $\mathrm{CH_3CH_2CH_2C^+H_2}$ — a primary carbocation. That would be the direct product from butan-1-ol without rearrangement. But that is less stable than the secondary. So why would they match D with Q?
      • Perhaps the question means the first-formed carbocation? No, it says "most stable Carbocations formed".
      • Let's re-read the column II: P: $(CH_3)_2C^+CH(CH_3)_2$ (tertiary) Q: $CH_3CH_2CH_2C^+H_2$ (primary) R: $(CH_3)_2C^+CH_2CH_3$ (tertiary) S: $CH_3C^+HCH_2CH_3$ (secondary)
      • For butan-1-ol, the most stable carbocation after all possible rearrangements is tertiary (R). But then butan-2-ol also gives tertiary (R). That would make both B and D map to R, which is not an option.
      • So perhaps the intended interpretation is that we consider only the direct carbocation from loss of water, without rearrangement? But then A (primary) would give Q, not R. That doesn't match either.
      • Let's check the typical textbook knowledge:
        • 2-Methylbutan-1-ol: dehydration gives mainly 2-methyl-2-butene via a tertiary carbocation after a hydride shift? Actually, the most stable carbocation is tertiary: $(CH_3)_2C^+CH_2CH_3$ (R).
        • Butan-2-ol: dehydration gives mainly 2-butene via the secondary carbocation (S). Rearrangement to tertiary is not common because it would require a methyl shift.
        • 3,3-Dimethylbutan-2-ol: undergoes a methyl shift to give tertiary carbocation (P).
        • Butan-1-ol: dehydration gives mainly 1-butene? Actually, butan-1-ol gives a mixture, but the carbocation rearranges via hydride shift to secondary (S), and then possibly to tertiary? In practice, butan-1-ol gives 1-butene and 2-butene, with 2-butene being major. The secondary carbocation is the key intermediate. So the most stable carbocation is secondary (S).
      • So the matches are: A→R, B→S, C→P, D→S? But D cannot be S because S is already used for B. In the options, D is matched with Q in (C) and with P in (D).
      • Let's look at option (C): A=R, B=S, C=P, D=Q. That means D (butan-1-ol) gives a primary carbocation? That seems unlikely.
      • Option (D): A=Q, B=S, C=P, D=R. That means A gives primary? No.
      • Let's systematically check each option:
        • (A) A=S, B=R, C=Q, D=P → A secondary, B tertiary, C primary, D tertiary.
        • (B) A=Q, B=R, C=S, D=P → A primary, B tertiary, C secondary, D tertiary.
        • (C) A=R, B=S, C=P, D=Q → A tertiary, B secondary, C tertiary, D primary.
        • (D) A=Q, B=S, C=P, D=R → A primary, B secondary, C tertiary, D tertiary.
      • The most plausible based on typical carbocation stability:
        • A (2-methylbutan-1-ol) → tertiary (R) after hydride shift.
        • B (butan-2-ol) → secondary (S) (no rearrangement needed).
        • C (3,3-dimethylbutan-2-ol) → tertiary (P) after methyl shift.
        • D (butan-1-ol) → secondary (S) after hydride shift, but S is taken. However, note that butan-1-ol's secondary carbocation is the same as butan-2-ol's. So if the question expects distinct matches, perhaps they consider that butan-1-ol does not rearrange? That would give primary (Q). But that contradicts the fact that hydride shifts are very fast. …

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10. COMEDK 2023Set 2023-E1 markMCQ
    Q.Which one of the following will undergo Nucleophilic substitution, by $\mathrm{S}_{\mathrm{N}}{ }^1$ mechanism, fastest? (A) $\mathrm{Br}-\mathrm{CH}_2-\mathrm{CH}=\mathrm{CH}_2$ (B) $\mathrm{C}_6 \mathrm{H}_5 \mathrm{Br}$ (C) $\mathrm{CH}_3-\mathrm{CH}=\mathrm{CHBr}$ (D) $\mathrm{C}_6 \mathrm{H}_{11} \mathrm{Br}$

    ›Reveal solutionSolution

    $\mathrm{S_N1}$ is fastest for the substrate forming the most stable carbocation. Allyl bromide gives a resonance-stabilised allylic cation, so it reacts fastest; aryl/vinyl halides don't ionise (sp²-C–X, partial double-bond character), and cyclohexyl bromide is only an ordinary secondary halide.

    $\mathrm{S_N1}$ proceeds through a carbocation, so its rate tracks carbocation stability:

    • (A) $\mathrm{BrCH_2CH=CH_2}$ (allyl): ionises to the allylic carbocation $\mathrm{CH_2=CH\text{-}CH_2^+ \leftrightarrow {}^+CH_2\text{-}CH=CH_2}$, resonance-stabilised $\Rightarrow$ fastest $\mathrm{S_N1}$. …

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11. COMEDK 2023Set 2023-M1 markMCQ
    Q.$t$-butyl chloride preferably undergo hydrolysis by (A) $\mathrm{S}_{\mathrm{N}} 1$ mechanism (B) $\mathrm{S}_{\mathrm{N}} 2$ mechanism (C) any of

    (a) and

    (b) (D) None of the above

    ›Reveal solutionSolution

    $t$-butyl chloride is a tertiary halide. On ionisation it gives a highly stabilised tertiary carbocation, so hydrolysis proceeds through the unimolecular $\mathrm{S_N}1$ pathway.

    The mechanism of nucleophilic substitution depends on the substrate:

    • Tertiary halides favour $\mathrm{S_N}1$ because (i) the $3^\circ$ carbocation intermediate is strongly stabilised by hyperconjugation and inductive effects, and (ii) the three bulky methyl groups sterically hinder the backside attack required for $\mathrm{S_N}2$.
    • $\mathrm{S_N}2$ is favoured by primary (and methyl) halides where the carbon is accessible. …

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12. KCET 2022Set B-31 markMCQ
    Q.The major product obtained when ethanol is heated with excess of conc. H_2SO_4 at 443K is (A) ethane (B) methane (C) ethene (D) ethyne

    ›Reveal solutionSolution

    Excess conc. $\mathrm{H_2SO_4}$ at 443 K dehydrates ethanol intramolecularly to the alkene, ethene.

    Step 1 — Recognise the conditions.

    The two classic reactions of ethanol with conc. sulphuric acid are distinguished only by temperature and stoichiometry — this is exactly what the question tests:

    Conditions	Reaction	Product
    443 K (170 °C), excess conc. $\mathrm{H_2SO_4}$	intramolecular dehydration (elimination)	ethene, $\mathrm{CH_2{=}CH_2}$
    413 K (140 °C), excess ethanol	intermolecular dehydration (condensation)	diethyl ether, $\mathrm{C_2H_5OC_2H_5}$

    The stem says excess acid and 443 K → the elimination branch.

    Step 2 — The mechanism (E1), and why acid is needed.

    $\mathrm{-OH}$ is a terrible leaving group. The acid's job is to convert it into a good one:

    1. Protonation of the alcohol:

    $\mathrm{CH_3CH_2{-}OH} + \mathrm{H^+} \rightleftharpoons \mathrm{CH_3CH_2{-}\overset{+}{O}H_2}$

    The $\mathrm{-OH}$ has become $\mathrm{-OH_2^{+}}$, i.e. water — an excellent leaving group.

    2. Loss of water (slow, rate-determining) to give the carbocation:

    $\mathrm{CH_3\overset{+}{C}H_2} + \mathrm{H_2O}$

    3. Loss of a $\beta$-proton to a base ($\mathrm{HSO_4^-}$ or $\mathrm{H_2O}$), forming the $\pi$ bond:

    $\mathrm{CH_3\overset{+}{C}H_2}\;\longrightarrow\;\mathrm{CH_2{=}CH_2}+\mathrm{H^+}$

    Net:

    $\mathrm{CH_3CH_2OH}\;\xrightarrow[\;443\ \mathrm{K}\;]{\text{conc. } \mathrm{H_2SO_4}}\;\mathrm{CH_2{=}CH_2}\;+\;\mathrm{H_2O}$ …

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