Q.Write the products of the following reactions:

Showing the 12 most recent of 17 on this concept.

1. COMEDK 2026Set 2026-A1 markMCQ
   Q.An unsaturated organic compound $\left(\mathrm{C}_3 \mathrm{H}_6\right)$, undergoes the following series of reactions: Identify compound [D] (A) Cyclohexane (B) 2,3-dimethylbutane (C) Hexane (D) 2-methyl pentane

   ›Reveal solutionSolution

   The reaction sequence is propene → isopropyl alcohol (via acid-catalysed hydration) → isopropyl chloride (via Lucas reagent) → 2,3-dimethylbutane (via Wurtz coupling). The final product [D] is 2,3-dimethylbutane, which corresponds to option (B).

   The key to this problem is recognising that each step is a classic organic reaction with a well-known regioselectivity. The starting material is an unsaturated hydrocarbon with formula C₃H₆ — that is propene (CH₃–CH=CH₂). The sequence uses dilute H₂SO₄ (hydration), then Lucas reagent (HCl/ZnCl₂, a test for alcohols), then sodium in dry ether (Wurtz reaction). Let’s walk through each transformation.

   1. Step 1: Hydration of propene with dilute H₂SO₄ Propene reacts with dilute sulphuric acid via electrophilic addition. According to Markovnikov’s rule, the hydrogen adds to the less substituted carbon of the double bond, and the –OH group adds to the more substituted carbon.

   $\text{CH}_3\text{–CH=CH}_2 + \text{H}_2\text{O} \xrightarrow{\text{dil. H}_2\text{SO}_4} \text{CH}_3\text{–CH(OH)–CH}_3$

   The product is propan-2-ol (isopropyl alcohol). This is the major product because the carbocation intermediate (secondary) is more stable than the primary one. So compound [B] is isopropyl alcohol.

   2. Step 2: Reaction with Lucas reagent (HCl/ZnCl₂) Lucas reagent converts alcohols to alkyl chlorides. The reaction works best for tertiary alcohols (immediate cloudiness), secondary alcohols (slow), and primary alcohols (very slow). Isopropyl alcohol is a secondary alcohol; it reacts to give isopropyl chloride.

   $\text{CH}_3\text{–CH(OH)–CH}_3 + \text{HCl} \xrightarrow{\text{ZnCl}_2} \text{CH}_3\text{–CH(Cl)–CH}_3 + \text{H}_2\text{O}$

   So compound [C] is 2-chloropropane (isopropyl chloride).

   3. Step 3: Wurtz reaction with sodium in dry ether The Wurtz reaction couples two alkyl halides in the presence of sodium metal to form a higher alkane. Here, two molecules of isopropyl chloride react: $2\,\text{CH}_3\text{–CH(Cl)–CH}_3 + 2\,\text{Na} \xrightarrow{\text{dry ether}} \text{CH}_3\text{–CH(CH}_3\text{)–CH(CH}_3\text{)–CH}_3 + 2\,\text{NaCl}$ …

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2. COMEDK 2025Set 2025-A1 markMCQ
   Q.Which one of the following is the major product formed when the given reaction occurs? (A) (B) (C) (D)

   ›Reveal solutionSolution

   The reaction of a para-ethoxy styrene with excess HI at 373 K leads to cleavage of the ether (forming phenol) and Markovnikov addition of HI to the vinyl group, giving the product with an OH group and a –CHI–CH₃ side chain. The correct option is (D).

   ---

   The key here is to recognize that two independent reactions occur on the same molecule under the given conditions: an aromatic ether cleavage and an electrophilic addition to an alkene. The challenge is to predict the outcome of each correctly and then combine them.

   Concept & Intuition:

   HI is a strong acid and a source of iodide, a good nucleophile. At 373 K (about 100 °C), it does two things:

   1. Cleaves aryl alkyl ethers (like –OCH₂CH₃) via an Sₙ2 or Sₙ1 mechanism, yielding a phenol and an alkyl iodide. The aromatic C–O bond is strong, so the alkyl–O bond breaks instead.
   2. Adds to alkenes following Markovnikov’s rule: the proton adds to the less substituted carbon of the double bond, and iodide adds to the more substituted carbon.

   The molecule has both an ethoxy group and a vinyl group on opposite sides of the benzene ring. They react independently because they are separated by the aromatic ring.

   ---

   Step-by-step reasoning:

   1. Ether cleavage:

      The ethoxy group (–O–CH₂–CH₃) is an aryl alkyl ether. With excess HI at high temperature, the alkyl–oxygen bond is cleaved. The mechanism: protonation of the ether oxygen, then nucleophilic attack by I⁻ on the ethyl carbon (Sₙ2), giving ethanol (which further reacts to ethyl iodide) and leaving a phenol group (–OH) on the ring.

      Result: The top substituent becomes –OH.

   2. Addition to the vinyl group:

      The vinyl group (–CH=CH₂) is an alkene. HI adds across the double bond. According to Markovnikov’s rule, the hydrogen (H⁺) attaches to the terminal carbon (CH₂) because it is less substituted, forming a more stable carbocation (secondary benzylic) on the carbon attached to the ring. Then I⁻ attacks that carbocation. …

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3. COMEDK 2025Set 2025-E1 markMCQ
   Q.Identify the Carbonyl compound which will not be formed when hydration of Alkynes is carried out with dil. $\mathrm{H}_2 \mathrm{SO}_4 / \mathrm{Hg}^{2+}$ at 333 K . Acetone, Butanal, Ethanal, Butanone. (A) Acetone (B) Butanone (C) Butanal (D) Ethanal

   ›Reveal solutionSolution

   Hydration of alkynes with dil. H₂SO₄/Hg²⁺ follows Markovnikov’s rule, so terminal alkynes give methyl ketones and internal alkynes give a mixture; butanal, an aldehyde, cannot be formed because it would require anti-Markovnikov addition.

   The key concept here is Markovnikov’s rule applied to the hydration of alkynes. The Hg²⁺-catalyzed addition of water across a triple bond proceeds via a vinyl carbocation intermediate, and the more stable carbocation (with more alkyl substituents) determines the product. This means the oxygen ends up on the more substituted carbon, yielding ketones (except for ethyne, which gives ethanal). An aldehyde like butanal would require the –OH to attach to a terminal carbon, which is the less substituted position — that’s anti-Markovnikov and does not happen under these conditions.

   Let’s check each option:

   1. Acetone (propanone) – Formed from hydration of propyne (CH₃–C≡CH). The triple bond is terminal; water adds so that the –OH goes to the internal carbon (more substituted), giving an enol that tautomerizes to acetone. ✓ Possible.

   2. Butanone – Formed from hydration of 1-butyne (CH₃CH₂–C≡CH) or 2-butyne (CH₃–C≡C–CH₃). For 1-butyne, Markovnikov addition gives butanone; for 2-butyne, symmetrical addition also gives butanone. ✓ Possible. …

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4. COMEDK 2025Set 2025-M1 markMCQ
   Q.Both reactions

   (i) and

   (ii) give the same compound X as the major product. Identify X (i). 3-Methylbut-1-ene $+\mathrm{HCl} \rightarrow \mathrm{X}$ (ii). Neopentyl alcohol $+\mathrm{HCl}\left(\right.$ anh. $\left.\mathrm{ZnCl}_2\right) \rightarrow \mathrm{X}$ (A) $\left(\mathrm{CH}_3\right)_2-\mathrm{CH}-\mathrm{CHCl}-\mathrm{CH}_3$ (B) $\left(\mathrm{CH}_3\right)_2-\mathrm{CCl}-\mathrm{CH}_2-\mathrm{CH}_3$ (C) $\mathrm{CH}_3-\mathrm{CH}_2-\mathrm{CH}\left(\mathrm{CH}_3\right)-\mathrm{CH}_2 \mathrm{Cl}$ (D) $\left(\mathrm{CH}_3\right)_2-\mathrm{CH}-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{Cl}$

   ›Reveal solutionSolution

   Both reactions proceed through the same tertiary carbocation intermediate, leading to the same major product: 2-chloro-2-methylbutane. The correct option is (B).

   Concept & Intuition

   The key here is that two different starting materials—an alkene and an alcohol—can funnel into the same carbocation intermediate under acidic conditions. Markovnikov addition to the alkene gives the more stable carbocation, while the alcohol (neopentyl alcohol) undergoes a carbocation rearrangement (a 1,2-methyl shift) to escape the instability of a primary carbocation. Both paths converge on the same tertiary carbocation, which then captures chloride to form the major product.

   Step-by-step reasoning

   1. Reaction (i): 3-Methylbut-1-ene + HCl

      • The alkene is unsymmetrical: $\ce{CH2=CH-CH(CH3)2}$.
      • According to Markovnikov’s rule, the proton adds to the less substituted carbon (the terminal $\ce{CH2}$), placing the positive charge on the more substituted carbon (the internal one).
      • This gives a secondary carbocation: $\ce{(CH3)2CH-CH+ -CH3}$.
      • However, this secondary carbocation can undergo a 1,2-hydride shift to form a more stable tertiary carbocation: $\ce{(CH3)2C+ -CH2-CH3}$ (2-methylbutan-2-ylium).
      • Chloride ion then attacks this tertiary carbocation to yield 2-chloro-2-methylbutane: $\ce{(CH3)2CCl-CH2-CH3}$.

   2. Reaction (ii): Neopentyl alcohol + HCl (anh. ZnCl2)

      • Neopentyl alcohol is $\ce{(CH3)3C-CH2OH}$.
      • Under acidic conditions (Lucas reagent, $\ce{ZnCl2/HCl}$), the $\ce{-OH}$ group is protonated and leaves as water, generating a primary carbocation: $\ce{(CH3)3C-CH2+}$. …

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5. COMEDK 2024Set 2024-E1 markMCQ
   Q.Identify the product $[\mathrm{C}]$ formed at the end of the reaction below 1,1,2,2- Tetrabromopropane + 2 $\mathrm{Zn}_{(\mathrm{S})}$ / Ethanol $\rightarrow$ [B] $[\mathrm{B}]$ + 2 moles of $\mathrm{HBr} \rightarrow[\mathrm{C}]$ (A) 2, 2-Dibromopropane (B) 1, 1- Dibromopropane (C) 1,3-Dibromopropane (D) 1, 2-Dibromopropane

   ›Reveal solutionSolution

   The reaction proceeds via debromination to an alkyne (propyne), followed by double hydrobromination that adds two HBr molecules in Markovnikov fashion, yielding 2,2-dibromopropane. The correct option is (A).

   Concept & Intuition

   The starting material is 1,1,2,2-tetrabromopropane — a propane chain with four bromines on the first two carbons. The first step uses zinc dust in ethanol, a classic reagent for dehalogenation: two vicinal bromines are removed to form a carbon–carbon bond of higher order. With four bromines and two Zn, both pairs are removed to give a triple bond (an alkyne). Then two moles of HBr add across the triple bond. HBr adds to alkynes in a Markovnikov fashion, and the addition happens twice, placing both bromines on the more substituted carbon.

   ---

   Step-by-step reasoning

   1. Structure of the starting compound. 1,1,2,2-Tetrabromopropane:

   $\mathrm{CH_3–CBr_2–CHBr_2}$

   (C1 has two Br, C2 has two Br, C3 is a methyl group.)

   2. First reaction: debromination with Zn/ethanol. Zinc removes vicinal bromine atoms; with four bromines and two Zn, both C1–C2 bromine pairs are eliminated, converting the linkage into a triple bond. The product [B] is

   $\mathrm{CH_3–C}\equiv\mathrm{C–H}$

   that is propyne (methylacetylene).

   Tip

   The reaction of a vicinal tetrahalide with Zn is a standard way to make alkynes — each Zn removes two adjacent halogen atoms.

   3. Second reaction: addition of 2 moles of HBr to [B]. Propyne has a terminal triple bond. The first HBr adds by Markovnikov's rule — H to the terminal carbon, Br to the internal carbon: $\mathrm{CH_3–CBr{=}CH_2}$ …

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6. COMEDK 2024Set 2024-E1 markMCQ
   Q.Given below are 4 reactions. Two of these reactions will give product which is an equimolar mixture of the d and 1 forms. Identify these 2 reactions. [A] 2- Methylpropene + $\mathrm{HI} \rightarrow$ --------- [B] But-1-ene + $\mathrm{HBr} \rightarrow$ ----------- [C] 3-Methylbut-1-ene + $\mathrm{HI} \rightarrow$ ----------- [D] 3- Phenylpropene + $\mathrm{HBr}$ (Peroxide) $\rightarrow$ ----------- (A) C & A (B) D & B (C) B & C (D) A & D

   ›Reveal solutionSolution

   [!TLDR]

   Only the reactions that generate a new chiral carbon via a planar carbocation give the racemic (d/l) mixture; these are B (but-1-ene + HBr) and C (3-methylbut-1-ene + HI), so option (C).

   Concept

   From CBSE Class 12 (Haloalkanes / Stereochemistry): Markovnikov addition of HX proceeds through a planar carbocation. If the carbon bearing the halogen ends up attached to four different groups, it is a stereocentre; attack of X$^-$ from both faces of the planar cation gives equal amounts of the two enantiomers (a racemic, optically inactive d/l mixture).

   Solution

   [A] 2-Methylpropene + HI: $(CH_3)_2C{=}CH_2+HI\to (CH_3)_3C{-}I$ (tert-butyl iodide). The carbon holding I bears three identical methyls – not chiral, no d/l pair.

   [B] But-1-ene + HBr (Markovnikov): $CH_3CH_2CH{=}CH_2+HBr\to CH_3CH_2\underset{Br}{CH}CH_3$ (2-bromobutane). C-2 carries $H, Br, CH_3, C_2H_5$ – four different groups → chiral → racemic. ✓ …

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7. KCET 2023Set D-21 markMCQ
   Q.Compounds P and R in the following reaction are CH$_3$CHO $\xrightarrow{\text{(1) CH}_3\text{MgBr}}$ P $\xrightarrow{\text{conc. H}_2\text{SO}_4, \text{heat}}$ Q $\xrightarrow{\text{(i) B}_2\text{H}_6}$ R $\xrightarrow{\text{(ii) H}_3\text{O}^+}$ $\xrightarrow{\text{(ii) H}_2\text{O}_2, \text{OH}^-}$ (A) Position isomers (B) Functional isomers (C) Metamers (D) Identical

   ›Reveal solutionSolution

   Track the three steps: Grignard → propan-2-ol, dehydration → propene, hydroboration–oxidation (anti-Markovnikov) → propan-1-ol; the two alcohols differ only in where the –OH sits.

   Step 1 — Grignard addition gives P

   A Grignard reagent's carbanion-like carbon attacks the carbonyl carbon; acid work-up then gives an alcohol. Acetaldehyde (an aldehyde other than formaldehyde) yields a secondary alcohol:

   $\mathrm{CH_3CHO} \xrightarrow{\ \mathrm{CH_3MgBr}\ } \xrightarrow{\ \mathrm{H_3O^+}\ } \mathrm{CH_3\!-\!\underset{\textstyle |}{\overset{\textstyle OH}{C}}H\!-\!CH_3}$

   $\boxed{\mathbf{P} = \text{propan-2-ol}\ (\mathrm{CH_3CH(OH)CH_3})}$

   Step 2 — Acid dehydration gives Q

   Concentrated $\mathrm{H_2SO_4}$ with heat dehydrates the alcohol (E1, via a carbocation) to the alkene:

   $\mathrm{CH_3CH(OH)CH_3} \xrightarrow{\text{conc. } \mathrm{H_2SO_4},\ \Delta} \mathrm{CH_3\!-\!CH\!=\!CH_2} + \mathrm{H_2O}$

   $\boxed{\mathbf{Q} = \text{propene}}$

   (Propene is the only alkene possible here — the molecule is symmetric about C-2, so Saytzeff offers no choice.)

   Step 3 — Hydroboration–oxidation gives R

   $\mathrm{B_2H_6}$ adds across the double bond with the boron attaching to the less substituted (terminal) carbon — partly steric, partly because the addition is syn and concerted with $\mathrm{B}^{\delta+}\!-\!\mathrm{H}^{\delta-}$. Alkaline $\mathrm{H_2O_2}$ then replaces the C–B bond by C–OH with retention. The net result is anti-Markovnikov hydration:

   $\mathrm{CH_3CH{=}CH_2} \xrightarrow{\ \mathrm{B_2H_6}\ } \left(\mathrm{CH_3CH_2CH_2}\right)_3\mathrm{B} \xrightarrow{\ \mathrm{H_2O_2},\,\mathrm{OH^-}\ } \mathrm{CH_3CH_2CH_2OH}$

   $\boxed{\mathbf{R} = \text{propan-1-ol}}$ …

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8. COMEDK 2023Set 2023-M1 markMCQ
   Q.Product of the following reaction is (A) (B) (C) (D)

   ›Reveal solutionSolution

   [!TLDR]

   Intramolecular oxymercuration-demercuration lets the existing tertiary hydroxyl attack the mercurinium ion, giving a Markovnikov bridged bicyclic ether with the gem-dimethyl group intact.

   Concept

   Oxymercuration ($\text{Hg(OAc)}_2/\text{H}_2\text{O}$, then $\text{NaBH}_4$) adds -H and -OH across a C=C by a mercurinium-ion mechanism: Markovnikov orientation, no carbocation rearrangement. When a suitably placed hydroxyl exists inside the same molecule, that oxygen acts as the nucleophile instead of water, so a cyclic ether is formed intramolecularly (alkoxymercuration).

   Solution

   • The mercurinium ion forms on the ring double bond.
   • The tertiary $-\text{C(CH}_3)_2\text{OH}$ oxygen is positioned across the ring and reaches the more-substituted (Markovnikov) alkene carbon, opening the mercurinium ring intramolecularly. …

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9. KCET 2021Set B-21 markMCQ
   Q.Peroxide effect is observed with the addition of HBr but not with the addition of HI to unsymmetrical alkene because (A) H-I bond is strong that H-Br and is not cleaved by the free radical (B) H-I bond is weaker than H-Br bond so that iodine free radicals combine to form iodine molecules (C) Bond strength of HI and HBr are same but free radicals are formed in HBr (D) All of these.

   ›Reveal solutionSolution

   The peroxide effect requires a self-sustaining radical chain; with HI the weak H–I bond gives $\mathrm{I^\bullet}$ radicals that prefer to dimerise to $\mathrm{I_2}$ rather than add to the alkene, so the chain never propagates.

   1. The mechanism the peroxide effect depends on

   In the presence of a peroxide, addition of HX to an unsymmetrical alkene switches from ionic (Markovnikov) to a free-radical chain (anti-Markovnikov, Kharasch effect):

   Initiation

   $\mathrm{R{-}O{-}O{-}R}\ \xrightarrow{\ \Delta\ }\ 2\,\mathrm{RO^\bullet}\qquad \mathrm{RO^\bullet + H{-}X \longrightarrow RO{-}H + X^\bullet}$

   Propagation

   $\mathrm{X^\bullet} + \mathrm{CH_2{=}CH{-}R} \longrightarrow \mathrm{X{-}CH_2{-}\overset{\bullet}{C}H{-}R} \qquad(i)$

   $\mathrm{X{-}CH_2{-}\overset{\bullet}{C}H{-}R} + \mathrm{H{-}X} \longrightarrow \mathrm{X{-}CH_2{-}CH_2{-}R} + \mathrm{X^\bullet} \qquad(ii)$

   The chain survives only if both (i) and (ii) are exothermic. That is a thermodynamic tug-of-war set by the H–X and C–X bond strengths.

   2. Why HBr works

   • H–Br ($\approx 366\ \mathrm{kJ\,mol^{-1}}$) is weak enough that $\mathrm{RO^\bullet}$ abstracts H readily ⇒ $\mathrm{Br^\bullet}$ is generated.
   • $\mathrm{Br^\bullet}$ adds to the double bond exothermically (step i) and the resulting carbon radical abstracts H from HBr exothermically (step ii).

   Both propagation steps are downhill ⇒ the chain runs ⇒ peroxide effect is observed.

   3. Why HI fails

   The H–I bond is the weakest of the hydrogen halides ($\approx 297\ \mathrm{kJ\,mol^{-1}}$), so $\mathrm{I^\bullet}$ radicals are formed very easily. But the C–I bond that step (i) would create is also very weak, which makes the addition of $\mathrm{I^\bullet}$ to the alkene endothermic — it simply does not go. The iodine radicals therefore accumulate and take the only path open to them, recombination:

   $\mathrm{I^\bullet + I^\bullet \longrightarrow I_2}$ …

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10. KCET 2021Set B-21 markMCQ
    Q.The major product of the following reaction is $\mathrm{CH_2=CH-CH_2-OH} \xrightarrow{\mathrm{HBr\ (excess)}}$ product (A) $\mathrm{CH_3-CHBr-CH_2Br}$ (B) $\mathrm{CH_2=CH-CH_2Br}$ (C) $\mathrm{CH_3-CHBr-CH_2-OH}$ (D) $\mathrm{CH_3-CHOH-CH_2OH}$

    ›Reveal solutionSolution

    With excess HBr, allyl alcohol undergoes both substitution of the $-\mathrm{OH}$ and Markovnikov addition across the double bond, giving 1,2-dibromopropane.

    1. Identify the two reactive sites

    $\mathrm{\underset{C3}{CH_2}{=}\underset{C2}{CH}{-}\underset{C1}{CH_2}{-}OH} \qquad \text{(prop-2-en-1-ol, allyl alcohol)}$

    • an alcohol $-\mathrm{OH}$ (reacts with HBr by substitution),
    • a C=C double bond (reacts with HBr by electrophilic addition).

    The word "excess" in the question is the signal that both must be consumed — a single equivalent would only do one.

    2. Step 1 — Substitution of the –OH

    $-\mathrm{OH}$ is a poor leaving group, so HBr first protonates it to $-\overset{+}{\mathrm{O}}\mathrm{H_2}$ (an excellent leaving group, water), and bromide then displaces it:

    $\mathrm{CH_2{=}CH{-}CH_2{-}OH} \xrightarrow{\ \mathrm{H^+}\ } \mathrm{CH_2{=}CH{-}CH_2{-}\overset{+}{O}H_2} \xrightarrow{\ \mathrm{Br^-}\ } \mathrm{CH_2{=}CH{-}CH_2Br} + \mathrm{H_2O}$

    (The allylic position makes this substitution especially easy — the allyl cation is resonance-stabilised.) This intermediate, allyl bromide, is exactly option (B) — but it is only the half-way product, not the answer, because HBr is in excess.

    3. Step 2 — Markovnikov addition of HBr across the C=C

    The second equivalent of HBr adds to the remaining double bond. The $\mathrm{H^+}$ attacks so as to generate the more stable carbocation:

    $\mathrm{\overset{3}{C}H_2{=}\overset{2}{C}H{-}\overset{1}{C}H_2Br} + \mathrm{H^+}$

    • H adds to C3 (the terminal $\mathrm{CH_2}$, which already has more hydrogens — Markovnikov's rule) ⇒ the positive charge lands on C2, a secondary carbocation.
    • The alternative (H to C2) would give a primary cation — much less stable. …

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11. KCET 2021Set B-21 markMCQ
    Q.A hydrocarbon A (C$_4$H$_8$) on reaction with HCl gives a compound B (C$_4$H$_9$Cl) which on reaction with 1 mol of NH$_3$ gives compound C (C$_4$H$_{10}$N). On reacting with NaNO$_2$ and HCl followed by treatment with water, compound C yields an optically active compound D. The D is (A) $\mathrm{CH_3-CH(Cl)-CH_2-CH_3}$ (B) $\mathrm{CH_3-CH(OH)-CH_2-CH_3}$ (C) $\mathrm{CH_3-CH(NH_2)-CH_2-CH_3}$ (D) $\mathrm{CH_3-CH(H)-CH_2-CH_3}$

    ›Reveal solutionSolution

    Butene $\to$ 2-chlorobutane $\to$ butan-2-amine $\to$ (diazotisation + water) butan-2-ol, whose C-2 is a stereocentre — hence optically active.

    Step 1 — A: the hydrocarbon $\mathrm{C_4H_8}$.

    Degree of unsaturation $=1$, so A is a butene. Adding HCl gives $\mathrm{C_4H_9Cl}$ (B).

    Step 2 — B: the chloride.

    $\mathrm{CH_3\text{-}CH{=}CH\text{-}CH_3} + \mathrm{HCl} \longrightarrow \mathrm{CH_3\text{-}CH(Cl)\text{-}CH_2\text{-}CH_3}$

    (For but-1-ene, Markovnikov addition gives the same secondary product, 2-chlorobutane — so B is 2-chlorobutane either way. The chlorine sits on C-2, a secondary carbon.)

    Step 3 — C: the amine.

    $\mathrm{NH_3}$ (1 mol) displaces the halide (nucleophilic substitution):

    $\mathrm{CH_3\text{-}CH(Cl)\text{-}CH_2\text{-}CH_3} + \mathrm{NH_3} \longrightarrow \underbrace{\mathrm{CH_3\text{-}CH(NH_2)\text{-}CH_2\text{-}CH_3}}_{\mathbf{C},\ \text{butan-2-amine}} + \mathrm{HCl}$

    C is a primary aliphatic amine ($-\mathrm{NH_2}$ on a secondary carbon).

    Step 4 — D: diazotisation followed by water.

    $\mathrm{NaNO_2 + HCl}$ generates nitrous acid $\mathrm{HNO_2}$ in situ. A primary aliphatic amine forms a very unstable alkanediazonium salt which immediately loses $\mathrm{N_2}$; the resulting carbocation is captured by water:

    $\mathrm{R\text{-}NH_2} \xrightarrow{\mathrm{NaNO_2/HCl}} [\mathrm{R\text{-}\overset{+}{N}{\equiv}N}] \xrightarrow{\mathrm{H_2O},\ -\mathrm{N_2}} \mathrm{R\text{-}OH}$

    $\Rightarrow\ \mathbf{D} = \mathrm{CH_3\text{-}CH(OH)\text{-}CH_2\text{-}CH_3}\quad(\text{butan-2-ol})$

    This matches the figure, in which D is drawn with an $-\mathrm{OH}$ on C-2 and is labelled 'optically active'.

    Step 5 — Check the optical activity. …

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12. COMEDK 2021Set 2021-B1 markMCQ
    Q.An aliphatic alcohol [X] on heating with Conc. H2SO4 gives a compound [Y]. Compound Y when reacted with HBr and then with aqueous KOH yielded 2-Methylpropan-2-ol. What are the compounds X and Y ? (A) X = Isobutyl alcohol Y= Methylpropene (B) X= Tertiary butyl alcohol Y= But-2-ene (C) X= Sec-butyl alcohol Y= But-2-ene (D) X= n-butyl alcohol Y= But-1-ene

    ›Reveal solutionSolution

    Y must be 2-methylpropene (gives tert-butanol via HBr/KOH); X is isobutyl alcohol, which dehydrates to it.

    Retrosynthesis from 2-methylpropan-2-ol ((CH$_3$)$_3$C–OH):

    • Aq. KOH on tert-butyl bromide $\to$ tert-butanol, so the HBr adduct is (CH$_3$)$_3$C–Br.
    • Markovnikov HBr addition giving (CH$_3$)$_3$C–Br comes from methylpropene (CH$_3$)$_2$C=CH$_2$ $\Rightarrow$ Y = methylpropene. …

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