Q.An organic compound having molecular formula C3H7Br on treatment with aqueous KOH solution gave the compound (A). When the vapour of the compound (A) was passed over red hot copper at 300°C compound (B) was formed. The compound (B) on treatment with I2 and dil. NaOH, formed a yellow solid (C). Identify the compounds A, B and C.

Showing the 12 most recent of 19 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.[Case study] Depending on the molecule involved in controlling the rate of reaction, nucleophilic substitution reaction can be divided in two categories: nucleophilic unimolecular (SN1) and nucleophilic bimolecular (SN2). Alkyl halide reactivity towards SN1 and SN2 reactions depends on a number of variables, including steric hindrance, stability of the intermediate or transition state and solvent polarity. Primary alkyl halides, followed by secondary and tertiary alkyl halide are most favourable to the SN2 reaction mechanism. In the case of SN1 reactions, this order is reversible.

   (i) Which of the following is most reactive towards nucleophilic substitution reaction ?

   (a) CHCl3

   (b) CH2=CHCl

   (c) ClCH2CH=CH2

   (d) CH2CH=CHCl

   ›Reveal solutionSolution

   Allyl chloride is most reactive because ionisation of the C–Cl bond gives an allylic carbocation stabilised by resonance with the adjacent C=C double bond, favouring rapid SN1 substitution.

   Comparing the four halides:

   • CHCl₃ (chloroform): a trihalomethane; it is not a typical alkyl/allyl/vinyl halide substrate for facile nucleophilic substitution under these conditions — its C–H (not a good leaving-group carbon) and its multiple Cl on the same very electron-poor carbon make it comparatively unreactive here.
   • CH₂=CHCl (vinyl chloride): the C–Cl bond is attached directly to an sp² carbon of the double bond. This bond has partial double-bond character (due to conjugation of a chlorine lone pair with the π system) and is both shorter and stronger than a normal C–Cl bond; also, any positive charge generated at that carbon cannot be stabilised by the adjacent π system (it would need an empty orbital where the π bond already sits). Vinylic and aryl halides are therefore very unreactive towards nucleophilic substitution. …

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2. CBSE 2026Set ANNUAL1 markMCQ
   Q.[Case study, continued]

   (ii) Isopropyl chloride undergoes hydrolysis by

   (a) SN1 and SN2 mechanism

   (b) SN1 mechanism

   (c) SN2 mechanism

   (d) None of the above

   ›Reveal solutionSolution

   Isopropyl chloride is a secondary alkyl halide; secondary halides sit in the intermediate reactivity zone and can react by either SN1 (via a moderately stable secondary carbocation) or SN2 (moderate steric hindrance still allows backside attack), so both mechanisms operate, often simultaneously depending on conditions.

   Reactivity trend for nucleophilic substitution:

   • Primary halides are sterically unhindered, favouring SN2 (backside attack is easy) but their carbocation would be unstable, disfavouring SN1.
   • Tertiary halides are too sterically hindered for backside attack (SN2 is disfavoured) but readily form a stable tertiary carbocation, strongly favouring SN1. …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.Which of the following on heating with aqueous KOH, produces acetaldehyde?

   (a) $CH_3CH_2Cl$

   (b) $CH_2ClCH_2Cl$

   (c) $CH_3CHCl_2$

   (d) $CH_3COCl$

   ›Reveal solutionSolution

   Only the geminal dihalide $CH_3CHCl_2$ hydrolyses to an unstable gem-diol that collapses to the aldehyde; the other substrates give an alcohol, a diol, or a carboxylate instead.

   $CH_3CHCl_2$ (c), a gem-dihalide (both $Cl$ on the same carbon):

   $CH_3CHCl_2 + 2KOH(aq) \rightarrow CH_3CH(OH)_2 + 2KCl$

   The geminal diol $CH_3CH(OH)_2$ is unstable (two $-OH$ groups on the same carbon) and spontaneously eliminates water:

   $CH_3CH(OH)_2 \rightarrow CH_3CHO + H_2O$

   Net: $CH_3CHCl_2 + 2KOH \rightarrow CH_3CHO + 2KCl + H_2O$ — acetaldehyde.

   Why the others are wrong:

   • (a) $CH_3CH_2Cl$ + aq. $KOH$ undergoes simple nucleophilic substitution to give ethanol, $CH_3CH_2OH$ — not an aldehyde. …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.Which one of the following is the correct statement?

   (a) Alkyl halides are more reactive than aryl halides towards nucleophilic substitution reaction

   (b) Alkyl halides are less reactive than aryl halides towards nucleophilic substitution reaction

   (c) Nucleophilic substitution reaction proceeds through carbocation

   (d) Aryl halides cannot be prepared by electrophilic substitution to arenes

   ›Reveal solutionSolution

   Resonance donation of the halogen's lone pair into the aromatic ring strengthens and shortens the aryl $C-X$ bond and makes the ring electron-rich, so alkyl halides are far more reactive than aryl halides toward nucleophilic substitution.

   Why aryl halides resist substitution: the halogen's lone pair delocalises into the benzene ring by resonance, giving the $C-X$ bond partial double-bond character — it becomes shorter and stronger than a normal $C-X$ single bond, and harder to break. The electron-rich ring also repels an incoming nucleophile, and backside ($S_N2$-type) attack on the $sp^2$ carbon is sterically/geometrically hindered by the planar ring; an $S_N1$-type pathway would require a highly unstable phenyl cation, which does not form under normal conditions.

   Why the other statements are wrong:

   • (b) is the exact reverse of the truth. …

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5. CBSE 2025Set ANNUAL1 mark
   Q.How can the following conversion be carried out? (Give chemical equation only) — Bromoethane to propane nitrile

   ›Reveal solutionSolution

   Nucleophilic substitution of bromide by cyanide ion (alcoholic KCN) gives the nitrile.

   Bromoethane is heated with alcoholic potassium cyanide (KCN); the cyanide ion (a good nucleophile, attacking through carbon) displaces bromide in an $S_N2$ reaction:

   $C_2H_5Br + KCN \xrightarrow{\text{ethanolic}} C_2H_5CN + KBr$

   …

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6. CBSE 2024Set D1 markMCQ
   Q.C2H5Br + NaOH -> C2H5OH + NaBr is an example of which of the following types of reaction?

   (a) Electrophilic substitution

   (b) Nucleophilic substitution

   (c) Both (A) and (B)

   (d) None of these

   ›Reveal solutionSolution

   The nucleophile OH- attacks the electrophilic carbon of C2H5Br and displaces the leaving group Br-, giving C2H5OH. This is nucleophilic substitution.

   In C2H5Br the C-Br bond is polar; carbon carries a partial positive charge. The hydroxide ion (OH-) from NaOH is an electron-rich nucleophile that attacks this carbon and expels bromide as the leaving group:

   C2H5Br + OH- -> C2H5OH + Br-

   …

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7. CBSE 2024Set ANNUAL1 markMCQ
   Q.Arylhalides are less reactive towards nucleophilic substitution reaction as compared to alkylhalides due to :

   (a) The formation of less stable carbonium ion

   (b) Resonance stabilization

   (c) Longer carbon-halogen bond

   (d) Inductive effect

   ›Reveal solutionSolution

   Aryl halides resist nucleophilic substitution because resonance between the halogen's lone pair and the ring strengthens the C-X bond.

   In an aryl halide such as chlorobenzene, a lone pair on the halogen conjugates with the pi-electron system of the ring. This delocalisation:

   1. Gives the carbon-halogen bond partial double-bond character, making it shorter and stronger than a normal C-X single bond, so it resists heterolytic cleavage (needed for both SN1 and SN2 pathways).
   2. Increases electron density on the ring (especially ortho/para to X), which repels an approaching nucleophile.
   3. The carbon bearing the halogen is sp2 hybridised, holding the bonding electrons closer to carbon and further strengthening the bond. …

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8. CBSE 2023Set ANNUAL1 markMCQ
   Q.Which of the following will react faster in SN2 reaction?

   (a) 1-bromopentane

   (b) 2-bromopentane

   (c) 3-bromopentane

   (d) 2-bromo-2-methyl butane

   ›Reveal solutionSolution

   SN2 rate depends inversely on steric hindrance around the carbon bearing the leaving group; the primary halide has the least hindrance and reacts fastest.

   In an SN2 reaction, the nucleophile attacks the carbon from the side opposite the leaving group in a single concerted step, so the rate is very sensitive to steric crowding at that carbon: primary (1 degree) > secondary (2 degree) > tertiary (3 degree), and tertiary halides essentially do not undergo SN2. Among the choices, 1-bromopentane has its Br on a pr …

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9. CBSE 2023Set ANNUAL1 markMCQ
   Q.$S_N2$ mechanism proceeds through the intervention of

   (a) carbonium ion

   (b) transition state

   (c) free radical

   (d) carbanion

   ›Reveal solutionSolution

   $S_N2$ substitution is concerted (bond-breaking and bond-making happen simultaneously in one step) via a single pentacoordinate transition state, unlike $S_N1$ (carbocation intermediate) or radical-chain reactions.

   Mechanism of $S_N2$: The nucleophile ($Nu^-$) attacks the electrophilic carbon from the side directly opposite (backside of) the leaving group ($X$). As the new $Nu-C$ bond starts to form, the $C-X$ bond simultaneously starts to break — both events occur in a single, concerted step:

   $Nu^- + R_3C\text{-}X \rightarrow \Big[Nu\cdots C R_3 \cdots X\Big]^{\ddagger} \rightarrow Nu\text{-}CR_3 + X^-$

   The bracketed species is the transition state: a fleeting, high-energy, pentacoordinate arrangement (5 groups around carbon — the incoming nucleophile, the three original substituents, and the departing leaving group) that is never actually isolable, unlike a true intermediate. Because there is only one transition state and no intermediate, the reaction is second order overall (rate depends on both $[Nu^-]$ and $[R_3CX]$) and proceeds with inversion of configuration at carbon (Walden inversion), since the nucleophile ends up on the opposite face from where the leaving group departed.

   …

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10. CBSE 2022Set ANNUAL1 markMCQ
    Q.A primary alkyl halide would prefer to undergo:

    (a) $S_N2$

    (b) $S_N1$

    (c) Elimination

    (d) None of these

    ›Reveal solutionSolution

    Primary alkyl halides favour $S_N2$ because the carbon is sterically open to backside attack and a primary carbocation (needed for $S_N1$) is too unstable. Option (A).

    The two nucleophilic substitution pathways:

    • $S_N1$ — proceeds through a carbocation; favoured by tertiary halides (stable 3° carbocation).
    • $S_N2$ — one-step backside attack by the nucleophile; favoured when the carbon is sterically uncrowded, i.e. by primary halides. …

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11. CBSE 2021Set A1 markMCQ
    Q.SN2 mechanism proceeds via formation of

    (a) Carbocation

    (b) Transition state

    (c) Free radical

    (d) Carbanion

    ›Reveal solutionSolution

    SN2 goes through one concerted step with a pentacoordinate transition state, not an intermediate carbocation/carbanion.

    In the SN2 (Substitution Nucleophilic Bimolecular) mechanism:

    • The nucleophile attacks the carbon from the side opposite to the leaving group.
    • Bond making (Nu-C) and bond breaking (C-leaving group) occur simultaneously in one step. …

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12. CBSE 2020Set 56/3/11 markMCQ
    Q.Racemisation occurs in (A) $S_N2$ reaction (B) $S_N1$ reaction (C) Neither $S_N2$ nor $S_N1$ reactions (D) $S_N2$ reaction as well as $S_N1$ reaction

    ›Reveal solutionSolution

    Racemisation occurs when a chiral centre is converted to a planar intermediate that can be attacked from either face with equal probability. This happens exclusively in $S_N1$ reactions through the formation of a planar carbocation. The correct option is (B).

    The key to understanding racemisation lies in recognising what happens to the stereochemistry at a chiral carbon during nucleophilic substitution.

    The Stereochemical Fate of Chiral Centres

    When a nucleophile attacks a chiral carbon bearing a leaving group, the three-dimensional arrangement of groups around that carbon determines whether we retain, invert, or lose stereochemical information entirely.

    1. The $S_N2$ Mechanism and Inversion

    In an $S_N2$ reaction, the nucleophile attacks from the side directly opposite to the leaving group. This backside attack forces all three other groups attached to the carbon to flip to the opposite side, like an umbrella turning inside-out in the wind.

    The result? Complete inversion of configuration at the chiral centre—what we call Walden inversion. If you start with an $(R)$-enantiomer, you end with an $(S)$-enantiomer, and vice versa. The product is a single stereoisomer, not a mixture.

    Watch out

    A common mistake is thinking that because $S_N2$ changes configuration, it must produce a racemic mixture. Inversion gives you the opposite enantiomer cleanly, not both enantiomers in equal amounts.

    2. The $S_N1$ Mechanism and Carbocation Formation

    The $S_N1$ reaction proceeds through a two-step mechanism. First, the leaving group departs, taking both bonding electrons with it. This leaves behind a carbocation at what was formerly the chiral centre.

    Here's the crucial point: a carbocation is $sp^2$ hybridised and planar. The three groups attached to it lie in a flat plane, with an empty $p$-orbital perpendicular to that plane. The molecule has lost its chirality at this intermediate stage.

    3. Nucleophilic Attack on the Planar Carbocation

    When the nucleophile approaches this planar carbocation in the second step, it can attack from either face of the molecule with equal probability. There's no steric or electronic preference for one side over the other.

    • Attack from the top face regenerates one enantiomer
    • Attack from the bottom face produces the opposite enantiomer …

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