JEE Challenger
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Match Organic Reactions with Corresponding Reaction Mechanisms

Match the LIST-I with LIST-II

List-I (Reaction)List-II (Mechanism)A. Williamson SynthesisI. Electrophilic additionB. Friedel Craft ReactionII. Free radical substitutionC. Bromination of vinyl benzeneIII. Nucleophilic substitutionD. Chlorination of toluene in lightIV. Electrophilic substitution\begin{array}{|l|l|} \hline \text{List-I (Reaction)} & \text{List-II (Mechanism)} \\ \hline \text{A. Williamson Synthesis} & \text{I. Electrophilic addition} \\ \text{B. Friedel Craft Reaction} & \text{II. Free radical substitution} \\ \text{C. Bromination of vinyl benzene} & \text{III. Nucleophilic substitution} \\ \text{D. Chlorination of toluene in light} & \text{IV. Electrophilic substitution} \\ \hline \end{array}

Choose the correct answer from the options given below:

Options

A

A-III, B-I, C-II, D-IV

B

A-III, B-IV, C-II, D-I

C

A-III, B-IV, C-I, D-II

Correct
D

A-I, B-III, C-IV, D-II

Step-by-Step Solution

To determine the correct matching between List-I (Organic Reactions) and List-II (Reaction Mechanisms), let us analyze each reaction individually:

  1. A. Williamson Synthesis:

    • Reaction: An alkyl halide reacts with an alkoxide ion to form an ether: RX+RONa+ROR+NaXR-\text{X} + R'-\text{O}^- \text{Na}^+ \longrightarrow R-\text{O}-R' + \text{NaX}
    • Mechanism: The alkoxide ion acts as a strong nucleophile and attacks the primary alkyl halide via a bimolecular nucleophilic substitution (SN2\text{S}_\text{N}2) pathway.
    • Match: A \rightarrow III (Nucleophilic substitution)
  2. B. Friedel-Crafts Reaction:

    • Reaction: Benzene reacts with an alkyl halide or acyl halide in the presence of a Lewis acid catalyst (e.g., AlCl3\text{AlCl}_3) to yield an alkylbenzene or acylbenzene.
    • Mechanism: The Lewis acid generates a carbocation or acylium ion (electrophile), which attacks the electron-rich aromatic ring, leading to an aromatic electrophilic substitution (SEAr\text{S}_\text{E}\text{Ar}).
    • Match: B \rightarrow IV (Electrophilic substitution)
  3. C. Bromination of vinyl benzene:

    • Reaction: Vinyl benzene (styrene, C6H5CH=CH2\text{C}_6\text{H}_5-\text{CH}=\text{CH}_2) reacts with bromine (Br2\text{Br}_2).
    • Mechanism: The π\pi-electrons of the carbon-carbon double bond (C=C\text{C}=\text{C}) attack the electrophilic bromine molecule to form a bromonium ion intermediate, followed by nucleophilic attack of Br\text{Br}^-. This is a characteristic electrophilic addition reaction across an alkene double bond.
    • Match: C \rightarrow I (Electrophilic addition)
  4. D. Chlorination of toluene in light:

    • Reaction: Toluene reacts with chlorine gas in the presence of sunlight (hνh\nu): C6H5CH3+Cl2hνC6H5CH2Cl+HCl\text{C}_6\text{H}_5-\text{CH}_3 + \text{Cl}_2 \xrightarrow{h\nu} \text{C}_6\text{H}_5-\text{CH}_2\text{Cl} + \text{HCl}
    • Mechanism: Sunlight initiates the homolytic cleavage of Cl2\text{Cl}_2 to generate chlorine free radicals, leading to a halogenation of the side-chain methyl group via a free radical substitution mechanism.
    • Match: D \rightarrow II (Free radical substitution)

Correct Match:

  • A \rightarrow III
  • B \rightarrow IV
  • C \rightarrow I
  • D \rightarrow II

This corresponds to Option C.

Match Organic Reactions with Corresponding Reaction Mechanisms | Chemistry PYQ Solution - JEE Challenger