Q.Match the reactions given in Column I with the names given in Column II.
Column I:
(i)

Concept: These are name reactions from organic chemistry — each is a standard method for forming carbon–carbon bonds or substituting halogens.

Reasoning:

1. (i) Aryl halide + alkyl halide + Na gives an alkylarene. This is the Wurtz–Fittig reaction (cross-coupling of two different halides). → matches (b).
2. (ii) Two aryl halides + Na in ether gives a biaryl. This is the Fittig reaction (homocoupling of aryl halides). → matches (a). …

The key to solving this is recognising the reagent pattern and the product structure — each reaction has a classic signature that tells you which name it belongs to. Let’s walk through them one by one.

1. Reaction (i): $\mathrm{C_6H_5X + RX \xrightarrow{Na} C_6H_5R}$

   An aryl halide and an alkyl halide react with sodium metal to give an alkylbenzene. This is a cross-coupling between an aromatic and an aliphatic halide.

   That’s the hallmark of the Wurtz–Fittig reaction — it’s a variant of the Wurtz reaction where one partner is aromatic.

   → Matches with (b).

2. Reaction (ii): $\mathrm{2\,C_6H_5X + 2Na \xrightarrow{Ether} C_6H_5{-}C_6H_5 + 2NaX}$

   Two aryl halides couple in the presence of sodium to form a biaryl (diphenyl). No alkyl halide is involved.

   This is the Fittig reaction (sometimes called the Wurtz–Fittig reaction when alkyl halides are also present, but here it’s purely aromatic).

   → Matches with (a).

3. Reaction (iii): $\mathrm{C_6H_5N_2^{+}X^{-} \xrightarrow{Cu_2X_2} C_6H_5X + N_2}$

   A diazonium salt is decomposed by cuprous halide ($\mathrm{Cu_2X_2}$) to replace the diazonium group with a halogen. Nitrogen gas is evolved.

   This is the classic Sandmeyer reaction — the copper(I) halide catalyses the substitution.

   → Matches with (d).

4. Reaction (iv): $\mathrm{C_2H_5Cl + NaI \xrightarrow{dry\ acetone} C_2H_5I + NaCl}$ …

Concept: Named Reactions in Organic Chemistry (Aryl and Alkyl Halides)

This is a matching question based on recognising the reagent, conditions, and product pattern of standard named reactions.

Method: Pattern Recognition by Reagent & Product Type

Steps:

1. Identify the substrate — is it an alkyl halide, aryl halide, or diazonium salt?
2. Identify the reagent and conditions — Na/ether, Cu₂X₂, NaI/acetone, etc.
3. Match the transformation to the standard reaction definition.

Step-by-step matching:

(i) $\mathrm{C_6H_5X + RX \xrightarrow{Na} C_6H_5R}$

• Substrate: Aryl halide + Alkyl halide
• Reagent: Sodium metal
• Product: Alkyl benzene
• Pattern: Combination of Wurtz reaction (alkyl-alkyl) and Fittig reaction (aryl-aryl) → Wurtz-Fittig reaction
• Match: (b)

(ii) $\mathrm{2\,C_6H_5X + 2Na \xrightarrow{Ether} C_6H_5{-}C_6H_5 + 2NaX}$

• Substrate: Two aryl halides
• Reagent: Sodium in dry ether
• Product: Biphenyl (diaryl)
• Pattern: Coupling of two aryl halides → Fittig reaction
• Match: (a)

(iii) $\mathrm{C_6H_5N_2^{+}X^{-} \xrightarrow{Cu_2X_2} C_6H_5X + N_2}$

• Substrate: Diazonium salt
• Reagent: Cuprous halide (Cu₂X₂)
• Product: Aryl halide + N₂ gas
• Pattern: Replacement of diazonium group by halogen using copper catalyst → Sandmeyer reaction
• Match: (d)

(iv) $\mathrm{C_2H_5Cl + NaI \xrightarrow{dry\ acetone} C_2H_5I + NaCl}$

• Substrate: Alkyl chloride
• Reagent: NaI in dry acetone …

Here are the common mistakes students make when matching these reactions, along with how to avoid each.

Mistake 1: Confusing the Wurtz-Fittig and Fittig Reactions

• The Mistake: Students often mix up reaction (i) and (ii). They see sodium ($\mathrm{Na}$) and an aryl halide in both and assume they are the same type of reaction. Specifically, they might label (i) as the Fittig reaction or (ii) as the Wurtz-Fittig reaction.
• Why it happens: Both reactions use sodium metal and involve aryl halides. The key difference is the number of different halides used.
• How to Avoid:
  • Focus on the reactants.
    • Wurtz-Fittig (i): Involves two different halides: one aryl ($\mathrm{C_6H_5X}$) and one alkyl ($\mathrm{RX}$). The product is an alkylbenzene ($\mathrm{C_6H_5R}$).
    • Fittig (ii): Involves only one type of halide: an aryl halide ($\mathrm{C_6H_5X}$). The product is a biphenyl ($\mathrm{C_6H_5{-}C_6H_5}$).
  • Memory Trick: "Wurtz-Fittig" has W and F — think "With Friends" (two different partners: alkyl + aryl). "Fittig" is just F — think "Family" (same type: aryl + aryl).

Mistake 2: Forgetting the Catalyst in the Sandmeyer Reaction

• The Mistake: Students match reaction (iii) with the Sandmeyer reaction but forget the crucial role of the copper(I) halide ($\mathrm{Cu_2X_2}$). They might think any diazonium salt decomposition is a Sandmeyer reaction.
• Why it happens: The reaction looks simple: a diazonium salt ($\mathrm{C_6H_5N_2^{+}X^{-}}$) is converted to an aryl halide ($\mathrm{C_6H_5X}$). Students memorize the "diazonium to halide" part but miss the specific reagent.
• How to Avoid:
  • Memorize the specific reagent. The Sandmeyer reaction is defined by the use of a cuprous halide ($\mathrm{Cu_2X_2}$ or $\mathrm{CuX}$) as a catalyst.
  • Compare with similar reactions:
    • Sandmeyer: $\mathrm{C_6H_5N_2^{+}Cl^{-} \xrightarrow{Cu_2Cl_2/HCl} C_6H_5Cl}$
    • Gattermann reaction: $\mathrm{C_6H_5N_2^{+}Cl^{-} \xrightarrow{Cu/HCl} C_6H_5Cl}$ (uses copper metal, not its salt).
    • Simple substitution (without catalyst): $\mathrm{C_6H_5N_2^{+}Cl^{-} \xrightarrow{H_2O} C_6H_5OH}$ (gives phenol, not halide).
  • Exam Tip: If you see $\mathrm{Cu_2X_2}$ or $\mathrm{CuX}$ with a diazonium salt, the answer is Sandmeyer.

Mistake 3: Misidentifying the Finkelstein Reaction

• The Mistake: Students fail to recognize reaction (iv) as the Finkelstein reaction because they don't immediately see the "iodide for chloride" swap. …