Match List-I with List-II
| List-I Reagents and Substrates | List-II Name Reaction |
| (A) Alkyl chloride + NaI + dry acetone | (I) Wurtz reaction |
| (B) Alkyl halide + Na | (II) Friedel-Crafts reaction |
| (C) Aryl halide + methyl chloride + anhyd. AlCl3 | (III) Finkelstein reaction |
| (D) Aryl halide + Na | (IV) Fittig reaction |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) - (III), (B) - (I), (C) - (II), (D) - (IV)
Named reactions of haloalkanes and haloarenes are recognised from their substrate and reagent. The common ones fall into three groups:
Halogen exchange (alkyl halides): Finkelstein reaction — an alkyl chloride or bromide with NaI in dry acetone gives an alkyl iodide; Swarts reaction — an alkyl chloride or bromide heated with a metallic fluoride (AgF, \( \ce{Hg2F2} \), \( \ce{CoF2} \), \( \ce{SbF3} \)) gives an alkyl fluoride.
Coupling with sodium in dry ether: Wurtz reaction — two alkyl halides give an alkane; Fittig reaction — two aryl halides give a diaryl; Wurtz–Fittig reaction — an alkyl halide and an aryl halide give an alkylarene.
Friedel–Crafts reactions of haloarenes: with an alkyl chloride (alkylation) or an acyl chloride (acylation) and anhydrous AlCl3, an alkyl or acyl group enters the ring at the ortho and para positions.
(A) Alkyl chloride + NaI + dry acetone — This matches (III) Finkelstein reaction. The chlorine is exchanged for iodine. NaCl is insoluble in dry acetone and precipitates, which drives the reaction forward.
\( \displaystyle \ce{R-Cl + NaI} \xrightarrow{\text{dry acetone}} \ce{R-I + NaCl v} \)
(B) Alkyl halide + Na — This matches (I) Wurtz reaction. Two alkyl groups join to form an alkane with double the number of carbon atoms.
\( \displaystyle \ce{2R-X + 2Na} \xrightarrow{\text{dry ether}} \ce{R-R + 2NaX} \)
(C) Aryl halide + methyl chloride + anhyd. \( \ce{AlCl3} \) — This matches (II) Friedel-Crafts reaction. \( \ce{AlCl3} \) generates the electrophile \( \ce{CH3^{+}} \), which attacks the ring; the halogen directs it to the ortho and para positions, and the para product is major.
\( \displaystyle \ce{C6H5Cl + CH3Cl} \xrightarrow{\text{anhyd. }\ce{AlCl3}} \text{2-chlorotoluene} + \text{4-chlorotoluene (major)} + \ce{HCl} \)
(D) Aryl halide + Na — This matches (IV) Fittig reaction. Two aryl groups join to form a diaryl.
\( \displaystyle \ce{2C6H5X + 2Na} \xrightarrow{\text{dry ether}} \ce{C6H5-C6H5 + 2NaX} \)
Shortcut: The pair (C)–(II) appears only in Option (1).