Match List-I with List-II.
| List-I | List-II |
| Name Reactions | Substrates/ Reagents |
| (A) Hofmann bromamide degradation reaction | (I)Conversion of benzene diazonium salt into haloarene using Cu2X2/(HX, X = Cl or Br). |
| (B) Gabriel phthalimide synthesis | (II) Conversion of benzene diazonium salt into haloarene using Cu/(HX, X = Cl or Br). |
| (C) Gattermann reaction | (III) Preparation of primary amines by the action of primary alkyl halides on phthalimide in aq KOH. |
| (D) Sandmeyer reaction | (IV) Synthesis of primary amines from an acid amide. |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → (A) – (IV), (B) – (III), (C) – (II), (D) – (I)
The named reactions in this list fall into two groups:
Preparation of primary amines: Hofmann bromamide degradation — an amide with \( \ce{Br2} \) and NaOH gives a 1° amine with one carbon fewer; Gabriel phthalimide synthesis — phthalimide with KOH, then a primary alkyl halide, then alkaline hydrolysis gives a pure 1° amine.
Diazonium salt to haloarene: Sandmeyer reaction — cuprous halide, \( \ce{Cu2X2} \)/HX; Gattermann reaction — copper powder, Cu/HX. Both replace –\( \ce{N2^{+}} \) by Cl or Br.
(A) Hofmann bromamide degradation reaction — This matches (IV) Synthesis of primary amines from an acid amide. The C=O carbon is lost as carbonate, so the amine has one carbon atom fewer than the amide.
\( \displaystyle \ce{R-CONH2 + Br2 + 4NaOH -> R-NH2 + Na2CO3 + 2NaBr + 2H2O} \)
(B) Gabriel phthalimide synthesis — This matches (III) Preparation of primary amines by the action of primary alkyl halides on phthalimide in aq KOH. The potassium salt of phthalimide is alkylated by R–X, and hydrolysis releases only the primary amine.
\( \displaystyle \text{phthalimide} \xrightarrow{\text{KOH}} \text{K-phthalimide} \xrightarrow{\ce{R-X}} \text{N-alkylphthalimide} \xrightarrow{\ce{NaOH}\text{ (aq)}} \ce{R-NH2} \)
(C) Gattermann reaction — This matches (II) Conversion of benzene diazonium salt into haloarene using Cu/(HX, X = Cl or Br). Copper powder with the halogen acid replaces the diazo group.
\( \displaystyle \ce{C6H5N2^+X^-} \xrightarrow{\ce{Cu}/\ce{HX}} \ce{C6H5X + N2} \)
(D) Sandmeyer reaction — This matches (I) Conversion of benzene diazonium salt into haloarene using Cu2X2/(HX, X = Cl or Br). Cuprous chloride or bromide in the halogen acid replaces the diazo group.
\( \displaystyle \ce{C6H5N2^+X^-} \xrightarrow{\ce{Cu2X2}/\ce{HX}} \ce{C6H5X + N2} \)
Shortcut: The pair (A)–(IV) appears in Options (1) and (4), and (C)–(II) then leaves only Option (4).
Note: It seems that there is a mistake in the original NTA paper, which prints aq KOH in item (III) when it should read ethanolic KOH.