Match List-I with List-II
|
List-I (Reaction Name) |
List-II (Reaction) |
|
(A) Carbylamine reaction |
(I) |
|
(B) Gatterman Reaction |
(II) |
|
(C) Sandmeyer reaction |
(III) |
|
(D) Hofmann Bromamide reaction |
(IV) |
Choose the correct answer from the options given below.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(I), (B)-(IV), (C)-(II), (D)-(III)
The four named reactions are best remembered by their starting materials:
From diazonium salts: the \( \ce{N2+} \) group is replaced by Cl or Br. With copper(I) halide (\( \ce{Cu2Cl2}/\ce{HCl} \)) it is the Sandmeyer reaction; with copper powder and HCl it is the Gattermann reaction.
From primary amines: heating with chloroform and alcoholic KOH gives a foul-smelling isocyanide (carbylamine reaction, a test for primary amines).
From amides: \( \ce{Br2} \) and NaOH convert an amide into a primary amine with one carbon atom less (Hofmann bromamide degradation).
(A) Carbylamine reaction — This matches (I), the primary amine with chloroform and KOH giving the isocyanide RNC.
\( \displaystyle \ce{R-NH2 + CHCl3 + 3KOH} \xrightarrow{\text{heat}} \ce{R-NC + 3KCl + 3H2O} \)
(B) Gatterman Reaction — This matches (IV), benzenediazonium chloride with Cu/HCl giving chlorobenzene.
\( \displaystyle \ce{C6H5N2^+Cl^-} \xrightarrow{\ce{Cu}/\ce{HCl}} \ce{C6H5Cl + N2} \)
(C) Sandmeyer reaction — This matches (II), benzenediazonium chloride with \( \ce{Cu2Cl2}/\ce{HCl} \) giving chlorobenzene.
\( \displaystyle \ce{C6H5N2^+Cl^-} \xrightarrow{\ce{Cu2Cl2}/\ce{HCl}} \ce{C6H5Cl + N2} \)
(D) Hofmann Bromamide reaction — This matches (III), the amide with \( \ce{Br2} \) and NaOH giving the amine \( \ce{RNH2} \).
\( \displaystyle \ce{R-CONH2 + Br2 + 4NaOH -> R-NH2 + Na2CO3 + 2NaBr + 2H2O} \)
Shortcut: (A)-(I) is in Options (1), (2) and (3). The Gattermann reaction is the one with copper powder, (B)-(IV), which among these three is only in Option (3).