Question

Match List-I with List-II:\[\begin{array}{|c|c|} \hline \textbf{List-I (Reaction Name)} & \textbf{List-II (Reactant)} \\ \hline \text{(A) Friedlander's synthesis} & \text{(I) o-Aminobenzaldehyde} \\ \hline \text{(B) Doebner-Miller's synthesis} & \text{(III) Aniline} \\ \hline \text{(C) Hantzsch's synthesis} & \text{(IV) \(\beta\)-Ketoneester} \\ \hline \text{(D) Bischler-Napieralski's synthesis} & \text{(II) \(\beta\)-Phenylethylamide} \\ \hline \end{array}\] Choose the correct answer from the options given below:

• (A) - (I), (B) - (II), (C) - (III), (D) - (IV)
• (A) - (I), (B) - (III), (C) - (IV), (D) - (II)
• (A) - (I), (B) - (II), (C) - (IV), (D) - (III)
• (A) - (III), (B) - (IV), (C) - (I), (D) - (II)

Hint:

Each reaction in List-I requires a specific reactant from List-II. Ensure you understand the reagents and substrates used in each reaction.

Answer 1

The Correct Option is A

Step 1: Friedlander's Synthesis.
Friedlander's synthesis involves the reaction of o-aminobenzaldehyde with a suitable electrophilic reagent. Hence, it matches with (I) o-Aminobenzaldehyde.

Step 2: Doebner-Miller's Synthesis.
Doebner-Miller's synthesis involves the reaction of $\beta$-phenylethylamide with a suitable electrophile, so it matches with (II) $\beta$-Phenylethylamide.

Step 3: Hantzsch's Synthesis.
Hantzsch's synthesis involves the reaction of aniline with a carbonyl compound, so it matches with (III) Aniline.

Step 4: Bischler-Napieralski's Synthesis.
Bischler-Napieralski's synthesis involves the reaction of a $\beta$-ketoneester with an amine, so it matches with (IV) $\beta$-Ketoneester.

Step 5: Conclusion.
Thus, the correct matching is (A) - (I), (B) - (II), (C) - (III), (D) - (IV), which corresponds to option (1).

Final Answer: \[ \boxed{\text{(1) (A) - (I), (B) - (II), (C) - (III), (D) - (IV)}} \]