In the Claisen-Schmidt reaction to prepare dibenzalacetone from 5.3 g of benzaldehyde, a total of 3.51 g of product was obtained. The percentage yield in this reaction was _____%. [2025]
(60)

Moles of benzaldehyde initially taken
For 100% yield of the reaction, 2 mol of benzaldehyde give 1 mol of dibenzalacetone, so 0.05 mol of benzaldehyde give 0.025 mol of dibenzalacetone.
Actual moles of dibenzalacetone obtained
Percentage yield
Identify the structure of the final product (D) in the following sequence of the reactions:

Total number of hybridised carbon atoms in product D is _____. [2025]
(7)

Match List I with List II [2023]
| List I | List II | ||
| Reaction | Reagents | ||
| A. | Hoffmann Degradation | I. | Conc. KOH, |
| B. | Clemmensen reduction | II. | |
| C. | Cannizzaro reaction | III. | , NaOH |
| D. | Reimer–Tiemann Reaction | IV. | Zn–Hg / HCl |
Choose the correct answer from the options given below:
(A) – III, (B) – IV, (C) – II, (D) – I
(A) – III, (B) – IV, (C) – I, (D) – II
(A) – II, (B) – IV, (C) – I, (D) – III
(A) – II, (B) – I, (C) – III, (D) – IV
Match List I with List II [2023]
| List I | List II | ||
| Name of reaction | Reagent used | ||
| A. | Hell–Volhard–Zelinsky reaction | I. | |
| B. | Iodoform reaction | II. | |
| C. | Etard reaction | III. | |
| D. | Gattermann–Koch reaction | IV. |
Choose the correct answer from the options given below:
A–III, B–I, C–II, D–IV
A–I, B–II, C–III, D–IV
A–III, B–I, C–IV, D–II
A–III, B–II, C–I, D–IV
(1)
