Aldehydes, Ketones and Carboxylic Acids Set-2

Q1. Among the following carbonyl compounds, which will be reduced most rapidly by sodium borohydride ($NaBH_4$) in ethanol, considering electronic and steric factors?

Q2. Crotonaldehyde ($CH_3CH=CHCHO$) is treated with $NaBH_4$ in methanol. Which is the major organic product after aqueous work-up?

Q3. When benzaldehyde ($C_6H_5CHO$) and acetone ($CH_3COCH_3$) in molar ratio 2:1 are treated with dilute $NaOH$ followed by acidification, the major organic product obtained is:

Q4. Assertion (A): When benzaldehyde ($C_6H_5CHO$) is mixed with acetaldehyde ($CH_3CHO$) and treated with dilute $NaOH$, the Cannizzaro reaction of benzaldehyde is largely suppressed and the crossed aldol condensation product predominates.
Reason (R): Acetaldehyde is readily enolized under base to give an enolate which rapidly attacks benzaldehyde; this fast enolate‑electrophile reaction consumes benzaldehyde before significant hydride transfer (Cannizzaro) can occur, thus favoring crossed aldol.

Q5. An unknown aromatic carbonyl compound A (molecular formula $C_8H_8O$) gives a positive 2,4‑DNP test, a negative Tollen's test, produces a yellow precipitate with $Br_2$ in $NaOH$ (haloform test), and on vigorous oxidation with $KMnO_4$ yields benzoic acid. Which of the following is compound A?

Q6. Which of the following carbonyl compounds will undergo the Cannizzaro reaction when treated with concentrated NaOH?

Q7. Among the following, which compound will form the most stable enolate (i.e., has the most acidic $\alpha$‑hydrogen)?

Q8. 2‑Methylcyclohexanone on treatment with LDA at $-78^\circ\text{C}$ followed by methyl iodide gives predominantly:

Q9. A mixture of benzaldehyde ($C_6H_5CHO$) and acetaldehyde ($CH_3CHO$) is treated with dilute NaOH at room temperature. The major isolated organic product after workup will be:

Q10. Which one of the following compounds gives a positive Tollens’ test but a negative iodoform test?