Step 1: Part (b)(i): Structure of semicarbazone of acetone.
Acetone ($CH_3COCH_3$) reacts with semicarbazide ($H_2N-NH-CO-NH_2$) by nucleophilic addition-elimination: \[ CH_3COCH_3 + H_2N-NH-CO-NH_2 \rightarrow CH_3-C(=N-NH-CO-NH_2)-CH_3 + H_2O \] Structure of acetone semicarbazone: $CH_3-C(=N-NHCONH_2)-CH_3$. The $C=O$ group of acetone is converted to a $C=N-$ group, with the rest of the semicarbazide unit ($-NHCONH_2$) attached to the nitrogen.
Step 2: Part (b)(ii): Why alpha-hydrogen atoms are acidic in aldehydes and ketones.
Two reasons work together: (i) Inductive effect: The carbonyl group ($C=O$) is strongly electron-withdrawing. This pulls electron density away from the alpha-C-H bond through the sigma framework, weakening the C-H bond and making the alpha-H easier to lose as $H^+$. (ii) Resonance stabilisation of the enolate: After removal of alpha-H, the resulting carbanion (enolate) is stabilised by resonance: \[ -CH_2-C=O \rightleftharpoons -CH=C-O^- \] The negative charge is delocalised onto the electronegative oxygen, greatly stabilising the enolate ion. This extra stability of the product (enolate) is what makes the alpha-H acidic.
Step 3: Part (b)(iii): Factors governing reactivity towards HCN.
Nucleophilic addition of $CN^-$ to $C=O$ depends on: (i) Electronic factor: electron-withdrawing groups on/near the carbonyl carbon increase electrophilicity (more reactive). Electron-donating alkyl groups reduce electrophilicity (less reactive). (ii) Steric factor: bulky groups around the carbonyl carbon create steric hindrance, making nucleophilic approach more difficult.
Step 4: Apply factors to the three compounds.
$HCHO$ (formaldehyde): No alkyl groups, maximum electrophilicity, zero steric hindrance. MOST reactive. $CH_3CHO$ (acetaldehyde): One $-CH_3$ group, slightly less electrophilic, slight steric effect. Less reactive than formaldehyde. $CH_3COCH_3$ (acetone): Two $-CH_3$ groups, further reduced electrophilicity, more steric hindrance from both sides. LEAST reactive.
Step 5: Arrange in increasing order of reactivity towards HCN.
\[ CH_3COCH_3 < CH_3CHO < HCHO \] (From least reactive to most reactive.)
Step 6: Explain Clemmensen reduction and state the boxed answer.
Clemmensen reduction: $C=O \xrightarrow{Zn(Hg)/\text{conc. HCl}} -CH_2-$. Used to reduce ketones/aldehydes to alkanes in acidic conditions. \[ \boxed{\text{Reactivity towards HCN (increasing order): }CH_3COCH_3 < CH_3CHO < HCHO} \]