See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: The rate of nucleophilic addition to a carbonyl compound depends on two main factors: (1) the electrophilicity (positive character) of the carbonyl carbon, and (2) steric hindrance around the carbonyl carbon. Greater electrophilicity and lesser steric hindrance both increase the rate. Step 1: Analyze compound (a) - p-nitrobenzaldehyde. The nitro group is a strong electron-withdrawing group (EWG) at the para position. It withdraws electron density from the ring and hence from the CHO group via resonance and induction, making the carbonyl carbon more electrophilic. Among all the compounds, this has the highest reactivity toward nucleophilic addition. Step 2: Analyze compound (b) - benzaldehyde. No substituent on the ring. The benzene ring conjugated with CHO slightly deactivates the CHO compared to aliphatic aldehydes, but benzaldehyde itself is the reference aromatic aldehyde here. It is more reactive than the methoxy-substituted analog but less than the nitro-substituted one. Step 3: Analyze compound (c) - p-methoxybenzaldehyde. The OMe group is an electron-donating group (EDG) via resonance, which donates electron density into the ring and onto the carbonyl carbon through the conjugated system, reducing electrophilicity of the carbonyl carbon. This makes it least reactive among the aldehydes. Step 4: Analyze compound (d) - methyl ethyl ketone (CH3-CO-Et). Ketones are generally less reactive than aldehydes toward nucleophilic addition due to: (i) two alkyl groups donating electron density to the carbonyl carbon (reducing electrophilicity compared to aldehydes which have one H and one alkyl/aryl), and (ii) slightly more steric hindrance than aldehydes. However, aliphatic ketones are more reactive than aryl ketones because aryl groups delocalize the carbonyl's positive character through resonance. Step 5: Analyze compound (e) - phenyl cyclohexyl ketone (cyclohexane ring with C(=O)Ph). This is a diaryl/aryl-alkyl ketone with significant steric bulk (cyclohexyl group) and the phenyl group conjugated with the carbonyl (reducing electrophilicity). Both steric and electronic factors reduce reactivity. This is less reactive than the simple aliphatic ketone (d). Step 6: Rank the compounds: - (a) p-NO2-C6H4-CHO: highest reactivity (EWG + aldehyde) - (b) C6H5-CHO: second (aldehyde, no substituent effect) - (c) p-MeO-C6H4-CHO: lower than (b) due to EDG reducing electrophilicity - (d) CH3-CO-Et: ketone, aliphatic, moderate steric, less reactive than aromatic aldehydes but more reactive than aryl ketone - (e) Cyclohexyl-CO-Ph: least reactive (large steric bulk + aryl conjugation) Wait - comparing (b), (c), and (d): Aromatic aldehydes vs aliphatic ketones. Aldehydes are generally more reactive than ketones. So (b) > (d). And (c) though deactivated is still an aldehyde, so (c) vs (d) is the key question. p-Methoxybenzaldehyde, despite the EDG, is still an aldehyde. The OMe group reduces reactivity compared to benzaldehyde but a strongly deactivated aldehyde can still be more reactive than a ketone OR less reactive. Given the answer is a > b > c > d > e, the order places (c) above (d), meaning even the electron-donating methoxy group doesn't reduce the aldehyde below the aliphatic ketone reactivity. Then (d) > (e) because (d) is an aliphatic ketone while (e) is an aryl ketone with greater steric and electronic deactivation. Therefore the order is: a > b > c > d > e, which corresponds to answer choice (a). Why other options fail: - Option (b) a > b > d > c > e: places (d) above (c), but (c) is still an aldehyde and should be more reactive than ketone (d). - Option (c) a > d > e > b > c: incorrectly places ketones above aldehydes (b) and (c). - Option (d) a > b > e > d > c: incorrectly places aryl ketone (e) above aliphatic ketone (d). Therefore, the correct answer is A.