See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1 - Identify reaction pathway to P and Q: CH3-CH=CH2 undergoes hydroboration-oxidation (BH3/THF then H2O2/OH-). This is an anti-Markovnikov addition of water, placing the OH on the terminal (less substituted) carbon. Product P = CH3-CH2-CH2-OH (propan-1-ol, a primary alcohol). PCC oxidizes primary alcohols to aldehydes without over-oxidizing to carboxylic acids. So Q = CH3-CH2-CHO (propanal, an aldehyde). Step 2 - Identify reaction pathway to R and S: CH3-CH=CH2 undergoes oxymercuration-demercuration (Hg(OAc)2/H2O then NaBH4/HO-). This is Markovnikov addition of water, placing the OH on the more substituted (internal) carbon. Product R = CH3-CH(OH)-CH3 (propan-2-ol, a secondary alcohol). PCC oxidizes secondary alcohols to ketones. So S = CH3-CO-CH3 (propanone/acetone, a ketone). Step 3 - Determine relationship between Q and S: Q = CH3CH2CHO (propanal) molecular formula C3H6O. S = CH3COCH3 (propanone/acetone) molecular formula C3H6O. Both Q and S have the same molecular formula (C3H6O) but differ in the nature of the functional group: Q is an aldehyde and S is a ketone. Compounds with the same molecular formula but different functional groups are functional isomers. Step 4 - Eliminate other options: (a) Positional isomers differ in the position of the same functional group on the same carbon skeleton - Q and S have different functional groups, so not positional isomers. (b) Chain isomers differ in the carbon skeleton arrangement - both have the same straight 3-carbon chain, so not chain isomers. (c) Stereoisomers have the same connectivity but differ in spatial arrangement - Q and S have different functional groups and connectivity, so not stereoisomers. Therefore, the correct answer is D.