See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The compound is 2-methoxy-4-methylphenyl ether of glycolaldehyde. More precisely, it is (2-methoxy-4-methylphenoxy)-substituted compound where the side chain is -O-CH(CHO)-CH2OH. The side chain attached via the ring oxygen is: -CH(CHO)-CH2OH, meaning the carbon attached to the aromatic oxygen bears an aldehyde (CHO) and a -CH2OH group. Step 2 - Reaction with NaBH4 (A): NaBH4 is a mild reducing agent that selectively reduces aldehydes and ketones to alcohols. It does not reduce aromatic rings, esters, or carbon-carbon double bonds under normal conditions. The aldehyde group (CHO) in the side chain is reduced to a primary alcohol (CH2OH). So compound A is: ArO-CH(CH2OH)-CH2OH, i.e., the aromatic ether where the side chain is now -CH(CH2OH)-CH2OH (a 1,2,3-triol fragment minus the aryl ether). Step 3 - Reaction with O3 (B): Ozonolysis cleaves carbon-carbon double bonds. The benzene ring contains three C=C double bonds (as a conjugated system). Ozonolysis of the aromatic ring (aromatic ozonolysis) cleaves the ring. The substituted benzene ring with CH3O at position 1 (via the side chain oxygen), CH3O at position 2, and CH3 at position 4 undergoes ozonolysis to give the mozonide/ozonide intermediate B. Step 4 - Reaction with H2O (C): Hydrolysis of the ozonide (B) with water (reductive workup implied or oxidative) gives the carbonyl/alcohol cleavage products. The key fragment from the side chain -O-CH(CH2OH)-CH2OH after cleavage of the Ar-O bond during ring ozonolysis liberates HO-CH(CH2OH)-CH2OH = HOCH2-CHOH-CH2OH = glycerol (1,2,3-propanetriol). Step 5 - Why option (b) is correct: The side chain after NaBH4 reduction becomes -O-CH(CH2OH)-CH2OH. When the aromatic ring is ozonolyzed and then hydrolyzed, the aryl-oxygen bond is cleaved, releasing the fragment as HO-CH(CH2OH)-CH2OH which is glycerol: HOCH2-CHOH-CH2OH, matching option (b) CH2OH-CHOH-CH2OH. Step 6 - Why other options fail: (a) is a pyruvate ester with no relation to the side chain fragment. (c) glyceraldehyde would result if only one CH2OH were present (no reduction of CHO prior to ozonolysis). (d) malondialdehyde would require two aldehyde groups without any reduction step producing alcohols. Therefore, the correct answer is B.