Final answer:
The products of reductive ozonolysis for the given compound should be a ketone and a carboxylic acid, due to cleavage at the double and triple bonds. The closest matching option to the products formed from the minimal decrease in oxidation state for the carbon initially in the alkyne is CH₃C≡CCH₂COOH.
Step-by-step explanation:
The student's question relates to the products of reductive ozonolysis of the given molecule. Reductive ozonolysis is a reaction in which an alkene or alkyne is cleaved by ozone (O₃) and then reduced, usually by a reducing agent such as zinc in acetic acid or dimethyl sulfide.
The given molecule is CH₃C≡CCH₂CH=CHCOOH. Upon ozonolysis, this molecule would cleave at the double and triple bonds to give two products: a carboxylic acid from the oxidation of the aldehyde formed due to the ozonolysis of the double bond and a ketone from the ozonolysis of the triple bond. Thus, the correct product options are CH₃C≡CCHO and CH₃C≡CCH₂COOH.
However, the provided options seem to have a discrepancy because none of the options are a perfect match to the expected products of reductive ozonolysis of the given compound. The products should be a mixture of a ketone and a carboxylic acid, but it's important to note that in reductive ozonolysis, the aldehyde produced by cleavage of the double bond is further reduced to an alcohol.
Considering the given options, the closest answer based on minimal decrease in oxidation state for the carbon atom initially present in the alkyne would be option b) CH₃C≡CCH₂COOH since it's the product of the reduction of an aldehyde from the ozonolysis of the alkene to a carboxylic acid.