See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: The acetoacetic ester synthesis involves alkylation at the active methylene group (C-2) of ethyl acetoacetate (CH3COCH2COOC2H5) using base (NaOEt) to form a stabilized enolate, followed by alkylation with an alkyl halide. Step 1: Acetoacetic ester = ethyl acetoacetate: CH3CO-CH2-COOC2H5. The active methylene (C-2, between the two carbonyl groups) is deprotonated by NaOEt to give the enolate anion CH3CO-CH(-)-COOC2H5. Step 2: Alkylation with CH3I introduces a methyl group at C-2, giving CH3CO-CH(CH3)-COOC2H5 (ethyl 2-methylacetoacetate). Step 3: Treatment with NaOEt again deprotonates C-2 of CH3CO-CH(CH3)-COOC2H5. Although C-2 now bears one methyl substituent (and one H), it is still flanked by two carbonyl groups and can be deprotonated to form CH3CO-C(CH3)(-)-COOC2H5. Step 4: Alkylation with CH3CH2CH2Br (n-propyl bromide) introduces the n-propyl group at C-2, giving CH3CO-C(CH3)(CH2CH2CH3)-COOC2H5. This is a dialkylated acetoacetic ester with a quaternary C-2 bearing: acetyl (CH3CO), methyl (CH3), n-propyl (CH2CH2CH3), and ethyl ester (COOC2H5). This product corresponds exactly to option (a): H3C-C(=O)-C(CH3)(CH2CH2CH3)-COOC2H5. Why other options fail: - Option (b) lacks the acetyl (C=O) group on the carbon skeleton, suggesting a decarboxylation/reduction that was not performed. - Option (c) shows only one alkyl substituent (n-propyl) and retains an H at C-2, meaning the first methylation did not occur or the product was hydrolyzed/decarboxylated prematurely. - Option (d) shows two methyl groups and no n-propyl, indicating CH3I was used twice instead of once CH3I and once n-PrBr. Therefore, the correct answer is A.