See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: Alkylation of 1,3-dicarbonyl compounds under basic conditions. Step 1: Identify the substrate. Cyclohexane-1,3-dione is a 1,3-diketone. The methylene group at C2 (flanked by two carbonyl groups) is highly acidic (pKa ~5-7). KOH deprotonates this position readily to form a stabilized enolate. Step 2: Identify the electrophile. Allyl bromide (CH2=CH-CH2-Br) is a primary alkyl halide that undergoes SN2 reactions efficiently. Step 3: Determine the site of alkylation. In 1,3-diketones treated with base and an alkyl halide, C-alkylation at C2 is thermodynamically and kinetically favored over O-alkylation under these conditions (KOH in protic or mixed solvent at moderate conditions gives predominantly C-alkylation). The enolate carbon (C2) attacks the allyl bromide via SN2. Step 4: Product formation. The allyl group (-CH2-CH=CH2) attaches to C2 of cyclohexane-1,3-dione, giving 2-allylcyclohexane-1,3-dione. Both carbonyl groups remain intact, and no ring unsaturation is introduced. Step 5: Match to options. Option (b) shows cyclohexane-1,3-dione with a -CH2-CH=CH2 group at C2, with both C=O groups retained and no additional ring double bond. This matches the expected C-alkylation product. Why other options fail: - (a) shows O-alkylation product (enol ether with ring double bond), which is minor under KOH conditions. - (c) shows a product with a ring double bond adjacent to the allyl group, which would require elimination - not expected here. - (d) shows alkylation at C5 (the non-acidic methylene), which would not be activated by two carbonyls and is therefore not favored. Therefore, the correct answer is B.