See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Concept: Thermal decarboxylation of unsaturated carboxylic acids. When a beta,gamma-unsaturated carboxylic acid is heated, it undergoes decarboxylation via a cyclic (pericyclic) six-membered transition state — specifically a [1,5]-sigmatropic or retro-ene type mechanism — losing CO2 and producing an alkene. Step 1: Identify the substrate. The compound is CH3-CH=CH-CH2-CO2H. Numbering from the carboxyl carbon: C1=COOH, C2=CH2, C3=CH (double bond), C4=CH, C5=CH3. The double bond is between C3 and C4 (the gamma and delta positions relative to COOH), making this a beta,gamma-unsaturated acid (the double bond is between the beta carbon C3 and gamma carbon C4). Step 2: Mechanism of thermal decarboxylation of beta,gamma-unsaturated acids. These acids undergo a concerted retro-ene (six-membered cyclic transition state) decarboxylation upon heating. In the transition state, the carboxylic OH hydrogen is transferred to the gamma carbon while CO2 is lost, and the double bond migrates. The six-membered cyclic TS involves: O-H...C(gamma), with simultaneous C(alpha)-C(beta) bond formation (new double bond) and loss of CO2. Step 3: Apply to CH3-CH=CH-CH2-CO2H. The beta,gamma double bond is CH=CH (between C3-C4). In the retro-ene TS: the OH of COOH interacts with the gamma carbon (C4, the CH next to CH3). CO2 is expelled, H migrates from OH to C4 (the gamma carbon), and a new double bond forms between C2 (alpha) and C3 (beta). This gives CH3-CH2-CH=CH2 (1-butene, but-1-ene). Wait — let me recount: CH3(C5)-CH(C4)=CH(C3)-CH2(C2)-CO2H(C1). The double bond between C3-C4 is beta(C2)-gamma(C3) if we count beta as one carbon from carbonyl and gamma as two. Actually: alpha = C2 (CH2), beta = C3 (=CH-), gamma = C4 (CH=), delta = C5 (CH3). So the double bond is at beta-gamma position. In the retro-ene TS: H from OH goes to gamma carbon (C4), new sigma bond lost is C1-C2, CO2 leaves, new pi bond forms between C2 and C3... Actually the product alkene: after losing CO2, the fragment from C2 onward rearranges. The product is CH3-CH2-CH=CH2 (1-butene), which is option (c). Step 4: Why is (c) the major product? The concerted retro-ene mechanism of beta,gamma-unsaturated acids gives the terminal alkene (less substituted) as the kinetically controlled product through this specific cyclic TS. Options (a) and (b) would require different mechanisms or more substituted intermediates. Option (d) is only a 3-carbon alkene, which is incorrect. Option (a) CH3-CH=CH-CH3 (2-butene) would require allylic rearrangement. The direct retro-ene product is 1-butene (option c). Therefore, the correct answer is C.