See image — Aldehydes Ketones and Carboxylic Acids Chemistry Question

💡 Solution & Explanation

Step 1: Starting material is acetone (CH3-CO-CH3). Treatment with NaNH2 (a strong base) deprotonates the alpha carbon to form the enolate anion P: CH3-CO-CH2(-) Na(+). Step 2: P reacts with HC≡CH (acetylene). NaNH2 also deprotonates acetylene to give HC≡C(-) Na(+), but here the sequence shows NaNH2 first forms P from acetone, then HC≡CH reacts. Actually, NaNH2 deprotonates acetylene (pKa ~25) more readily than acetone's alpha carbon. Re-interpreting: NaNH2 converts HC≡CH to HC≡C(-) which then acts as a nucleophile. But the arrow shows acetone + NaNH2 → P, then P + HC≡CH → Q. More likely: NaNH2 deprotonates acetone alpha position to give enolate P, but the classical reaction here is: acetone reacts with NaNH2 to give the acetylide from acetylene. Let me reconsider the standard sequence. Standard interpretation: The starting ketone is acetone. NaNH2 deprotonates acetylene (added implicitly or the solvent), but the diagram shows acetone → P with NaNH2, then P + HC≡CH → Q. Most likely P is the sodium enolate of acetone, and Q forms by addition of acetylide to acetone. Actually the standard reaction is: HC≡CH + NaNH2 → HC≡C(-) Na(+) (sodium acetylide), then this adds to acetone (nucleophilic addition) to give Q = HC≡C-C(CH3)2-O(-) Na(+) (alkoxide). Step 3: H3O(+) workup of Q gives R = HC≡C-C(CH3)2-OH (2-methylbut-3-yn-2-ol), a propargylic tertiary alcohol. Step 4: R + 1H2/Pd (one equivalent of H2 with Lindlar's catalyst or similar) reduces the triple bond to a cis double bond, giving S = H2C=CH-C(CH3)2-OH (2-methylbut-3-en-2-ol), a tertiary allylic alcohol. Step 5: S treated with Al2O3 at elevated temperature (D = heat) causes dehydration (E1 or E2 elimination of water). The tertiary alcohol S = (CH3)2C(OH)-CH=CH2 undergoes dehydration. Al2O3/heat is a classic dehydration condition. The OH is on a tertiary carbon adjacent to a vinyl group. Dehydration removes OH and an adjacent H. The available beta-H positions: the methyl groups give CH2=C(CH3)-CH=CH2 (isoprene, 2-methylbuta-1,3-diene) via loss of H from one methyl group, giving a conjugated diene. This is option (d). Why other options fail: - (a) contains a carbonyl group, but no oxidation step is present after the alcohol stage. - (b) is compound S itself (the alcohol before dehydration), not T. - (c) is 2-methylbut-2-ene, which would require non-conjugated dehydration and does not match the preferred product from Al2O3 dehydration of a beta-hydroxy alkene (which favors the conjugated diene). - (d) isoprene (2-methylbuta-1,3-diene) is the product of dehydration of 2-methylbut-3-en-2-ol over Al2O3, consistent with forming the thermodynamically stable conjugated diene system. Therefore, the correct answer is D.