See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Concept: Bromine addition to alkenes proceeds via a bromonium ion intermediate, resulting in anti addition of the two bromine atoms across the double bond. This makes the reaction stereospecific. Step 2 - Reaction of fumaric acid (P, trans): Fumaric acid has the two COOH groups trans to each other. Anti addition of Br2 to the trans alkene: the two COOH groups are on opposite faces. Anti addition means the two Br atoms add from opposite faces. Starting from the trans configuration, anti addition gives a product where the two COOH groups end up on the same side (erythro) after considering the Newman/Fischer projection. Specifically, anti addition to fumaric acid (trans) gives the dl-pair (racemic mixture) of the threo dibromosuccinic acid... Let us be careful: For fumaric acid (trans-HOOC-CH=CH-COOH), anti addition of Br2 gives the (2R,3R) and (2S,3S) enantiomers — this is the racemic mixture. In Fischer projection terms, with COOH at top and bottom, the Br atoms end up on opposite sides — this is the threo racemate. Wait, let me reconsider using standard definitions. Erythro: like substituents on same side in Fischer projection; threo: like substituents on opposite sides. Step 3 - For fumaric acid (trans, P): anti addition. Draw Fischer projection with COOH at top and bottom. In fumaric acid the H and COOH are trans on the double bond carbon. Anti addition of Br2: both Br atoms add from opposite faces. The product from fumaric acid via anti addition is the meso compound (2R,3S)-2,3-dibromosuccinic acid. In Fischer projection: COOH top, Br on left at C2, Br on right at C3 (or vice versa) — the two Br atoms are on opposite sides, and the two COOH are at top and bottom. This corresponds to erythro (same substituents on same side in Fischer for the H atoms) — actually the meso form. The meso dibromosuccinic acid is the erythro isomer (Br and H on same side when drawn in Fischer with COOH at ends). So X (from fumaric, anti addition) = erythro (meso) isomer. Step 4 - For maleic acid (cis, Q): anti addition of Br2. Maleic acid has both COOH groups cis. Anti addition to cis alkene gives the (2R,3R) and (2S,3S) pair — the racemic (dl) mixture. In Fischer projection, the two Br atoms appear on the same side — this is the threo (racemic) isomer. So Y (from maleic, anti addition) = threo (racemic) isomer. Step 5 - Evaluate statements: Statement I: reactions are stereospecific — TRUE, because anti addition gives a specific stereochemical outcome depending on the geometry of the starting alkene. Statement II: (X) is erythro and (Y) is threo — TRUE based on the analysis above (X from fumaric = erythro/meso; Y from maleic = threo/racemic). Statement III: (X) is threo and (Y) is erythro — FALSE (opposite of II). Statement IV: each gives a mixture — FALSE, each gives a specific stereoisomer (stereospecific). Step 6 - Therefore statements I and II are correct, corresponding to option (a). Therefore, the correct answer is A.