See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the substrate: The compound is Ph-CH2-CH(D)-CH3, where the central carbon bears a deuterium (D), a hydrogen (implicitly), a benzyl group (PhCH2-), and a methyl group. The central carbon (C2) already has four different substituents: D, H (wait, actually C2 has D, CH3, PhCH2, and... let us re-examine). C2 bears: D, CH3, CH2Ph, and H — making it a chiral center with four different groups. Step 2 - Identify sites of free radical bromination under Br2/hv: Free radical bromination is selective for the most stable (weakest C-H bond) position. The benzylic position (PhCH2-) is the most activated site for radical bromination due to resonance stabilization of the benzylic radical. The C2 position also has a C-H (and C-D) bond, but benzylic C-H is preferentially abstracted. Step 3 - Bromination at the benzylic carbon (C1, PhCH2-): The benzylic carbon (Ph-CH2-) has two equivalent hydrogens. Abstraction of one H gives a benzylic radical at C1: Ph-CH•-CH(D)-CH3. This radical carbon (C1) becomes a new chiral center upon bromine capture (it will have: Ph, Br, H, and -CH(D)CH3 as substituents). Step 4 - Existing chirality at C2: C2 is already a stereocenter (bearing D, H, CH3, CH2Ph). The radical at C1 is planar (sp2), so Br• can attack from either face with roughly equal probability, generating both R and S configurations at C1. Step 5 - Two stereocenters in the product: The product Ph-CHBr-CH(D)-CH3 has two stereocenters: C1 (Ph, Br, H, -CH(D)CH3) and C2 (D, H, CH3, -CHBrPh). The original C2 stereochemistry is retained (radical reaction does not disturb C2). Since C1 is formed with equal probability of R and S (planar radical), we get two products: (R,C2-config) and (S,C2-config). These two products are non-superimposable, non-mirror-image stereoisomers of each other — they are diastereomers (one specific enantiomer at C2 combined with R at C1, and the same enantiomer at C2 combined with S at C1). They are not a racemic mixture because they are not enantiomers of each other; they differ at only one of two stereocenters. They are not meso because the molecule lacks the requisite internal symmetry. They are not constitutional isomers because they have the same connectivity. Step 6 - Why not racemic mixture: A racemic mixture would require two enantiomers (mirror images). The two products here share the same configuration at C2 but differ at C1, so they are diastereomers, not enantiomers. Step 7 - Why not meso: Meso requires an internal plane of symmetry; the presence of deuterium and the unsymmetrical substitution pattern prevents this. Therefore, the correct answer is A.