See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1: Identify the structure. The compound shown is bicyclo or tricyclic structure consisting of three fused cyclobutane rings (a [3]triangulane-like or linearly fused tricyclobutane system) with a bromine substituent at the central carbon connecting the ring systems. Step 2: Identify stereocenters. The central carbon bearing Br is attached to four different groups: Br, H (implied), and two different cyclobutane-ring portions. We need to assess whether this carbon and any other carbons are stereocenters. Step 3: Analyze the structure more carefully. The molecule appears to be spiro[3.3]heptane or a related tricyclic system with Br at a junction. Looking at the drawing: three cyclobutane rings are linearly fused, with Br on the central carbon of the middle ring. The central carbon with Br has: Br, H, and two ring substituents (left bicyclobutane portion and right cyclobutane portion). These two ring portions are different if the molecule is not symmetric at that point, making the carbon a stereocenter. Step 4: Count stereocenters and determine stereoisomers. The structure has the Br-bearing carbon as a stereocenter. Additionally, the ring junction carbons between the cyclobutane rings may also be stereocenters. With 2 stereocenters and no plane of symmetry making a meso form, we get 2^2 = 4 stereoisomers, but if one pair is meso, we get 3 (2 enantiomers + 1 meso). The optically active ones exclude meso compounds. Step 5: Determine meso possibilities. If the molecule has an internal plane of symmetry in one configuration (meso form), that stereoisomer is optically inactive. With 2 stereocenters: possible stereoisomers are (R,R), (S,S), (R,S), and (S,R). If (R,S) = (S,R) due to symmetry (meso), then total distinct stereoisomers = 3: one meso (optically inactive) + one pair of enantiomers (both optically active). Thus, optically active isomers = 2. Step 6: Why other options fail. Option (a) 0 is wrong because the stereocenter with Br is genuine and non-symmetric configurations exist. Option (b) 1 is wrong because enantiomers come in pairs. Option (d) 3 is wrong because the meso form is not optically active, leaving only 2 optically active isomers. Therefore, the correct answer is C.