See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Step 1 - Identify the starting material: The structure shown is 1-(methylene)cyclobutane with a CH3 group and an H (on wedge) at the carbon bearing the exocyclic double bond. More precisely, the molecule is a cyclobutane ring where one carbon bears: an H (wedge, meaning it is coming out of the plane), a CH3 group, and an exocyclic =CH2 group. This makes that ring carbon a stereocenter in the product after hydrogenation. Step 2 - Hydrogenation reaction: Addition of H2 over Pd catalyst performs catalytic hydrogenation of the exocyclic C=CH2 double bond. The =CH2 becomes -CH3, so the product is 1,1-dimethylcyclobutane... wait, let me reconsider the structure. The ring carbon has H (wedge), CH3, and =CH2. After hydrogenation the =CH2 becomes -CH3. The product carbon now has: H, CH3, CH3, and two ring carbons - that would be 1,1-dimethylcyclobutane with no stereocenters, which cannot give stereoisomers. Step 3 - Reinterpretation: The H shown on wedge is on C1 of cyclobutane, and C1 also bears CH3, while the =CH2 is at C1. Alternatively, the molecule may be (1-methylcyclobutyl)methylene - i.e., the ring carbon C1 has H (wedge, existing stereocenter), CH3, a ring bond, and the =CH2. Upon hydrogenation of the double bond, the =CH2 becomes CH3, generating a NEW stereocenter at the exo carbon. Now there are TWO stereocenters: C1 of the ring (with existing H wedge and CH3) and the newly formed CH3-bearing carbon. Since H2/Pd can deliver H2 from either face of the double bond to the new carbon, two diastereomers are formed (the existing stereocenter at C1 is fixed by the wedge H, and the new center can be R or S), giving a pair of diastereomers rather than enantiomers because the two stereocenters are not mirror images of each other in a symmetric sense. Step 4 - Why not racemic mixture: A racemic mixture requires equal amounts of enantiomers (non-superimposable mirror images). Here, because one stereocenter is already fixed (the cyclobutane C1 with wedge H), the two products formed by adding H2 to either face of the exo double bond are diastereomers, not enantiomers. Step 5 - Why not meso: Meso requires identical substituents and internal plane of symmetry canceling chirality. That condition is not met here. Step 6 - Why not constitutional isomers: Constitutional isomers differ in connectivity. Both products have the same connectivity, differing only in spatial arrangement. Therefore, the correct answer is B.