See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Step 1: Determine x - the number of planes of symmetry for p-bis(chloromethyl)benzene (1,4-bis(chloromethyl)benzene). The molecule is a benzene ring with -CH2Cl groups at the 1 and 4 (para) positions. To find planes of symmetry, we consider all mirror planes that map the molecule onto itself. Plane 1: The plane of the benzene ring itself (the molecular plane). Both CH2Cl groups lie in or are symmetric about this plane. This is a valid plane of symmetry. Plane 2: A plane perpendicular to the benzene ring that passes through C1, C4, and the two CH2Cl groups (the axis connecting the two substituents). This plane bisects the ring symmetrically and maps the left half onto the right half. Plane 3: A plane perpendicular to the benzene ring that passes through C2-C5 (the axis perpendicular to the C1-C4 axis, bisecting the molecule between the two CH2Cl groups). This maps the two halves of the molecule onto each other, swapping the two CH2Cl groups which are identical. Thus x = 3 planes of symmetry. Step 2: Determine y - the number of meso isomers of 1,2-dichlorocyclopentane. 1,2-dichlorocyclopentane has two stereocenters (C1 and C2). The possible stereoisomers are: cis-1,2-dichlorocyclopentane and trans-1,2-dichlorocyclopentane. The cis isomer (with both Cl on the same face) has an internal plane of symmetry bisecting the C1-C2 bond and passing through C3, C4 (approximately), making it a meso compound. The trans isomer exists as a pair of enantiomers (R,S configurations). Therefore, there is 1 meso isomer of 1,2-dichlorocyclopentane, so y = 1. Step 3: Calculate x + y = 3 + 1 = 4. Therefore, the correct answer is 4.