See image — Isomerism and Stereochemistry Chemistry Question

💡 Solution & Explanation

Concept: In cumulenes with an even number of cumulated double bonds (like butatriene, C=C=C=C with 3 double bonds), the two terminal CH groups lie in perpendicular planes. Geometrical (cis/trans) isomerism arises when the terminal carbons each bear two different substituents, and the relationship between the end groups determines whether the compound is a distinct stereoisomer. Key point about the arrows l1 and l2: l1 in structure (I) spans the entire length of the cumulene chain (all four carbons, three double bonds), while l2 in structure (II) spans only part of the chain (appears to cover fewer bonds/carbons). The arrows represent the conjugated pi-system delocalization length or the relevant bond length being compared. Step 1: In a 4-carbon cumulene (butatriene system, 3 cumulated double bonds), the molecule has an even number of double bonds (3 is odd, meaning the terminal planes are perpendicular). Wait - with 3 cumulated double bonds (odd number), the end groups are in perpendicular planes, giving axial chirality or atropisomerism, not classical geometric isomerism in the alkene sense. However, with an even number of double bonds the end planes are parallel and cis/trans isomerism is possible. Step 2: The key reasoning for why I and II are geometric isomers relates to the effective conjugation length. l1 represents the full delocalized length in isomer I (larger, spanning all bonds), while l2 in isomer II is shorter because the arrangement of substituents interrupts or shortens the effective conjugation. Therefore l1 > l2. Step 3: The fact that l1 > l2 is the structural/electronic basis that distinguishes the two geometric isomers - the different spatial arrangement of substituents (CH3/H vs H/CH3 at the terminal carbons) leads to different effective bond lengths or delocalization extents, confirming they are indeed geometrical isomers with measurably different l values, specifically l1 > l2. Why other options fail: - (a) l1 = l2: If the lengths were equal, there would be no distinction between the two structures as geometric isomers based on this criterion. - (c) l2 > l1: The arrow in (I) clearly spans a greater portion of the chain than in (II). - (d) Cannot be compared: They are both bond lengths/delocalization lengths in analogous systems and can be compared. Therefore, the correct answer is B.