See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

This question involves matching molecules to their rotational free energy barriers (barriers to rotation about a specified bond or axis). Concept: The rotational free energy barrier depends on steric and electronic factors. Larger steric bulk around the bond increases the barrier. A negative barrier implies the gauche/orthogonal conformation is more stable than the planar one (no real barrier in the traditional sense; the perpendicular geometry is preferred). Step-by-step reasoning: (a) The molecule is a highly substituted bi-cyclohexadienone (a biaryl-type with two bulky ring systems, each bearing tert-butyl groups, phenyl substituents, and ketone groups). These bulky ortho-like substituents create significant steric strain. The barrier is high but not the highest among the options. This matches (q) 88.3 kJ/mol — a substantial but measurable barrier consistent with atropisomerism in sterically hindered biaryl quinones. (b) The molecule is a biphenyl with CPh2 groups para to the biaryl bond. The para substituents do not contribute to steric hindrance about the central biaryl bond, and the biphenyl itself with no ortho substituents has a relatively low rotational barrier. This matches (r) 21 kJ/mol — a low barrier typical of unhindered or lightly hindered biphenyls. (c) The molecule has a polycyclic core with four peri-chlorine atoms (Cl,Cl/Cl,Cl) flanking the central bond and large =CPh2 exocyclic groups. The four chlorines in peri positions create extreme steric compression around the central bond. This results in a situation where the ground state is already highly strained, and the transition state for rotation may actually be lower in energy than expected — or more accurately, the system is so strained that the rotational energy profile gives a negative free energy barrier (the perpendicular conformation is the ground state and the planar one is the transition state, or there is no classical barrier). This matches (s) Negative barrier. (d) The molecule is stilbene (PhHC=CHPh) with rotation about the C=C double bond indicated. The barrier to rotation about a C=C double bond is very high because it requires breaking pi-bond overlap. The barrier for rotation about the C=C double bond in stilbene is approximately 180 kJ/mol. This matches (p) 180 kJ/mol. Why other options fail: - (a) cannot be 180 kJ/mol because that level of barrier corresponds to C=C pi bond rotation, not a biaryl single bond. - (b) cannot be 88.3 kJ/mol because para substituents don't hinder rotation about the central bond. - (c) cannot be 88.3 kJ/mol because the extreme steric strain from four peri-chlorines leads to a negative barrier scenario. - (d) cannot be 21 kJ/mol because single-bond rotation barriers are ~21 kJ/mol, not double-bond rotation. Therefore, the correct answer is {"a": ["q"], "b": ["r"], "c": ["s"], "d": ["p"]}.