See image — GOC and Organic Chemistry Basics Chemistry Question

💡 Solution & Explanation

Step 1 – Identify the intermolecular forces at play. Boiling point depends on the strength of intermolecular attractions. Stronger attractions → higher boiling point. For non-polar or weakly polar molecules, London dispersion forces (LDF) dominate. Polar molecules may also have dipole–dipole interactions. Step 2 – Analyze compound B (hydrocarbon alkene, b.p. = 67°C). Compound B is a simple hydrocarbon (2-methyl-2-pentene). It has only C–H and C=C bonds. Its electrons are relatively polarizable, and molecules can stack/interact via LDF. There are no strongly electronegative atoms at the periphery, so molecules can approach each other closely, maximizing dispersion interactions. Step 3 – Analyze compound A (perfluorinated alkene, b.p. = 58°C). Compound A is heavily fluorinated. Fluorine is the most electronegative element, and in C–F bonds the electron density is pulled strongly toward fluorine. This means the outer surface of molecule A is covered with partially negative fluorine atoms (lone pairs, negative partial charges). Although C–F bonds are very polar individually, the overall molecule may be relatively symmetric (reducing net dipole), but more importantly, the peripheral F atoms are all electron-rich. Step 4 – Explain why F atoms reduce intermolecular attraction. When molecules of A approach each other in the liquid phase, the negatively charged fluorine atoms on one molecule repel the negatively charged fluorine atoms on the neighboring molecule. This inter-molecular repulsion between the δ– F atoms prevents molecules from getting as close to each other as hydrocarbon molecules can. Reduced proximity means weaker London dispersion forces. Additionally, fluorine's electrons are tightly held and less polarizable than hydrogen's electrons (C–H), so LDF in fluorocarbons are inherently weaker per unit surface area. Step 5 – Compare molecular weights (check if MW favors A having higher b.p.). Despite A having a much higher molecular weight than B (which would normally increase LDF and raise b.p.), the repulsive effect of the peripheral fluorines and their low polarizability more than compensates, resulting in A having a lower boiling point. Step 6 – Conclusion. The highly electronegative fluorine atoms surround the periphery of molecule A. In the liquid state, these F atoms carry partial negative charges and repel each other between adjacent molecules, weakening intermolecular attractions and lowering the boiling point relative to B. Therefore, the correct answer is In A, highly electronegative F-atoms are present at the periphery. In liquid term these F-atoms will repel each other due to partial negative charge and thus A will have lower b.pt..