Rank the following compounds from lowest boiling point to highest boiling point.

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Question restatement: Rank the following compounds from lowest boiling point to highest boiling point. The answer field indicates the ranking: C < B < A. Answer options: a. B < C < A; b. C < A < B; c. C < B < A; d. A < B < C.

To compare boiling points, consider the key factors that influence BP in organic compounds: intermolecular hydrogen bonding, molecular weight, and molecular structure/branching which affects surface area and dispersion forces.

Option a (B < C < A): If this were the correct ranking, it would imply B has the lowest BP, followed by C, with A having the highest. We must examine whether B truly would have a lower BP than C. If B can form stronger hydrogen bonding or has greater molecular weight than C, this ordering would be inconsistent. In many amide series, a tertiary amide (no N–H) tends to have a lower BP than a secondary or primary amide due to reduced hydrogen bonding, assuming similar molecular weights. If B contains an N–H and C does not, this option would be unlikely because C would be expected to have a lower BP than B, not higher.

Option b (C < A < B): This would place C as the lowest BP, A in the middle, and B as the highest. If C is a tertiary amide lacking N–H, its hydrogen bonding capability is reduced, often giving a lower BP; however, if B is also an amide with fewer hindrances or a larger surface area and higher molecular weight, B could show higher BP. Yet this ordering would require A to have more hydrogen bonding or heavier mass than B, which is not typical if B is the more substituted amide; this makes the sequence less plausible without additional structural details.

Option c (C < B < A): Here C is lowest, B is middle, A is highest. This aligns with a common trend in amide series where the tertiary amide (no N–H) like C has the lowest BP due to absence of hydrogen bonding, B has some hydrogen bonding and moderate mass, and A has the highest BP due to larger molecular weight and increased hydrogen-bonding potential from N–H along with greater dispersion forces from its possibly bulkier substituents. If A is a primary or secondary amide with substantial alkyl substitution, both hydrogen bonding and dispersion forces would contribute to a higher BP than B and C, making this sequence reasonable.

Option d (A < B < C): This would imply A has the lowest BP and C the highest, which contradicts the general understanding that a more hydrogen-bonding-capable amide (with N–H) and/or greater molecular size tends to raise BP. If C is a tertiary amide with no N–H, it would typically have a lower BP, not the highest, so this ordering is inconsistent with expected trends.

Overall reasoning: Among these three amide variations, the presence or absence of N–H hydrogen bonding significantly affects BP. A compound with an N–H (capable of H-bonding) and larger hydrophobic surface area tends to have a higher boiling point due to stronger intermolecular attractions. A compound lacking N–H (tertiary amide) will generally have a lower boiling point because it forms fewer hydrogen bonds and often has slightly reduced polarity, though mass and surface area can offset this to some extent. Between B and C, the one with an N–H (and possibly greater weight or surface area) would typically exhibit a higher BP than the tertiary amide C. Therefore, C (lowest BP) < B (middle) < A (highest BP) is a plausible order.

Note: If any specific details about exact masses or conformations were provided, the ranking could adjust accordingly, but with the common structural trends described, option C < B < A is a reasonable analysis.

Rank the following compounds from lowest boiling point to highest boiling point.

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