Anilinium is the conjugate acid of Aniline (1) and cyclohexylammonium is the conjugate acid of cyclohexyl amine (II). Which would you pick to be the stronger acid and why? NH2 NH2 I II A) Anilinium is the stronger acid because aniline is the weaker base B) Cyclohexylammonium is the stronger acid because cyclohexyl amine is the weaker base C) Anilinium is the stronger acid because cyclohexyl amine is the weaker base D) Cyclohexylammonium is the stronger acid because aniline is the stronger base 17. Which of the following pairs of structures represents a pair of resonance structures? + & + + & + OH HO N N + & + B. A. C. D. All of them do
The Correct Answer and Explanation is:
For question 16, the correct answer is A) Anilinium is the stronger acid because aniline is the weaker base.
To determine which conjugate acid is stronger, we consider the relative basicity of their conjugate bases. Aniline is an aromatic amine where the nitrogen lone pair is partially delocalized into the benzene ring through resonance. This delocalization reduces the availability of the lone pair for protonation, making aniline a weaker base. On the other hand, cyclohexylamine lacks such resonance stabilization, so its nitrogen lone pair is more available to accept a proton, making it a stronger base. A weaker base corresponds to a stronger conjugate acid, which is why anilinium is more acidic than cyclohexylammonium.
The presence of the aromatic ring in aniline draws electron density away from the nitrogen, further stabilizing the conjugate acid (anilinium) once it donates a proton. In contrast, cyclohexylammonium lacks this stabilization. Thus, the equilibrium of proton transfer favors deprotonation of anilinium more readily, confirming its stronger acidity.
For question 17, the correct answer is D) All of them do.
Resonance structures are alternate ways of drawing a molecule where the arrangement of electrons differs, but the placement of atoms remains the same. In Pair A, the shifting positions of the double bonds within the aromatic system reflect classic resonance within a benzene-like ring. Pair B involves redistribution of charges and electron pairs involving a hydroxyl group and a nitrogen-containing ring system, which fits the criteria for resonance. Pair C shows alternate locations for a charge and double bond within a ring, which again reflects electron delocalization without altering the atomic connectivity. Since all three pairs exhibit electron redistribution with consistent atomic frameworks, they all qualify as resonance structures.
