A common error that students make is to account for stereochemistry on every question, regardless of the context. One such context where stereochemistry is not always applied is resonance. Let's look at the different resonance structures in q.25:

Resonance refers to the delocalization of electrons between atoms within a molecule. Thus, the specific atoms involved in this delocalization matter, not just the overall appearance. Here, the negative charge is distributed across two carbons and a nitrogen.

Now, let's compare the middle resonance structure with Option A:

Here, the middle resonance structure looks identical to Option A. The structures are superimposable via a 180 degree rotation about the vertical axis. But be careful, these structures are not the exact same since the position of their atoms differ. The two carbons that can carry the charge are highlighted in green and blue (see above). The middle resonance structure carries the charge on the green carbon. When we rotate this structure to match Option A, the charge is still carried on the green carbon. In contrast, Option A carries the charge on the blue carbon.

No matter how you rotate the two molecules in space, the charge still resides on different atoms. This holds true for any resonance structures that are superimposable (see below). Although we can rotate one structure to perfectly match the other, the charge still resides on fundamentally different atoms, making them distinct resonance structures.

*Note: As a quick rule of thumb, assume that the molecule is never rotated or moved in questions about resonance structures.

There are two trends that describe the effect of branching on the melting point of hydrocarbons:

These trends seem to contradict. After all, a molecule with extensive branching, such as neopentane, will have a high degree of symmetry. In short, for highly branched molecules, Trend #2 trumps Trend #1. To understand this, we must look at the structure of solids: crystal lattices.

Crystal lattices are highly ordered structures composed of repeating arrangements of molecules. These crystal lattices are held together by various intermolecular forces (e.g. London Dispersion forces). These crystal lattices become stronger when molecules pack more tightly together, which maximizes the strength and number of intermolecular forces. Highly symmetrical molecules can pack more tightly together, and due to their symmetry, allow interactions to form in several directions. Thus, symmetrical molecules form stronger crystal lattices which require more energy (e.g. heat) to disrupt, resulting in higher melting points.

Consider the following image of blocks from the popular video game Tetris. These blocks provide a rough analogy for the crystal lattice arrangement of hydrocarbons.

To determine the most acidic proton in this molecule, we must compare the stability of the conjugate bases formed after deprotonation. The most acidic proton is the one whose conjugate base is the most stable. A useful framework for evaluating acidity is CARDIO (Charge, Atom, Resonance, Dipole induction, and Orbital hybridization). While CARDIO is generally applied in that order, it is important to consider the extent and quality of each stabilizing factor rather than treating the acronym as an absolute ranking.

At first glance, Hc appears to be a strong candidate because deprotonation produces a conjugate base that can be resonance stabilized. The negative charge can be delocalized throughout the ring system, making Hc a benzylic and allylic proton. However, the presence of resonance alone does not guarantee significant stabilization. We must also consider the quality of the resonance structures. In this case, the negative charge is distributed primarily over carbon atoms, and carbon is not particularly electronegative. As a result, the resonance stabilization is relatively weak. Furthermore, deprotonation at Hc disrupts the aromaticity of the ring, which introduces additional instability into the conjugate base.

This image outlines the general acidity of each hybridization of carbon, and we can see a very large increase in acidity for sp hybridized carbons compared to sp2 hybridized carbons.

So although the resonance offered by Hc is slightly stabilizing compared to that of an alkene, it is not very significant especially when compared to the stability offered by the greater s character of the carbon on He.

He is attached to an sp-hybridized carbon of a terminal alkyne. When He is removed, the resulting acetylide ion is stabilized through orbital hybridization. An sp-hybridized carbon contains 50% s-character, compared with 33% for an sp² carbon and 25% for an sp³ carbon. Because electrons in an s orbital are held closer to the nucleus, greater s-character increases the effective electronegativity of the carbon and allows it to stabilize a negative charge more effectively.

So although the resonance offered by Hc is slightly stabilizing compared to that of an alkene, it is not very significant, especially when compared to the stability offered by the greater s-character of the carbon on He.

This question highlights an important limitation of applying CARDIO mechanically. Although resonance (R) often has a greater influence on acidity than orbital hybridization (O), the effectiveness of the stabilization must always be considered. In this molecule, the relatively weak resonance stabilization and loss of aromaticity associated with Hc are outweighed by the substantial stabilization provided by the sp-hybridized carbon of He.

He is the most acidic proton in the molecule. While Hc benefits from some resonance stabilization, the conjugate base is not strongly stabilized because the negative charge remains on carbon atoms and aromaticity is disrupted. The conjugate base formed by deprotonation of He is significantly more stable due to the high s-character of the sp-hybridized carbon, making He the most acidic proton.