The calculated heat of formation for \((E)-\mathrm{CH}_{3} \mathrm{CH}=\mathrm{CHCH}_{2} \mathrm{CH}_{2}^{+}\) is \(227 \mathrm{kcal} / \mathrm{mol}(949 \mathrm{~kJ} / \mathrm{mol})\) and for \((E)-\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}=\mathrm{CHCH}_{2}^{+}\)is \(221 \mathrm{kcal} / \mathrm{mol}(926 \mathrm{~kJ} / \mathrm{mol})\). Draw the two cations showing the three-dimensional perspective and explain why one cation is more stable than the other.

The heat of formation of a molecule is the energy change that occurs when the molecule is formed from its constituent atoms in their standard states. For cations like the ones given in the exercise, the heat of formation helps us determine their stability. The lower the heat of formation, the more stable the molecule is.
The exercise compares two cations: \((E)-\mathrm{CH}_{3} \mathrm{CH} = \mathrm{CHCH}_{2} \mathrm{CH}_{2}^{+}\) with a heat of formation of 227 kcal/mol and \((E)-\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH} = \mathrm{CHCH}_{2}^{+}\) with a heat of formation of 221 kcal/mol.
The cation with the lower heat of formation, 221 kcal/mol, is more stable. Thus, \((E)-\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH} = \mathrm{CHCH}_{2}^{+}\) is more stable than \((E)-\mathrm{CH}_{3} \mathrm{CH} = \mathrm{CHCH}_{2} \mathrm{CH}_{2}^{+}\).

The three-dimensional structure of a molecule gives us an idea of how atoms are spatially arranged, which affects the stability and reactivity of the molecule.
In the exercise, we have two cations:

• \((E)-\mathrm{CH}_{3} \mathrm{CH} = \mathrm{CHCH}_{2} \mathrm{CH}_{2}^{+}\)
• \((E)-\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH} = \mathrm{CHCH}_{2}^{+}\)

The difference between these two cations lies in the position of the double bond and the positive charge. Drawing these structures in 3D perspective helps visualize how atoms' placement can influence the molecule's overall stability. For instance, a carbocation at a more substituted position is generally more stable.
This allows us to see that the second cation, with the positive charge closer to a double bond, results in more stabilizing interactions.

The substitution effect refers to how the substituents (such as hydrogen or alkyl groups) surrounding a functional group (like a carbocation) influence the molecule's properties. The stability of carbocations is drastically impacted by substitution levels around the positively charged carbon.
In our specific cations:

• \((E)-\mathrm{CH}_{3} \mathrm{CH} = \mathrm{CHCH}_{2} \mathrm{CH}_{2}^{+}\)
• \((E)-\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH} = \mathrm{CHCH}_{2}^{+}\)

\((E)-\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH} = \mathrm{CHCH}_{2}^{+}\) is more substituted near the carbocation because the positive charge is closer to electron-donating groups (next to the double bond), making it more stable. The higher degree of substitution facilitates increased electron density around the positive charge, reducing its overall energy.

Electron-donating groups (EDGs) are groups of atoms that can donate electron density through inductive or resonance effects, which can stabilize a carbocation.
In the case of carbocations, EDGs stabilize the positive charge by decreasing its electron deficiency. In our examples:

• \((E)-\mathrm{CH}_{3} \mathrm{CH} = \mathrm{CHCH}_{2} \mathrm{CH}_{2}^{+}\)
• \((E)-\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH} = \mathrm{CHCH}_{2}^{+}\)

\((E)-\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH} = \mathrm{CHCH}_{2}^{+}\) is more stabilized due to the electron-donating effect of the nearby double bond, which increases electron density near the carbocation, thereby reducing its energy and increasing stability.
Understanding how EDGs work helps explain why certain carbocations have lower heats of formation, making them more stable.