Problem

Hydrocarbons are generally considered to be nonpolar or weakly polar at best, characteri...

Hydrocarbons are generally considered to be nonpolar or weakly polar at best, characterized by dipole moments that are typically only a few tenths of a debye. For comparison, dipole moments for molecules of comparable size with heteroatoms are commonly several debyes. One recognizable exception is azulene, which has a dipole moment of 0.8 debye:

Optimize the geometry of azulene using the HF/6-31G* model and calculate an electrostatic potential map. For reference, perform the same calculations on naphthalene, a nonpolar isomer of azulene. Display the two electrostatic potential maps side by side and on the same (color) scale. According to its electrostatic potential map, is one ring in azulene more negative (relative to naphthalene as a standard) and one ring more positive? If so, which is which? Is this result consistent with the direction of the dipole moment in azulene? Rationalize your result. (Hint: Count the number of π electrons.)

Step-by-Step Solution

Request Professional Solution

Request Solution!

We need at least 10 more requests to produce the solution.

0 / 10 have requested this problem solution

The more requests, the faster the answer.

Request! (Login Required)


All students who have requested the solution will be notified once they are available.
Add your Solution
Textbook Solutions and Answers Search
Solutions For Problems in Chapter 15