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### Course: Organic chemistry>Unit 9

Lesson 3: Aromatic stability

# Aromatic stability IV

Aromaticity of the cyclopentadienyl anion and cycloheptatrienyl cation. Created by Jay.

## Want to join the conversation?

• At about , how do you know there is 3 bonding and 4 anti bonding instead of the other way around?
• When constructing molecular orbitals using a Frost circle, one of the vertices is always drawn in the centre pointing down. In an aromatic compound there will only be one arrangement where all of the p orbitals are in phase. This arrangement has maximal orbital overlap so will be the lowest energy bonding orbital (in the pi system). This single lowest-energy MO is represented by the vertex at the bottom of the Frost circle (it will always stand alone without any degenerate orbitals). Having a vertex at the bottom of the Frost circle (as opposed to a side of the polygon at the bottom) means that there are 3 bonding and 4 anti-bonding MOs.
• Is there ever a situation where you calculate the pi electrons at the start and when you get to the frost circle, they don't create a 'full outer shell'? Or is the Frost circle just a way to demonstrate BMO etc, but in reality you don't need it to work out if a molecule is AROMATIC?
• The Frost circle is just an easy visual way to work out if a molecule is aromatic without having to do the mathematical calculations.
• What is pKa and what does it numeric value suggest?
(1 vote)
• Ka is the value of the equilibrium constant for the ionization of an acid in water. The greater the Ka, the more of the acid ionizes, and the stronger the acid. Most common weak acids have Ka values ranging from 10^-2 to 10^-14. A Ka value of 10^-2 means that an acid is stronger than one with a Ka of 10^-12.
pKa is the negative logarithm (to the base 10) of Ka. Thus the pKa values of these acids range from 2 to 14. The lower the value of pKa, the stronger the acid. So an acid with pKa = 2 is stronger than an acid with pKa = 14.
• () A pKa value of 16 indicates a weak acid, doesn't it?
• in the very first example... arent those hrydrogens on the sp2 double bonded carbons acidic... i thought that becoz sp2 orbitals are more acidic than sp3 orbitals??
• Quite correct! Normal sp3 bonded H atoms have a pka ~ 51, while sp2 bonded H atoms have pKa ~44, and acetylenic (sp bonded) H atoms have pKa ~25. But in cyclopentadiene the sp3 H atoms have pKa ~16. They are 10^9 times more acidic than even an alkyne H.
The reason is that the cyclopentadienide ion (C5H5-) is exceptionally stable because it is aromatic. So it is much easier to lose the sp3 bonded H (the activation energy is much less), and the H is therefore more acidic than you might expect.
• I've read in my textbook that Hükel's rule doesn't work with polycyclic aromatic compounds does anyone know why?
(1 vote)
• For the last figure, at , why does the does the top carbon have a pi orbital. Isn't it only the electrons involved with the pi bond?
(1 vote)
• The top carbon is positively charged, so is sp2 hybridized. Thus, it has an unhybridized p orbital. This p orbital has the necessary orientation to be involved in resonance structures with the double bonds in the ring. Thus, through resonance, the p_ orbital on the carbocation is part of the _pi system.
• do frost circles only work with cyclic compounds?
(1 vote)
• It's even more limited than that. They work only for planar cyclic conjugated compounds.
• if having a high pKa value indicates that it is a weak acid, doesnt that mean that the molecule is not willing to give up a hydrogen very easily to reach this apparently very stable conformation? so why is it constantly referred to as "so acidic" in the video?
(1 vote)
• It is very acidic relative to other hydrocarbons. When Jay first talks about this he says "extremely acidic for a hydrocarbon".

Typical pKa values:
alkanes pKa ~50
alkenes are pKa ~45.

Various sources list the pKa of cyclopentane as being ~45 to >60 and I can't find anything for cyclopentene.

So the acidity of cyclopentadiene about 10^29 times greater than for a typical alkene!