Subjects

Sections | |||
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Aromaticity | 8 mins | 0 completed | Learn |

Huckel's Rule | 10 mins | 0 completed | Learn Summary |

Pi Electrons | 5 mins | 0 completed | Learn |

Aromatic Hydrocarbons | 15 mins | 0 completed | Learn |

Annulene | 17 mins | 0 completed | Learn |

Aromatic Heterocycles | 20 mins | 0 completed | Learn Summary |

Frost Circle | 15 mins | 0 completed | Learn Summary |

Naming Benzene Rings | 13 mins | 0 completed | Learn Summary |

Acidity of Aromatic Hydrocarbons | 10 mins | 0 completed | Learn Summary |

Basicity of Aromatic Heterocycles | 11 mins | 0 completed | Learn Summary |

Ionization of Aromatics | 19 mins | 0 completed | Learn |

Additional Practice |
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Physical Properties of Arenes |

Resonance Model of Benzene |

Aromatic Heterocycle Nomenclature |

Cumulative Aromaticity Problems |

Polycyclic Aromatic Hydrocarbon Nomenclature |

Concept #1: Counting pi electrons

**Transcript**

In this video, we’re going to become experts on counting pi electrons. Technically, a pi electron is any electron found within an unhybridized p-orbital. That's the technical definition.

However, for the purposes of this video, I think it’s just easier if we memorize a few different types of electrons that are pi electrons. That’s going to cover our bases. We can learn that double bonds, radicals, and cations all contribute different amounts of electrons to pi electrons. Double bonds and anions contribute two pi electrons each. Radicals contribute one pi electron. Cations because of the fact that they’re empty orbitals, they contribute zero electrons.

Simply the job of counting of pi electrons is as easy as counting with your fingers, how many electrons you see by counting up with this method.

I’m going to have you guys do six compounds and figure out the pi electrons for those six. Go ahead and start off with the first one here and use the rule that I gave you to count up all the pi electrons you see and then we'll go ahead and do the answer.

Example #1: Count the pi electrons present

**Transcript**

The answer for the first molecule is 6 because I've got 2 from this double bond, I've got 2 from this bond, I've also got another 2 from this double bond and I've got zero from this cation. You add it all together and you get 6 Pi electrons. Easy, move on to the next one.

Example #2: Count the pi electrons present

**Transcript**

The answer for this next one was 10 Pi electrons because I've got 2, 2, 2, 2, 2. That equals 10 Pi electrons. Too easy, let's move on to the next one.

Example #3: Count the pi electrons present

**Transcript**

The answer for this next compound was 4 pi electrons because I had 2 electrons coming from my double bond, and then I had another 2 coming from my anion which we said contributed 2 pi electrons as well, that would give us 4 pi electrons total.

Aright, now in order to the answer the next three questions we're going to have to get a handle on this tricky notation. So if you see hydrocarbon ring with just a charge in the middle, that means you've got a lazy organic chemist on your hands but hey this is an acceptable notation and you do need to learn what it means. Basically if you just see a charge in the middle that means that we're going to assume, so I'm just going to put here by definition this means that we're going to assume alternating double bonds and then the space left over or the carbon left over, that atom left over would possess that charge so that means that this molecule, that five of a is actually a short cut or a short hand for a five carbon diene with a positive charge on one carbon.

So that being said go ahead and try to figure out how many pi electrons that has and then use that information to then draw the correct structures of the following two so that you can get the right answers for those as well. You got this.

Example #4: Count the pi electrons present

**Transcript**

Alright, so the correct answer for this one was 4 now that we know how to draw it because you had 2, 2 and zero giving us a total of 4 pi electrons. Go ahead and try the next one.

Example #5: Count the pi electrons present

**Transcript**

So the correct structure of this molecule was a triangle with simply one double bond and a negative charge in the space afterwards because there was only enough space for one double bond total. This means that this molecule would have a total of 4 pi electrons. Let's move on to the last question.

Example #6: Count the pi electrons present

**Transcript**

The answer for this last one was 8 pi electrons. How did we get the answer? Because the negative charge in the middle meant that we had to assume double bond, double bond, double bond with a negative charge here I can just cross that out and we would get 8 pi electrons from the 3 double bonds and the anion which would contribute electrons 7 and 8 so this would be 8 pi electrons total. Now I know that you're not really completely sure how important this pi electron thing is but it's a huge deal we're talking about in aromaticity so now that you know how to count them, let's go ahead and figure out why they're going to be so important for aromaticity.

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Concept #1: Counting pi electrons

Example #1: Count the pi electrons present

Example #2: Count the pi electrons present

Example #3: Count the pi electrons present

Example #4: Count the pi electrons present

Example #5: Count the pi electrons present

Example #6: Count the pi electrons present

How many lone pairs are found in the structure of vitamin C?

Ramelteon is a hypnotic agent used in the treatment of insomnia: (e) How many lone pairs are present in this structure?

Cycloserine is an antibiotic isolated from the microbe Streptomyces orchidaceous. It is used in conjunction with other drugs for the treatment of tuberculosis. (e) How many lone pairs are present in this structure?

Cycloserine is an antibiotic isolated from the microbe Streptomyces orchidaceous. It is used in conjunction with other drugs for the treatment of tuberculosis. (f) Identify each lone pair as localized or delocalized.

Determine the number of pi electrons in each molecule below

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