Organic II - Reaction of Aromatic Compounds

In summary, the conversation discusses the use of Cl as a leaving group and drawing resonance structures to understand the stability of the intermediate formed. The presence of a methyl group on the bottom ring affects the reaction and leads to the observed product. The use of propylmagnesium bromide as a Grignard reagent is also mentioned, but its exact role is not fully understood.
  • #1
StefanoFIU
3
0

Homework Statement



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The Attempt at a Solution


I want to recognize the Cl as a good leaving group. After leaving, I get the ionized form of the chloroazirine reactant. Then, I want to draw some resonance structures to move the charge around the ring.
sadasd-1.jpg



I do not understand the role of propylmagnesium bromide (Grignard reagent?) or what to do next. Also I do not see the the reason why the "not observed" is not observed. Am I correct up until this point? I appreciate any help, preferably any tips that could lead me to the answer, thank you!
 
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  • #2
Now I'm thinking that the methyl on the bottom ring exerts an effect on the entire compound forcing the propyl chain to attach to the bottom carbon of the chloroazirine (demonstrated in the major product).
 
  • #3
The compound is asymmetrical. The methyl group is an electron-releasing group which stabilizes the intermediate formed, in this case the ionized azirine (formed upon removal of the Cl leaving group). Thus, the 3rd resonance structure is more stable than the first and second structures. Therefore, we should not expect to form the "not observed" product.

If the above is true, then my only problem left is to understand why propylmagnesium bromide was chosen for this reaction.
 

Related to Organic II - Reaction of Aromatic Compounds

1. What is the difference between aromatic and aliphatic compounds?

Aromatic compounds contain a benzene ring and have a delocalized pi electron system, making them more stable and less reactive. Aliphatic compounds do not have a benzene ring and have more localized pi bonds, making them more reactive.

2. How do you predict the products of electrophilic aromatic substitution reactions?

The products of electrophilic aromatic substitution reactions are determined by the electrophile (electron-seeking) attacking the aromatic ring and replacing a hydrogen atom. The final product will have a new substituent attached to the ring.

3. What is the purpose of using a Lewis acid catalyst in Friedel-Crafts reactions?

A Lewis acid catalyst, such as aluminum chloride, helps facilitate the reaction by forming a complex with the electrophile and making it more reactive towards the aromatic ring. It also helps regenerate the catalyst for future reactions.

4. How do you determine the orientation of products in electrophilic aromatic substitution reactions?

The orientation of products is determined by the directing effects of substituents on the aromatic ring. Electron-donating groups direct the incoming electrophile to the ortho and para positions, while electron-withdrawing groups direct it to the meta position.

5. How does the acidity of the benzene ring affect the reactivity of aromatic compounds?

The acidity of the benzene ring, influenced by substituents, can affect the reactivity of aromatic compounds. A more acidic ring will make it easier for an electrophile to attack and undergo substitution. This can be seen in reactions such as nitration and sulfonation.

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