Organic II Help!

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ImaJerseyGirl

WesternU CVM c/o 2014
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Hi everyone,

I would greatly appreciate some insight to two orgo problems. Both are from my online hw which we only get 3 chances to submit. One answer I submit which I believe was right, was wrong and I do not understand why.
The second problem, I don't know what is going on!

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The first one, for some reason, isn't working for me....so I can only offer assistance with the second.

I want to start off by saying I LOVED orgo...got A's without trying. So feel free to PM me for future help.

The second one (with the benzene ring and other chlorinated molecule) is a basic chlorination. Using AlCl3, you essential attach the non-chlorinated molecule to the site of the Cl in the other. In this case, the benzene is attached to the carbon where the Cl was. I can take you through the mechanism if needed, but you pretty much form an AlCl4 hybrid which detaches and allows the benzene to attack the site.

Let me know if this helps at all....
 
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yeah, first one isn't working for me either. Try posting it again. But yeah, the second one is pretty much what kayakman said. You form a transient molecule with the AlCl3 salt and the substituted cyclohexane ring at the chlorine site. The pi bonds of the benzene ring then act as a nucleophile displacing aluminum tetrachloride (-1), and you have the product they show. I wouldn't say it's a chlorination though, because you are not adding Cl2.
 
Twelvetigers- Yeah sometimes this stuff is kinda Ew lol. I like chemistry alot (and orgo-no math!), I just dont like getting screwed on the exams when there are 1200 ppl in the course...

Kayakman- Hello fellow Jersian! Where are you from in the Great Garden State?
My understanding is this: The Cl attatches to the AlCl3. The C takes a partial +, the now -AlCl4 breaks off. Along comes the benzene and one of the pi bonds attack the carbocation (where Cl was). Now its sp3, the H breaks off and the product is 1, what was drawn, 2, HCl, and 3, AlCl3.
I entered the the product online and it said the answer was wrong. The only thing I could think of was maybe rearrangement? Or now that C is sp2 since the H broke off?
Friedel-Crafts is ruining my chemistry enjoyment!:laugh:

Thanks for your help.
This is the other question: What is the product? I don't know where the electrons are going!
 
number 1 isn't that hard. The AlCl3s chlorine with the positive charge will attract the electrons from the the ethly group. This forms an ethyl group with a positive charge. The conjugated rings on the benzyl group will act as a nucleophile and attack the positive charge on the ethyl group, forming the compound I have attached. Hope this helps.
 
For those interested in the answers to those two lovely questions, here's what finally earned points:

The Friedel Crafts has the benzene attached to the carbon with the methyl group, no charges.

The second one that has the t-butyl as a para director just needed me to draw a "C+" in stead of just "+" on the carbon, my bad. So I had the right answer, just not in computer terms. Go figure.
+pissed+
But thank you!
 
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