Predict the major products formed when benzene reacts (just once)...

Question

Predict the major products formed when benzene reacts (just once) with the following reagents.
(g) 1-chloro-2,2-dimethylpropane + AlCl3

Accepted Answer

All right. Hi everyone. So for this question, it's asking us, what is the major product when told Queen reacts only one time with ter pudo chloride in the presence of catalytic A L C L three. So let's get started by drawing out our told us so told Queen is a benzene ring, looks a little wonky, but it works. And attached to our benzene ring is a methyl group. So let's tell you in. And we are reacting to valuing with turd butto chloride. So we have chlorine and a church bugle group attached. And we are reacting this molecule with ALCL three. So let's get started. Right. Notice how our church bugle chloride is an actual um group. Right? It's an acc you'll molecule, I guess you could say. And so this is an example of what we call frito Crafts ocular ation. Because recall it in Friedel Crafts calculation, a hydrogen of a given aromatic system is going to be replaced by an actual group. In this case we're going to replace that hydrogen with a turbulent group. So the question then becomes, where is it going to add? Right, because we have so many different possibilities. But recall that the fact that we have a substitue in on our aromatic system is going to change um where our group is going to add to our benzene ring in this case. So methyl groups recall just like any other acute group is an example of what we call an activating group. So activating groups are named as such because they donate electrons to the aromatic system because they are less electro negative. So they don't want to keep it themselves. They donate it instead. So an activating group is also what we call my bad. An activating group is what we call or so slash para directing. So activating groups are what we call Ortho and para directors. So what does that mean? Right, I want to remind you that if our medical group is attached to carbon one, then the Ortho position is carbon to the meta position is carbon three. And our para position is carbon four. So if we have an Ortho and para directing group then are all kill group in this case, archer bugle is going to react either on carbon too Or on Carbon four. So the question is asking for the major product. So which one is it going to be? And notice how the turk bugle group is kind of big. Right? It's a very bulky um all kill group. So if we were to add it to the Ortho position, there would be substantial Sterritt hindrance with our um with our method group, right? Because they're going to be very close to each other and they're going to crash against each other all the time. So that's not really favored instead because we have such a bulky um such a bulky alco group here, then the para product is going to be favored. It's going to be favored because our substitute prints are on opposite sides of the ring. so they interact with each other less and there is less hysteric hindrance. And with that being said, our para product is our major product, even though I want to point out that, um, are both Ortho product is still possible, just in a much lesser degree. So thanks for watching. And I hope you found this helpful. Yeah.

Predict the major products formed when benzene reacts (just once) with the following reagents. (g) 1-chloro-2,2-dimethylpropane + AlCl3

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