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4. Consider the molecule azepine (shown below). azepine Do you think azepine would be aromatic, anti-aromatic, or neither? Why? Draw any p-orbitals and fully justify your answer! with &T present in this molecule, it foluo the (Yr9m e- rule which Provey its anti-aromatic
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media%2F76b%2F76b162ea-a066-4750-9192-95I have most of the last one finished but I need help with E, G, and F
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e. If you took a 1:1 mixture of both products and treated with Br2 at 0 oC, what ratio of 1,2:1,4 adducts would you expect? Why?

Ans: The ratio of 1,2: 1,4 would be same (1:1) at 0 oC. Because the reaction is under kinetic control, so both products would be more stable. And also other reason is thermodynamic product is more stable product than kinetic product always, so it won’t change at 0 oC.

f. If you took a 1:1 mixture of both products and treated with Br2 at 40oC, what ratio of 1,2:1,4 adducts would you expect? Why?

Ans: Ans: The ratio of 1,2: 1,4 would be 0:100. The 1,2-adduct would completely convert to 1,4-adduct at 40 oC.

The reason is kinetic control product unstable at 40 oC. So it would change to more stable thermodynamic product; 1,4- adduct.

g. What is the rate determining step (slow)? Draw below

Ans: Rate determining step is proton transfer step which is forming carbocation. Similar to SN1reaction, forming of carbocation step is rate determining step .

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