Destroyer: OC #83

[Mrbubbles]

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Ok u have 1-butene, when u add Br2, (X2), u get anti-addition. The first Br breaks the DB, a "transition state forms (like a triangle), and then u can have the Br attack the less substituted C.

But if u have Methanol ( a base), wouldn't it attack the Less substitted carbon????. in the answer, it has it attack the more substuited C ring.

i thought acids do that. same with question number 80.

basically, my queston is, doesn't a base attack the less subsituted Ring or Oxirane?? if so, why doesnt' questions 80 and #83 show it? please help =(

 

[TeamGuo]

it's Sn1 my friend. Since we are using methanol = not methoxide which is not negatively charged = only has lone pairs due do the work = so it will not do Sn2.

If it was Sn2 yes it will attack the primary C, but here you have the Br leave first then methanol will go attach.

 

[TeamGuo]

and number 80 = it's not base catalyzed = it's acidic catalyzed, epoxide gets protonated then methanol attacks the more substituted C, remember it's based on the catalyst not the reagent itself.

If acid catalyzed = favor more substituted
If base catalyzed = less substutited.

 

[Mrbubbles]

and number 80 = it's not base catalyzed = it's acidic catalyzed, epoxide gets protonated then methanol attacks the more substituted C, remember it's based on the catalyst not the reagent itself.

If acid catalyzed = favor more substituted
If base catalyzed = less substutited.

wow thank you! it makes sense! ok more more question, why in #80, does the 2nd step give a terminal alkene? doens't it want more subsituted, (why not 2-butene? instead of the answer 1-butene)

 

[TeamGuo]

wow thank you! it makes sense! ok more more question, why in #80, does the 2nd step give a terminal alkene? doens't it want more subsituted, (why not 2-butene? instead of the answer 1-butene)

That is because we are using a bulky base, since it's so bulky it will not form the internal alkene.

Bulky = 400 lb person = does not like to be crowded.

 

[Mrbubbles]

genius! thank you again!

 

[topdent1]

it's Sn1 my friend. Since we are using methanol = not methoxide which is not negatively charged = only has lone pairs due do the work = so it will not do Sn2.

If it was Sn2 yes it will attack the primary C, but here you have the Br leave first then methanol will go attach.

I was also confused by this question..infact I still am. Once we form the triangle (what the hell is this called??? Epoxide??) with the Br on top, you say it goes through Sn1. In order for it to go through Sn1, there would obviously be a need for carbonation. If a carbonation does form, where would it sit? C1 or C2 or both? I don't think the triangle is a very stable molecule so it why would it go through sn1?

 

[TeamGuo]

I was also confused by this question..infact I still am. Once we form the triangle (what the hell is this called??? Epoxide??) with the Br on top, you say it goes through Sn1. In order for it to go through Sn1, there would obviously be a need for carbonation. If a carbonation does form, where would it sit? C1 or C2 or both? I don't think the triangle is a very stable molecule so it why would it go through sn1?

that portion of the mechanism is NOT sn1. That's just bromination portion where you attach Br anti to each other on 2 adjacent carbons. Sn1 we are talking about is to get rid of one of the Br. Don't get confused with the Br2 reaction and Sn1.