3 Destroyer ?s: E2, 1,2/1,4 addition, & Ka

[nicUD]

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My 1st question is 177 ochem. The part of this problem I dont understand is why does 2-chlorobutane turn into butene instead of 2-butene when reacting with potassium t-butyl alcohol? I realize that this is an E2 rxn but since the halide is secondary why wouldn’t it try to form the most stable alkene?

My second question deals with 105 ochem. It says that 1,2 addition is under kinetic control and 1,4 addition is under thermodynamic control. I really don’t understand if thats important or how that changes things etc..

Lastly, in 132 gen chem. The question says HN3 has a Ka= 2x 10^-5, and says if HN3 was put into soln what would be true? The answer is [N3-]=[H+] at equilbrium. I understand how to set up the Ka equation, but how do we know those concentrations are equal at equilib??

Thanks & good luck to all my fellow august DATers!

 

[dentist2be2011]

My 1st question is 177 ochem. The part of this problem I dont understand is why does 2-chlorobutane turn into butene instead of 2-butene when reacting with potassium t-butyl alcohol? I realize that this is an E2 rxn but since the halide is secondary why wouldn’t it try to form the most stable alkene?

My second question deals with 105 ochem. It says that 1,2 addition is under kinetic control and 1,4 addition is under thermodynamic control. I really don’t understand if thats important or how that changes things etc..

Lastly, in 132 gen chem. The question says HN3 has a Ka= 2x 10^-5, and says if HN3 was put into soln what would be true? The answer is [N3-]=[H+] at equilbrium. I understand how to set up the Ka equation, but how do we know those concentrations are equal at equilib??

Thanks & good luck to all my fellow august DATers!

1. t-butyl is very bulky. Due steric hindrance, a less substituted alkene is formed.

2. 1,2 addition is under kinetic control because it occurs at low temperature, and an unstable product is formed at a rapid rate. (kinetics is relate to speed). 1,4 addition occurs at high temperature, a more stable product is formed. Think thermodynamic and stability (deltaG=-RTlnK). Increase temperature decreases deltaG, thus more stable.

3. I think it is because HN3 a monoprotic weak acid you, when it dissociates, the concentration of base equals to concentration of acid. However, it doesn't dissociate fully, that's why the Ka is low. But the disassociated acid contributes to the equal amount of acid and base.

Please correct me if I'm wrong.🙂

 

[rnbaxter]

1) I'm sure just because it is a bulky strong base that is why your getting the least substituted elimination, in reality you probably get a 60 40 mixture.

2) 1,2 addition occurs at low temperature (0C), therefore this is controlled kinetically. If you want a 1,4 addition, you perform the reaction at high temps (>40C), therefore this is controlled thermodynamically. If you create your 1,2 addition and let the product sit out. What happens? Well room temperature is about 25C?, so the initial product will eventually turn into the 1,4 product.

3) You would get a better explanation if I could see the other choices to explain while they were wrong. But HN3 is a weak acid, so at equilibrium [H3O+ or H+]=[A-]

 

[Jia7]

I had a hard time understanding the last one too. But when they put it in soluton just write the equation out and it will make it easier to understand. Here,

HN3 + H2O <--> H3O+ + N3-

Concentration of H3O+ and N3- is equal in the above equation.