Destroyer O. Chem #80

[predentn]

Advertisement - Members don't see this ad

For the elimination step, why isn't 2-butene formed preferentially over 1-butene? I would have thought that 2-butene would be formed in greater concentration because it is the more substituted alkene and thus more stable.

 

[fancymylotus]

For the elimination step, why isn't 2-butene formed preferentially over 1-butene? I would have thought that 2-butene would be formed in greater concentration because it is the more substituted alkene and thus more stable.

I swear I've repeated myself 1000 times about this one. E2 with big bulky bases gives Hoffman products.

 

[shoombol tallah]

If you've answered it 1000 times then look at the post and move along swiftly. Noone asked you to "repeat" yourself. Anyways about the question because the base used in problem 80 is bulky ---(CH3)3C0-K+ then the least substituted alkene will form. However if the base used would have been C2H5O- K+ then the most substituted alkene product would have formed. Just remember bulky bases like potassium t-butoxide always form less substituted alkenes. Look at road map 3 for a great example of this and the two possible products. If you have other questions feel free to PM me.

 

[fancymylotus]

If you've answered it 1000 times then look at the post and move along swiftly. Noone asked you to "repeat" yourself. Anyways about the question because the base used in problem 80 is bulky ---(CH3)3C0-K+ then the least substituted alkene will form. However if the base used would have been C2H5O- K+ then the most substituted alkene product would have formed. Just remember bulky bases like potassium t-butoxide always form less substituted alkenes. Look at road map 3 for a great example of this and the two possible products. If you have other questions feel free to PM me.

Hokay, lets fill up the forums with repeat questions without using the convenient search function that SDN provides for us.
🙄