elimination/substitution part II =)

[Tina324]

1.
2-bromobutane + (CH3)3CO- ---> CH2=CHCH2CH3
what happened to zaitseff's rule of creating a more substituted alkene? Is it because the base is too bulky and has to eliminate the most terminal H?

2.
2-bromo, 2-methyl butane + -CN/heat --> E1 mechanism
Is this reaction E1 as opposed to SN1, only because of the heat?

Bonus question:
3. What reaction is this? and what the heck does the NaH do?
2-hydroxy butane + 1.NaH/2.CH3I in THF ---> 2-methoxy butane
...never seen this reactino in my life =/

thanks so much guys. i reallly reallly appreciate the help!

---

Members don't see this ad.

 

[tranv117]

1) with a big bulky base like t-butoxide, it will not undergo Zaitsev elimination rule but rather Hoffman. Theres two alpha protons to pull off, just think of it as if theres a big bulky base, it is much harder for it to go in and pull off the proton on carbon 3. So it'll pull off proton on carbon 1.

2) pretty much because of the heat. The trick is anytime you see heat, the reaction prefers elimination.

3) NaH is a base. It'll deprotonate the H off alcohol leaving you an alkoxide. It then turns into a nucleophile grabbing the electrophile methyl group in methyl iodide.