I don't get these two elimination reactions

[2PacClone23]

I don't get why one goes a certain way, while the second goes totally different.


The first reaction makes more sense than the second. What gives?

---

Members don't see this ad.

 

[OUDent88]

For E2 reactions normally it will form an alkene with the more substituted carbon according to zaitsev rule like in the bottom reaction. But the top reaction has a t-butyl group which is a bulky base reacting with a tertiary halide and that will form the anti-zaitsev product also known as the hoffman product. Keep in mind that this rule only applies when the bulky base is reacting with a tertiary halide.

If the bulky base is reacting with a secondary halide then it just follows the zaitsev rules like normal but make sure it is trans because it is more stable than the cis formation.

 

[OUDent88]

Another note on the second reaction since you said you were more confused with that one...you have a strong base reacting with a tertiary halide in heat. Adding heat to a reaction will usually always favor elimination because its entropic. The double bond that was formed in the bottom reaction could have easily formed on the other side instead since the carbon on either side of the methyl groups are equal to each other meaning they both have anti-coplanar hydrogens which are essential for and E2 reaction to occur.

Hope this helps

 

[mestrada9447]

The first reaction is an E2 elimination, but when you react a
Tertiary substrate with a BULKY BASE (like t-butoxide) it will form the Hoffman product, making the least substituted double bond.
(This also happens when there are no anticoplanar H available)


The second reaction is reacted with Alcohol (which can be considered both a weak nucleophile and a weak base)
and KOH (which can be a strong base or a protic solvent)
Either way you look at it the reaction will be an elimination that will form the most substituted double bond.

 

[UCSD1984]

Im sure the previous responses answered it in full detail, but for the sake of the DAT, I remembered this simply as; Bulky Base = less substituted alkene. Strong unbulky = most substituted alkene. Another simple yet tricky thing to watch out for is making sure there is an anti-coplanar hydrogen to be deprotonated.

Sent from my DROID3 using Tapatalk

 

[2PacClone23]

So these are both E2 reactions correct?

 

[DontCallItACmBk]

So these are both E2 reactions correct?

The bottom one looks borderline because of the heat and polar protic solvent, but KOH is a strong base, so I would say E2 because of the strong base

 

[mestrada9447]

The bottom one looks borderline because of the heat and polar protic solvent, but KOH is a strong base, so I would say E2 because of the strong base

I agree.