[hankb]

2-iod-2-methylbutane reacts Worth OH^- creating an alkene, H₂O and I^-.


Step 1: OH^- reacts with a H in one of the carbons connected to the middle carbon, making H₂O.

Step 2: The carbon atom now have only 3 bonds, therefore it creates a double bond to the centre carbon.
It also has a negative charge, so it releases I^-

Is this wrong?
I think the alkene is 2-methyl-but-(1 or 2)-en, however I am not certain about how the process work..

[discodermolide]

Don't think an E1 uses basic conditions. I would search this forum. This question has been asked and answered many times.

[Babcock_Hall]

hankb, What is the first step in a typical E1 mechanism?  Check your textbook, or check on-line.

[hankb]

Thanks for the replies. I found this example: http://chemwiki.ucdavis.edu/Organic_Chemistry/Reactions/E1_Reaction

Step 1: I leaves on it's own, steals one electron away from C(so it becomes I^-), forming the carbocation.
Step 2: OH^- comes into play, stealing a H⁺. C now has an extra electron, and forms a double bond.

Is this correct?
Does this mean the original hydrocarbon is unstable and will break apart one way or another anyway?

[Babcock_Hall]

I would say that the iodine atom departs with both electrons to become the iodide anion.  In the same vein, I would say that when a base (not necessarily hydroxide ion) removes a proton from the carbocation, both of the electrons are free to form the double bond between the two carbon atoms.

[PhDoc]

You did find a very nice example online that fails to understand the definition of the term pKa along with a certain equilibrium that will be at play when you mix a base with an acid other than its conjugate acid.

Methoxide is derived from methanol, pKa 15.5. Water has a pKa of 15.7, so we can expect exchange of protons under these conditions.

MeO-   +   H2O   ------>    MeOH   +   HO-

I'd be careful with this problem. Just because it's on a UC-Davis website doesn't mean your professor believes it's correct. The area of SN1, SN2, E1, E2 is where authors and professors <<<disagree most>>>.

What I'd also like you to realize is that this flowchart also came from University of California, albeit in Los Angeles.

http://voh.chem.ucla.edu/vohtar/spring06/classes/30A/pdf/Flowchartfor.pdf

According to the chart, the product for this reaction is one of E2 elimination. Also, Vollhardt & Schore (Berkeley) indicate this as E2.

For this particular topic, what I or anyone else other than your professor tells you is irrelevant. Your professor grades your exam. It's in your best interest to talk with him/her about it - NOT your TA.

[orgopete]

In a workshop, I saw the reaction of tert-butyl halides with NaOH written on the board and I thought they were going to be E2 eliminations. Then we did the kinetics, it was E1. Ah, hold on there. It was really a solvolysis reaction and the base was quite dilute and there monitor the pH. When enough HX had been generated, the indicator turned color.

This is always the problem with these reactions. How should we interpret them. For example, when I wrote the problems, I might ask for the elimination products. A second rule that I applied was that if NaOH (or any reagent) were present, it had to be present in kinetically significant amounts. You can't expect an E1 elimination reaction in 10M NaOH nor expect an E2 elimination in 0.001M NaOH.

If this was similar to the effect of solvent on SN1/E1 reactions, then it will be an E1 reaction (because the base was too dilute to cause a reaction to occur by its presence). Concentrations can be an important factor in answering some questions. I just don't know there can be any guidance for how it will be applied.

Note the similarity to the discussion of this reaction: http://www.chemicalforums.com/index.php?topic=63955.0. NaOMe is present, but seemingly not involved in the kinetics of the reaction.