Joseph K. Myers
4-6-04
The elimination reaction mechanism forms an alkene from an alkyne. This is done by elimination of a substituent and one additional hydrogen atom, forming a double carbon-carbon bond.
This mechanism competes with SN1 and SN2 reactions. Conditions not favorable to those reactions will increase ratios of E2 products. For this reason, E2 can be made to produce a dominant product, and so it is a useful synthesis technique. Generally, a bulky, strong base is good for E2 synthesis to avoid competing pathways, such as SN2.
It is important to discuss some of the ideas for E2 reactions.
E2 stands for elimination bimolecular. This means that the reaction is concerted, and there is one intermediate. Both reactants are involved in the rate-determining step. Two molecules are participating in the transition state.
A good leaving group should be a weak base. The base which draws away the hydrogen atom should be a strong base.
In the removal step, the hydrogen being removed and the leaving group should be 180 degrees opposite each other, or "anti-periplanar."
Attached on separate paper.
(Table 01)
| compound | formula | molecular weight (g) | mp (deg-C) | bp | d (g/cm3) |
|---|---|---|---|---|---|
| 2-methyl-1-butene | C5H10 | 70.133 | -137.53 | 31.2 | 0.6504 |
| potassium hydroxide | KOH | 56.105 | 406 | 1327 | 2.044 |
| 1-propanol (propyl alcohol) | C3H8O | 60.095 | -124.39 | 97.2 | 0.7997 |
| 2-methyl-2-butene | C5H10 | 70.133 | -133.72 | 38.56 | 0.6623 |
| t-butyl alcohol | C4H10O | 74.1224 | 25.5 | 82.2 | 0.786 |
| 2-bromo-2-methylbutane* | C5BrH11 | 151.0459 | 108 | 1.182 | |
| potassium t-butoxide* | C4H9OK | 112.21 |
(*More data for these compounds could not be found in the reference material.)
(Table 02)
With KOH:
| compound | potassium hydroxide | 2-bromo-2-methylbutane |
|---|---|---|
| equivalence | 1 | 1 |
| mmol | 99.1 | 19.56 |
| grams | 5.56 | 2.955 |
| mL | n/a (25 mL 4 M) | 2.5 |
With potassium t-butoxide:
| compound | potassium t-butoxide | 2-bromo-2-methylbutane |
|---|---|---|
| equivalence | 1 | 1 |
| mmol | 22.28 | 19.56 |
| grams | 2.50 | 2.955 |
| mL | n/a (25 mL 1 N) | 2.5 |
The process is the same for the two different bases, except in step one of the flowchart.
1) Prepare 4 M KOH in 25 mL 1-propanol.
Or, if you are using potassium tert-butoxide, then prepare a 1 N solution of this in 25 mL 1-propanol.
2) add 2.5 mL 2-bromo-2-methylbutane.
3) Reflux for 45 min. (Lubricate joints because of strong base.)
4) Cool 10-15 min.
5) Fractional distillation.
6) Collect fraction below 45-deg-C.
Details: use two condensers together. Switch water tubes to the vertical condenser for reflux, and to the horizontally-sloping condenser for fractional distillation. Keep the receiving flask in an ice bath while collecting the distillate.
The solution was powdered milk-looking after reflux.
By accident, the reaction with KOH was carried to a head temperature of above 50 degrees C.
(Table 03)
This is for the E2 elimination with KOH:
| compound | experimental yield (g) | theoretical yield | %y | bp (exp.) | bp (lit.) |
|---|---|---|---|---|---|
| 2-methyl-2-butene | 1.0 | 1.37 | 73.0 | 44-47 | 38.56 |
(Show calculations: %y = ME/MT * 100 = 1.0/1.37 * 100 = 73.0.)
This is for the E2 elimination with potassium t-butoxide:
| compound | ME (g) | MT | %y | bp (exp.) | bp (lit.) |
|---|---|---|---|---|---|
| 2-methyl-1-butene | 0.50 | 1.37 | 36.5 | 39-42 | 31.2 |
(Show calculations: %y = ME/MT * 100 = 0.50/1.37 * 100 = 36.5.)
Note that the large, bulky base encourages formation of the less highly substituted product because the bulk of the base increases steric hindrance in the pathway to the more highly substituted product. This blocks some production of the more highly substituted product, even though it would be more stable.
A small base favors the 2-methyl-2-butene product because that is more stable (as shown by its higher melting point and boiling point).
We learned in this laboratory not only some of the details of elimination reactions, but also how to control the product by favoring a specific E2 pathway, by use of different reagents.
Contact with too much surface area, and transfer between too many containers can affect the yield of a small reaction. This may be a source of error and account for lower product formation with potassium t-butoxide.
With potassium hydroxide, the experiment was fairly successful, and in part this may be due to more efficient preparation and mixing of the original compounds for the experiment. Also, the larger yield may partly be due to increased impurity of the distillate because of a higher head temperature, although this is mitigated by the fact that the product also had a higher boiling point.
1. Gilbert JC, Martin SF. Experimental Organic Chemistry. 2002. 3rd Edition. p313-323.