Experiment 6: Elimination Reactions Investigation of the distribution of alkene

Question

Experiment 6: Elimination Reactions Investigation of the distribution of alkene products from E1 and E2 reaction pathways.

HO

substitution

Cl

elimination

elimination

ratio?

This experiment will investigate the distribution of alkene elimination products from both E1 and E2 reaction pathways. 2-Methyl-2-butanol will be the starting material for the E1 reaction, and 2-Chloro-2-methylbutane will be the starting material to for the E2 reaction. Be careful! The two starting materials have similar names, don’t mix them up! In this experiment you’ll be working with a partner. Set up the E2 as a team, bring the reaction to reflux and then do the E1 reaction while the E2 “cooks”. Be sure to share your data before leaving the lab. NOTE: The products of this reaction are volatile, keep samples in an ice bath at all times or you will lose your product.

Background:

Elimination reactions often have more than one possible product. While mixtures are very common, one product is usually produced in a larger proportion than the others. Zaitsev’s Rule states that the product having the most substituted double bond will be the dominant product. More substituted double bonds are more stable and are thus favored.

There are two major reaction pathways by which a leaving group may be eliminated to form an alkene. In the E1 mechanism, the reaction rate depends only on the concentration of substrate, that is the reaction displays first order kinetics. This means that the slowest step of the mechanism must involve only the substrate molecule. In the E2 mechanism, the rate of the reaction depends both on the concentration of base and on the concentration of substrate, that is the reaction displays second order kinetics. In this case the slowest step of the mechanism must involve both a molecule of substrate and a molecule of base.

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Heating Under Reflux

Refluxing a reaction is a convenient way to maintain a constant, elevated reaction temperature. The temperature is the boiling point of the solvent that the reaction is dissolved in. The reaction flask is fitted with a condenser (see Figure 1) which cools the vapors from the boiling solvent so that they drip back into the solution and a constant volume of solvent is maintained. The condenser is usually cooled by circulating water. The water must go in through the bottom of the condenser and out through the top so that the condenser fills with water. Once the condenser is filled, only a slow trickle of water is required. A faster flow of water can cause the tubing to pop off the condenser and spray water all over.

Always use a boiling stone or a magnetic stirrer to keep the boiling solution from “bumping.” It is essential that the cooling water be flowing before the heating has begun. If the water is to remain flowing overnight, it is important to fasten the rubber tubing securely with wire to the condenser.

If the heating rate has been correctly adjusted, the liquid being heated under reflux will travel only partly up the condenser tube before dripping back into the flask. You may be able to see a line of vapor in the condense at the condensation point. Below the condensation point, solvent will be seen running back into the flask; above it, the condenser will appear dry. This point should be less than half way up the condenser. The temperature of a reaction in a refluxing mixture will be approximately the boiling point of the solvent used for the reaction.

Figure 1: Reflux apparatus with additional air condenser for conver- sion to distillation apparatus.

Pre-Lab Questions:

1. If both reactions obey Zaitsev’s rule, which product should be the major one?

2. In this experiment we use two closely related compounds to demonstrate both E1 and E2 reactions. Could we use the same two compounds to do experiments that demonstrate SN1 and SN2? Why or why not?

Figure 2: Fractional distillation apparatus.

Ice Bath

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Procedure: E2 Reaction - Dehydrochlorination of 2-Chloro-2-methylbutane

Since the products will be distilled directly out of the reaction vessel, place an air condenser between the flask and the water condenser to make it easier to convert this reaction setup to a distillation apparatus. Fit a 25 mL round bottomed flask with an air condenser then a water condenser as shown in Figure 1. After the period of heating at reflux, the solution will be cooled and the reflux condenser will be removed and attached to a distillation head that is placed onto the air condenser. The setup for fractional distillation is shown in Figure 2. If necessary, review distillation from Experiment 2.

Place 1.5 g (23 mmol) of potassium hydroxide (this number is corrected for the fact that KOH pellets are about 15% water by weight) and 15 mL of propanol in the 25 mL round- bottomed flask along with a magnetic stir bar. Warm the solution using a sand bath and stir the mixture until the potassium hydroxide has dissolved. Quickly cool the flask to room temperature, using an ice-water bath, and cautiously add 1.3 g (12 mmol, 1.5 mL) of 2-chloro- 2-methylbutane. Replace the condenser and heat the mixture for one hour, maintaining a gentle reflux. During the period of heating potassium chloride precipitates from solution.

Remove the heating source from the reaction flask and allow the mixture to cool to room temperature. Equip the flask for fractional distillation, as shown in Figure 2. Remove the reflux condenser from the top of the air condenser and replace it with a distillation head fitted with a thermometer. Attach the reflux condenser to the outlet of the distillation head. Immerse the receiving flask in an ice-water bath. Distill the product mixture, collecting all distillate boiling below 45 °C (2-methyl-l-butene, bp 31 °C; 2-methyl-2-butene, bp 38 °C). Keep the recieving flask in an ice bath at all times to prevent evaporation of your products Analyze your product by NMR and gas chromatography. Be sure to keep your samples on ice as you wait for the GC. If necessary, review distillation and gas chromatography in experiments 2 and 3.

Cl

KOH

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Procedure: E1 Reaction - Dehydration of 2-Methyl-2-butanol

In this experiment, the product is removed from the reaction mixture by fractional distillation as the reaction proceeds. The flask containing the reaction mixture is connected to a fractional distillation setup. (Figure 2)

Place 2.0 mL (1.6 g) of 2-methyl-2-butanol in a 25-mL round-bottomed flask containing a magnetic stir bar. Carefully add 10 mL of 6 M sulfuric acid to the material in the flask. Caution: Sulfuric acid will cause burns if it comes into contact with your skin. If you get acid on yourself, tell your TA and wash it off immediatly. If you spill any acid, tell your TA and clean it up immediately. Gently heat the mixture and when distilling begins, regulate the heating so that the temperature of the vapor distilling over does not exceed about 45 °C. The distillation rate should be about 1 to 2 drops per second. Collect all distillate boiling below 45 °C (2-methyl-1-butene, bp 31 °C; 2-methyl-2-butene, bp 38 °C).Keep the recieving flask in an ice bath at all times to prevent evaporation of your products Analyze your product by NMR and gas chromatography. Be sure to keep your samples on ice as you wait for the GC. If necessary, review distillation and gas chromatography in experiments 2 and 3.

In your lab report:

Report your product ratios based on NMR, show calculations and peak assignments. Deter- mine the ratio based on integration of the alkenyl hydrogens. Remember to account for the different numbers of hydrogens being integrated in the two products.

Report your product ratios based on GC, show calculations and peak assignments. Did both reactions obey Zaitsev’s rule?
Write mechanisms for each reaction.

One reaction is significantly more selective than the other. Why do you think this might be? Hint: Is one of these reactions reversible? How might that effect the product ratio?

HO

substitution

Cl

elimination

elimination

ratio?

to be Using a previous lab desiachinthe sae beow thet describes the carried out Lab number Lab came tin Reference Procedure

Explanation / Answer

E1 reactions are unimolecular whereas E2 reactions are bimolecular. By extension,

RE1=k[LG?] and

RE2=k[LG?][NUC]

where the notations stand for leaving group and nucleophile.

As a consequence of the preceding, E2 reactions usually proceed with a strong nucleophile (e.g. base), whereas E1 reactions are fine with a weaker nucleophile (e.g. base).

Mechanistically, E2 reactions are concerted (and occur faster), whereas E1 reactions are stepwise (and occur slower and at a higher energy cost, generally).

Due to E1's mechanistic behavior, carbocation rearrangements can occur in the intermediate, such that the positive charge is relocated on the most stable carbon. Factors influencing this include hyperconjugation and resonance. As a result of this, E1 usually won't occur on primary substrates.

Moreover, when competing with SN1, E1's rate is more sensitive and will dominate when supplied with heat.

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