Experiment 4 - Isolation of Caffeine from Tea

introduction

• coffee contains the stimulant caffeine

• a cup of coffee contains between 80 and 125 mg of caffeine, while a cup of tea contains 30 to 75 mg of caffeine

introduction

• caffeine belongs to a large class of compounds called alkaloids

• nicotine, cocaine, morphine, and strychnine are also alkaloids

• these chemicals are of plant origin, contain basic nitrogen, and often have a bitter taste and physiological activity 

introduction

• in this experiment, caffeine will be isolated from tea leaves

• caffeine can be extracted from tea with hot water 

introduction

• the extraction of water-soluble components of tea leaves effects the separation of the caffeine from most of the other naturally occurring substances found in tea leaves.

• the main component of tea is cellulose, the structural material of all plant cells

• cellulose is virtually insoluble in all temps of water and thus is separated from the caffeine in the initial extraction 

introduction

• tannins are another water-soluble component of tea leaves

• tannins include a large group of phenolic compounds with molecular weights between 500 and 3000

introduction

• there are two main types of tannins found in tea:

• 1) one group includes tannins that are esters of gallic acid and glucose

• they have structures in which some of the hydroxyl groups in glucose have been esterified by digalloyl

• when these tannins are extracted into hot water, some of them are partially hydrolyzed to form free gallic acid and the sugar

introduction

• The other major group of tannins found in tea are condensation polymers of catechin of varying size

• Thus the tannins of tea are extractable into hot water, along with caffeine.

• Fortuitously, it is possible to separate the caffeine from the tannins (and their hydrolysis products, glucose and gallic acid) with ease.

introduction

• The tannins (because of their phenolic groups) and gallic acid (because of its carboxyl group) are both acidic.

• If a base (such as sodium carbonate) is added to the aqueous extract of tea these acids are converted to their sodium salts, which are highly soluble in water.

introduction

• Caffeine can be extracted from the basic tea solution with an organic solvent, but the sodium salts of gallic acid and the tannins will remain in the aqueous layer.

• (Glucose will also remain in the aqueous layer, since it is highly soluble in water due to its many hydroxyl groups).

introduction

• The brown color of an aqueous extract of tea is due to flavonoid pigments, chlorophylls, and their respective oxidation products.

• Although chlorophylls are soluble in the organic solvent, most other substances aren’t

• thus, the organic extract of the basic aqueous extract of tea should contain nearly purse caffeine

procedure 1

• Place 100 mL of distilled water into a 250 mL beaker.

• Add a few boiling chips, cover the beaker with a watch glass and heat the water to boiling.

• Once the water is boiling, add 20 mL of this boiling water to a 100 mL beaker and immediately place a tea bag into the hot water so that it lies flat on the bottom of the beaker and is covered completely with the hot water

procedure 1

• Cover with a watch glass and keep the tea bag immersed in the hot water for 10 minutes.

• During this extraction period, it is helpful to push down GENTLY on the tea bag with a test tube intermittently, while keeping the beaker covered at all other times to inhibit evaporation.

• Keep the remaining water in the 250 mL beaker simmering on the hot plate so that you can replace any evaporated water from the tea solution.

procedure 1

• Pour the extract into your 35 mL. large test tube.

• Tilt the beaker so that the tea bag is in the upper corner (farthest away from the pouring spout), so that the extract can drip out of the tea bag. Roll and press gently with the test tube to help to squeeze much of the water out of the tea bag.

• BE CAREFUL not to break the tea bag!

procedure 1

• Extract the tea bag two more times (with about 5 mL of hot water for 5 minutes each time).

• After the last extraction, squeeze as much water as you can out of the tea bag.

• Dissolve 1.5 g sodium carbonate in the hot liquid (it may not all dissolve).

procedure 1 

• Take only as much ethyl acetate (organic solvent) as you need.

• When the liquid is cool, extract it three times with approximately 4 mL of ethyl acetate (4 mL is about two "squirts" from a 2 mL Pasteur pipette)

• Ethyl acetate is a low boiling point liquid and must be treated accordingly.

• For example, the warmth from your hand on a sealed container containing ethyl acetate is sufficient to build up a significant vapor pressure).

procedure 1:

• For each extraction, add about 4 mL ethyl acetate and stopper the test tube with a cork. (Do NOT use a rubber stopper, since it will start to dissolve in ethyl acetate)

• Ensure a good seal, by moistening the cork with a bit of water first.

• CAREFULLY swirl the test tube 15 times. (NOTE: If you are not careful, and mix the layers too vigorously, the ethyl acetate and the aqueous tea extract will form an intractable emulsion.

• This will drastically decrease your yield.

procedure 1

• After thoroughly mixing the test tube, allow the ethyl acetate layer to settle.

• Transfer it to a clean test tube with a Pasteur pipette.

• Try not to transfer any of the aqueous solution along with the ethyl acetate layer.

• If any drops of the dark aqueous solution are visible in your pooled ethyl acetate extracts, transfer the extract to another clean test tube using a clean Pasteur pipette.

procedure 1

• dry the ethyl acetate extract with anhydrous magnesium sulfate

• decant or filter the dried extract into a clean and dry pre-weighted sample tube

• (note - plastic sample containers cannot be placed in the heated sand-back)

procedure 1:

• gently evaporate off the ethyl acetate (BP 40 degrees celsius) in the fume hood, using the heated sand-baths

• be careful that the caffeine doesn’t sublime

• leave the sample container open for one week so the product dries off completely

procedure 2: after one week

• determine the weight of your crude caffeine, using the three-place balance

• perform a TLC on a small portion of your sample against a standard caffeine sample for qualitative analysis

pre-lab question 1 and answer

• Give a general equation for the hydrolysis of the tannins that are esters of gallic acid and glucose.

• It is not necessary to balance this equation, but indicate the reactants and products by general structure.

• answer: look at page 37 in lab manual

pre-lab question 2 and answer

1. Why was the sodium carbonate added? (Hint: there are at least three reasons.)

• as a base, it converts the tannins and gallic acid to salts so the they are more soluble in water and thus, the caffeine can be extracted into the organic phase in a purer form

• a base also ensures that the caffeine in non-protonated so that it will be more organic soluble for the extraction

• salting out - by saturating the aqueous phase with any salt, the extraction efficiency of the organic solvent for an organic solute will be increased

pre-lab question 3 and answer

1. The crude caffeine isolated from tea has a yellowish or greenish tinge. Why?

• due to small amounts of flavonoid pigments and chlorophyll that are also extracted

pre-lab question 4 and answer

• An alternative procedure for removing the tannins and gallic acid is to heat the tea leaves in an aqueous mixture containing calcium carbonate. Calcium carbonate reacts with the tannins and gallic acid to form insoluble calcium salts of these acids. If this procedure were used, what additional step (not done in this experiment) would be needed in order to obtain an aqueous tea solution?

• a filtration step would be required

pre-lab question 5 and answer

• In a typical extraction of a solute from one solvent to another, once the two immiscible liquids have separated into two distinct layers, an equilibrium is established such that the ratio of the concentrations of the solute in each solvent is a constant. This constant is called the partition (or distribution) coefficient and is defined by: K=C2/C1

• where C, and C, are the equilibrium concentrations of the solute in solvent 1 (the solvent in which the solute is initially dissolved) and solvent 2 (the solvent which is doing the extracting), respectively.

  The extraction of 8.0 g of a particular organic compound dissolved in 100 ml of water (solvent 1) into diethyl ether (solvent 2) proceeds with a distribution coefficient of 5. Demonstrate (show all calculations) that three 50 ml extractions yields 7.8 g of the compound extracted into the ether, whereas one 150 mL extraction yields 7.1 g. 

• LOOK AT PAGE 38 FOR ANSWER