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Electronic Spectra of Molecules: The Absorption of UV and Visible Light

By Walt Volland-----
Bellevue Community College--

Description

Wavelength absorbed

Phenolphthalein

Equipment

Procedure

Chromatography

Data & report

Description

Absorption or emission of ultraviolet or visible light by a molecule depends on electron transitions between molecular orbital energy levels, just as absorption or emission of electromagnetic radiation by an atom is determined by electron transitions between different energy levels in the atom and the DEs for those transitions. Molecular spectra follow rules analogous to the rules for atomic spectra: energy is absorbed only when the amount of energy provided matches the difference in energy, DE, of 2 energy levels. When an electron goes from a higher to a lower energy state, a photon of definite wavelength and frequency is emitted. Every atom or molecule has a characteristic electronic spectrum depending on its characteristic DEs. Because of a molecule's greater complexity, we can often construct a molecule that will give a particular spectrum, rather than having to just accept the spectra available as we do with atoms. This possibility arises because of the interdependence of molecular orbital energy level values for the molecule, molecular shape, bonding, and distribution of electron density within the molecule. Energy transitions in phenolphthalein indicator, plant pigments, and isolated pi systems will be examined to illustrate this.

Molecular orbital theory provides a model for the way electromagnetic radiation interacts with molecules. For example, an electron in the pi bonding molecular orbital (MO) of an alkene can be excited to a pi antibonding MO. This is described as a transition.

For an isolated pi bond the energy separation, DE, between the pi bonding and pi antibonding MO's is large; ultraviolet light with its large energy and short wavelength is needed to excite the pi electron. Molecular orbital theory predicts that the energy difference, DE, between levels will decrease if the double bond is conjugated with another double bond. (Conjugated double bonds have one single bond separating them. They are coplanar and pi electrons can move through out the pi system.) Cogjugation exists when series of alternating double and single bond. This means that the molecule has a single bond between the two double bonds. The term "conjugated" is used in chemistry to refer to a series of alternating single and double bonds. The predicted decrease in DE for conjugated structures is also observed in experiments.

Here is a general rule that describes the effect of double bonds conjugation on the energy absorbed by the pi system. The greater the number of conjugated multiple bonds in a compound, the longer the wavelength of the light that the compound will absorb. This translates into a smaller energy requirement for the transition from the highest occupied MO to the lowest unoccupied MO. This idea is illustrated schematically in the following diagram for alkenes where the transition is for . As the conjugation increases, electrons are delocalized over more space and the system is more stable.

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 Compound name

 Condensed formula

 Wavelength absorbed

 Ethene, ethylene

 CH2::CH2

 171 nm

 

1,3-butadiene

 

CH2::CHCH::CH2

 

217 nm

 

Trans 1,3,5-hexatriene

 

CH2::CHCH::CHCH::CH2

 

274 nm

 

b-carotene

 

see structure below

 

425 nm

The plant pigment b-carotene has this line structure.

There are 11 double bonds in the alternating double bond-single bond system in b-carotene. The double bonds are essentially coplanar and the electrons are delocalized over a 22 carbon chain.

The wavelength absorbed by b-carotene is in the blue region of the visible spectrum. When we look at b-carotene, no "blue wavelengths" reach our eyes because the carotene molecule is absorbing these. The rest of the wavelengths in the visible region (minus the blue) reach our eyes and carotene looks orange to us. Remember, the color of light absorbed by the molecule is not the color we see. It is the one we don't see!

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Observed Color of Compound

 

Color of Light Absorbed

 

Approximate Wavelength of Light Absorbed

 

Green

 

Red

 

700 nm

 

Blue-green

 

Orange-red

 

600 nm

 

Violet

 

Yellow

 

550 nm

 

Red-violet

 

Yellow-green

 

530 nm

 

Red

 

Blue-green

 

500 nm

 

Orange

 

Blue

 

450 nm

 

Yellow

 

Violet

 

400 nm

Absorption or emission of ultraviolet or visible light by a molecule depends on electron transitions between molecular orbital energy levels, just as absorption or emission of electromagnetic radiation by an atom is determined by electron transitions between different energy levels in the atom and the DEs for those transitions. Molecular spectra follow rules analogous to the rules for atomic spectra: energy is absorbed only when the amount of energy provided matches the difference in energy, DE, of 2 energy levels. When an electron goes from a higher to a lower energy state, a photon of definite wavelength and frequency is emitted. Every atom or molecule has a characteristic electronic spectrum depending on its characteristic DEs. Because of a molecule's greater complexity, we can often construct a molecule that will give a particular spectrum, rather than having to just accept the spectra available as we do with atoms. This possibility arises because of the interdependence of molecular orbital energy level values for the molecule, molecular shape, bonding, and distribution of electron density within the molecule. Energy transitions in phenolphthalein indicator, plant pigments, and isolated pi systems will be examined to illustrate this.

Phenolphthalein is another example of the effect of molecular shape on delocalization of pi electrons. Phenolphthalein molecule is colorless in acid solution and its pi system is nonplanar. The electrons in the double bonds are "localized. The excitation energy is large and in the ultraviolet range. The molecule loses protons in basic solutions and forms a minus 2 anion. This negative ion is flat and the pi system is conjugated. In the planar ion the pi electrons in the eleven double bonds can delocalize over the 3 rings, the COO-1 carboxyl group and the oxygens. The anion imparts a color to solutions because the energy separation between the pi and pi antibonding levels is smaller and matches light in the visible range.

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In the first part of this experiment you will observe this difference in color for phenolphthalein and its anion.

In the second part of this experiment you will extract chlorophyll from green leaves and then use chromatography to separate chlorophyll a from chlorophyll b. You may also see the separation of carotene and other plant pigments. You will be asked to explain the different colors of the 2 chlorophylls by examining their structures and identifying the type of change in molecular orbital energy levels caused by the difference in structure.

Paper chromatography is a separation technique that anyone who ever spilled coffee or tea on a piece of paper has seen. The solvent wets the paper, creeps along carrying solutes along with it. The different dissolved materials move with the solvent, but at different rates because the paper attracts the solutes differently. You will use this behavior in a systematic way to separate pigments in chlorophyll. You will make an extract of leaf pigments by soaking plant leaves in a mixture of cold acetone and ethanol. The extract appears to be green, but other pigment colors may be masked by the strong green tint. You will use paper chromatography to separate any pigments in the extract. The separation occurs because the pigments have different solubilities in the mobile solvent and different absorption affinities (attractions) to the chromatography paper. The separation process will give a chromatogram.

 

 Equipment and Supplies

Al foil

2 beakers, 100 mL

graduated cylinder, 10 mL

medicine cup

3 Beral type pipets

erlenmeyer flask, 50 mL

Beral type pipet with narrow stem

glass jar about 7 inches tall (about 1 quart)

2 pencils

string

scissors

1-3 g fresh spinach leaves

ice

phenolphthalein indicator solution

acetone

vinegar

ethanol

5M NaOH

petroleum ether

chromatography paper

plastic wrap

2 rubber bands

tape

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Procedure

Please read the entire procedure before starting the experiment.

Wear your goggles at all times while doing the experiment!

Gather your equipment and chemicals. Cover your work area with Al foil or waxed paper to catch spills and spatters.

Preparation for Extracting Chlorophyll

Do this first, then do the phenolphthalein part of the experiment. Finally, finish up with the chlorophyll chromatography.

Collect 1 to 3 grams of fresh spinach leaves. If you can't get spinach, other fresh green leaves (turnip greens, broccoli leaves) will work, but your chromatogram may be slightly different from the one described later in this experiment. Any dark green leaves should allow you to extract the 2 chlorophyll pigments.

Make an ice bath by placing a little cold water and some ice in a beaker.

Remove the containers of acetone and ethanol from your kit. Do not open yet.

Set the 2 containers in the ice bath so the acetone and the ethanol get cold. Do not let the water get into the acetone or ethanol containers.

Phenolphthalein Molecule and Anion Colors

Pour about 2 mL of water into one of the medicine cups from your kit. The reference marks on the cup are adequate; you do not need to measure the water with your graduated cylinder.

Add one or 2 drops of the phenolphthalein indicator solution to the water in the medicine cup. You can do this with one of your Beral type pipets if there is no "built in" dropper in your phenolphthalein container.

Use a different Beral type pipet to add 2 drops of vinegar to the medicine cup. Stir with your stirring rod. Note and record the color of the solution in the medicine cup.

USE CARE WHEN POURING THE NaOH. IT IS CAUSTIC.

If the 5.0 M NaOH solution gets into your eyes, rinse the eyeball for 15 minutes in warm running water (force your eyelids apart with your fingers to keep your eye open and the eyeball exposed to the running water). Contact your physician for advice if your eye continues to hurt or "burn".

Use a third Beral type pipet to add 2 or 3 drops of 5 M NaOH to the contents of the medicine cup. Stir with your stirring rod (no need to wash the rod between steps). Note and record the color of the mixture in your cup

Discard the solution in the medicine cup by pouring it into the drain followed by cold water.

Preparation of Pigment Extract

Work in a well ventilated area away from any open flames or hot surfaces (stove, space heater, water heater, etc.). Acetone and ethanol convert to vapor very easily, especially the acetone. Both the liquid and the vapor are flammable!

Label your erlenmeyer flask "pigment extract mixture".

Cut the leaves into tiny pieces using your scissors. The smaller the pieces, the better the contact between pieces of leaf material and the extraction solvent.

Place the small leaf pieces in the erlenmeyer flask and add 5 mL of cold ethanol.

Then add 5 mL of cold acetone. There is no need to wash your graduated cylinder between the ethanol and acetone.

Quickly cover the erlenmeyer flask with a piece of plastic. Hold the plastic wrap snugly against the neck of the flask with a rubber band. This helps keep the solvent from evaporating while the pigments are being extracted from the leaves.

Dump the water out of the beaker. Keep the ice. Set the erlenmeyer flask in the ice in the beaker.

Store both in a cool, well ventilated area for about 24 hours. Do not put the beaker and erlenmeyer in your refrigerator. Store away from food, hot surfaces, and open flames.

The solvent will extract (dissolve) pigments from the leaf material. Then you can do the chromatography to separate the different pigments.

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Chromatography

Assemble a chromatography chamber. You need a glass jar that is about 7 inches tall, a piece of plastic wrap to cover the top of the jar, and a large rubber band to go around the neck of the jar to hold the plastic wrap tightly in place. A recycled jelly jar, instant coffee jar, or just about any quart jar will do.

Obtain a piece of chromatography paper that measures about 3 cm by 15 cm. Handle only the edges. Try not to touch the surface. Oils from your fingers can alter the absorptive properties of the paper and the separation may not happen normally.

Use a ruler and a pencil to draw a line one centimeter from the bottom 3 cm edge of the paper. DO NOT USE INK! Ink will dissolve in the solvent and ruin the separation process. This is the line where you will place a spot of the extract.

Lightly mark the edges at one-centimeter distances along the edges of the paper, lengthwise. This helps you keep track of the leading edge of your solvent (the solvent front) as it "climbs" the strip of paper.

Use the tip of your scissors to poke a small hole about 1/2 cm from the top 3 cm edge of the paper. Insert a piece of string or thread through the hole. Tape the ends of the string to the middle of a pencil or pen. The paper should hang freely from the pencil when you hold up the pencil.

Hang the paper in the center of your glass jar by placing the pencil across the mouth of the jar.

Rotate the pencil to wind the string around it until the bottom of the paper just clears the bottom of the glass jar. The paper should not touch any part of the jar, not even the bottom.

Tape the wound-up string in place on the pencil so it can't unwind when you let go of it. Remove paper from jar by lifting the pencil. Apply the "pigment extract mixture" as directed below.

 

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This animation illustrates what happens in the separation of a two component mixture. The solvent front is marked by the upper black line. There is a green component that moves faster than the yellow component in the mixture. The Rf value for the green is closer to "1" than the Rf value for the yellow component.

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Take the plastic wrap off the beaker of "pigment extract mixture". Use a Beral type pipet with a very narrow stem to draw up a very small quantity of pigment extract.

Place one drop of the pigment extract on or just slightly above the pencil line on the chromatography paper. The pigment extract spot should be about 3 to 5 mm in diameter which is the size of this spot .

 

 

Let the spot dry for at least 30 seconds.

Repeat this procedure to apply 4 or 5 more drops directly on top of the first drop. ALLOW THE PIGMENT TO DRY BETWEEN APPLICATIONS ! If you apply a large amount at one time or if the spot is still wet from the previous drop, the spot will spread over an unmanageable area and give poor separation results. You need to have several drops of pigment on the paper to make the colors of minor parts of the mixture visible in your finished chromatogram.

Place the plastic wrap cover back on the erlenmeyer flask of pigment extract. You may need to repeat the chromatography separation and need this extract for the duplicate trial.

Prepare a mixture of 1 mL acetone and 9 mL petroleum ether. This is the solvent for the chromatography separation.

Place enough acetone-petroleum ether solvent in your "chamber" to give a depth of 1/2 cm in the bottom of your glass jar. The liquid level must be lower than the pencil start line across the paper.

Place the "loaded" chromatography paper into the chamber. Lower the pencil until the bottom of the paper just touches the solvent in the bottom of the jar. You want the bottom of the paper to get wet with solvent. You do not want the spot of pigment extract to touch the solvent at this time.

Cover the top of the chamber with plastic wrap; use a rubber band to hold the wrap in place.

Watch the solvent climb the paper.

Gauge the speed of movement of the solvent front. The progress of the solvent will be quick at first. The pigment extract spot will start to smear out as the solvent carries the pigments away from the start line and the chromatogram "develops".

Allow the chromatogram to develop for 10 minutes, or until the solvent front comes within 1 cm of the top edge of the paper. Remove the paper from the chamber by lifting the pencil.

If you let the chromatogram develop too long, the solvent front will actually come to a stop. Solvent molecules will be evaporating from the wet paper as fast as they are "climbing" onto the paper at the bottom, so the front will stop rising. However, solvent molecules will still be moving up the paper, washing the pigment along. The different components of the pigment mixture will start to over-take one another and "pile up", undoing the separation you achieved.

Allow the paper to dry. Use pencils to sketch the outlines of the colors on the paper. Begin with the color closest to the start line and label the different color regions. The first pigment spot should be yellow-green; this is chlorophyll b. The blue-green pigment is chlorophyll a. You should also see 2 yellow-orange xanthophylls in the middle. The darker orange spot near the top is carotene. Xanthophylls and carotenes are members of the carotenoid class of pigments.

The chromatogram will fade on exposure to light because of light induced oxidation reactions. If you want to preserve it, store it in a dark place.

 

 

Data and Conclusions

Phenolphthalein

What is the color of the phenolphthalein molecule in acidic solution?

What is the color of the anion in a basic solution?

 

Why do you see a color for the anion in basic solution?

 

Which pi system will have a smaller DE for the transition, the phenolphthalein molecule or the anion?

 

What is the approximate wavelength absorbed by the anion of phenolphthalein?

 

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Chromatography results

These illustrations for chlorophyll a and chlorophyll b do not show the hydrogen atoms because they would obscure the carbon skeleton. The carbon, nitrogen, oxygen, and magnesium atoms are shown. Single and double bonds are indicated. There are 12 double bonds in a series of alternating double and single bonds. This is a low energy, conjugated pi system in chlorophyll a.

Chlorophyll b has an aldehyde group in place of a CH3 group in chlorophyll a. The rest of the structure is the same for the 2 molecules.

 

How many double bonds are in the alternating pi system for chlorophyll b?

 

Explain the difference in color for chlorophyll a and chlorophyll b in terms of the number of double bonds in the 2 molecules, the wavelength absorbed and the size of the DE that corresponds to this wavelength.

 

Why is it necessary to use a closed chamber when you develop the chromatogram of the plant pigment extract?

 

The spots that appear in the chromatogram of the plant pigment extract do not always completely separate. The trailing edge of one color may overlap with the leading edge of the one behind. If the chromatography paper were rotated by 90o and another development period carried out with the solvent now moving at right angles to its original direction, would better separation be possible? Explain your answer.

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All rights reserved 1999 Walt Volland

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