Question

Purpose: To observe how contaminants can accumulate in organisms within a food web. Materials: "shaker" cup, 9 small cups (minnows), 3 medium cups (sunfish), 1 large cup (osprey), 20 M&Ms of one color (represent producers), 10 M&Ms of a second color (represent producers with DDT)—Feel free to substitute for other candy, plastic beads, dried beans, etc. If you have 30 of the same item, you can always mark 10 of them with paint/marker/etc. to represent the second color. (You can also use all the same size cups if needed—or even arrange the M&Ms into small piles if you don't have cups—just be sure to keep clear separation of which piles represent which organisms so you don't get confused.) Introduction: Many chemicals produced by industries end up in our waterways through dumping or runoff. Many of these chemicals can accumulate in the cells of organisms that live in or ingest the water (bioaccumulation). These chemicals then move up through the food chain and can be damaging or fatal to many types of organisms. Although the initial level of these chemicals might be low, the concentration of these chemicals can increase as they move up food chains. This is known as biomagnification. DDT (dichlorodiphenyl-trichloroethane) was the first commonly used insecticide. DDT was relatively inexpensive to manufacture and had long-lasting effects. DDT enters aquatic environments by attaching itself to the surfaces of plankton or accumulating in the cells of these organisms. DDT is harmful to many organisms as it is a neurotoxin that attacks the nervous system of animals. DDT also has adverse effects on many seabirds because it prevents proper eggshell production in birds, which results in very brittle eggs that are then easily broken. The specific food chain we will be focusing on contains organisms common in lake and river ecosystems in North America: algae → minnows → sunfish → osprey Procedure: Place all 30 of the M&Ms into the "shaker" cup to represent the producer population. 1. In the data table, record the amounts of DDT per producer (algae) in the "Trial 1" column. (10 contaminants per 30 producers total = 10/30 = .33) 2. Give the shaker cup a good shake. 3. Now simulate minnows eating some of the producers. Each minnow eats 3 algae, so randomly remove 3 M&Ms from the "shaker" cup and place them into one of the small cups (minnows). 4. Repeat step 3 for the remaining eight small cups (minnows). In your data table, record the number of DDT particles present within the groups of 9 minnows in the "Trial 1" column. (The number will be out of—divided by—9.) 5. Now simulate the sunfish eating minnows. Each sunfish needs to eat 2 minnows. Randomly select 2 small cups (minnows) and empty the contents into one of the medium cups (sunfish). This will represent the DDT particles contained in the minnows being ingested by the sunfish that eat the minnows. Repeat for the remaining medium-size cups (sunfish). In your data table, record the number of DDT particles contained within the group of 3 sunfish in the "Trial 1" column. (The number will be out of—divided by—3.) 6. Finally, simulate an osprey eating sunfish. One osprey needs to consume 2 sunfish. Randomly select 2 medium-size cups (sunfish) and empty the contents into the large cup (osprey). In your data table, record the amount of DDT in the osprey in the "Trial 1" column. (The number will be out of—divided by—1, so in this case, just the total number counted.) 7. Empty the contents from all cups back into the "shaker" and repeat all steps two more times, recording the data in "Trial 2" and "Trial 3" columns. 8. Calculate the average amount of DDT for each organism. To do this, calculate the percentage for each organism in each trial by dividing. For each organism, add the percentages together and divide the total by 3 for the average PERCENTAGE. Multiply the average percentage by the total number of organisms for the average DDT particles per individual organism. So for example, if you recorded for minnows: 3 particles out of 9 in trial 1, 4 out of 9 in trial 2, and 5 out of 9 in trial 3 You will have: Organism | Trial 1 | Trial 2 | Trial 3 | Average (DDT/organism) Minnow | 3/9 = .33 | 4/9 = .44 | 5/9 = .56 | .44 .33, .44, and .56 are the amounts of DDT particles per individual in each trial. Now, calculate the average DDT particles per organism: .33 + .44 + .56 = 1.33 (divide by 3 to average, so 1.33/3 = .44) Amount of DDT Per Organism Organism | Trial 1 | Trial 2 | Trial 3 | Average (DDT/organism) Algae Minnow Sunfish Osprey Analysis Questions: 1. What happened to the amount of DDT per organism as you move up the food chain? 2. Which organism contained the largest concentration of DDT? 3. In what ways is DDT harmful to osprey? In addition to the organisms from our simple example above, the organisms in the following relationships are also part of a more complex food web containing these same organisms. Zooplankton eat algae Minnows eat zooplankton, algae, and insect larvae Largemouth bass eat sunfish and minnows Sunfish eat minnows and insect larvae Osprey eat largemouth bass Insect larvae eat zooplankton and algae Snapping turtles eat minnows and sunfish 4. Based on the information above, what is one other organism besides osprey that you would expect to have high concentrations of DDT? Explain why. 5. If the osprey population were to decrease due to DDT, what are two other populations that would be affected and how? The following diagram illustrates a typical marine food web. Mercury, a toxic heavy metal, can also accumulate in the fatty tissue of organisms. 6. Which organism from the food web above would you predict to have the highest level of mercury? Explain your answer. 7. Of the following types of fish, which would most likely be the safest to eat? Explain your answer. Mackerel | Snapper | Mako Shark Bluefish

Purpose: To observe how contaminants can accumulate in organisms within a food web.

Materials: "shaker" cup, 9 small cups (minnows), 3 medium cups (sunfish), 1 large cup (osprey), 20 M&Ms of one color (represent producers), 10 M&Ms of a second color (represent producers with DDT)—Feel free to substitute for other candy, plastic beads, dried beans, etc. If you have 30 of the same item, you can always mark 10 of them with paint/marker/etc. to represent the second color. (You can also use all the same size cups if needed—or even arrange the M&Ms into small piles if you don't have cups—just be sure to keep clear separation of which piles represent which organisms so you don't get confused.)

Introduction:
Many chemicals produced by industries end up in our waterways through dumping or runoff. Many of these chemicals can accumulate in the cells of organisms that live in or ingest the water (bioaccumulation). These chemicals then move up through the food chain and can be damaging or fatal to many types of organisms. Although the initial level of these chemicals might be low, the concentration of these chemicals can increase as they move up food chains. This is known as biomagnification.

DDT (dichlorodiphenyl-trichloroethane) was the first commonly used insecticide. DDT was relatively inexpensive to manufacture and had long-lasting effects. DDT enters aquatic environments by attaching itself to the surfaces of plankton or accumulating in the cells of these organisms. DDT is harmful to many organisms as it is a neurotoxin that attacks the nervous system of animals. DDT also has adverse effects on many seabirds because it prevents proper eggshell production in birds, which results in very brittle eggs that are then easily broken.

The specific food chain we will be focusing on contains organisms common in lake and river ecosystems in North America:
algae → minnows → sunfish → osprey

Procedure:
Place all 30 of the M&Ms into the "shaker" cup to represent the producer population.

1. In the data table, record the amounts of DDT per producer (algae) in the "Trial 1" column. (10 contaminants per 30 producers total = 10/30 = .33)
2. Give the shaker cup a good shake.
3. Now simulate minnows eating some of the producers. Each minnow eats 3 algae, so randomly remove 3 M&Ms from the "shaker" cup and place them into one of the small cups (minnows).
4. Repeat step 3 for the remaining eight small cups (minnows). In your data table, record the number of DDT particles present within the groups of 9 minnows in the "Trial 1" column. (The number will be out of—divided by—9.)
5. Now simulate the sunfish eating minnows. Each sunfish needs to eat 2 minnows. Randomly select 2 small cups (minnows) and empty the contents into one of the medium cups (sunfish). This will represent the DDT particles contained in the minnows being ingested by the sunfish that eat the minnows. Repeat for the remaining medium-size cups (sunfish). In your data table, record the number of DDT particles contained within the group of 3 sunfish in the "Trial 1" column. (The number will be out of—divided by—3.)
6. Finally, simulate an osprey eating sunfish. One osprey needs to consume 2 sunfish. Randomly select 2 medium-size cups (sunfish) and empty the contents into the large cup (osprey). In your data table, record the amount of DDT in the osprey in the "Trial 1" column. (The number will be out of—divided by—1, so in this case, just the total number counted.)
7. Empty the contents from all cups back into the "shaker" and repeat all steps two more times, recording the data in "Trial 2" and "Trial 3" columns.
8. Calculate the average amount of DDT for each organism. To do this, calculate the percentage for each organism in each trial by dividing. For each organism, add the percentages together and divide the total by 3 for the average PERCENTAGE. Multiply the average percentage by the total number of organisms for the average DDT particles per individual organism.

So for example, if you recorded for minnows:
3 particles out of 9 in trial 1, 4 out of 9 in trial 2, and 5 out of 9 in trial 3
You will have:
Organism | Trial 1 | Trial 2 | Trial 3 | Average (DDT/organism)
Minnow | 3/9 = .33 | 4/9 = .44 | 5/9 = .56 | .44
.33, .44, and .56 are the amounts of DDT particles per individual in each trial.
Now, calculate the average DDT particles per organism:
.33 + .44 + .56 = 1.33
(divide by 3 to average, so 1.33/3 = .44)

Amount of DDT Per Organism
Organism | Trial 1 | Trial 2 | Trial 3 | Average (DDT/organism)
Algae
Minnow
Sunfish
Osprey

Analysis Questions:
1. What happened to the amount of DDT per organism as you move up the food chain?
2. Which organism contained the largest concentration of DDT?
3. In what ways is DDT harmful to osprey?

In addition to the organisms from our simple example above, the organisms in the following relationships are also part of a more complex food web containing these same organisms.
Zooplankton eat algae
Minnows eat zooplankton, algae, and insect larvae
Largemouth bass eat sunfish and minnows
Sunfish eat minnows and insect larvae
Osprey eat largemouth bass
Insect larvae eat zooplankton and algae
Snapping turtles eat minnows and sunfish

4. Based on the information above, what is one other organism besides osprey that you would expect to have high concentrations of DDT? Explain why.
5. If the osprey population were to decrease due to DDT, what are two other populations that would be affected and how?

The following diagram illustrates a typical marine food web. Mercury, a toxic heavy metal, can also accumulate in the fatty tissue of organisms.

6. Which organism from the food web above would you predict to have the highest level of mercury? Explain your answer.
7. Of the following types of fish, which would most likely be the safest to eat? Explain your answer.
Mackerel | Snapper | Mako Shark
Bluefish

Added by Ramon M.

EXISTING SUBSTITUENTS DIRECT THE POSITION OF ELECTROPHILIC AROMATIC SUBSTITUTION An existing group on a benzene ring can direct the position of a new group added by an electrophilic aromatic substitution (EAS) reaction. Ortho-para directors predominantly direct the incoming group to an ortho or para position. (FIGURE CANNOT COPY) G is an ortho-para director. Ortho product Para product Meta directors predominantly direct the incoming group to a meta position. (FIGURE CANNOT COPY) G is a meta director. Meta product These directing effects stem the initial substituent's electron-donating or electron withdrawing properties. Electron-donating and electron-withdrawing effects are caused by resonance, induction, or both. Of these two, when both effects are possible, resonance plays a greater role than induction. As we shall also see, electron-donating groups activate (enhance) the reactivity of a ring toward EAS, and electron-withdrawing groups deactivate (reduce) EAS reactivity. A Ortho-Para Directors The unshared electron pair of an existing ring substituent causes electrophilic aromatic substitution to take place at the ortho and para positions. Para substitution usually predominates over ortho due to steric effects. For example, acetanilide undergoes nitration primarily at the para position, with only a trace of the ortho product formed. (FIGURE CANNOT COPY) Acetanilide $p$ -Nitroacetanilide $(90 \%)$ o-Nitroacetanilide (trace) Strongly activating groups can cause substitution at both the ortho and para positions. For example, aniline and phenol both react with bromine to produce products that are fully substituted at their ortho and para positions. Amino and hydroxyl substituents are strong activating groups. (FIGURE CANNOT COPY) 2,4,6 -Tribromoaniline $(-100 \%)$ $2,4,6-$ Tribromophenol $(-100 \%)$ We can account for the predominance of ortho and para versus meta substitution, as well as the greater reactivity of aniline and phenol relative to acetanilide, by considering resonance structures for the arenium ions that would result from attack on aniline by a generic electrophile $\mathrm{E}^{+}$ at the ortho, meta, and para positions, as shown here. Ortho Attack (FIGURE CANNOT COPY) Meta Attack (FIGURE CANNOT COPY) Para Attack (FIGURE CANNOT COPY) Relatively stable contributor As we see here, four reasonable resonance structures can be written for the arenium ions resulting from ortho and para attack, whereas only three can be written for the arenium ion that results from meta attack. This observation, in itself, suggests that the ortho-and para-substituted arcnium ions should be more stable. Of greater importance, however, are the relatively stable structures that contribute to the hybrid for the ortho- and para-substituted arenium ions. In these structures, nonbonding pairs of clectrons from nitrogen form an additional covalent bond to the carbon of the ring. This extra bond-and the fact that every atom in each of these structures has a complete outer octet of electrons makes these structures the most stable of all of the contributors. Because these structures are unusually stable, they make a large - and stabilizing - contribution to the hybrid. This means, of course, that the ortho- and para-substituted arenium ions themselves are considerably more stable than the arenium ion that results from the meta attack. As a result, electrophiles react at the ortho and para positions very rapidly. Use resonance theory to explain why the hydroxyl group of phenol is an activating group and an ortho-para director. Illustrate your explanation by showing the arenium ions formed when phenol reacts with a Br' ion at the ortho, meta, and para positions. We can account for the lower reactivity of acctanilide relative to aniline and phenol on the basis of resonance involving the carbonyl group in acctanilide. In acctanilide the unshared electron pair on the nitrogen can be delocalized into the carbonyl of the amide functional group, thus diminishing availability of the electron pair for contribution to the resonance structures of the arenium ion. In aniline or phenol, this resonance effect is not possible due to absence of the carbonyl group. Thus, the amide group of acctanilide is less activating than the amino group of aniline or the hydroxyl group of phenol. (FIGURE CANNOT COPY) Electron donation to the ring by resonance is reduced when there is an alternative resonance pathway away from the ring. Phenol reacts with acetic anhydride in the presence of sodium acetate to produce the ester phenyl acetate: (FIGURE CANNOT COPY) Phenol Phenyl acetate The $-\mathrm{OCOCH}_{3}$ group of phenyl acetate, like the $-\mathrm{OH}$ group of phenol (Practice Problem 15.6 ), is an ortho-para director. (a) What structural feature of the $-$ OCOCH $_{3}$ group explains this? (b) Phenyl acetate, although undergoing reaction at the ortho and para positions, is less reactive toward electrophilic aromatic substitution than phenol. Use resonance theory to explain why this is so. Halogen atoms are also ortho-para directors by resonance, but they deactivate the ring toward electrophilic aromatic substitution by their electronegativity. Their unshared clectron pairs stabilize the arenium ion by resonance, but this stabilization is limited by the mismatch in atomic size and orbital overlap between carbon and the halogens (except for fluorine), versus the more effective resonance orbital overlap between carbon and oxygen or nitrogen. Thus, halogens are deactivating because their electronegativity destabilizes the arenium ion, yet they are ortho-para directors as a result of resonance. Alkyl groups are ortho-para directors, as well. Alkyl groups stabilize the arenium ion through inductive release of electrons, similar to the way an ordinary carbocation (e.g., a carbocation that is not conjugated like an arenium ion) is stabilized by adjacent alkyl groups. We find, for example, that when toluene undergoes electrophilic substitution, most of the substitution takes place at its ortho and para positions. When we nitrate toluene with nitric and sulfuric acids, we get mononitrotoluenes in the following relative proportions: (FIGURE CANNOT COPY) o-Nitrotoluene $(59 \%)$ $p$ -Nitrotoluene $(37 \%)$ $m$ -Nitrotoluene $(4 \%)$ Of the mononitrotoluenes obtained from the reaction, $96 \%$ ( $59 \%+37 \%$ ) have the nitro group in an ortho or para position. Only $4 \%$ have the nitro group in a meta position. Predominant substitution of toluene at the ortho and para positions is not restricted to nitration reactions. The same behavior is observed in halogenation, sulfonation, and so forth. To understand the predominance of ortho and para substitution in toluene, let's consider the arenium ions that would result from attack on toluene by a generic electrophile $\mathrm{E}^{+}$ at the ortho, meta, and para positions. The trifluoromethyl group is a meta dircctor. Explain why this is so using resonance structures for the arenium ions that would result upon ortho, meta, and para bromination of trifluoromethylbenzene. (TABLE CANNOT COPY) Label each of the following aromatic rings as activated or deactivated based on the substituent attached, and state whether the group is an ortho-para or meta director. (FIGURE CANNOT COPY) STRATEGY AND ANSWER: If a substituent donates electron density it will activate the ring and cause ortho and para substitution. If a substituent withdraws electron density it will deactivate the ring and cause meta substitution (except for halogens, which are electron withdrawing but cause ortho-para substitution). (a) Activated; an ether is an ortho-para dircctor; (b) deactivated; the ester carbonyl is a meta dircctor; (c) activated; the single-bonded oxygen of the ester is directly bonded to the ring, and therefore it is an ortho-para dircctor; (d) activated; the amide nitrogen is an ortho-para director; (e) deactivated; however, the halogen is ortho-para director through resonance; (f) deactivated; the sulfonate group is a meta director. Predict the major products formed for each of the following. If the major products would be a mixture of ortho and para isomers, you should so state. (a) Toluene is sulfonated. (c) Nitrobenzene is brominated. (b) Benzoic acid is nitrated. (d) Isopropylbenzene reacts with acetyl chloride and AlCl_s.

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