Solutions: Environmental Chemistry - a Global Perspective 4Th Edition
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Solutions: Environmental Chemistry - a global perspective 4th Edition Chapter 20 Synthetic organic chemicals PROBLEMS/SOLUTIONS 1. Consider the structure of the pesticide metolachlor and indicate structural features that would be involved in binding with organic matter in the soil. Solution Metolachlor would be associated with the humic material within the soil. The aromatic ring portion of the structure would likely be associated with similar ring structures within the humic material (for pi-bond stacking interactions) and the chain portions of the molecule could align with more hydrophilic structures where the lone pairs of oxygen, nitrogen and chloride may H-bond and allow for other van der Waals type of interactions. Experimental results confirm that metolachlor absorbs weakly to moderately to soil and the leaching of metolachlor from soil is high to medium. Volatilization of metolachlor from poorly adsorbing moist soil to the atmosphere may be significant especially if assisted by solar heating and high winds. Depending on the nature of soil and climatic conditions, the average field t1/2 of metolachlor in soil 30 to 90 days. (reference http://www.speclab.com/ using the ‘Interactive List of Services’ link leading to the ‘searchable database’ Search metolachlor, and choose 3rd result: http://www.speclab.com/compound/c5121845.htm) 2. Compare the chemical forces with respect to the retention of the pesticides dieldrin and malathion by soil organic and mineral phases. Solution Soil organic matter would bind dieldrin more strongly, since dieldrin has a very pronounced hydrophobic portion within its structure. Chlorinated hydrocarbons are strongly lipophilic. The oxygen at one end of the molecule could also play a role in hydrogen bonding to soil organic matter. Therefore, van der Waals type 170 Solutions: Environmental Chemistry - a global perspective 4th Edition Chapter 20 Synthetic organic chemicals forces would act to retain dieldrin within the soil. The tendency to bind to soil organic matter is reflected in the large log KOW value (5.4). Malathion on the other hand is more hydrophilic and would tend to associate to a lesser degree with the organic phase of the soil. The log KOW is somewhat smaller (2.75). The polar portions of the malathion molecule might interact to some degree with polar and ionic groups with the mineral portion of the soil, but will also interact favourably with water. In summary, the overall retention of dieldrin by the soil is expected to be greater than retention of malathion. 3. A rainstorm causes the water to penetrate 5 cm into a soil containing 3.6% OM. Qualitatively predict the relative extent of downward movement of the herbicides aldicarb and trifluralin if they have been applied to the soil just before the rain begins. Solution Aldicarb is in a class of pesticides known as carbamates. Therefore it has the following typical properties: -1 -1 KOC = 50 – 150 mL g Kd = 0.5 – 1.5 mL g Rf = 0.7 – 0.4 high mobility soil -1 Specifically the t1/2 = 7 d and KOC = 17 mL g with GUS = 2.34 Trifluralin is a much less mobile compound. soil -1 Specifically the t1/2 = 83 d and KOC = 8000 mL g with GUS = 0.66 Qualitatively, based solely on the comparison of the KOC values (a small value favours leachability), aldicarb would move down the soil further than trifluralin. Trifluralin would likely be substantially immobilized. 4. The presence of a halogen atom at an odd-numbered carbon, with respect to a primary substituent site on a benzene ring, reduces the stability of the compound in comparison to a situation where the halogen is on an even-numbered carbon. Explain. Solution Electrophilic substitution A substituted ring is classified as being activating if it is more reactive than benzene and deactivating if the ring is less reactive than benzene. The halogens in being classified in this way end up in a class by themselves. They are known as being deactivating, but ortho and para-directing. Other ortho and para- directors include NH2, OH, OCH3, and CH3. (all activating). 171 Solutions: Environmental Chemistry - a global perspective 4th Edition Chapter 20 Synthetic organic chemicals Meta directors include NO2, CN, SO3H and CHO to name a few and are all deactivating. When there are two substituents are on a benzene ring, they may be oriented in such a way that one position reinforces that of the other. When they oppose each other in terms of directive effects (i.e. ortho, para or meta directive) it becomes more difficult to predict which products will predominate. But generally, strongly activating groups will win out over deactivating or weakly activating groups. As well, a second generalization which answers the question above is that there is often little substitution between two groups that are meta to each other. So the molecule containing the halogen in the meta position, 2 or 4 (from a primary substituent) will be less reactive than those with a halogen in positions 1, 3 or 5 (ortho and para) from a primary substituent. This generalization however, may be complicated with having the primary constituent being activating ortho, para or deactivating meta or deactivating ortho, para (i.e. the primary constituent is a halogen) compared to the halogen. Nucleophilic substitution In nucleophilic substitution for para and ortho attacks (at the halogen position) relative to the primary substituent, an especially stabilized carbocation results over that of the meta attack. As a consequence of the more stabilized carbocation, the para and ortho isomers react faster than the meta isomer. Thus, the presence of a halogen atom at an odd numbered carbon, with respect to a primary substituent site on a benzene ring reduces the stability of the compound in comparison to a situation where the halogen is on an even numbered carbon. 5. The half-life (t1/2) for the hydrolytic degradation of the carbamate insecticide carbaryl is reported to be 31 days at 6ºC and 11 days at 22ºC. Calculate the activation energy for the reaction. Write an equation for the hydrolysis of this pesticide. Solution carbaryl t1/2 = 31 d at 6ºC & t1/2 = 11 d at 22ºC Calculate the activation energy (Ea) for the reaction. Assume that the hydrolysis reaction follows pseudo first order kinetics. -Ea/RT k = Ae t1/2 = ln 2 / k 172 Solutions: Environmental Chemistry - a global perspective 4th Edition Chapter 20 Synthetic organic chemicals Determination of activation energy -2.6 -2.8 -3 -3.2 ln k -3.4 -3.6 -3.8 0.003350.00340.003450.00350.00355 0.0036 1/T k at 6ºC is 0.0224 d-1 & k at 22ºC is 0.0630 d-1 Rearrange the Arrhenius equation by taking the natural log (ln) of both sides to give: ln k = ln A – Ea / RT or written in the form of a straight line (y = mx + b) leads to: m = –5350 K –Ea 1 ln k = -------- x ------ + ln A R T y = m x + b Plot ln k vs. 1 / T : the slope of the line is m = –Ea / R (shown above) Using the data from above (assuming a straight line) slope = Δln k / Δ(1/T) slope = –5350 = –Ea / R therefore Ea = 44 484 J = 44.5 kJ The activation energy for the reaction is 44.5 kJ and the pre-exponential factor A = 4.75 x 106 d-1. An equation for the hydrolysis is shown below. 173 Solutions: Environmental Chemistry - a global perspective 4th Edition Chapter 20 Synthetic organic chemicals O O C NH CH3 OH + H O CH NH + + CO 2 3 2 2 carbaryl 6. At 22ºC, the organophosphorus compound diazinon has a degradation t1/2 of 80 days in river water and 52 days in the same water after filtration. In contrast, the triazine cyprazine has corresponding t1/2 values of 190 days and 254 days. Suggest reasons why filtration increases t1/2 in one case and decreases it in the other. Solution Organophosphorus compound – diazinon t1/2 = 80 d (in river water) & t1/2 = 52 d (in filtered river water) and triazine compound – cyprazine t1/2 = 190 d (in river water) & t1/2 = 254 d (in filtered river water) Diazinon has a moderate affinity for organic matter based on the log KOW value of 3.3 for this organophosphate. As such the removal of particulate organic matter through filtration of the river water would eliminate the stabilizing effect provided by interactions between the diazinon and OM. This would allow for a more rapid degradation of diazinon in the filtered river water. This accounts for the decrease in the measured t1/2 from 80 days to 52 days. For diazinon, hydrolysis has been reported to be slow at pH > 6, but may be significant in some soils. Hydrolysis may be a significant fate process with reported t1/2 of 31 days at pH 5, 185 days at pH 7.4, and 136 days at pH 9.0 at 20ºC. In distilled water at pH 6 at room temperature t1/2 was 14 – 21 days. Major products of hydrolysis are 2-isopropyl-4-methyl-6-hydroxypyrimidine and diethyl thiophosphoric acid or diethyl phosphoric acid. (reference http://www.speclab.com/ using the ‘Interactive List of Services’ link leading to the ‘searchable database’ Search diazinon, and choose 4th result: http://www.speclab.com/compound/c959988.htm) 174 Solutions: Environmental Chemistry - a global perspective 4th Edition Chapter 20 Synthetic organic chemicals X OH N N N N + H2O + HX (X = Cl, OCH3, SCH3) N R' R" R' N R" Cyprazine – 6-Chloro-N-cyclopropyl-N’-(1-methylethyl)-1,3,5-triazine-2,4-diazine. While organic matter can stabilize some molecules, it can enhance hydrolysis in other cases.