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 t 1/2 of metolachlor in soil 30 to 90
days.
(reference using the ‘Interactive List of Services’ link leading to the ‘searchable
database’ Search metolachlor, and choose 3rd result: />
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

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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:
KOC = 50 – 150 mL g-1

Kd = 0.5 – 1.5 mL g-1

Specifically the t1/2soil = 7 d and KOC = 17 mL g-1

Rf = 0.7 – 0.4

high mobility


with GUS = 2.34

Trifluralin is a much less mobile compound.
Specifically the t1/2soil = 83 d and

KOC = 8000 mL g-1

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 paradirectors include NH2, OH, OCH3, and CH3. (all activating).

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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.
k = Ae-Ea/RT

t1/2 = ln 2 / k

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Determination of activation energy

-2.6
-2.8

ln k

-3
-3.2
-3.4
-3.6

-3.8
0.003350.00340.003450.00350.003550.0036
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:
–Ea
1
ln k = -------- x -----R
T


y

=





m

x


+ ln A


+

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.

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O
O C NH CH3
+ H2O


OH
CH3NH2

+

+ CO2

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

and

t1/2 = 80 d (in river water)
& t1/2 = 52 d (in filtered river water)

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 K OW 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 t 1/2 was 14 – 21 days. Major
products of hydrolysis are 2-isopropyl-4-methyl-6-hydroxypyrimidine and diethyl thiophosphoric acid or
diethyl phosphoric acid.
(reference using the ‘Interactive List of Services’ link leading to the ‘searchable
database’ Search diazinon, and choose 4th result: />
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X
N

N
N

R'

OH
N

+ H2O
R"


R'

N
N

+ HX

(X = Cl, OCH3, SCH3)

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. Atrazine is a
related triazine pesticide. Its rate of hydrolysis has been found to significantly increase upon the addition of
a small amount humic matter, indicating that atrazine hydrolysis could be catalyzed. For example, the t 1/2
of atrazine at 25ºC and pH of 4 was 244 days without an additive and 2 days with the presence of 2%
humic acid. At 25ºC, a 5 mg.L-1 solution of fulvic acid (naturally occurs in soils and most surface waters,
including rivers) resulted in t1/2 of 35, 174, 398 and 742 days at pHs of 2.9, 4.5, 6.0 and 7.0, respectively.
Hydrolysis of atrazine followed first order kinetics producing hydroxyatrazine as the transformation
product. One could speculate that there is a similar effect of OM on cyprazine hydrolysis.
(reference using the ‘Interactive List of Services’ link leading to the ‘searchable
database’ Search cyprazine, and choose the only result />
7.

Predict chemical degradation processes that might occur for the herbicide metribuzin.

Solution
Degradation processes for metribuzin.
If metribuzin is released to the atmosphere, degradation of vapour phase metribuzin by reaction with

photochemically produced hydroxyl radicals (estimated t1/2 of 11 hrs) will be important. Metribuzin can be
removed from air via rainfall and particulate phase metribuzin may be removed from air via dry deposition.
If released to soil, biodegradation will be the primary fate process. Metribuzin is moderately adsorbed (K OC
of 95) on soils with high clay and organic content by a H-bonding mechanism. Adsorption decreases with
an increase in soil pH. Little leaching occurs on soils with high organic content, but metribuzin is readily
leached in sandy soils. The soil t1/2 is in the range of 14-60 days. In water, biodegradation may be
important based on soil studies. Activity of soil micro-organisms, higher temperatures and aerobic
conditions increase the rate of metribuzin biodegradation. Slow hydrolysis (t 1/2 of 90 days ) may also be an

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important removal mechanism in conjunction with biodegradation. Degradation of metribuzin takes two
forms – deamination, or loss of the NH2 group, and oxidation resulting in displacement of the –SCH3 by
forming a diketo compound.
(reference using the ‘Interactive List of Services’ link leading to the ‘searchable
database’ Search metribuzin, and choose the 6th result />
8.

The values of log KOM for the insecticide lindane and the herbicide 2,4-D are 2.87 and 1.54,
respectively. For each pesticide, suppose rain water dissolves 12 ppb from the plants and surface soil
and then carries it downward. If the soil has 2.3% organic matter, calculate the equilibrium
concentration for each pesticide in the soil.

Solution

Lindane (C6H6Cl6) M.W. = 290.83 g mol-1, Caq = 12 ppb = 4.13 x 10-8 mol L-1
Let Cs (mol g-1) represent the concentration of the substrate in the soil.
KOM = COM / Caq and Cs = fOM x KOM x Caq where Cs has the units of mol g-1; Caq has units of mol L-1;
fOM is the fraction of organic matter in the soil; and KOM is the partitioning constant for the specific organic
substrate.
Log KOM = 2.87, therefore KOM = 741.3
Cs = 0.023 x 741.3 x 4.13 x 10-8 mol L-1
= 7.04 x 10-7 mol g-1
For 1.0 g of soil, 7.04 x 10-7 mol of lindane is present, which is equal to 205 µg of lindane.
The equilibrium concentration in the soil can be expressed as 205 μg g-1 (ppm).
2,4-D (C8H6Cl2O3) M.W. = 221.04 g mol-1, Caq = 12 ppb = 5.43 x 10-8 mol L-1
Let Cs (mol g-1) represent the concentration of the substrate in the soil.
KOM = COM / Caq and Cs = fOM x KOM x Caq where Cs has the units of mol g-1; Caq has units of mol L-1;
fOM is the fraction of organic matter in the soil; and KOM is the partitioning constant for the specific organic
substrate.
Log KOM = 1.54, therefore KOM = 34.7
Cs = 0.023 x 34.7 x 5.43 x 10-8 mol L-1
= 4.33 x 10-8 mol g-1
For 1.0 g of soil, 4.33 x 10-8 mol of 2,4-D is present, which is equal to 9.6 µg of 2,4-D.
The equilibrium concentration in the soil can be expressed as 9.6 μg g-1 (ppm).

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9.

Synthetic organic chemicals


Using Additional Resources 4, look up the structures and log K OW values of the ‘Dirty dozen’.
Describe the general correlation that exists between log KOW and the number of chlorines in the
individual molecules.

For classroom discussion, no solution provided. Use for a class assignment and discussion. Depending
on student numbers, I usually assign each student only one and ask for more physical data and have them
present these in class.
10.

Calculate the amount of toluene (log KOW = 2.69) that would bioconcentrate in a 1.0 kg fish using the
following data and approximations. The lake in which the fish lives is contaminated with toluene at a
level of 10 µg L-1, due to having a gasoline station and marina in close proximity for many years.
Equilibrium has been attained and the fish proximate a composition is 9% fat, 4% bone and other nonabsorbing material, 7% air, and 80% aqueous. Assume that the fat is equivalent to octanol. How might
this result compare to a biomagnification value for this same fish?
Repeat this hypothetical calculation with one of the ‘Dirty dozen’ persistent organic pollutants having
a log KOW = 5.0 using a concentration of 10 ng L-1 (note much lower concentration). How does this
result compare to the result obtained for toluene?

Solution
The air component requires Henry’s Law constant for toluene (KH = 0.271 mol L-1a mol-1 Lw )
We will treat the fat component as if it were octanol (and use KOW = 490)
The aqueous component is assume to be equal to the aquatic environment (10 µg L-1)
We will assume the bone and other components are non- partitioning and that the density of the fish is 1.2
kg L-1.
The mass of toluene that will bioconcentrate (partition) into the fish can be calculated as follows:
Mass of toluene = 0.07 x (1.0 / 1.2) L x 0.271 mol L-1a mol-1 Lw x 10 µg L-1
+ 490 x 10 µg L-1 x 0.09 x (1.0 / 1.2) L
+ 10 µg L-1 x 0.80 x (1.0 / 1.2) L
= 0.16 µg + 367.5 µg + 6.67 µg
= 374 µg

It is clear from this calculation that the majority of the toluene is concentrated into the fat of the fish with
the overall average of toluene in the fish being 374 ppb.
Biomagnification would cause these levels to be even higher, as more toluene would accumulate as a result
of the fish feeding on other organisms that have also accumulated toluene at elevated levels.

a

Composition described in terms of classes of substances present is generally referred to as proximate composition,
because the classes or groups, for example fats or minerals, are those first arrived at in the process of analysis; in
proximate analysis the groups are measured as such, rather than as individual proteins or specific minerals. Food and
Agriculture Organization of the United Nations, FAO Corporate Document Repository
accessed November 2016. We will use the proximate
composition of our fish to be fats, minerals (bone), water, air and other.

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Using a generic POP value of log KOW = 5.0 and only considering the fat and aqueous components we
obtain:
Mass of a generic POP = 100 000 x 10 ng L-1 x 0.09 x (1.0 / 1.2) L
+ 10 ng L-1 x 0.80 x (1.0 / 1.2) L
= 75000 ng + 6.67 ng
= 75 µg
The significant size of the KOW value clearly dominates this calculation and significantly increases the
concentration of the POP in the fish, despite it being in the water at a concentration three orders of

magnitude lower than toluene.

11. The Equation (20.52) for the groundwater ubiquity score (GUS) has been modified from that given in
the reference to use KOM values in place of KOC values. Show that the two forms of the equation are the
same.
Solution
GUS = log10(t½soil) x (3.77 – log10(KOM))

Equation 20.52

GUS = log10(t½soil) x (4.00 – log10(KOC))

from reference

KOC is an alternative parameter for describing partitioning of a species between organic matter and water,
assuming that organic carbon makes up 60% of organic matter.
OM = 1.7 x OC
and
KOC = 1.7 x KOM
Using chlorpyrifos as an example, the KOM value is 3600, so KOC = 1.7 x 3600 = 6120 ≈ 6100
Since it is the log of these two values used in the two equations given above, the constant (4.00) must also
be corrected by log 1.7 = 0.23 (4.00 – 0.23 = 3.77)
Example calculation of GUS for Chlorpyrifos:
using Equation 20.52

GUS = log10(t½soil) x (3.77 – log10(KOM))
= log10 (54) x (3.77 – log10(3600)
= 1.73 x (3.77 – 3.556)
= 0.37


using reference eqn.

GUS = log10(t½soil) x (4.00 – log10(KOC))
= log10 (54) x (4.00 – log10(6100)
= 1.73 x (4.00 – 3.785)
= 0.37

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12. The organophosphorus pesticide phosmet is used as an insecticide for treating many orchard crops.
Consult the Extoxnet website (Additional Resource 4), and using the representative half-life in soil,
calculate the GUS score. What is the significance of this score and what are the limitations in its
interpretation?
Solution
The search result for phosmet at the Extoxnet website gave 10 results, and using result 8, National Pesticide
Information Centre – OSU Extension Pesticide Properties Database ( />accessed November 2016) the half-life was reported as 19 days and KOC = 820.
KOM = KOC / 1.7 = 820 / 1.7 = 482
GUS =
=
=
=

log10(t½soil) x (3.77 – log10(KOM))
log10(19) x (3.77 – log10(482))

1.28 x (3.77 – 2.68)
1.4

This calculation places the pesticide phosmet just into the limited leachability category (1.75 to –6.00). Due
to the nature and variability of the natural environment and conditions that exist within, this calculation
may not adequately describe specific situations. In other words, other conditions may make this pesticide
more or less leachable than what the GUS value has predicted.

13. A triazine pesticide is present in groundwater. In this particular environment, the Kd value for the
pesticide is 6.5 mL g-1. The soil has a particle density of 2.7 g mL-1 and a porosity of 44%. The lateral
flow rate of the groundwater is 0.75 m d-1. Estimate how far the pesticide will move in one year.
Solution
The water will move 0.75 m d -1 x 365 d y-1 = 273.75 m y-1
Following Example 20.2, and using Equation 20.51
Rf = 1 / (1 + 6.5 mL g-1 x 2.7 g mL-1 x [(1-0.44)/0.44]) = 0.04284
The distance the pesticide will travel is:
273.75 m y-1 x 0.04284 = 11.7 m y-1
The pesticide will travel about 12 m in one year, assuming it doesn’t degrade.

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