INTRODUCTION
With the possible exception of some inorganic
and elemental substances, no chemical is environmentally inert.
One of the principle advantages claimed for organophosphate and
carbamate pesticides, as opposed to such ìhardî organochlorine
materials as DDT and dieldrin, is the ability of the former to
break down in only a few hours or days after application, as opposed
to months or years for the latter. Breakdown may take place
chemically and photo-chemically in the environment, or biologically
through metabolism by plants, microorganisms, and animals. Furthermore,
most breakdown falls into the simple categories of oxidation,
reduction, hydrolysis and the like.
Among the organophosphate and carbamates,
oxidation and hydrolysis are the most common purely chemical environmental
reactions. Rainfall, atmospheric water vapor, soil moisture,
and air are the reagents. In some cases sunlight provides energy
which causes the reactions to occur at more rapid rates than
they would in the dark. To predict environmental stability then,
simple laboratory tests of hydrolysis rates are often carried
out in dilute aqueous solution. The disappearance of a chemical
by this procedure has at least an indirect bearing on its persistence
in environmental waters1 and in soil, and this could have an impact
on the choice of formulation technology and storage procedures.
Furthermore, in disposal of unwanted or accidentally spilled
materials, hydrolysis under various conditions offers a convenient
means of decontamination2.
A common organophosphate, such as parathion,
hydrolyzes in basic media by one of the following two pathways:
The P=S may be oxidized to the oxygen analogue,
which is quite rapidly hydrolyzed; or the P=S compound may undergo
direct (but slower) hydrolysis itself. An example of the relative
hydrolysis rates of P=S and P=O compounds, and the experimental
determination of rate constants, half lives, and activation energies
is given by Gomaa et al3.
In the present experiment, the rates of
hydrolysis of three representative organo-phosphate insecticides
will be determined under the "artificial" conditions
of high pH and aqueous alcohol solvent. These conditions are vigorous
enough to cause complete hydrolysis of parathion, one of the
more stable organophosphorus compounds, in a few hours2. The
goal of the experiment is to illustrate laboratory techniques
appropriate to the determination of hydrolysis rates, and to allow
a comparison of stability of related chemicals. Though the conditions
are artificial, the rates may be used as a first approximation
to predict relative persistence under some environmental conditions.
References
1. J.W. Eichelberger and J.J. Lichtenberg.
Persistence of pesticides in river water. Environ Sci Technol,
5:541, 1971.
2. D.P.H. Hsieh et al. Decontamination
of noncombustible agricultural pesticide containers by removal
of emulsifiable parathion. Environ Sci Technol, 6:826,
1972.
3. H.M. Gomaa, I.N. Suffet, S.D. Faust.
Kinetics of hydrolysis of diazinon and diazoxon. Residue
Reviews, 29:171, 1969.
4. N.M. Gomaa, S.D. Faust. Chemical hydrolysis
and oxidation of parathion and paraoxon in aquatic environments.
In: ìFate of Organic Pesticides in the Aquatic
Environmentî ACS Advances in Chemistry series 111, Chapter
10, 1972.
Procedure
The procedure is based on the relative
hydrolysis of three organophosphorus insecticides, methyl and
ethyl parathion and diazinon, but it is applicable to many chemicals.
Allow the prepared 50 % aqueous solution
that is 1 N in NaOH to equilibrate in a water bath at
room temperature for 15 minutes; measure and record the bath temperature.
Prepare 7 screw-top 15 ml calibrated test tubes by adding ca
4 ml of saturated aqueous sodium chloride solution, 3.0 ml of
ethyl acetate, and place a piece of label tape to each tube.
From the standard solution of mixed pesticides
in ethanol (standard contains diazinon, ethyl parathion , and
methyl parathion), transfer 1.0 mg diazinon, 1.5 mg ethyl parathion,
and 2.0 mg methyl parathion into a 10 ml volumetric flask using
a 1.0 ml pipette. After the pesticides have been added, add the
1 N NaOH water: ethanol solution just to the mark, mix
well, and place in the water bath.
Remove a 1.0-ml aliquot within 3 minutes
of mixing and place it into one of the test tubes; this sample
should be considered t=0 minutes. At recorded time intervals following
the t=0 sample (e.g. 3, 8, 15, 30 and 60 and 90 minutes) remove
1 ml samples using the same pipet and place them in the screw-top
test tubes containing the saturated salt solution and ethyl acetate.
Extract by inverting each test tube slowly for 2 minutes. Allow
the layers to separate, and remove aliquots from the organic fraction
(top or bottom layer?) for direct injection into the HPLC.
Analysis
Analysis for the amount of parent compound
remaining will be performed by HPLC employing a reverse phase
C18 column, a water:acetonitrile mobile phase, and a UV detector
set at the wavelength of your choice. Samples are quantitated
using a regression equation that you will generate from known
standards.
To familiarize yourselves with the instrument
and to obtain a standard curve, inject 2.5, 5, 7.5, and 10 ml
of the mixed standard solution provided; for your samples inject
10 ml.
Use your time efficiently! One
partner should start the HPLC work immediately while another
prepares the hydrolysis reaction mixtures.
Calculations
Obtain the peak area (or peak height) for
the unreacted methyl and ethyl parathion and diazinon for each
sample, and place these in a table with the time the aliquot was
taken. Calculate the concentration of the parent compound in
the reaction flask, and determine C/Co values for each time point.
Plot the natural log of the fraction remaining (C/Co) vs. time
on regualr graph paper; place all three curves in the same graph.
Obtain the 1st order rate constants (kobs) from the resulting
plots and Calculate t1/2 (half-lifes) (include units).
Of course doing this with any available computer program such
as Cricket Graph (Macintosh) greatly facilitates this process.
Addendum: First Order Reactions
A first order reaction is defined as a
chemical reaction in which the rate depends only on the first
power of the concentration of a single reacting species at a given
temperature. In this case, the reaction rate of hydrolysis is
dependant only on the temperature of the reaction mixture and
the concentration of the organophosphate pesticide. The hydroxide
ion concentration is in large excess; therefore it remains essentially
constant throughout the experiment. If the volume is constant,
which it is in this experiment, and the concentration of the reacting
speciesóthe organophosphate pesticideóis represented
by ìcî, then we can write the first order rate law
as:
The rate constant (k) has the units of reciprocal
time (t-1).
Since experimental results obtained in the study of reaction rates are usually values of ìcî (or some quantity such as peak area related to ìcî) found at various times, the data can be more readily dealt with by integrating the first order rate law from t0 to t1 and c0 to c1. If the initial concentration at some time t0 is c0, then at a later time t the concentration will have fallen to c. Integration gives:
If a plot of ln c vs. t gives a straight
line the reaction can be said to be first order and can be represented
in the form of a linear regression, y = mx + b, where m is the
slope and b is the y-intercept.
Rearranging gives a simpler form:
This equation should be used for your graphing
purposes. The slope is then equal to -k.
Pseudo forst order reactions, derivation
of half-lives, leaving group ability, etc. will be discussed during
the lab period (be prepared).
Lab Report
In the discussion section of your lab report
include answers or discussion relevent to the following questions.
1. Give the equations you used to calculate
kobs and t1/2, and define each term.
2. Write a balanced chemical equation for
the hydrolysis of one of the compounds.
3.
a. If the conc. of base was doubled in
the reaction, how does it affect the observed rate constant? The reaction rate?
b. How does temperature affect the rate constant? Why?
c. If the reaction were done in water
instead of 50% ethanol, do you think the reaction rate would be different? Would it be
slower or faster? Why?
4. Why does the solution turn yellow when
the base and standards are mixed?
5. Why is saturated NaCl used in the workup
procedure?
6. What mechanistic reason can you give
for the difference in hydrolysis for the three compounds?