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LS1a Fall 2014
Questions
Answers are in the back
Additional Practice ICE #2
I. Basic Concept Questions
1. Consider the following single-step reaction mechanism:
a. Write the rate expressions for the forward and reverse reactions.
b. If the reaction has already reached equilibrium, how would the rate of the
reverse reaction be affected immediately after doubling the concentration
of AB? Briefly explain your answer.
c. If the reaction has already reached equilibrium, how would the rate of the
forward reaction be affected immediately after tripling the concentration of
AB? Briefly explain your answer.
d. As the temperature increases, do the reaction rates in the forward and
reverse directions increase, decrease, or stay the same? Briefly explain your
reasoning.
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e. Initial reaction rates were measured for the reaction at two different
temperatures. Both measurements were made with initial conditions such
that [A2] = 6.30 M, [B2] = 2.6 M and [AB] = 2.8x10-2 M. At 500 K, the
observed initial forward reaction rate was 4.5x10-19 M/s. At 400 K, the
observed initial forward reaction rate was 6.4x10-20 M/s. Based on these
measurements, calculate the forward rate constant at 400 K and the forward
rate constant at 500 K. Be sure to include units in your answer.
2. Draw a reaction coordination diagram for a three-step reaction with the following
characteristics:
a. Products have higher energy than starting materials but lower energy than
both intermediates
b. The first intermediate is higher in energy than the second intermediate
c. G‡Step 1  G‡Step 2  G‡Step 3
Legibly label the axes, products, reactants (starting material), transition states,
intermediates, Grxn, and G‡ for each step.
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3. Shown below are six mechanisms. Draw the products of each reaction. Identify
all atoms that are acting as nucleophiles by drawing a box around them. Identify
all atoms that are acting as electrophiles by drawing a circle around them.
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4. For each of the reactions shown below, draw the transition state that is
associated with the indicated mechanism. Indicate partial charges where
appropriate. It is not necessary to draw the products of the reaction.
5. Draw the transition states and the products of each of the reactions shown
below.
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6. For each of the reactions shown below, draw the arrow pushing mechanism on
the starting materials that are associated with the mechanism as described by
the transition state.
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II. Applied Concept Questions
7. Scientists have developed a number of laboratory methods to covalently connect
separate protein fragments. One such method is done via a two-step reaction, of
which Step 1 is shown below. In addition to generating an intermediate that
covalently attaches these two protein fragments, Step 1 also produces a
byproduct molecule.
a. (9 points) It was discovered that the rate of this reaction is strongly
dependent upon pH. This reaction was conducted separately in solutions of
pH = 7.0 and pH = 5.0 under otherwise identical conditions. The
concentration of the byproduct was monitored and plotted versus time in the
graph below.
Assuming that only one of the two mechanisms shown below can occur,
which one is consistent with these data? Briefly explain your answer and
include an explanation for why a shift in pH would affect the rate of the
reaction in the observed fashion. Be sure to discuss the rate expression for
this step in your answer.
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b. (4 points) Based on the mechanism you selected in part (a), draw the
structures of the intermediates in the space provided above.
c. (4 points) In the space below, draw the structure of the transition state
associated with Step 1 of this reaction. Label partial charges where
appropriate.
d. (4 points) Step 2 of the reaction begins with the intermediate you drew in
part (b) and proceeds through the transition state shown below to produce
the final product. On your drawing of the intermediate, draw the arrow
pushing mechanism that leads from the intermediate to this transition state.
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e. (4 points) Several of the thermodynamic values for this reaction have been
experimentally determined:
i.
ii.
iii.
iv.
ΔG‡ of step 1: 29.6 kJ/mol
ΔG‡ of step 2: 17.7 kJ/mol
ΔGrxn of step 1: -2.2 kJ/mol
ΔGrxn of step 2: -7.4 kJ/mol
Draw a reaction energy diagram showing both steps of this reaction on the
axes shown below such that the relative differences between the energies of
all starting materials, products, intermediates, and transition states are
consistent with the data given. There is no need to label specific values on
the axes, and it is not necessary to draw the graph to scale. Label starting
materials, products, intermediates, and transition states in your drawing.
8. Penicillin is an antibiotic that inhibits transpeptidase, a key enzyme for many
bacteria. Shown below are penicillin and the groups in the transpeptidase
enzyme that are involved in the inactivation of the enzyme.
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a. Draw in all the lone pair elections on the above diagram (Sulfur has an
octet).
b. Below is the mechanism by which penicillin inhibits transpeptidase. Using the
arrow pushing mechanisms shown, draw the expected products.
c. Label the electrophiles and nucleophiles directly involved in the reaction
mechanism on the above diagram.
9. The second step of gluconeogenesis is a one step reaction that is catalyzed by
phosphoenolpyruvate carboxykinase (PEPCK). Its mechanism is shown below.
a. Draw the transition state associated with this reaction.
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b. Draw the products of the reaction.
10. N-ethylmaleimide (NEM) is a reagent that is commonly used to label cysteines in
laboratory experiments. NEM reacts with cysteine via a two-step mechanism. The
reactants and both transition states are shown below.
 Draw the structures of the intermediates and products in the spaces
provided.
 Using arrow pushing notation, indicate the mechanism that leads from the
reactants to transition state 1 and from the intermediates to transition state
2.
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1. a.
Answers
rate forward  k for [ A2 ][ B2 ]
ratereverse  k rev [ AB ] 2
b. Because the reverse reaction rate is related to AB by the square of its
concentration, doubling the concentration of AB increases the rate of
the reverse reaction by a factor of 4 (22).
c. The forward reaction would not change immediately because it does
not depend on the concentration of the products.
d. Both the forward and reverse rates will increase as the temperature is
increased. An increase in temperature increases the kinetic energy of
all molecules, which increases both the collision frequency as well as
the fraction of molecules that will collide with enough energy to form
products or reactants, both of which affect reaction rate. The elevated
temperature increases the rate of the reaction by increasing the value
of the rate constant.
Equally valid answer: temperature increases both reaction rates since
𝒌 = 𝑨𝒆−(𝑬𝒂/𝑹𝑻) .
e. Ratefor = kfor * [A2][B2]
For 500 K:
4.5x10-19 M/s= kfor *(6.3 M)(2.6 M)
kfor(500K) = 2.7x10-20 M-1·s-1
For 400 K:
6.4x10-20 M/s= kfor *(6.3 M)(2.6 M)
kfor (400K) = 3.9x10-21 M-1·s-1
2.
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3.
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4.
5.
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6.
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7.a.
Mechanism 2 is consistent with these data. The reaction takes place at
a faster rate when the pH is higher. Mechanism 2 requires the sulfur on
protein B to be deprotonated, whereas Mechanism 1 requires the sulfur
to be protonated. The concentration of deprotonated sulfur increases
as the pH increases. Since an increased concentration of deprotonated
sulfur is associated with an increased rate, deprotonated sulfur must
be the reactant in this reaction.
Since only the deprotonated sulfur-protein B species is involved in the
reaction, the rate expression for this reaction is:
Rate = k*[protein A] [protein-B-deprotonated].
As the concentration of deprotonated protein B increases, as it does at
higher pH’s, the rate of the reaction also increases.
[Also accept answers that argue that as pH increases, the sulfur
becomes more negative, which makes it more nucleophilic, thus
enhancing the rate of the reaction. Having a more nucleophilic sulfur
lowers the activation energy and therefore increases the rate constant.
We will cover this more in depth later in the semester.]
b.
c.
d.
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d.
8. a.
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b.
c.
9a.
electrophiles labeled as ‘e’ and nucleophiles as ‘n’.
Electrophiles and nucleophiles usually address the atoms directly
involved in the formation of new bonds with a nucleophile. The terms
usually do not refer to an atom receiving or donating electrons to an
existing bond.
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b.
10.
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